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MCAT Content / AAMC MCAT Prep Physics Question Pack Solutions

AAMC QPack Physics [Web]

Physics Question Pack: Passage 1

1) To answer this question, we want to know how long it takes the original number of americium nuclei to reach a certain threshold: 3.75 x 106 nuclei. In other words, we want to know how long it will take to get down to this number of nuclei. We’re going to need to reference the passage to find the half-life of radioactive americium.

We have a portion of the passage here. It says most units are initially fueled with 60 million nuclei of radioactive Americium 241 (half-life 430 years). We’re going from 60 million to 3.75 x 106 nuclei and want to know how many years that will take. We can rewrite 60 million as 60×106 nuclei as well. We’re simply manipulating the numbers here. We move our decimal point from the end of our 60 million 6 places to the left. Moving a decimal point left means we add to the exponent. We have the same base 10 and multiplier now: 106. We can divide 60 by 3.75 to find the fraction of the initial amount in our final amount. 3.75 goes into 60 16 times.

So ultimately to find the number of half-lives, set two equations equal to one another. X is our number of half lives:

(1/x)2 = (1/16) 

Take the square root of both sides. We get 4 as our number of half-lives. We want to convert the number of half-lives to actual years that have passed. We multiply 430 years by 4 and get 1,720 years. 

This was a math problem where we solved for an exact value. Quick glance at our answer choices show they are not close to one another. We didn’t do any rounding or approximation, so we can compare all 4 of our answer choices at once. Our calculation matches answer choice A: 1720 years.

2) To answer this question, we’re going to have to explain the differences and similarities between different isotopes of the same element. Isotopes of the same element have the same atomic number. The atomic number of an element is the number of protons in the element’s nucleus. This stays consistent, and a change in the number of protons will correlate with a different element. The number of neutrons differs for different isotopes, meaning atomic weight will also vary, because it’s influenced by the number of neutrons. 

  1. atomic number. Atomic number is the number of protons in the atom’s nucleus and the number of electrons. This number influences bond formation, and stays the same for both isotopes. I’m liking this answer for the time being; it matches our breakdown of the question.
  2. number of neutrons. This number can vary between isotope like I mentioned, so answer choice A remains superior.
  3. mass number. Mass number is dependent on the number of neutrons and protons found in the nucleus. If the number of neutrons varies, the mass number will vary also. Answer choice A remains our best option.
  4. atomic weight. This is similar to answer choice C. Atomic weight is dependent on the number of neutrons and protons found in the nucleus. If the number of neutrons varies, the atomic weight will vary also. We’re left with our best answer, answer choice A.

3) To answer this question, we’re going to need to use the passage to find specific details about ionization detectors. We’ll have to solve for the electric field value using the numbers provided in the passage. 

We have part of the passage here. We’re told two parallel electrodes are separated by 3 cm, and have a 5-volt potential difference across them. 

Electric field is given as volts per meter (V/m) or Newtons/Coulomb

We can convert our 3 cm to meters using dimensional analysis. 3 cm multiplied by a factor (1 meter/100 cm). We get 0.03 meters. 

Write out the equation: Our electric field is potential difference divided by the distance between our two points.

5V / (0.03 M) = 166.7 V/m

This is a math problem where we solved for an exact value. Quick glance at our answer choices show they’re not close to one another, so we can compare all four values at once. Answer choice D is consistent with our calculation.

4) To answer this question, we’re going to need to use the passage to find specific details about the photoelectric detectors. Then we’re going to have to solve for frequency by relating the information we’ve been given. 

We’re given wavelength and intensity of the light beam here. Our main light equation is:
C=wavelength x frequency 
We’re also told our C, speed of light, is 3×10^8 m/s

Passage tells us λ is 6.0 x 10^-7 m
-Even if we didn’t know which number was our lambda, we can look at units and see it is our wavelength. We can isolate frequency: C/lamda = frequency

Plug in the numbers we’re given and solve for frequency. 
(3.0 x 10^8 m/s) / (6.0 x 10^-7m) = f 

To more easily estimate, when we’re dividing numbers in scientific notation, we can divide the integers first, and then deal with our base 10 and exponents. 3 divided by 6 is 0.5. 10^8 over 10^-7 is 10^15. To get this number in proper scientific notation, we move the decimal point one place to the right. Remember LARS-if we move a decimal point to the right, we subtract from the exponent. We get a final answer of:

f=5×10^14 Hz

This was another math problem where we solved for an exact value. Let’s compare all of our answer choices at once. We see there are only two answer choices with the correct exponent on our base 10 (C & D). That means we can eliminate answer choices A and B. Answer choice D matches our calculation exactly, so we’ll stick with option D as our best answer.

5) While this question is tangenetially related to the information in the passage, we should be able to answer this question using our general knowledge. Destructive interference occurs when two waves are shifted by half a wavelength. Think of one wavelength as one revolution. So, 360 degrees would be one wavelength and 180 degrees would be half a wavelength.

  1. 0 degrees. This means there is no difference in phase between the two waves. We said we’re expecting half a wavelength, or 180 degrees. This contradicts what I said in our breakdown of the question.
  2. 90 degrees. Again, this contradicts our breakdown. This would be a quarter wave difference in phase between the two waves, not a half wave difference.
  3. 180 degrees. This matches our breakdown exactly. Destructive interference occurs when two waves are shifted by half a wavelength. Let’s hold on to this answer choice, we can eliminate answer choices A and B.
  4. 360 degrees. This is the same as answer choice A essentially. What do we know about questions on the MCAT? There’s only one correct answer. If we have two numerical answer choices that are essentially the same, we know they won’t be the correct answer. We’re left with our correct answer, answer choice C, 180 degrees.

6) The answer for this question is going to come from the passage-we’re going to need specific details to find the frequency of the initial light used in the detector so we can compare with our new frequency. This is also a call back to question #4 where we actually solved for this frequency. 

We previously set frequency times wavelength equal to the speed of light to solve for a frequency of 5×10^14 Hz. This was in question #4. 

We are given new wavelength, and the question stem tells us λ is 2.0 x 10^-7 m

Use our equation again:

C=wavelength x frequency 

We’re know our C, speed of light, is 3×10^8 m/s, wavelength is 2.0 x 10^-7 m, and we’re solving for frequency. Dividing both sides by wavelength gives us a frequency of 1.5 x 10^15 Hz. Is this our final answer? No it’s not. We have to find the difference in frequencies, so we subtract. 

We want to have our numbers multiplied by the same base with the same exponent. We can rewrite our initial frequency as 0.5 x 10^15 Hz. Remember LARS, we moved the decimal point one place to the left, so we added to the exponent. 

Subtract the frequencies to find the difference as 1.0 x 10^15 Hz

This was a math problem where we solved for an exact value. Let’s compare all of our answer choices at once. Answer choice D matches our calculation exactly.

 

Physics Question Pack: Passage 2

7) We’re asked about the three waveforms in Figure 1, and want to determine which has the shortest period. That means we’re going to need to reference Figure 1 in the passage, and then we’ll need to need to relate the figure to the period of the waves. 

The first harmonic always has the longest wavelength. Subsequent harmonics – second, third, fourth…will steadily decrease their wavelength. If the first harmonic has the longest wavelength, its fundamental frequency is the shortest there because wavelength and frequency are inversely proportional.

We can conclude the frequency increases with increasing harmonics, and the third harmonic will have the highest frequency and first harmonic will have the lowest frequency.

Period and frequency are inversely proportional:
frequency = 1/T
Where T = period

If frequency is highest at one point, the period will be its lowest. We established frequency increases with increasing harmonics. Since the third harmonic has the highest frequency, the period here will be the shortest.

We can also take a different approach. Period is the time it takes to complete one cycle of the wave. Look at the figure to answer the question:

  • Figure 1a is all three harmonics
  • The third harmonic’s period is shorter than those of harmonic 1 and harmonic 2
  1. first harmonic this is the opposite of what I said in the breakdown. The first harmonic has the lowest frequency and the longest period.
  2. second harmonic. The second harmonic is shorter than the first harmonic. Answer choice B is superior to answer choice A and our best option for the time being.
  3. third harmonic. The third harmonic has a shorter period than both the first and second harmonic. That means answer choice C is our best answer. 
  4. Answer choice D says the waveform in figure 1c. The waveform in Figure 1c has the same period as the first harmonic. This also contradicts our figure and breakdown of the question. We can eliminate answer choice D and we’re left with our correct answer choice, answer choice C.

8) We’re going to need to reference Figure 1 in the passage. We want to find the point where all three curves intersect. We have to be able to actually interpret the curves in Figure 1a and determine what the intersection point means.

We have our figure here, and we can highlight the second position where our three curves intersect. First intersection is at very beginning of time axis, and third intersection is all the way to the right side of the time axis. 

The second position where the three curves intersect in Figure 1a is along the time axis, when all three curves are at zero displacement. What’s displacement? Displacement is the overall change in position of the object from start to finish and isn’t affected by the accumulation of distance traveled during the path from start to finish. All three curves are intersecting with one another (and the time axis) and at zero displacement.

  1. in phase. That only happens when two waves have the same frequency and are perfectly aligned. Not the case here, so let’s see if we can find a better answer.
  2. out of phase. When waves are out of phase, they’ll reach a zero value at different x-values. Again, not the case here. 
  3. at zero displacement. This matches our breakdown. All three curves intersect along the x-axis and at zero displacement. That means answer choice is our best answer so far.
  4. at maximum displacement. Maximum displacement would be the maximum change in position from the x-axis in this case. Not the case either. We’re left with our correct answer. Answer choice C.

9) We’re going to need to reference Figure 2 in the passage. We’re also going to use external information to relate frequency and period for the two harmonics. 

We have our figure here, and we’re focused on 2a specifically. We’re told frequency of the first harmonic is 100 Hz. 

Frequency refers to how often something happens. Period refers to the time it takes something to happen. 

Period (T) and frequency (f) are related by T= 1/f for a tone (OR f = 1/T) First harmonic frequency is 100 Hz. Going by our figure, the second harmonic has a frequency of 200 Hz.

1/frequency = 1/200Hz = 0.005 seconds

This was a math problem in which we did no rounding or approximation. None of the answer choices are particularly close to one another, so we can compare all of our answers to one another at once. Our correct answer is answer choice A.

10) We’re going to need to reference Figure 1 in the passage. We can answer this question just by looking at the figure and comparing the amplitudes of the three different harmonics.

We have Figure 1a here. The amplitude is the distance from rest to crest. Harmonic 1 gets slightly further away from the time axis than the other two harmonics, so it has the greatest amplitude. The second and third harmonic have slightly smaller relative amplitudes that look to be equal.

a.

Answer choice A matches our breakdown exactly. We said the first harmonic has the greatest amplitude, while the second and third have slightly smaller, but equal amplitudes.

b.

Answer choice B is the opposite of our breakdown. First harmonic has the largest relative amplitude, not the smallest. We can eliminate answer choice B.

c.

Answer choice C shows the first and third harmonics having the same relative amplitude, but this contradicts Figure 1. Answer choice A remains our best answer for the time being.

d.

Answer choice D again shows the first and third harmonics having the same relative amplitude, but this contradicts Figure 1. We’re left with our correct answer, answer choice A. 

11) To answer this question, we have to compare the properties of this hypothetical fourth harmonic with the third harmonic. That means we’re going to need to reference Figure 1 in the passage. This question is fairly open-ended, but glancing at our answer choices, we want an answer that either compares amplitude or frequency. 

We have our Figure above. The amplitude is the distance from rest to crest. The second and third harmonic have slightly smaller relative amplitudes than the first harmonic, but look to be equal. We can’t predict from this information whether the amplitude will be different for a fourth harmonic. 

Let’s recall some information from earlier questions. The first harmonic always has the longest wavelength. Subsequent harmonics – second, third, fourth…will steadily decrease their wavelength. If the first harmonic has the longest wavelength, its fundamental frequency is the shortest there because wavelength and frequency are inversely proportional:

frequency = 1/wavelength

Therefore, we can conclude the frequency increases with increasing harmonics. The fourth harmonic will have the highest frequency.

  1. lower amplitude. We cannot predict whether the fourth harmonic is going to have a lower or higher amplitude. This is not a great answer choice, so let’s keep looking for something better.
  2. higher amplitude. This is similar to answer choice A because we cannot predict whether the fourth harmonic is going to have a lower or higher amplitude.
  3. lower frequency. This goes against what I said in the breakdown of the question. We actually expect the opposite to be true. 
  4. higher frequency. This is consistent with our breakdown. We said the fourth harmonic will have the highest frequency of the first four. That means we can stick with our best answer, answer choice D: higher frequency.

12) Once again, we’re going to need to reference Figure 1 in the passage. Similar to our last question, this question is also fairly open-ended. We can glance at the answer choices to see what type of answer the testmaker is asking for. We’re comparing the period of the waveform to the first, second, and third harmonics. 

We have our figure here. Period is the time it takes to complete one cycle of the wave. Looking at Figures 1a and 1c, we see the wave hits the time axis as often as the first harmonic. The second harmonic hits the axis twice in the same time, while the third harmonic hits the axis three times in the same time. The period of the waveform in Figure 1c is the same as the period of the first harmonic.

  1. same as the period of the first harmonic. This is exactly what we said in the breakdown, so I am going to hold on to this answer. It’s a viable option if we don’t find anything better.
  2. same as the period of the second harmonic. The second harmonic hits the axis twice in the same time the waveform hit the axis once. Answer choice A remains the best answer choice.
  3. same as the period of the third harmonic. The third harmonic hits the axis three times in the same time the waveform hit the axis once. Answer choice A is still our best option.
  4. sum of the periods of the first, second, and third harmonics. The period of the waveform is the same as the period of the first harmonic, not the sum of the three periods of the first three harmonics. We checked answers B-D, but none of them were better options than answer choice A. We’re sticking with our correct answer, answer choice A: same as the period of the first harmonic

 

Physics Question Pack: Passage 3

13) To answer this question, we’ll have to know about relative velocities, and how movement can affect frequency.

First thing we mentioned is relative velocity. The jet moving away at 300 m/s from a stationary receiver will show a certain frequency. To replicate the frequency for a jet moving at 600 m/s, we have to travel in the same direction as the jet at 300 m/s to account for the difference. We’re dealing with an electromagnetic wave, so all that matters is the velocity difference between source and observer. 

Said differently: a jet moving 600m/s while you move in the same direction as the jet at 300m/s will make you observe a doppler shift as if the jet were moving only at 300m/s away from you while stationary.

  1. A receiver moving at 900 m/s in the opposite direction as the jet. This is not consistent with my breakdown of the question. We want to move in the same direction as the jet and have a velocity difference of 300 m/s so we replicate the frequency.
  2. A receiver moving at 300 m/s in the opposite direction as the jet. This is against our breakdown too. We want an answer choice that mentions moving in the same direction as the jet.
  3. A stationary receiver. A stationary receiver doesn’t account for the velocity difference in the two jets.
  4. A receiver moving at 300 m/s in the same direction as the jet. This matches our calculated velocity and direction. We accounted for the 300 m/s difference. A receiver moving at 300 m/s in the same direction as the jet will detect the same change in frequency. We can stick with answer choice D as our best answer.

14) To answer this question, we will pick one of the four graphs that demonstrates the relationship between speed of the transmitter away from the receiver and change in wavelength. We’re going to use trends from in the tables given in the passage to answer our question.

We want to see what happens to our change in wavelength as speed away from the receiver varies.

Column 1 of table 2 shows an increase of speed away from the receiver. What happens to change in wavelength? In Column 3 we see a linear increase.

Table 3 shows speeds toward the receiver, so we should see opposite results. We have an increase in speed in Column 1. We have an expected decrease in wavelength in Column 3. Why is it a decrease here? Because the jet is coming toward the receiver, so this is consistent with Table 2 as well. Both tables are telling us the same thing. As the speed of the jet moving away from the receiver increases, the change in frequency also increases. We have a linear increase.

a.

Answer choice A looks exactly like our prediction. Speed increases, and we see a linear increase in the change in wavelength as well.

b.

Answer choice B shows an increase in the change in wavelength, all at the same speed. This contradicts our tables and our prediction. The change in frequency varies after changes in speed.

c.

Answer choice C shows an increase in change in wavelength as speed decreases. This is opposite our breakdown of the question. We can eliminate this answer choice.

d.

Answer choice D is the exact opposite of our breakdown. We identified the trend from our tables, but this answer is showing an increase in change in wavelength as speed decreases. That happens when speed increases. We can eliminate answer choice D. We’re left with our correct answer, answer choice A

15) To answer this question, we want to know what happens to the distance between adjacent peaks of transmitted waves, which is another way of saying wavelength. This question is testing your knowledge of the Doppler effect. 

The jet is moving away from the receiver and increasing speed. The waves are being spread out by the motion of the jet, and the waves will show an increased wavelength and decreased frequency. As the jet moves away, the waves appear “stretched out” so distance between the peaks will appear to increase therefore lower frequency. If instead the jet was moving toward the receiver, the waves would be “compressed” so a higher frequency and smaller wavelength. In this case, with the jet flying away, we expect an increased wavelength.

  1. it decreases. This would be the case if the jet were flying toward the receiver. This contradicts our breakdown of the question. 
  2. it remains constant. This contradicts what we know about the Doppler effect. We’re going to see the wavelength increase because of the jet speeding away. 
  3. it increases. This matches our breakdown exactly. There’s not a ton of subjectivity to this question, so we can eliminate answer choices A and B, and we’re going to keep C as our best answer choice so far.
  4. it changes, but is not dependent on the speed. This is a contradiction as well. The jet moving away and increasing speed is what causes the wavelength to increase. We can eliminate answer choice D. We’re left with our correct answer, answer choice C. 

16) We want to explain some of the data found in the passage. We’re going to have to use our general knowledge to explain the relationship between frequency and wavelength. We’ll also think about the differences between sound and radio waves. 

Changes in frequency and wavelength are dependent on variation in speed. The velocity of radio waves is much higher than that of sound. A small change in velocity is therefore going to have a much bigger effect on frequency and wavelength when dealing with sound waves. The same change in the radio waves would be a relatively smaller change.

  1. Sound waves travel more slowly. This matches our breakdown. We said sound waves have a smaller velocity normally, so any change in velocity will cause a larger relative change in frequency and wavelength
  2. Sound waves have a much higher frequency. Note something interesting here about answer choices B and C. B and C are practically saying the same thing. Frequency = 1/wavelength. Higher frequency = shorter wavelength. Wavelength and frequency are inversely proportional. Both answer choices are saying practically the same thing, but every MCAT question can only have one correct answer. If we have two answers that are the same (for all intents and purposes), then we know we will be able to eliminate them.
  3. Sound waves have a much shorter wavelength. We have the same reasoning as answer choice B here. If we have two answers that are the same, then we know we will be able to eliminate them. Answer choice A remains the best option. 
  4. Interference in the atmosphere affects sound waves much more. This is out of scope. We didn’t discuss interference in the passage or the question stem. We can eliminate answer choice D. We’re left with our correct answer, answer choice A.

17) Let’s think about what’s happening in this situation. We have a receiver in one jet that’s pointed at a second jet that’s emitting radio waves. Both are traveling at the same speed. We’re going to have to find the relative motion between the two jets and solve for change in frequency.

Let’s focus on the fact the jets are both moving at the same speed. What does that mean? There is no relative motion between the two. For one of the systems, the other is relatively “at rest”. There will be no change in frequency. The change in frequency detected will be zero because there is no frequency shift.

Alternatively, if something is emitting sound and moving away, the waves appear “stretched out” so distance between the peaks will appear to increase therefore lower frequency. If it’s moving towards you, the waves are being “compressed” so it’s a higher frequency. In this case, no change.

In this case, we already solved for our correct answer. This is not a subjective answer. We said there’s no relative motion between the two jets, so there is no frequency shift. The change in frequency detected at the receiver will be answer choice A: 0 Hertz

18) To answer this question, we have to understand the relationship between wavelength, frequency, and the Doppler effect.

The wavelength of the hydrogen in the star’s light is slightly shorter than the wavelength in the laboratory. Frequency is inversely proportional to the wavelength. This means that the frequency of the arriving star light is higher than normal. Increased frequency means that the source is moving toward us. A deceased frequency would mean the opposite. Think of it this way. The classic example is the ambulance siren. It has a higher pitch, or a higher frequency, when the ambulance moves toward us. That’s what we see here.

  1. The star is moving away from Earth. This contradicts our breakdown. This would be the case if the wavelength of the star’s light was higher.
  2. The wavelength of light that the star is emitting changes constantly. This wavelength is not constantly changing, but rather we’re seeing a discrepancy in the value because the star is coming toward the earth.
  3. The frequency of light that the star is emitting changes constantly. Reasoning here is going to be similar to answer choice B so I’ll reiterate. These variables are not constantly changing, but rather we’re seeing a discrepancy in the values because the star is coming toward the earth. We can eliminate both answer choices B and C. 
  4. The star is approaching Earth. This matches our breakdown exactly. Increased frequency means that the source is moving toward us. Sounds like the best option of the 4 listed. We keep our correct answer choice, answer D.

 

Physics Question Pack: Questions 19-23

19) This is a question that’s fairly easy to visualize. Think about a soft drink to which you add ice. Part of some cubes are submerged, while parts of them may float above the soft drink. What are the forces acting on the cube? We have gravitational force down, while the buoyant force is in the opposite direction. That buoyant force (FB = Vρg) is what is allowing the cube to float. We can set the forces equal to one another:

(mass of the ice cube) x (g) = (volume cube submerged) (density of soft drink) (g)

The key here is that we have “g” on both sides of our equation. If we have 1/6th of g on both sides, then both sides of the equation are affected equally. What does that mean for our question? Even with gravitation force being 1/6 that of earth, the cube would not change how much it floats. 

This question was a math problem and we did no rounding or approximating. We solved for an exact value, so we can compare all of our answer choices at once. Only answer choice B is consistent with our breakdown.

20) Let’s look at the data we have in our question stem. It’s comparing depth and pressure, and we want to know the effect on pressure if density is doubled. Like many math questions, knowing the relationship between variables ends up being the key to getting the correct answer.

First thing we notice is if we increase depth by 5 cm, pressure increases by 200 N/m2, but at 5 cm we see an additional 50 N/m2. What does that mean? We have atmospheric pressure of 50 N/m2 that has to be added, and for our calculations, pressure at these various depths for the initial liquid is actually the amount listed in the table minus 50 N/m2 atmospheric pressure. Density and pressure are directly related, so if we double density, we also double pressure. At 10 cm, we have (450 N/m2 – 50 N/m2) * (double density) = a pressure of 800 N/m2. Once again, we can add the 50 N/m2 atmospheric pressure back. That gives us a pressure of 850 N/m2

This question was a math problem and we did no rounding or approximating. We solved for an exact value, so we can compare all of our answer choices at once. Only answer choice C is consistent with our breakdown. We might incorrectly get answer choice A if we thought density and pressure were inversely related. We might incorrectly get answer choice B if we believed density did not affect pressure.

21) Let’s run through some content here and then attack our specific question. Refraction: when light ray changes its direction in a new medium depending on the incident angle and the changes in speed. Snell’s Law: n1sinθ1 = n2sinθ2

A ray of light changes direction when it passes from one medium to another. We follow the arrows in our question stem and see the light go from air to the medium. The angle of incidence (towards the surface, the incidence ray) and angle of refraction (of the refracted ray) are measured relative to a perpendicular to the surface at the point where the light ray crosses it. This perpendicular line is called the normal line – if light enters a medium with a higher density, it bends towards the normal line like we see in our image. In this case, we have a higher index of refraction of the medium (index of refraction of air is ~1.0) and θ > α

We’re told we have parallel surfaces; the light ray is going to be reflected perfectly and will exit the medium at the same angle. In other words, the reflected angle is going to leave the medium at the same angle, α. That means θ = θ’. The only answer choice that reflects everything we said in our breakdown is our correct answer, answer choice B. 

22) To answer this question, we have to consider what we know about half-lives. Half-life is the time taken for the activity of a given amount of a radioactive substance to decay to half of its initial value. In our question stem, activity would go from an initial 240 counts/min and halves to 120 counts/min, then would half again to the 60 counts/min. We’re able to go through two half lives in 24 minutes. That means each individual half-life is 12 minutes.

  1. 4 min this answer incorrectly divides the initial activity by the activity after 24 minutes to get a numerical value of 4.
  2. 12 min this answer choice matches our breakdown exactly. We can also go about this a different way. We can determine the factor by which the activity decreases: (240 counts/min)/n = 60 counts/min. n=4. We can use 2x = n (with x being the number of half-lives and n being the factor by which the activity has been reduced) to determine how many times smaller the activity has gotten. 2n = 4, so n=2. Two half-lives over 24 minutes means one half life is 12 minutes. 
  3. 24 min. This would be the correct answer if we were asked about two half-lives.
  4. 48 min. We might incorrectly get this answer if we doubled the 24 minutes instead of halving the 24 minutes in our breakdown. We can stick with answer choice B as the best answer.

23) We can visualize this experiment. The student has a beaker that’s held at copper’s exact melting temperature. They quickly light a Bunsen burner and hold it for a split second under the beaker. Not a very long time, but just a quick exposure to the flame so we’re adding a slight bit of energy. We want to know what happens to the metal. Something I want you to keep in mind: it’s currently in the solid state. A lot of energy still needs to be added to convert it all to liquid, despite it being at the correct temperature. Phase changes require some energy. In theory, if we kept the flame longer, the copper would all turn to liquid as temperature increases. But the split second is not long enough for this to happen. But what about the part of the copper that’s right next to the flame? That part will feel the flame directly and liquify. Think of cooking food-the part on the outside will get warm quickly, and we’re often worried about making sure the center gets enough heat and cooked thoroughly. 

  1. A small amount of the metal will turn to liquid, with the temperature remaining the same. This is exactly what I brought up in the breakdown of the question. The part of the metal that is next to the hot flame will liquify momentarily, but there’s not enough exposure to turn the entire sample to liquid. Temperature will not increase until we reach the point where the sample is liquid and continues to be heated. 
  2. All the metal will turn to liquid, with the temperature remaining the same. This could happen with longer exposure, but the test-maker explicitly says the Bunsen burner is used for a fraction of a second. The metal won’t turn to liquid that quickly.
  3. The temperature of the metal at the top of the beaker will increase. The opposite is likely true, if anything. The Bunsen burner hits the bottom of the beaker and that small bit of energy is transferred. We don’t expect temperature to change, but especially not at the top of the beaker. 
  4. The temperature of the whole mass of the metal will increase slightly. Until we have the phase change with additional energy added to the metal, we don’t expect temperature to vary. We can stick with answer choice A as our best answer.

 

Physics Question Pack: Passage 4

24) To answer this question, we’re going to need to know about electric fields. Quick glance at our answer choices shows we want to know in relation to L, our distance between electrodes, or R, the electrical resistance. 

Electric field is potential difference divided by the distance between our two points. It’s given as volts per meter (V/m). And the distance “L” is 0.01m. Potential difference is approximately equal to battery voltage, or 50 volts in the example. So, in order to decrease electric field, we either want to decrease the potential difference, or battery voltage. Or we want to increase the distance between our electrodes.

The battery effective voltage is given as V_eff = V – IR where R is the internal resistance of the battery. While changing R is going to affect electric field, change in L (the distance) is going to be an exact factor. 

  1. Increasing L by a factor of 2. This would decrease the electric field by a factor of 2 as well. The distance, L, is in the denominator, and increasing it by a factor of 2 will decrease the electric field. Let’s hold on to this answer choice. 
  2. Decreasing L by a factor of 2. This would have the opposite effect of what we need. Decreasing the denominator in our ratio is actually going to increase the electric field between the electrodes. 
  3. Increasing R by a factor of 2. This would decrease the electric field, but not as much as increasing L by a factor of 2. Increasing L by a factor is a direct change in the electric field, and has a larger effect. We can eliminate answer choice C
  4. Answer choice D says decreasing R by a factor of 2. This has the opposite effect of what we want. This would increase the electric field between the electrodes. We can also eliminate answer choice D. We’re left with our correct answer, answer choice A.

25) We want to compare the kinetic energy of an ejected electron that goes from cathode to anode with its potential energy immediately after being ejected from the cathode. We’re told the photon has an energy slightly greater than the work function of the cathode. We’re going to need to know about potential vs kinetic energy.

When the electron is released from the cathode, it has potential energy, and little kinetic energy. It isn’t moving yet, or it’s moving very slowly. As the electron moves toward the anode, the potential energy is turned into kinetic energy. Remember, energy is conserved so by the time it reaches the anode, the electron has its highest kinetic energy and little potential energy. The question also says the electron is ejected by a photon with an energy that’s slightly greater than the work function. For our purposes, this energy is going to be mostly negligible. The photon gave just enough energy to eject the electron. Most of the energy is just going to be the potential energy converted to kinetic energy, and these are about equal.

  1. it will be 2 times as large. This is not the case. Energy is conserved, and while the final energy might be slightly higher, it’s a very small difference. Let’s keep comparing
  2. it will be approximately equal. This is exactly what we said in our breakdown. Both the initial potential energy and final kinetic energy should be approximately equal. Energy is conserved. This is a better answer than answer choice A, we can eliminate that answer choice.
  3. it will be ¼ as large. This sounds like an arbitrary number, and we know the energies will be approximately the same because energy is conserved. We can eliminate answer choice C.
  4. it will be 0. The potential energy will be converted to kinetic energy, and the final kinetic energy will not be zero. We can eliminate answer choice D. We’re left with our correct answer choice, answer choice B. The kinetic energy and potential energy will be approximately equal.

26) To answer this question, we’re going to need to know the definition of work function. That’s the minimum work or energy needed to free an electron from a solid. So, if we have an increase in the number of photons incident on the cathode that have an energy above the work function, what do we expect to happen? More electrons are freed, and more electrons are ejected.

  1. number of electrons ejected. This is consistent with our prediction. When the number of our photons with energies above the value of the work function increases, the number of electrons ejected increases. Let’s keep answer choice A and keep comparing.
  2. potential energy of each ejected electron. The potential energy of each individual electron should not increase, but rather the number of electrons will increase. We can eliminate answer choice B. 
  3. magnitude of the electric field between the electrodes. Electric field will not increase with an increased number of incident photons. We’re only increasing the number of electrons. This contradicts our breakdown, we can eliminate answer choice C. 
  4. speed of the electrons at the anode. Another question mentioning an increase in speed or energy. This isn’t the case. Only the number of electrons increases if we increase the number of incident photons. We can eliminate answer choice D. We’re left with our correct answer, answer choice A.

27) To answer this question, we can check the figure in the passage to give us a better visual. We’re also going to use general knowledge to explain the movement of electrons and movement of a test charge in an electric field.

We have our figure and excerpt from the passage here. As you’re practicing, I don’t recommend going back to the passage often, unless it’s for specific details. The reason I reference the passage fairly often in these solutions is so you can clearly see my thinking and what I’m referencing from the passage. 

We can see where the electron is ejected near the left side of the vacuum tube, from the cathode. We have an electric field between our two electrodes. We’re told the negative side of the negative terminal of the batter is connected to the cathode, the positive terminal to the anode. 

Our charged electron is released in the electric field and feels a force. Remember, force is dependent on mass and acceleration. We expect the electron to accelerate across the electric field toward the anode.

  1. It is stationary until collisions propel it toward the anode. The electron is a test charge, it’s introduced to an electric field between our two electrodes, and it accelerates toward the anode. It’s not stationary, so this contradicts our breakdown.
  2. It moves with constant speed toward the anode. We said the electron feels a force in the electric field. This by definition means it will accelerate, and not be in constant motion. This contradicts our breakdown as well. 
  3. It accelerates toward the anode. This matches our breakdown exactly. We said the electron feels a force in the electric field. It accelerates toward the anode. This is our best answer so far, so we can eliminate answer choices A and B.
  4. It exits through a side of the vacuum photodiode. The electron’s not going to exit through the side, nor do we know if it has the ability to do so. This answer choice is irrelevant, so we can keep our best answer choice, answer choice C. The electron will accelerate toward the anode.

28) We can check the figure in the passage once again to give us a better visual. Quick glance at our answer choices shows we’ll have to know about voltage and circuits to compare our options. 

We have our figure. Typically, electrons travel from Anode to Cathode. Current travels from Cathode to Anode (movement of positive charge).

In this case, electrons are ejected from the cathode via the photon beam-remember the passage said the photoelectric effect causes electrons to be ejected. The anode can replace its “lost” electrons. As a result, this movement of electrons generates a current that flows through the circuit. There was no current until the electrons start getting ejected. Our voltage and potential difference are approximately equal, and aren’t going to change here.

  1. The voltage across the electrodes reverses polarity. What did we say in our breakdown? Voltage and potential difference aren’t going to be changed. Reversed polarity would also damage the battery in our experiment. Not a great option to start.
  2. The voltage difference between the electrodes increases. This again contradicts our experiment. Voltage difference doesn’t increase in this situation. Let’s try and get a better option.
  3. Current flows through the circuit. The electrons ejected from the cathode will generate a current that flows through the circuit. This is a better answer choice than answer choices A and B, we can eliminate those options.
  4. The total resistance of the circuit increases. Resistance isn’t going to increase in this situation. We have the electrical resistance of the resistor, which isn’t going to increase. This contradicts our breakdown, and the passage as well. We’re left with our correct answer, answer choice C.

29) We can go back to the passage to see the effect of the photons on our vacuum photodiode detector. Once we do that, we’ll need to deduce the effect of the increased frequency. We’ll have to know how frequency is related using our general knowledge.

We have the first part of our passage here, and we want to highlight a few things. Our second sentence says there is an ejection of electrons from a metal plate when photons are absorbed by the metal. So, a photon hitting the cathode will eject an electron.

Later in the excerpt, we’re given the equation for the energy of a photon, E=hf. An increase in frequency is going to increase the energy. 

We’re told kinetic energy of the ejected electron is going to be equal to the photon’s energy minus the work function, or the binding energy. The photon’s energy is partly used to break the electron away from the material, and the rest is kinetic energy. The work function stays constant here. Meaning by increasing the energy of the incoming photon, the kinetic energy is going to increase. Because mass stays constant, velocity increases.

  1. decrease the number of photons ejected. Looking at answer choices A and B, both of these mention a change in the number of photons ejected. But what did we say? Every single photon hitting the cathode will still eject only one electron. This contradicts the passage and our breakdown of the question. We’ll likely end up eliminating answer choices A and B.
  2. increase the number of photons ejected. We’re using the same reasoning as answer choice A to eliminate this answer choice.
  3. decrease the speed of the ejected electrons. We said there’s an increase in kinetic energy, and increase in velocity. This answer choice goes against our breakdown.
  4. increase the speed of the ejected electrons. This one matches our breakdown exactly. Increasing the frequency of each photon will increase the speed of the ejected electrons. We can stick to our correct answer choice, answer D. 

 

Physics Question Pack: Passage 5

30) To answer this question, we want to know the effect on electrical force if the speed of the charged particle is increased. We can answer this question using general knowledge. We have to define electric force, and see if it is impacted by speed.

We are dealing with electric force. We can write out our equation dealing with electric force as:

E=F/q

E is electric field (that’s given in Newtons/Coulomb or volts/meter). In this case, charge was given in coulombs in the passage, so electric field is Newtons/Coulomb. 

F is the electric force (that’s given in newtons). This is the variable we’re focused on. We want to see the impact of speed on electric force.

q is the charge, given in Coulombs. Our units check out on both sides of our equation-Newton’s over Coulombs.  

Given this equation and these units, how does speed affect electrical force? Going by our equation, speed’s not going to affect electrical force. Increasing speed by a factor of 2 won’t affect any of the variables in our equation. And why did we break out the units of each variable? It’s an extra level of insurance. None of the units are units for speed. We want an answer choice that mentions there will be no change with an increase in speed. 

We can compare all of our answer choices at once in this situation. We said there’s going to be no change in force. There’s only one answer choice that reflects this. Answer choice B says the electrical force will remain the same. Answer choices A, C, and D all reflect a change in electrical force, these are all incorrect.

31) We can go back to the passage to reference our figure. We’re going to need to know how capacitance varies, and we can compare all 4 of our answer choices. We recall from that passage that we’re dealing with parallel plates, so we’ll want to point out relationships between capacitance and other variables in parallel plate capacitors.

We have our figure on the top left, we have our four answer choices on the top right. What do we know about the capacitor? The capacitor is fully charged, and there’s a potential difference between its plates. We also have a parallel plate capacitor. Remember we said we had to know our content outline? Parallel plate capacitors are the first subtopic under capacitance. We can wirte out the equation for the capacitance of a parallel plate capacitor:

C=Eo x A/d

A is going to be area, and d is the distance between the plates. So, a larger area means more charge can be stored. Does that make sense? The bigger the plates are, the more charge they can store—because the charges can spread out more. 

A larger distance between the plates means less charge can be stored. Does that make sense? It should. The closer the plates are together, the greater the attraction of the opposite charges on them. So, Capacitance should be greater for smaller distance. Ultimately, we want an answer that mentions an increase in the area of the plates. 

Quick glance through our answer choices. A and B don’t have to do with geometric factors. That’s what ultimately determines capacitance for parallel plates. Answer choice C says increase separation between plates. Looking at our equation, that would have the opposite effect. Capacitance and distance have an inverse relationship. Increasing distance would decrease capacitance. Lastly, answer choice D says increasing the area of the capacitor plates. That’s exactly what we expect from our equation. We’ll pick answer choice D as our best option: Increasing the area of the capacitor plates.

32) To answer this question, we need to know the force on a particle in terms of the variables given. We also need to know our physics equations and units to be able to convert as necessary. 

The force on a particle with a charge q and electric field E is given by:

F=qE. We want our force to be in terms of our given mass, m. We can substitute for F on the left side of our equation. Force = mass x acceleration. So ma = qE.

We want to solve for acceleration, a. We isolate acceleration and we’re given a= qE/m

We solved this question by knowing the equation. How can we solve this question using units? Electric field can be given in volts/meter or Newtons/Coulomb. Newtons/Coulomb is equal to electric force (in newtons) over charge (in Coulombs). E=F/q. This can be rearranged to F=qE, and we can solve for acceleration the same way. 

Why do we go through this problem with two different methods? So we can see the importance of two things: knowing the content, and being able to manipulate units. You have to be careful with units, especially in a math-heavy section like physics.

Glancing at our answer choice, we want our acceleration in terms of three variables, E (electric field), q (charge), and m (mass). We already isolated our acceleration and can pick the corresponding correct answer, answer choice A.

33) To answer this question, we’re going to have to know about equivalent capacitances, and capacitors in series. Glancing at our answer choices, we’re going to be comparing the capacitance of the new circuit with that of the original circuit. 

Equivalent capacitance of two capacitors connected in parallel is the sum of the individual capacitances. If we add another capacitor in series to the circuit in the passage, and the capacitor is identical, each will contribute and equal capacitance. Now, does the capacitance change? The capacitance itself is not going to change. Each individual capacitor is only going to need to contribute half of the capacitance, and the sum of the individual capacitances is going to equal the initial, total capacitance. So, the equivalent capacitance of both the new circuit and the original circuit are going to be ½ as great as the capacitance of the original capacitor’s capacitance. That means we can pick answer choice A because it matches our breakdown exactly. There’s no subjectivity to this question, so no need to spend a lot of time going through the other answer choices.

34) We can answer this question using general knowledge, but we may want to reference our figure once again. We’re going to have to know about parallel-plate capacitors, and the properties of each plate. Keep in mind our particle is negatively charged.

We have our figure on the top left, we have our four answer choices on the top right. The particle in our figure is negatively charged in this case, and it’s traveling through the parallel plate capacitor. Typically, capacitors will have a negative plate and a positive plate. So first of all, which way do we expect the particle to travel? We expect the negative particle to travel toward the positive plate. The negative charged particle is going to be attracted to the oppositely charged, positive plate. Quick glance at our answer choices, we have two options that mention the particle traveling toward the positive plate. We have to narrow our choices down further. One answer mentions moving with a constant speed, while the other mentions accelerating. Now, we actually touched on this a few questions ago as well. The force on a particle with a charge q and electric field E is given by: F=qE.

Force is also equal to mass x acceleration. So, because there is a force on the particle, the particle is going to be accelerating toward the positive plate.

  1. It moves with constant speed toward the positive plate. We already looked at our answer choices a bit during our breakdown. We can look at A and B together. Both of these mention a constant speed, and we know that’s not the case. The negative particle is accelerating toward the positive plate.
  2. It moves with constant speed toward the negative plate. The reasoning here is going to be similar to answer choice A. 
  3. It accelerates toward the positive plate. This matches our breakdown, and sounds like the best option here. We can officially eliminate answer choices A and B. 
  4. It accelerates toward the negative plate. This is the opposite of our breakdown. The particle does accelerate, but the particle itself is negatively charged, so it will accelerate toward the oppositely charged, positive plate. We can eliminate answer choice D. We’re left with our correct answer, answer choice C. 

 

Physics Question Pack: Passage 6

35) To take this question a step further, we want to explain why the positive ions were considered fixed, while the electron sea moved back and forth. That means we want to identify and use the difference between the positive ions and the electron sea to explain why the ions can be considered fixed.

We have to look no further than their relative sizes. Which do we expect to be able to move and change direction more quickly: an ion or electrons? Electrons are thousands of times smaller than even the smallest ions, so they’re able to move much more easily and rapidly than the ions. Even if the ions are feeling the opposite charge, they’re not able to oscillate and move as quickly as the electrons.

  1. The ions are bound together with strong nuclear forces. This wasn’t really discussed in the passage, and it’s not very relevant to answering this question. Even if the ions are bound by strong, nuclear forces that doesn’t explain why there’s no movement relative to the electron oscillations.
  2. An ion is much more massive than an electron. This is consistent with our reasoning in our breakdown of the question. We said the ions are much more massive-several thousand times more massive in many cases. This makes any movement negligible and seemingly nonexistent relative to the oscillations of the electrons. That’s why the ions can be considered fixed. This option is superior to answer choice A. 
  3. The ions experience no force when the electron sea is displaced. This contradicts the passage and our breakdown. The ions still experience a force from the charge difference. The electrons oscillate so much more rapidly as a result of this charge difference, that it appears the ions are fixed. It’s not because they don’t experience the force. We can eliminate answer choice C.
  4. Coulomb’s law prohibits the motion of the ions. This contradicts the passage as well. The charges repel and that’s what causes the oscillations in the first place. There’s negligible motion in the positive ions relative to the motion of the electrons, and that’s why the ions can be considered fixed. We can also eliminate answer choice D. We’re left with our correct answer, answer choice B. 

36) We’ll have to reference the passage to look at Figure 1A. We’ll use our general knowledge to explain the potential energies at different points in the oscillation. 

We’re looking for maximum potential energy. There are two types of energy that we’re going to be seeing in this figure: kinetic and potential. Kinetic energy of a mass is given by K =  1/2 mv2 and potential energy is given by U =  1/2 kx2 where x is displacement. All of the energy is kinetic energy in Figure 1B where displacement is zero. All of the energy is potential energy in Figures 1A and 1C when the electron sea hits the end of its movement in one direction and is preparing to swing in the other direction. Maximum potential energy occurs in Figure 1A and 1C, while maximum kinetic energy occurs in Figure 1B. This corresponds to answer choice D.

37) We want to know the frequency for plasma oscillations given a specific density. We’re going to need the passage to find the formula for frequency. We’re going to solve using the number we’re given in the question stem. 

We have an excerpt from the passage here that gives us our formula for frequency. The question stem gave us a density, n, of 1018 m-3. Look at the last sentence in the excerpt. It says the approximation on the right side of the equation gives the frequency in hertz, when n is expressed in m^-3. That means we can use the approximation given to us in the passage. Frequency is equal to roughly 9.0n^1/2.

We can plug in our value for n, and we now have 1018 raised to the ½ power. This means we multiply our exponents together. If we’re raising an exponent to a different power, multiply the exponents. The exponents 18 times ½ gives us 9. So, we’re left with a frequency equal to approximately 9.0 x 10^9 Hz.

We want to be careful here. We said there was an approximation in calculating our frequency. But when we look at our answer choices here, they’re all off by at least a thousand-fold. The exponents are all 3 or greater away from one another. We can find our calculated answer, answer choice C.

38) We won’t need the passage to solve for wavelength. We just have to know the relationship between speed, frequency, and wavelength. This is almost like a standalone question. 

We can write out the relationship between our variables as:

Speed=wavelength x frequency

In this case we’re given both speed and frequency. We can write out our equation by plugging in these numbers:

3.0 x 10^7 m/s = 1.0 x 10^9 Hz x wavelength.

Divide both sides by frequency. These are relatively big numbers, so how do we divide two numbers in scientific notation? We’re can turn this into two fractions instead of trying to solve the entire problem at once. So instead this becomes:

(3.0/1.0) and (10^7 / 10^9) When dividing exponents on the same base, we’re going to subtract the bottom exponent from the top exponent on our base 10.

We get 3.0 x 10^-2 meters. Quick glance at our answer choices shows our answer is either going to be in meters or centimeters. That means we can use dimensional analysis here. Multiply by a factor (100 centimeters/1 meter). Multiplying by 100 cancels out our 10^-2. So, we’re left with 3.0 centimeters. We want an answer that is equal to 3.0 centimeters or 0.03 meters.

We can compare all of our answer choices at once here since we calculated for an exact value. We’re left with our correct answer, answer choice A.

39) We’ll have to reference the passage to look at Figures 1A and 1B once again. We’ll have to use our general knowledge to explain the differences. Because this is an open-ended and broad question, we can quickly glance at our 4 options to know exactly what we’re looking for. We want to compare energy differences, or any changes in power and turbulence. 

We actually covered the energy differences a few questions ago. There are two types of energy that we’re going to be seeing in this figure: kinetic and potential. Kinetic energy of a mass is given by K =  1/2 mv2 and potential energy is given by U =  1/2 kx2 where the x is displacement. All of the energy is kinetic energy in Figure 1B where displacement is zero. All of the energy is potential energy in Figure 1A; the electron sea hits the end of its movement in one direction and is preparing to swing in the other direction. So, transitioning from position A to B means potential energy is transformed to kinetic energy. We do not have any references to power or turbulence here. 

  1. Kinetic energy is transformed into potential energy. This is the opposite of what is happening. This change happens from position B to position A. This contradicts our breakdown of the question, and the passage itself. 
  2. Potential energy is transformed into kinetic energy. This is consistent with our breakdown. We have maximum potential energy in position A. As it moves to position B, that potential energy is transformed into kinetic energy. This is a better answer than answer choice A.
  3. Power is dissipated as heat. Like we mentioned in our breakdown, there’s no reference to power dissipating as heat. We have a change from potential to kinetic energy. Power dissipation is more common in electrical devices. We can eliminate this answer choice.
  4. Turbulence brings the electron sea to rest. This is another answer that contradicts our breakdown, and the passage. The electron sea isn’t brought to rest. It overshoots the equilibrium position in Figure 1B and continues oscillating. We can eliminate this answer choice, so we’re left with our correct answer, answer choice B.

 

Physics Question Pack: Questions 40-44

40) This is a classic Doppler effect question from AAMC. The Doppler effect is an alteration in the observed frequency of a sound due to motion of either the source or the observer. The observer moving toward a source receives them at a higher frequency, and the person moving away from the source receives them at a lower frequency. Similarly, if the source is coming toward an observer, the observer receives them at a higher frequency and vice versa.

  1. rotating. This is out of scope. When we think Doppler effect, we usually think different loudness or frequency.  
  2. louder than it actually is. As the source is moving away from the observer, we expect the observer will experience the sound at a lower frequency. 
  3. lower in frequency than it actually is. This is consistent with our breakdown and the visual I’ve added above. 
  4. higher in frequency than it actually is. This is the opposite of what we expect. We expect the observer has the impression that the sound source is lower in frequency than it actually is.

41) Key things we want to note from the question stem: we have an organ pipe closed at one end, and n increases by +2 for consecutive wavelengths. We’re told resonances occur at L =  nλ/4, so we can plug in the necessary numbers here for the wavelengths given in the question stem (8m and 4.8m). We have two equations and we can multiply both sides by 4 to isolate n:

L = n(8 m)/ 4 → 4L = n(8 m)

L = ((n + 2)(4.8 m)) / 4 → 4L = (n+2)(4.8 m)

We can set the two equations equal to one another and solve for n (because both are equal to 4L):

n(8 m) = (n+2)(4.8 m):

8n=4.8n + 9.6:

3.2n = 9.6:

n=3

Plugging this value of n into one of the equations, we can solve for L:

L = (3)(8 m) / 4 = 6 m 

This question was a math problem and we did no rounding or approximating. We solved for an exact value, so we can compare all of our answer choices at once. We can pick our correct answer, answer choice C. Like many math problems on the MCAT, you’re not being tested on how quickly you can add and subtract numbers. Instead, AAMC wants to make sure you know how to manipulate numbers and understand relationships between variables.

42) This is another math problem that involves being able to manipulate numbers and understand the relationship between different values given to us. We’re given the threshold of pain as 120 decibels and asked to solve for the intensity (we’re given the equation in the question stem).

Set the threshold of pain equal to the equation given in the question stem:

120 = 10 log10(I/I0)

Isolate the logarithm on one side:

12= log10(I/I0)

Solving for I/I0 gives us 1012. We already know from the question stem that I0 is equal to 10–12 W/m2. Plus that value into our fraction:

I/(10–12 W/m2) =1012

I is equal to 100 W/m2

This question was a math problem and we did no rounding or approximating. We solved for an exact value, so we can compare all of our answer choices at once. Our answer choices are not particularly close to one another. We can pick our correct answer, answer choice B.

43) The net work done is going to be the sum of the force acting along the path multiplied by the distance travelled. This question comes down to knowing the equation for work and comparing the two situations described in the question stem. Work is going to be given by mgΔh, regardless of the direction we’re going. Moving the weight from A to B or from B to A is going to yield the same absolute value of work, given by the same equation. The value of m and g stays constant in our equation. Δh is going to remain the same as well because we’re moving from the same point A to point B and point B to point A. The only answer choice that matches this breakdown is answer choice C.

44) First thing we want to note is the difference in charges. We have the small negatively charged particle. We’re concerned with the motion of this particle. This negatively charged particle is placed near a fixed positively charged particle Q. The positively charged particle is fixed, while we want to note the motion of the negatively charged particle. Opposite charges will attract, while same charges will repel each other. We anticipate the negatively charged particle will be attracted to Q and will accelerate toward Q.

  1. It accelerates away from Q. This is the opposite of what we expect. If the non-Q charged particle was instead positively charged, this could be a correct answer.
  2. It accelerates toward Q. This matches what we said in our breakdown. Opposite charges will attract, while same charges will repel each other. We anticipate the negatively charged particle will be attracted to Q and will accelerate toward Q.
  3. It moves with constant speed away from Q. We anticipate the charged particle will move toward Q.
  4. It moves with constant speed toward Q. While we do expect the negatively charged particle to move toward Q, we don’t expect a constant speed. The attractive force will cause the particle to accelerate toward Q. We can stick with answer choice B as our best answer.

 

Physics Question Pack: Passage 7

45) To answer this question, we’re going to need to know how to solve for work, and we’re also going to need to know specific details and values given in the passage.

We have an excerpt from the passage here, it tells us the amount of force applied on the object, and the distance the object was propelled upwards. What’s our equation for work? It’s the force applied over a distance: Work = Fx d. 

In this case, Force is given as 19.6 newtons. Distance is given as 0.5 meters. Multiplying the two gives 9.8 newtons times meters. That unit should look familiar, it’s a unit of energy; that’s what makes up a joule. Our work is going to be given in joules.

We came up with an exact value and units for our calculated value, so we can start there. Which of our answer choices match our calculated units? Answer choices B and C. 

Answer choice D is in newtons and is missing the “meters.” Answer choice A is a mass multiplied by units for acceleration. In other words, Newtons. So, answer choices A and D have the same, incorrect units and we can eliminate those answer choices. Let’s go through our remaining choices for our correct answer. Answer choice C corresponds to our correct answer choice: 9.8 joules.

46) We can go back to the passage to see where our unknown variable is mentioned, and we’re going to use general knowledge to make any necessary connections. We want to focus on the relationship between variables. Looking at our 4 options, we want in terms of meters, velocity, and “g” or gravity.

We have our passage excerpt up top. We have our 4 answers down below. 

We’re told our unknown variable is peak height above the launch point. The peak height above the launch point is when the object is no longer rising, or when it’s switching from rising to falling. 

Has the force on our object changed? We have a projectile, meaning the only force on the object is gravity. That’s not going to change. All of its kinetic velocity has switched over to potential energy. Remember, we had an initial velocity of v. Initial kinetic energy is given by ½ mv^2. Final potential energy of the projectile is given by mgh. Energy is conserved, so we can actually set these two energies equal to one another. Remember, we said when the object was initially at rest and propelled upwards, we had kinetic energy. At the peak height, the object isn’t moving for the split second, and it has all potential energy. We can set our equations equal to one another. The masses cancel out. So we have ½ v^2 is equal to gh. We can isolate “h”-that’s what we’re solving for. We get h = 1/2v^2 x 1/g  or  V^2/ 2g. 

We came up with an exact value and units for our calculated value. We can keep our correct answer, answer choice B: V2/2g

Just a reminder, for physics questions, units are super important! So is the ability to manipulate units. We didn’t deal with any specific values here, but rather we needed to be able to manipulate equations. 

47) We’re going to need specific details and values from the passage, but we’ll need general knowledge to relate wavelength to our other variables. Quick glance at our answer choices shows our answer is going to be given in meters.

We have our passage excerpt here. We’re asked the wavelength at the peak height, h. What do we know about this height? At the peak height, the object isn’t moving for that split second, and it has all potential energy. That means we’d see the object at its maximum height after rising, and right below it begins falling. In this case, there is no Doppler shift. The object is no longer moving in relation to the microphone, so we can solve for wavelength without taking that nonexistent movement into account. 

What’s our main equation for wavelength? Speed is equal to wavelength x frequency. We have our speed of sound given as 340 m/s. We have frequency is given as 170 Hz. We can solve for wavelength in meters. Isolate wavelength and we have 2.0 meters. This is the correct units for our answer choices as well. We can keep our correct answer, answer choice C: 2.0 meters.

48) This question is going to be similar to our last two questions, meaning we’ll have to reference the passage, but we’ll ultimately answer the question using general knowledge.

We have our passage excerpt here, and we’ve dealt with this part of the passage a few times already in this question set. We’re asked about the frequency shift, or the Doppler effect in this question. What did we say about the flight of the projectile in our previous question? A force is applied and the object is propelled upwards and begins to slow down until it reaches a peak height. At that point, it changes direction and begins to fall toward the ground and the microphone on the ground. 

We need to find the magnitude of the frequency shift from 170 hertz. Meaning negative and positive signs aren’t going to matter. As the object is initially propelled and it rises up, we have a negative frequency shift that continues to decrease in magnitude as the object also slows down. When the object reaches the peak height, it’s no longer moving in relation to the ground, and frequency shift is zero. It begins to fall immediately after, and at that point, the shift starts to increase again in magnitude-this time it’s positive.

  1. It falls to zero, then increases. This matches our breakdown exactly. We have our frequency shift fall to zero when the object reaches peak height. Then after it changes direction and falls, the frequency shift increases in magnitude.
  2. It is constant throughout the flight. Unlike answer choice A, this answer contradicts what we said in our breakdown of the question. We said we have a negative value as the object initially rises, and a positive value as it falls back down. There’s zero shift at the peak height.
  3. It rises continuously. The magnitude of the frequency shift does not rise continuously. The magnitude actually drops to zero initially before rising again.
  4. It falls continuously. The magnitude of the frequency shift initially falls to zero, but then when the object begins to fall, it increases. We’re left with our correct answer, answer choice A.

 

Physics Question Pack: Passage 8

49) To answer this question, we’re going to need to reference the passage, and particularly Figure 2. We’re going to be changing a variable that has an effect on the focal length of the convex mirror. We’re also going to be using external information because we need to know the relationship between the different variables listed in our answer choices and focal length of the convex mirror.

We have an excerpt from the passage on the top, and we have Figure 2 right below. I also included the 4 answer choices we’re considering. 

We have to identify which of the options will affect focal length of the convex mirror. Right away, we can look at answer choices C and D and notice they are not changing the mirror at all. We have our figure right above that confirms this. The lens is not going to affect the focal length of the mirror. 

The way we approximate the focal length of a spherical mirror: f = R / 2. So, the focal length of the mirror does depend on the radius of the curvature answer choice B looks promising. Let’s consider answer choice A. There is no refraction here. The figure shows the light rays are reflected back along the incoming path, so the index of refraction is not going to affect the focal length either.

We keep our correct answer choice, answer choice B. We can eliminate answer choices A, C, and D because changing those will not change the focal length of the convex mirror in our figure.

50) We can answer this question by writing out Snell’s Law.

Snell’s law relates the angles of incidence and refraction. This question is simply asking to relate the two angles.

Snell’s law states that the ratio of the sines of the angles of incidence and refraction is equivalent to the reciprocal of the ratio of the indices of refraction:

Sin theta A / sin theta 1  =  n1/nA

We can multiply both sides by the denominators to get our angles on separate sides of the equation. We’re left with:

Sin theta A x nA = Sin theta 1 x n1

This answer is going to be simply comparing the formulas given with our breakdown. Answer choice C matches our formula exactly and it shows Snell’s Law. 

51) To answer this question, we have to know about diverging lenses, and how they differ from converging lenses. With a converging lens, we see the rays all come together at a point on the other side of the lens. A diverging lens will have divergent rays out. Note, their names explain the rays that result from going through.

  1. converging rays. This contradicts our breakdown of the question. We said the rays would all converge if we still had our converging lens. We actually saw this in the figure in the passage as well. We’ll likely end up eliminating this answer choice.
  2. parallel rays in and out. We covered this also in our breakdown. We said we’ll have divergent rays out, not parallel rays out. Let’s keep looking for a better option.
  3. reflected rays diverging. We’ve got refracted rays as they hit the lens and pass through. We’re not reflecting them in a different direction. 
  4. diverging rays. This matches our breakdown exactly. There was not much subjectivity here. We came into the answer choices knowing exactly what we were looking for, and it’s answer choice D: diverging rays.

52) Optically dense material by definition means the material will slow down a wave that’s moving through the material. We want to explain this definition. We’ll have to use our knowledge about how photons work, and how this may affect how quickly they travel.

Photons travel in a vacuum at a speed of c (3×10^8 m/s) or the speed of light in a vacuum. But, when a light wave travels through a medium, energy is absorbed and electrons with then atoms are set into vibrational motion. Energy is transported through the medium, and the process of being absorbed and reemitted by the atoms of the medium cause speed to decrease.

  1. is absorbed and re-emitted by the atomic structure of the optically dense medium. This is exactly what we said happens. Photons will interact with matter through exciting electrons. This ultimately slows down light waves. This answer is a strong frontrunner for the time being.
  2. is absorbed and re-emitted by the nucleus of the material in the optically dense medium. We said photons will interact with electrons, not nuclei. This contradicts what we said in our breakdown of the question, so we can eliminate this answer choice. Answer choice A remains superior. 
  3. bounces around randomly inside of the optically dense medium before emerging. This answer choices mentions light bounces around randomly. However, that motion is not random. This contradicts our breakdown of the question as well. We can eliminate this answer choice.
  4. loses amplitude as it passes through the optically dense medium. This answer choice is out of scope. Amplitude has to do with the intensity of the light wave, not the speed or wavelength. We’re left with our correct answer, answer choice A

 

Physics Question Pack: Passage 9

53) We’ll want to reference the figure in the passage to see switch S and the effect of closing it to the left. We’re going to have to understand the inner workings of capacitors and charge to actually answer our question. 

We have an excerpt from our passage and Figure 1 shown above that. The switch, S, is closed to the left, meaning charge is accumulating on the capacitor. The capacitor is going to be fully charged soon, but we have to explain why that charge can’t accumulate indefinitely. We said we have to understand the charging of the capacitor, so let’s visualize this situation. We can simplify this by saying we have positive charges beginning to accumulate on one of the parallel plates, while negative charges accumulate on the other. What do we say about these charges? Opposites attract, and like charges will repel one another. So as more positive charges accumulate on a single plate, for example, it gets increasingly difficult to keep adding more positive charges. Same thing goes for the negative plate. 

  1. the variable resistor inhibits the current flow. The variable resistor is involved when the capacitor discharges, not this charging process. This answer is out of scope.
  2. the battery continually loses charge. This answer is also not really answering why charge can’t accumulate indefinitely. The amount of charge that the capacitor can hold is much going to be the limiting factor here-the battery running out of charge isn’t going to come into play
  3. successive charges brought to the plates are repelled by charges accumulated earlier. That sounds like our breakdown of the question. We said as more positive charges accumulate on a single plate, it gets increasingly difficult to keep adding more positive charges. Same thing goes for the negative plate. This sounds like the best option so far. We can eliminate answer choices A and B.
  4. the fixed resistor loses energy to heat. Another answer choice that isn’t addressing the ability of the charge to accumulate indefinitely. We didn’t mention this in our breakdown, nor was it discussed in the passage. We can eliminate this answer choice, so we’re left with our correct answer, answer choice C.

54) We’ll want to reference the figure in the passage and we’re going to focus on the capacitor discharging. Quick glance at our answer choices shows that our answer will have to do with either of the two resistances mentioned in the passage. We’ll use our general knowledge, specifically our knowledge of resistors and Ohm’s law.

The passage says R is adjusted so that the discharge current is constant during the discharge time. We have to be careful with the verbiage and the key terms here. The charge stored in a capacitor is proportional to the voltage for a given capacitance. As we discharge the capacitor, the charge on the capacitor is reduced, and voltage reduces. Next thing we note, is we’re going to focus on capital R resistance to answer our question. OHM’s law states that electric current is proportional to voltage and inversely proportional to resistance. V=IR, voltage equals current times resistance. 

We already established voltage is being reduced. If we’re keeping current constant, resistance must be decreased to compensate. We’re looking for an answer that says capital R is decreased.

  1. the resistance R must be continually increased. This is the opposite of our breakdown. We said resistance R is going to continually decrease.
  2. the resistance R must be continually decreased. This answer choice matches our breakdown exactly. Resistance is continually decreased to compensate because voltage is being reduced and current is kept constant. 
  3. the resistance r must be continually increased. We said in our breakdown that our answer is going to focus on capital R resistance. Answer choices C and D depend on lower case r resistance, so we can eliminate these answer choices.
  4. the resistance r must equal R. Reasoning here is going to be the same as answer choice C. We’re left with our correct answer, answer choice B.

55) The three elements listed were mentioned in our passage, but we don’t necessarily need the passage to answer this question. The answer is going to come exclusively from knowing about these three elements from our content. Capacitors and batteries can store energy. That leaves resistors. Resistors control the flow of current. We mentioned fixed and variable resistors in the passage, but neither can store energy. That means only options I and III store energy. The only answer choice that matches options I and III is answer choice C. 

56) We’re focusing on the variable resistor. We want to know the resistance at the beginning of the discharge process, when the battery is full. We’re going to revisit our passage to get specific details about our resistor and the voltage of the battery. Then we’re also going to use Ohm’s Law again to solve for resistance. Quick glance at our answer choices shows that our units are going to be in Ohms.

We said we’re going to use Ohm’s Law. Ohm’s law states that electric current is proportional to voltage, and inversely proportional to resistance. V=IR, voltage equals current times resistance. 

We want to know the voltage at the beginning of the discharge process. There are going to be 12 volts initially across the fully charged capacitor-the same as the battery voltage. Next, what is our current? It’s constant during the discharge time, meaning it’s the same at the beginning and throughout the discharge. Look at the Y-axis of the graph, it’s current. The value of current in our calculation is going to be 0.002 amps. We can plug these values into our equation and solve for resistance. 12 volts divided by 0.002 amps gives us 6000 ohms

This question was a math problem and we did no rounding or approximating. We solved for an exact value, so we can compare all of our answer choices at once. Our answer choices are not particularly close to one another. We can pick our correct answer, answer choice D.

 

Physics Question Pack: Questions 57-61

57) A few things we want to note: the block is moving at a constant speed, so there’s no acceleration. Net force is going to be zero on the block. The weight W is equal to the force exerted by the ground on the object which accounts for the vertical forces. Horizontally, we have frictional force F and T(cos θ) equal to one another. Why do we know they’re equal? Because we said the sum of the horizontal forces is zero. We can stick with answer choice D as our best answer. 

58) First thing we’ll do is recall the equation that relates pressure and density:

P = ρgh

g is going to be identical for both equations, but the density of mercury is 14x that of water, while the height of mercury, h, is ½ that of the container of water. The pressure of the mercury is therefore: (14)(1/2)(Pressure of water).

The pressure of mercury can be written as 7 x Pw. This question was a math problem and we did no rounding or approximating. Our answers are not very close to one another, and we solved for an exact value. We can compare all of our answer choices at once and pick the correct answer: Answer choice B.

59) We’re given an intensity level difference between Sounds A and B. We want to know how many times greater the intensity level of Sound B is compared to that of Sound A. 

A decibel is a ratio of the observed amplitude, or intensity level to a reference, which is 0 dB. The equation for intensity in dB is: dB = 10Log(IA/I0). In this specific case, we know we have an intensity level of 20 dB greater in Sound B. We can set this as our intensity in the equation and get:

20 dB = 10Log(IB/IA)

Solving for IB/IA gives us 102, or 100 times greater. This question was a math problem and we did no rounding or approximating. Our answers are not very close to one another, and we solved for an exact value. We can compare all of our answer choices at once and pick the correct answer: Answer choice D.

60) Let’s think about what we know about Doppler effect and the variables involved:

We can use the above equation and consider all four of our answer choices at once. We do need to know the speed of sound in the medium and the frequencies of the emitted and observed sounds. The distance between the source and observer is not required to determine the relative speed of approach of a sound source. Answer choice D is our correct answer.

61) First thing we do whenever we see light going through multiple mediums is think of Snell’s Law.

Snell’s Law: n1sinθ1 = n2sinθ2 where n1 will be the index of refraction of air (1.0) and n2 is the index of refraction of the glass (which we’re solving for). We can write out our equation and isolate for our variable in question, as well as add the angles given to us in the question stem: 

n2= n1 x sinθ1/sinθ2

n2=n1 x sin(45°)/sin(30°) we also know n1 is 1.0 for air, so n2 = sin(45°)/sin(30°)

We can plug in the values given in the question stem for the sine of these angels to solve for n2. We get a value of 1.414. This question was a math problem and we did no rounding or approximating. We can compare all of our answer choices at once and pick the correct answer: Answer choice B.

 

Physics Question Pack: Passage 10

62) To answer this question, we’re going to need the figure in our passage, but we’re also going to have to understand angles of reflection and incidence. We’ll use the law of reflection.

We have an excerpt from our passage here. We have Figure 1 below. We’re told some of the light returning from the array is reflected, at a right angle, to a viewing screen by the beam splitter. The law of reflection for the case of our question states that when a ray of light reflects off a surface, the angle of incidence is equal to the angle of reflection. Both angles are measured from the path of the light to the normal. Because the light is reflected at a perfect right angle, the reflected ray (we can call it theta 1) would have to be at 45 degrees relative to the normal, but also equal to the angle of incidence (which we can call theta 2). So to find the angle of the beam splitter relative to the incident ray. We set theta 1 + theta 2=90 degrees. 2 theta equals 90 degrees because we said theta 1 and theta 2 are equal. Solving for theta, we get an angle of 45 degrees

This question was a math problem and we did no rounding or approximating. Our angles are given in intervals of 15 degrees, so our answers are not very close to one another. We solved for an exact value, so we can compare all of our answer choices at once. Answer choice C is our value of theta: 45 degrees

63) This is something that was touched on in the passage, but we’re not going to need the passage to show the path for a ray of light passing through a retroflecting bead. We’re going to have to know about normal lines, and we’ll have to know how light rays are reflected. We want to pay attention to the fact that the index of refraction of the bead is greater than that of air according to the question stem. 

First thing we want to reiterate is the fact that the index of refraction of the bead is greater than that of air. That means the ray should bend toward the normal. It’s going from a lower refractive index to a higher refractive index. The normal line is the line perpendicular to the surface of the bead where the light is entering. I’m going to draw out a visual here to hopefully help us go through what’s happening a bit more clearly: 

If we draw the bead as a circle with a light entering, we can draw the normal line as a line that’s perpendicular to the point of entry. As the ray enters the bead, going to a higher refractive index and the ray is bending toward the normal line. It bends slightly left in this case. We’re also told in the passage that light returns on a path that’s no longer than the diameter of the bead. So, we don’t expect it to deviate much from a straight line toward the further end of the bead, reflecting, and exiting the bead. We can draw this out and also draw the normal again. When the ray goes from the bead to air, it now bends away from the normal. This is consistent with our passage because the light ray enters the bead, and is reflected back. 

  1. Answer choice A shows the ray not bending at all. Instead, we should expect it to bend toward the normal like our drawing. Another quick thing we mentioned, light returns on a path that’s no longer than the diameter of the bead. This entire reflection looks to be too convoluted and possibly longer than that diameter. The light ray does enter and reflect back though. 
  2. Answer choice B shows the ray bending, but this shows it bending away from the normal, not toward it like our prediction. We can eliminate this answer choice. 
  3. Answer choice C looks a lot like our breakdown, only it doesn’t show the normal like we drew. Our ray enters the bead, bends toward the normal which we can see in our prediction, it reflects and exits the bead. Let’s keep this answer choice though because it’s superior to A and B.
  4. Answer choice D shows the ray bending away from the normal again. Be aware of the arrows in this answer choice. All of the other answer choices have a ray entering the bead on the right side of the figure. This one has the ray entering the bead on the left side of the figure. We can eliminate answer choice D as well. We’re left with our correct figure, answer choice C.

64) We’ll need to relate that index of refraction to the ratio of the angles, theta. Another question that’s tangentially related to the topics that are covered in the passage, but we can use Snell’s law to answer this question. Snell’s law relates the angles of incidence and refraction. This question is simply asking to relate the two angles 

Snell’s law states that the ratio of the sines of the angles of incidence and refraction is equivalent to the reciprocal of the ratio of the indices of refraction:

Sin theta A / sin theta B  =  nB/nA

That means if we change the ratio of the indices of refraction, we’re also going to be changing the ratio of the angles as well. The light’s going to be refracted to a different place-our image is going to move.

We also know that if we have a larger index of refraction, light bends more toward the normal. We saw this earlier in the passage actually. If you change the index of refraction, the projected image is going to move. The light bends more toward the normal.

  1. It will move. This matches our breakdown exactly. We used Snell’s Law to relate the relationship between the angles of incidence and refraction. That showed us our projected image moves.
  2. It will become larger. Changing the index of refraction and changing our angle isn’t going to change the size of the image. We can only do this my changing the pinhole itself.
  3. It will become smaller. Same reasoning here as we used in answer choice B. We can eliminate both answer choice B and answer choice C.
  4. It will become more clear. This answer is out of scope. Our image is not going to get clearer by changing the index of refraction. This doesn’t address what’s happening in the question stem. We’re left with our correct answer, answer choice A. The pinhole image will move.

65) We’re going to need to reference the passage for information about the retroflecting beads and relating to diameter. We’ll also need to use our general knowledge to relate the variables given in our question stem to wavelength to ultimately solve the question. 

Here we have an excerpt from the passage, it says each light ray is returned on a path no farther than the diameter of a bead from the source ray. Let’s use that as the distance traveled by the light ray. First thing we want to do though, is solve for wavelength. What’s our main equation when dealing with wavelengths, frequency, and speed?

C=frequency x wavelength. Our C is our speed of light, 3 x 10^8 m/s. Our frequency is given as 10^15 hertz. When dividing exponents on the same base, we’re going to subtract the bottom exponent from the top exponent on our base 10. So, our 10^8 divided by 10^15 means we leave our base 10 the same, and we subtract the bottom exponent from the top exponent, leaving 10^-7. Our 3.0 doesn’t change, so solving for wavelength gives us. 3.0 x 10^-7 meters.

We want to know the number of waves that will fit in our diameter, so we divide once again. We divide the diameter by wavelength. Let’s turn this into two fractions: 5.0/3.0 and (10^-5)/(10^-7). First part is approximately 1.67. Second, we divide exponents the same way. Negative 5 minus negative 7 gives us a positive two. Combine and we get our number of waves to be 1.67 x 10^2.

This question was a math problem and we did a bit of approximating. We approximated 1 and 2/3 to be 1.67, but the exponent on our 10 remains as 2. Glancing at our answer choices, our answers are all roughly one hundredfold away from one another in magnitude, so we can use our approximation without having to solve for a more precise number.

We can compare all of our answer choices to our calculated value at once. Answer choice B is closest to this calculated value, and even ends up matching our value after rounding.

 

Physics Question Pack: Passage 11

66) We’re asked to describe the direction of motion in the presence of an external magnetic field, for all three of the types of radioactive decay discussed in the passage. We’re going to need to know the properties of each of our three rays, and that’s going to help us determine the direction of motion.

We have an external magnetic field, which may influence the direction of our alpha, beta, and gamma rays. The magnetic field is going to exert a force that’s given by the equation F=qE. Our q is the electric charge, and we have electric field E. So, we know force is proportional to electric charge, so we can use that. What did we say about charges in our three situations? Alpha particles carry a positive charge, beta particles carry a negative charge, and gamma rays are neutral. That means we’re expecting the force on the alpha and beta rays to be in opposite directions. There’s no force on the gamma rays, because gamma rays are neutral, so there’s no electric charge.

  1. They all travel straight. This doesn’t sound like our breakdown of the question. We said alpha and gamma rays will feel a force, but we do expect the gamma rays to travel straight.
  2. They are all bent in the same direction. This also contradicts our breakdown. We said alpha and beta rays will feel an opposite force because of their opposite charges. Gamma rays won’t feel that same force and won’t bend. This is slightly better than answer choice A though, which didn’t have any of our rays bending. We can now eliminate answer choice A.
  3. Gamma rays travel straight; alpha and beta rays are bent in the same direction. This is getting closer to our breakdown. We said alpha and beta rays will feel an opposite force because of their opposite charges. They won’t bend in the same direction. Gamma rays won’t feel that same force and won’t bend, so that part of this answer is consistent with our breakdown. This is better than answer choices A and B. 
  4. Gamma rays travel straight; alpha and beta rays are bent in opposite directions. Bingo, this matches our breakdown exactly. We said alpha and beta rays will feel an opposite force because of their opposite charges. Gamma rays won’t feel that same force and won’t bend. That’s exactly what answer choice D is saying. We’re sticking with answer choice D as our correct answer. 

67) We’re given a beginning and intermediate species in a series, and we need to determine the possible sequence of particle emissions between the two.

We can look at this question two ways. We’re told the specifics about the three types of radioactive decay in the passage, but the information we need is also from knowing the content. In this case, we can use our general knowledge, but also reference our passage. During the exam, you want to take the most efficient method. We want to be thorough here, however, so in this case we’ll do a hybrid of the two. We’ll have to know about atomic masses, atomic numbers, and obviously the three types of decay. Quick glance at our answer choices show we’re dealing with only alpha and beta decay, not gamma rays. Also, each answer choice contains at least one alpha and one beta in its sequence.

We have the equations given to us in the passage right below our question stem. First thing we want to note in the question stem is the difference in atomic mass and electric charge. The atomic mass of our intermediate product is 4 units smaller than the beginning species. There is no net change in electric charge. 

Now below: Alpha particles are equivalent to a helium (He) nuclei, made of two protons and 2 neutrons. Each nucleon has a mass of 1 atomic mass unit. What’s the overall charge? +2 because of the protons. 

Beta radiation is the streams of electrons. Electrons have a -1 charge, but negligible mass. So, let’s write out these changes in each one: 

Alpha decay is going to increase the atomic mass unit by 4 and charge by +2. Beta decay is going to decrease electric charge -1 and not affect mass.

Let’s start with alpha decay. Following that reaction, we’ll have a mass of 234, just like our end goal, but we will have a +2 charge relative to our desired intermediate. We can follow with two beta emissions which will keep atomic mass the same, but bring charge back to the desired 92 once again. So, we’re predicted an answer choice with a sequence that includes two beta emissions, and one alpha.

  1. Beta, beta, beta, beta We can consider this an incorrect answer choice. Four beta emissions would give us a mass of 238 and charge of 88. Neither is consistent with what’s mentioned in the question stem.
  2. Alpha, beta, beta, beta We can consider this an incorrect answer choice. Alpha emission gives us the proper atomic mass, but we have three beta emissions which would put our charge at 91 instead of our desired 92.
  3. Alpha, alpha, beta, beta We can also consider this an incorrect answer choice. Two alpha emissions give us an atomic mass of 230. Two alpha emissions and two beta emissions give us an electric charge of 94.
  4. Alpha, beta beta This matches our breakdown exactly. Alpha emission gives us an atomic mass of 234. Single alpha emission and two beta emissions gives us a charge of 92. Answer choice D is going to be our best answer choice.

68) To answer this question, we’re going to need to know how half-lives work. We’re simply relating the half-life and activity levels given in the question stem to one another.  

Half-life is the time taken for the activity of a given amount of a radioactive substance to decay to half of its initial value. 

We’re told we have a sample with activity of 100 millicuries after 24 hours. We’re also told half-life is 15 hours, so we’re going to work backward from our 100 millicuries after 24 hours. 

We can go a half-life back, which would leave us at 9 hours, but double the 100 millicuries. After 9 hours, we can say activity is 200 millicuries. Now, another 15 hours would double the millicuries to 400 millicuries. But we don’t wait an additional 15 hours, we’re only dealing with 9. So that means activity will be somewhere between 200 and 400 millicuries. 24 hours is between 1 and 2 half-lives

So, our estimated activity is going to reflect that by being between 200 and 400 millicuries. Let’s see if our estimate is good enough for now, or if we should dig deeper into an exact value. The only answer choice that falls between 200 and 400 millicuries is answer choice B: 300 millicuries.

69) This is very similar to a previous question we went over in this question set, so we shouldn’t have any trouble reasoning this out. We’re dealing with alpha and beta decay once again, so we can pull up our visual.

Alpha decay is going to decrease the atomic mass units by +4 and charge by +2. Beta decay is going to decrease electric charge -1 and not affect mass.

We have 6 alpha emissions and 4 beta emissions. The sum of the 6 alpha emissions is going to be 24 atomic mass units and a positive 12 charge. The sum of the 4 beta emissions is a net decrease in charge of -4. The overall change we’ll see from the beginning of the series to the end 24 atomic mass units, and a net change in charge of +8. The new atomic mass units should be 232 minus 24, which is 208. The new electric charge is going to be 90 minus 8, which is 82. So our prediction is going to be the element lead. We have an atomic mass of 208.

We found an exact atomic mass and element in this situation, so we can actually compare all of our answer choices at once. The only answer choice that matches our breakdown is answer choice A. Answer choices B-D either have incorrect atomic masses or show the incorrect element. 

 

Physics Question Pack: Passage 12

70) To answer this question, we want to compare the densities of the three balls given in the passage. We can go back to the passage to find the necessary information, then we’ll use our general knowledge to actually work out the densities.

As you’re practicing, I don’t recommend going back to the passage often, unless it’s for specific details. The reason I reference the passage fairly often in these solutions is so you can clearly see my thinking and what I’m referencing from the passage. 

Here, I want to focus on two things: It says the balls all have the same volume, then it says Ball 1 floats in the water with a part of it above the surface, Ball 2 is suspended in the water, and Ball 3 rests on the bottom of the tank. So, all three balls have the same volume, but are at different levels, and that’s because of density differences. Density is the mass per unit volume of an object. Because volume is identical, Ball 3 has the highest mass because it sank to the bottom of the water tank and is resting on the tank floor. Ball 2 is suspended in the water, so it has the 2nd highest density. Ball 1 floats in the water, meaning it has the lowest density.

  1. D1 < D2 < D3. Answer choice A says the density of D1 is less than D2 is less than D3. This matches our breakdown exactly. Ball 3 has the highest density, it was the ball that sank to the bottom of the tank. Ball 2 had the 2nd highest density, it was suspended in the water. Ball 1 has the smallest density, it was floating on top of the water.
  2. D1 < D2 = D3. Answer choice B says the density of D1 is less than D2, which equal to D3. This answer choice is a little tricky. The very first part of this answer choice is correct. The density of D1 is less than D2. However, think back to the passage. The author explicitly says Ball 2 is suspended, but Ball 3 is resting on the floor. That tells us Ball 3 is denser than Ball 2. Those densities won’t be equal
  3. D1 = D2 < D3. Answer choice C says the density of D1 is equal to that of D2, which is less than D3. This is inconsistent with our breakdown of the question. 
  4. D1 = D2 > D3. Answer choice D says the density of D1 is equal to that of D2, which is greater than D3. Again, this is inconsistent with our breakdown of the question. We can stick with answer choice A as our best answer. 

71) Keep in mind here, we’re trying to find the approximate pressure difference between two balls: Ball 2 and Ball 3. We’re told a distance difference between the balls. We’re going to need the density of the water in the tank, and the value of g to solve for our answer. Although some of this should be general knowledge, we can get those from the passage to be thorough. We’ll ultimately solve for the difference in pressure using general knowledge, and our answer is given in N/m2.  

We have the major detail we needed from our passage right above, which is the density of the water. We’re told in the question stem that Ball 2 is 20 centimeters above Ball 3. We’re looking for an approximate answer according to our question stem, and we can assume the water is static around the balls. In static fluids, pressure increases with depth. We need to determine the overall effect of this change in depth. The change in pressure is going to be equal to density, times g, times change in depth. We often just deal with rho x g x h. In this case, we want to know about the difference in pressures, so let’s plug in our values:

Our change in pressure is going to be equal to a rho of 1000 kg/m^3 (our density) multiplied by g (9.8 m/s^2) multiplied by the difference in depth (20 centimeters).

We want to keep our units consistent, so let’s use dimensional analysis. We use a factor, there is 1 meter per 100 centimeters) This gives us a difference in depth of 0.2 meters. When we solve for our difference in pressure, we get a numerical value of 1960 and units, we’re multiplying a distance, and a mass (kg) over volume (meters3) by units for acceleration. In other words, we have Newtons from our mass and acceleration. And that is divided by the meters squared that does not cancel out. We have a calculated value of 1960 N/m2.

We have 4 answers that are all rounded to the nearest whole number and given in scientific notation. We can rewrite our calculated value as 1.96 x 10^3 N/m2, or 2.0 x 10^3 N/m2

Because we solved a math problem to get to this answer, we can compare all of our answer choices at once. None of the answer choices are extremely close to one another, they’re all either tenfold or even hundredfold away from one another. So, our slight rounding is consistent with answer choice C, 2.0 x 10^3 N/m2

72) We’re given the density of Ball 1, and we’re going to need to compare with the density of water to determine how much of the ball will be below the surface of the water and how much will be above. We can answer this question using the information given in our question stem and comparing to the density of water given in the passage. We’re also going to need to use buoyancy and Archimedes’ principle from our content outline. Looking at our answer choices, we want our answer in terms of fractions.

We said we’re dealing with buoyancy and we have the density of Ball 1 as 8.0 x 10^2 kg/m^3, which was given to us in the passage. When we’re dealing with buoyancy. The fraction of an object’s volume that’s submerged is given by the ratio of its average density to that of the fluid: ρobject/ρfluid. That means we’re going to be dividing the density of ball 1 by the density of the fluid, the water.

How do we divide two numbers in scientific notation? We’re going to turn this into two fractions instead of trying to solve the entire problem at once. So instead this becomes:

(8.0/1.0) and (10^2/10^3)

We can still round to simplify our numbers, so our first fraction becomes 8.0 divided by 1.0. First division problem ends up being 8.0. We move on to our 2nd fraction. When dividing exponents on the same base, we’re going to subtract the bottom exponent from the top exponent on our base 10. So our 10^2 divided by 10^3 means we leave our base 10 the same, and we subtract the bottom exponent from the top exponent, leaving 10^-1. So 8.0 x 10^-1 is equal to 0.8 or 4/5. That means 80% of the ball is submerged. The remaining 20% of the ball, or 1/5 is above the surface of the water.

Similar to our last question, we solved for an exact value in this math problem. Our answer is going to correspond to answer choice D: 1/5.

Another thing we want to notice here is answer choice A. Why might we have picked correct answer choice A? If we accidentally answered how much of the ball was submerged, not how much was above the surface of the water. This shows the importance of reading the question carefully! 

73) We can answer this question using details given in our question stem. And we refer back to our passage as well for Bernoulli’s equation which gives us the relationship between pressure and velocity in fluids.

These questions are fairly common where we have water coming out of a punctured surface. We can use the equation for projectile motion as the final velocity in Benoulli’s equation. The thinking here is, we know energy is conserved, so we can set potential energy equal to kinetic energy:

mgh=1/2mv^2

Cancel Mass & multiply both sides by 2:

2gh=v^2

square root both sides

root 2gh=v

Let’s plug in the values we’ve been given. G is 9.8 m/s^2 and h is 5 meters. We can round “g” to 10, and we have v is equal to 10 m/s.

Another question where we have a math problem and we did a little bit of approximating by rounding “g” up. What that means is we overestimated our answer slightly. What we also want to notice is our question stem asks that we find the approximate speed of the water. You’re expected to approximate on the MCAT. Going through our answer choices, answer choice A matches our calculated value. But we said we rounded, is that going to be an issue? No-because we said that we overestimated our answer slightly. Our actual answer would be slightly smaller than 10 m/s which is already the smallest answer choice listed. This gives us confidence in our answer choice, so we can eliminate answer choices B-D. 

 

Physics Question Pack: Questions 74-78

74) The work done on a system by a constant force is the product of the component of the force in the direction of motion times the distance through which the force acts. Easiest way to look at this is looking at the mass M. The force and distance in the context of M is going to be equal to the work done by the force F:

(mass of M: 4-kg) x (g: ~10 m/s2) x (distance: 5 m) = 200 kg*m2/s2 = 200 J

Note the common sine and cosine values are given to you as distractors here. It’s common you’ll get extra information in the question stem that you don’t end up using to answer the question. Make sure you are able to convert units and you understand the relationship between variables. Just being able to do that can get you an extra 2-3 questions correct!

This was a math problem, and we did minimal rounding or approximating. We can pick our correct answer from the choices listed here, that’s answer choice D.

75) To answer this question, we’ll have to look at all four of the variables listed in the question stem and determine which property will remain the same before and after reflection. Big key here is it’s bouncing off an imperfect reflector moving toward the source. If you’re not strong with sound content, make sure to check the content outline on our website: Jackwestin.com/resources/mcat-content/sound. 

  1. Speed. Regardless of bouncing off the reflector, the medium here isn’t going to change, so the speed of sound does not change. 
  2. Intensity. This is where we want to pay close attention to the question stem. The test-maker says the sound bounces off an imperfect reflector. That means the reflector absorbs some energy, and intensity will be affected. Answer choice A remains the best option. 
  3. Frequency. This ties into answer choice B. There’s a good chance the reflector absorbs some energy. In other words, the sound wave loses energy and affects frequency which is related to energy. Frequency is affected, while speed is not, so we can stick with answer choice A as our best option.
  4. Wavelength. I mentioned in answer choice C, frequency is affected. Affecting frequency also affects wavelength. We can eliminate answer choice D. Answer choice A is our best option. 

76) This question boils down to visualizing the scenario in the question stem. We have static equilibrium, so the net force is going to be zero (the test-maker says we have equilibrium). AAMC actually brings up a great example of a see-saw. Imagine a force 0.6 m to the left of the center of the see-saw, and a force 0.4 m to the right of the see-saw. We’ll have to solve for the torque on both sides: (Force) x (distance) x (sin θ). In this case, both sides we have a downward force, so we won’t have to worry as much about sin θ. Let’s plug some values into the torque equation:

Left side we have: Force x (0.6 m)

Right side we can substitute mass x acceleration for force, so we have: (10–7 kg x 10 m/s2) x (0.4 m).

Multiply out the right side and set both sides equal to one another because we’re in equilibrium:

(0.6 m) F = (4.0 x 10–7 N/m)

F=6.67 x 10–7 N

We did a small bit of rounding here when dealing with the constant “g”. It’s typically easy to round up g from 9.8 to 10.0 m/s2 for any MCAT math, which is what I did here. That means our approximated, calculated force is slightly larger than the force in reality. Looking at our answer choices, only one answer choice matches what we’re looking for, that’s answer choice B.

77) A ray of light changes direction when it passes from one medium to another. The angle of incidence (towards the surface, the incidence ray) and angle of refraction (of the refracted ray) are measured relative to a perpendicular (normal) to the surface at the point where the light ray crosses it. In the context of this question, the angle of incidence is the same as the known angle of reflection.

For a ray at a given incident angle, a large change in speed causes a large change in direction, and thus a large change in refraction angle. This is expressed in Snell’s Law:

n1sinθ1 = n2sinθ2

In the image below, the angle of incidence and the angle of refraction are respectively θ1 and θ2. In this specific question, we need to know θ2 to solve for n2 (refractive index of the liquid) in Snell’s law.

  1. Angle of incidence. I mentioned this is the same as the known angle of reflection. 
  2. Angle of refraction. This is the θ2 we’re looking to solve for to be able to find n2.
  3. Refractive index of air. This is known information.
  4. Wavelength of the light. This does not help us find the angle of refraction in this case.

78) This question just comes down to knowing the equation for an electric field and how different variables are related to one another. Near a point charge, the strength of an electric field can be calculated as E = kq/r2, where k is a constant, q is the charge we’re interested in for this question, and r is the distance from the point charge. Let’s focus on E = kq/r2. The question stem mentions we triple the charge q. By doing so, we also triple the other side of the equation; we triple E. The only answer choice consistent with this is the 1:3 ratio shown in answer choice B. 

 

Physics Question Pack: Passage 13

79) We want to explain how the work done by friction as it travels down the ramp is affected by increasing the initial speed of the box. We’re going to have to determine the work done by friction on the box and see which variables affect this work variable. 

Work is the transfer of energy by a force acting on an object as it’s displaced. We’re concerned with the work done by friction on the box. The work done on the box by a constant force (in this case friction). That’s going to be the product of the component of the force in the direction of motion times the distance through which the force acts. That can be written out as:

W=Fdcos(θ)

W is work, F is the magnitude of the force on the system, d is the magnitude of the displacement of the system, and θ is the angle between the force vector F and the displacement vector d. We’re asked the change in work given an increase in initial speed. We don’t see speed in any of these variables, and it’s not going to affect this equation or the amount of work. That means we want an answer that says changing initial speed won’t affect work. There’s only one answer choice here that reflects that: Answer choice B.

80) The answer is ultimately going to come from our general knowledge. But it’s also going to be a passage-based question. Figure 2 compared velocity and time, so we need to go back to our figure and see what the specific area mentioned in the question stem represents. Quick glance at our answer choices shows that we’re going to find a physical quantity related to the box in the passage.

Velocity is a vector quantity. It denotes the rate of change of position with respect to time. We can also look at units for our given quantities in Figure 2. The y axis is velocity in meters/second. The x-axis is time in seconds. When we multiply both of these variables to get that area, we have a unit in meters. This means the area in that quadrilateral is going to be in meters. It’s going to have to deal with position, or distance.

  1. Average speed of the box. The average speed of the box would come from analyzing the y-axis only, not the area of the quadrilateral. This contradicts our breakdown.
  2. Average acceleration of the box. If instead of multiplying our variables, we divided our velocity by time, then we would be finding acceleration. That wasn’t the case when we were finding the area under the quadrilateral. This also contradicts our breakdown.
  3. Distance traveled by the box. This matches our breakdown of the question. We said the area represented by the quadrilateral is distance, usually given in meters. This is our best option of the three answer choices we’ve gone through. 
  4. Work done on the box. We actually went over work in our previous question: Work=Fdcos(theta)

W is work, F is the magnitude of the force on the system, d is the magnitude of the displacement of the system, and θ is the angle between the force vector F and the displacement vector d. Speed doesn’t affect work, so it’s not represented by the quadrilateral. We can eliminate answer choice D. We’re still going to stick with our correct answer, answer choice C

81) In other words, we want to find an expression that represents acceleration from Point B to Point C. We’ll reference our passage again to see what each of the variables in our question stem represent. Quick glance at our answer choices tells us that we want to find an answer in terms of g, I, and the coefficient of friction. Those three variables are in every answer choice. 

As you’re practicing, I don’t recommend going back to the passage often, unless it’s for specific details. The reason I reference the passage fairly often in these solutions is so you can clearly see my thinking and what I’m referencing from the passage. 

We have our figure above. We can see Point B and Point C in Figure 1. We want to find the expression for acceleration as we move from Point B to Point C. The passage mentions the box reaches Point B, which is at the bottom of the ramp. And after reaching point B, the box travels on a flat surface and stops at point C. Note that ‘I’ is the unit vector in the x-direction. Let’s solve for the acceleration vector of the box. We know we have friction that opposes the motion of the box as it travels from Point B to Point C. 

When the box is moving, the magnitude of kinetic friction (fk) is = -μkhNi. N is the normal force of the surface on the box. In this case, we want our answer in terms of “g” also. We know upward normal force is equal to downward gravity force, so we can substitute mg as the normal. So, our force is equal to -μkhmgi. Force is equal to ma, which allows us to cancel mass on both sides and solve for our desired acceleration. Acceleration is therefore equal to -μkhgi.

We solved for an exact value for our acceleration vector expression. We can compare all of our answer choices at once to find the expression that matches this expression. Answer choice D matches what we came up with in our breakdown.

82) We want to know how work changes if the angle of the ramp changes from 45 degrees to an increased angle. We can reference our passage again to see Figure 1 and the original angle of the ramp. We’re have to understand the relationship between work and the angle of the ramp from our general knowledge.

We have our figure here, and we’re told the ramp is designed at a 45-degree angle. Remember our question wants to know the effect on work if the angle is increased. The formula for work which we’ve used a few times in this question set already is:

Work=Fdcos(θ)

W is work, F is the magnitude of the force on the system, d is the magnitude of the displacement of the system, and θ is the angle between the force vector F and the displacement vector d. We’re going to also relate this to the normal force and friction as the angle is changed.

Friction’s maximum is proportional to the normal force, which is mg x COS(θ). Cosine is highest when angle is zero, lowest when angle is 90 degrees. So, increasing the angle decreases the normal force, which decreases the maximum possible frictional force, and ultimately decreasing work.

Let’s visualize whether this makes sense. If we increase the angle to 90 degrees, what happens? The box falls straight down, with no friction whatsoever. Because COS(90) is zero. So, we’re expecting work will ultimately decrease.

  1. W will decrease, because the normal force to the surface of the ramp will decrease. Right away we can tell this matches our breakdown of the question. We said increasing the angle of the ramp to the horizontal will decrease work because the normal force decreases between the box and the ramp will decrease.
  2. W will not change, because the coefficient of friction between the box and the ramp will not change. We said work will decrease because of the change in normal force. The reasoning here is correct, but the first part of the answer is a contradiction. Answer choice A remains the best option so far.
  3. W will not change, because the gravitational force is always constant and the length of the ramp is not changed. The first part of the answer is a contradiction, even though the reasoning in the second half is accurate-even though we didn’t discuss the length of the ramp. We can eliminate answer choice C.
  4. W will increase, because the height from which the box falls increases. This is the opposite of our breakdown. We said W decreases, not increases, so we can eliminate answer choice D. We’re left with our correct answer, answer choice A.

 

Physics Question Pack: Passage 14

83) We can answer this question using our general knowledge. We know the relationship between velocity of blood flow and the total cross-sectional area of the artery. We simply have to use this relationship and explain what happens when cross-sectional area is halved. 

The velocity of blood flow varies inversely with the total cross-sectional area of blood vessels. The equation is given by F=VxA where F is flow, v is velocity, and A is the cross-sectional area. So as the total cross-sectional area of the vessel increases, the velocity of flow decreases. As the total cross-sectional area of the vessel decrease, like in our question here, the velocity increases. So right away we know the speed is going to be faster. Next, we know we have an inverse relationship. If the cross-sectional area is halved, that means the speed of the blood is going to be doubled.

  1. it’s ¼ as fast. This contradicts the breakdown of our question and what we said about the relationship between cross-sectional area and speed. If the cross-sectional area decreases, speed is going to increase. Let’s keep comparing.
  2. it’s ½ as fast. Another contradiction. We’re expecting the speed of blood flow to increase. And just a quick aside, notice how important it is just to know the relationship between variables. Just by knowing our two variables are inversely related, we can eliminate two answer choice right away. A 50/50 shot is much better than a 1 in 4 shot at getting the right answer. Let’s keep comparing.
  3. it’s 2 times as fast. This matches our breakdown of the question. We’re expecting speed is doubled, which is exactly what this answer choice says. This is our best option so far. 
  4. it’s 4 times as fast. This contradicts the breakdown of the question. We do have an increase in speed, but it’s only doubled, not 4 times as fast. Answer choice C remains the best answer choice.

84) We can answer this question using our passage, and then ultimately using our knowledge of the variables discussed in the passage to get to our answer.

We’re doubling the strength of the magnet, meaning our magnetic field (B) is doubled. We know magnetic force and magnetic field are related by Fm=q x B. 

If we have double magnetic force, we can compare with electric force. We’re told at equilibrium, Fe is equal to Fm. So, let’s break all of this down, we have an increase in magnetic field. That, in turn, means increase in magnetic force and ultimately electric force. Because we’re told an equilibrium condition occurs when Fe=Fm, we’re also expecting Fe, the electric force, increases. We’re told in the passage, the voltage across the artery (V) is measured by the meter in figure 1 and is equal to Ed. So an increase in electric field means an increase in voltage across the artery also. 

  1. The electric field will reverse polarity. This isn’t discussed in the passage, nor do we expect this to happen with a stronger magnet. This answer choice is irrelevant and contradicts our breakdown of the question.
  2. The electric field will decrease. We’re expecting an increase in electric field, not a decrease. This also contradicts our breakdown of the question.
  3. The voltage will increase. We said voltage is going to increase as we have the electric field increase. That means we can hold on to this answer choice because it’s our best answer choice so far.
  4. Blood will flow faster. We’re not expecting blood is going to flow faster. We’re given an equation for the velocity that relates magnetic field and voltage. But we’re expecting voltage and magnetic field both increase, and not one or the other. Velocity itself isn’t going to be changed so we can eliminate answer choice D. We’re left with our correct answer, answer choice C.

85) We want to know the volume flow rate of blood given a specific velocity and diameter. We’re able to solve this question using our general knowledge. We touched on this in a previous question actually where we related speed and cross-sectional area. We’ll be doing the same thing here, only with the specific values we’ve been given in the question stem. Quick glance at our answer choices shows we’re dealing with numbers in scientific notation, and they all have the same units: m3/s.

We said in our previous question that the velocity of blood flow varies inversely with the total cross-sectional area of blood vessels. The equation is given by F=VxA where F is flow, v is velocity, and A is the cross-sectional area. We’re given velocity and diameter, and we need to solve for flow. This means we want to relate diameter and cross-sectional area. How do we do that? We know the formula for the area of a circle is pi r squared. 

However, we have the diameter, not the radius. Let’s convert. We said A = πr squared. We know radius is ½ of diameter, so we can substitute: A=π((1/2)d) squared. So A = (π/4) x d squared.

We can plug in numbers and multiply.

We have velocity as 0.2 m/s multiplied by the area we just solved for, and we can plug in diameter of 1.0 x 10^-2 meters. We can rewrite velocity in scientific notation as 2.0 x 10^-1 m/s. We moved the decimal point one place to the right, so we subtract one from the exponent on our base 10 (mnemonic LARS). Next, let’s deal with our diameter squared. If we’re raising an exponent to a different power, multiply the exponents. So we have:

2.0 x 10^-1 m/s x (pi/4) x (1.0 x 10^-4). 

Multiply the integers together first, then we add exponents while keeping the same base. So we have 0.5pi x 10^-5 m^3/s.

Or 5.0 x 10^-6 using LARS once again.

This is a math problem that took a few steps to convert units and manipulate exponents. But ultimately, we didn’t need to round or approximate to solve for our final answer, so we can compare all of our answer choices at once. Answer choice B matches the breakdown of our question.

86) We’re able to solve this question using our general knowledge. We want to relate direction of the magnetic force to the direction of velocity and magnetic field. We can do that using the right-hand rule and what we know about force direction. 

Magnetic force is perpendicular to magnetic field and charge velocity. We can use the right-hand rule to visualize the direction of magnetic force, but in any case, it’ll be perpendicular to both velocity and magnetic field. 

What’s the right-hand rule? To determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v (velocity). The fingers in the direction of B (magnetic field vector), and the perpendicular to the palm points in the direction of F.

  1. Parallel to both the direction of v and the direction of B. This is the opposite of our breakdown of the question. We said that both v and B are perpendicular to the direction of the magnetic force on the ion, not parallel.
  2. Parallel to the direction of v and perpendicular to the direction of B. This is better than answer choice A, but not quite the same as our breakdown. We want the answer choice to say perpendicular, not parallel. 
  3. Perpendicular to the direction of v and parallel to the direction of B. The reasoning here is going to be the same as answer choice B. These are both better than answer choice A, but not quite the same as our breakdown. We want both to say perpendicular, not parallel.
  4. Perpendicular to both the direction of v and the direction of B. This is the answer choice we were looking for. The direction of the magnetic force is perpendicular to both the direction of v and the direction of B. This answer choice is better and answer choices B and C. We’re left with our best answer, answer choice D.

 

Physics Question Pack: Passage 15

87) We can reference the apparatus in the passage, but we’re ultimately going to use our general knowledge to explain the movement happening in the apparatus. We know energy is conserved, so we’re going to explain the energy of the sliding block and where it came from. Quick glance at our answer choices shows our ultimate kinetic energy is going to come from an initial kinetic energy or gravitational potential energy.

We have part of our figure here that shows our apparatus. We know that as mass is added to the hook, we eventually reach and overcome the threshold, and the block begins to slide. The kinetic energy of this sliding block had to come from somewhere. There was no movement initially in the apparatus, so the energy comes from potential energy, not kinetic energy. Potential energy is the energy difference between the energy of an object in a given position and its energy at a reference position. So, after we reach the threshold mass, the mass on the hook begins to fall toward the floor and the block slides across the wooden board. That potential energy of the mass is decreasing, but the kinetic energy of the block is increasing. We can say the kinetic energy of the sliding blocks comes from the gravitational potential energy of the mass.

  1. kinetic energy of the string. This contradicts our breakdown of the question. We went through an experiment in the passage and there was no initial kinetic energy. The string is simply moving along with the mass and the wooden block. There’s no initial kinetic energy; let’s keep comparing.
  2. kinetic energy of the board. This also contradicts our experiment from the passage. We said there is no initial kinetic energy, only potential energy. Let’s keep comparing and trying to find a superior answer choice.
  3. gravitational potential energy of the block. This answer deals with gravitational potential energy, so we’re getting closer to our breakdown. But that potential energy comes in the form of our mass, not the block. The block is simply resting on the wooden board and there is static friction preventing the block from sliding as smaller masses are added to the apparatus. This doesn’t quite match our breakdown, but we can still eliminate answer choices A and B. Both of those mention kinetic energy, and both are incorrect.
  4. gravitational potential energy of the mass on the hook. This matches our breakdown of the question. The mass on the hook has potential energy because it’s a certain height above the floor. As the mass falls, this height becomes less and the energy is transformed into kinetic energy. That’s where the kinetic energy of the sliding block comes in. So, we can now eliminate answer choice C and keep our correct answer, answer choice D.

88) We can reference Table 1 given to us in the passage and see what we can find about the relationship between static friction and base area.

We have Table 1 above, and we’re analyzing the relationship between static friction and base area. 2nd column lists all of the block base areas, and 4th column lists the threshold mass. What do we know about the threshold mass? From set to set, these numbers don’t vary, but within the set they do based on the block material. The wooden blocks all have the same threshold mass, the stone blocks have the same threshold masses, and the steel blocks all have identical threshold masses. And why do we have this discrepancy? It’s because of the varying coefficients of friction.

Now, the question we’re concerned with, is the relationship between static friction and base area. We see base area is changing, but does the threshold mass change? No, it doesn’t. Besides some very small changes that can be attributed to human error and rounding errors, the threshold masses don’t change as the base areas change. The same mass is added to the hook to move the blocks, regardless of the base area. We can say there isn’t a direct relationship between static friction and base area.

  1. Static friction is independent of base area. This sounds like what we’re looking for from the breakdown of our question. We had a change in base area, but our threshold mass didn’t change. The same mass was needed to move the blocks, regardless of base area. So, static friction is independent of base area.
  2. Static friction is directly proportional to base area. This contradicts the table in the passage. As base area increases, our threshold mass didn’t change. We can eliminate this answer choice. 
  3. Static friction is directly proportional to the square of base area. This contradicts the table in the passage. Once again, as base area increases, our threshold mass didn’t change. We can also eliminate this answer choice. 
  4. Static friction is inversely proportional to base area. This also contradicts the table in the passage. We don’t see a change in static friction as the base area varies. We can eliminate answer choice D as well. We’re left with our correct answer, answer choice A.

89) We want to note how the speed of the block varied after it begins to slide on the wooden board. And the note here says tension and kinetic friction force were constant in magnitude. So that’s going to be important as the block slides. We can answer this question using general knowledge. We’re going to have to relate kinetic and static friction. We’ll have to analyze how speed varies with time. If we think of that in units, we want to know how m/s varies with seconds, or relate this sliding to acceleration.

Let’s imagine what’s happening. The block begins to slide, and the mass on the end of the hook falls. Potential energy is converted to kinetic energy at that point. There’s a constant acceleration as the mass falls toward the floor, and the block accelerates and slides along the board. This means acceleration is constant, but there is variance in speed itself. The speed is constantly increasing, but it’s increasing linearly, hence the constant acceleration.

  1. It was constant in time. This contradicts our breakdown. We said acceleration is constant, not velocity. Velocity is going to vary linearly, because acceleration is constant
  2. It increased exponentially with time. This also contradicts our breakdown of the question, but is a better answer than answer choice A. We can eliminate answer choice A. We’re expecting speed varies in time and increases, but not exponentially. Acceleration is constant, so speed varies and increases linearly. 
  3. It was first constant, then increased linearly with time. Another contradiction. Speed isn’t constant, just acceleration is constant. The last part of the answer is technically correct. Speed does increase linearly with time, but it does so the entire time. This is a better answer than answer choice B, though. We can eliminate answer choice B.
  4. It increased linearly with time. Here’s the answer choice we were looking for. This matches our breakdown. Speed increases linearly with time. We can eliminate answer choice C and stick with our correct answer choice, answer choice D.

90) This is a wordy question, so we can break this down a bit at a time. The researchers develop a new methodology to measure static friction-they placed the blocks on the wooden board and raised the end of the board to make the block begin to slide. We want to know how the researchers can solve for static friction this way. Our answer’s going to come from general knowledge. When a block is placed on the wooden board at a smaller angle, the block won’t slide down because the force caused by gravity acting on the object is less than the resistance from friction. The object isn’t moving, so this resistance is static friction. We’ll have to determine the forces acting on the block when sliding begins.

When our block is on the inclined board, we’ll have a frictional force going up the plane, keeping the block stationary. As the angle increases, we have two additional forces we want to think about. We have our usual mg downwards, but we have mg cos theta that’s opposite of the normal force. We have mg sin theta that’s opposite the frictional force. We can find the coefficient of static friction by using the ratio of force up the board, compared with the normal. Or mg sin theta over mg cos theta. Ultimately that yields tangent theta determining the coefficient of static friction and influencing the force of the static friction.

  1. Time it took for the block to slide down the board. The time it takes the block to slide doesn’t have to do with the static friction force on the block. We don’t know the length of the block, or what this time represents. The value won’t be helpful in our calculations regardless.
  2. Distance the block slid down the board before coming to rest. This is similar to our previous answer choice. The distance traveled won’t tell us about the force on the block. But we ultimately need to know the angle of the board to horizontal to find the friction force on the block.
  3. Mass of the board. The mass of the board on which the block is sliding isn’t going to determine the friction force on the block. The mass of the board is irrelevant in this case. We’re focused on the angle of the board, and the block itself.
  4. Angle of the board with respect to the horizontal. This matches our breakdown exactly. We said the angle of the board with respect to the horizontal will determine the static friction force on the block. We can now eliminate answer choices A-C. We’re left with our correct answer, answer choice D.

 

Physics Question Pack: Questions 91-95

91) This question comes down to understanding buoyant force and the related equation. The buoyancy force on a completely submerged object is given by the formula:

FB = Vρg 

where V is the volume of the object, ρ is the density of the fluid, and g is gravitational acceleration. For reference, Specific gravity=Density of material/Density of water.

We know V and g will remain the same in both equations, so we can write out:

FB/ρ = Vg

We can set FB/ρ equal to one another to solve for the approximate specific gravity of the unknown liquid.

FB/ρ = (5 N)/0.7 = (12 N)/specific gravity of unknown

Solving for the specific gravity of the unknown gives us 1.68. This was a math problem in which we did no rounding or approximation. None of the answer choices are particularly close to one another, so we can compare all of our answers to one another at once. Our correct answer is answer choice C. 

92) To answer this question, we have to consider 𝑣=𝑢+𝑎𝑡

Final velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v. In this specific example, we’re reducing g to g/6, so acceleration (a) is also reduced by 1/6. To compensate for this decrease in acceleration, we must increase time (t) by 6. The only answer choice that is consistent with this increase is answer choice A.

93) 

  1. Sound travels more slowly in a solid than in air. The opposite is actually true. Sound travels more quickly in a solid than in liquids and gases.
  2. The frequency of sound is lower in a solid than in air. Frequency is not going to change in the solid. Let’s keep comparing. 
  3. Part of the sound energy is reflected by the solid. This is exactly what is happening in this situation. We have the sound absorbed or reflected when we have a solid (like a wall) between the source of the sound and the listener. The intensity of the sound heard is less.
  4. The wavelength of sound is shorter in a solid than in air. We know frequency does not change, while sound travels faster in the solid. That means wavelength is actually going to be longer in a solid than in air. We can eliminate this answer choice as well. Answer choice C is our best answer.

94) This is purely a content question. The heat of fusion of a solid is the energy it takes for the substance to transform into a liquid (melting). As energy is added to a system, the bonds in the solid gain energy and reach a point where they can break and undergo a phase change. 

  1. Melting of a solid. This matches the definition we came up with in our breakdown.
  2. Sublimation of a solid. Sublimation is the transition of a solid to a gas. 
  3. Boiling of a liquid. Boiling is the transition of a liquid to a gas.
  4. Condensation of a gas. Condensation is the transition of a gas to a liquid. Answer choice A is our correct answer.

95) This is a classic torque question from AAMC. This question boils down to visualizing the scenario in the question stem. We have equilibrium, so the net force is going to be zero (the test-maker says we have equilibrium). We’re going to have to think of distances in relation to the suspension point (30 cm). On one side of the suspension point, we have an unknown mass at the 10-cm mark (20 centimeters from the suspension point). On the other side of the suspension point, we have to account for the 0.2-kg mass hanging at 80 cm (50 centimeters from the suspension point) and the weight of the meter stick at 50 cm (20 centimeters from the suspension point). Torque is given by (Force) x (distance) x (sin θ). In this case, we can rewrite force as mg, and at equilibrium the forces are all vertical, so sin θ=sin 90=1. We can rewrite as (mg) (distance) for the three torques. g will be the same, so we can eliminate from the three formulas as well. The torque on each side of the suspension will equal the torque on the other side.

Side 1: (unknown mass) (0.2 m) 
Side 2: (0.2-kg) (0.5 m) + (0.5-kg) (0.2 m) = 0.2 kg*m

Set the two sides equal to one another:
0.2 m x (unknown mass) = 0.2 kg*m

Divide both sides by 0.2 m to get the unknown mass of 1.0 kg, which corresponds to answer choice D.

 

Physics Question Pack: Passage 16

96) We’re going to need to reference the passage for the formula of fundamental frequency. We have to understand the relationship between frequency and tension in the formula and use that to answer our question. 

Above I have included two small excerpts from the passage: one from the first paragraph and I also have included the equation for fundamental frequency. 

We want to adjust tension in a way that raises fundamental frequency by a perfect fifth. In the passage, we said the fundamental frequency of each string is 2/3 that of the next higher frequency string. But in this case, we’re going the other way. We’re going to need to raise frequency by a perfect fifth, so we’re expecting our factor to be 3/2. We had 2/3 going from higher to lower, now we have 3/2 going from lower to higher. We still have our perfect fifth.

Now, looking at our equation, to increase frequency, tension also has to increase, but we also see that ½ exponent outside of tension. What does that mean? We’re taking the square root. So that means to get a 1 to 1 change between tension and frequency, we would have to square the increase in tension. So, to increase frequency by 3/2, we need to increase tension by 3/2 squared. That would account for the ½ exponent. We can write that out here. 3/2 squared is equal to 9/4. So when tension is increased by a factor of 9/4, the tension is raised by the square root of 9/4, or 3/2.

This was a math problem, and we did minimal rounding or approximating. We can pick our correct answer from the choices listed here, that’s answer choice D: 9/4.

Just a quick aside following this question. You just read through a 4-paragraph passage that barely mentioned any numbers or equations, but in our first question we focused on those few numbers and equations mentioned in the passage. What does that tell you? We want to read the passage carefully, but we want to understand the most important points. This was a physics passage and topic, and what does almost every physics passage go over? Math and equations. Those are almost always the most important topics, and that shows here in our first question.

97) We’re going to use the equation for our fundamental frequency once again, to determine which of our answer choices would be conducive to a lower-toned instrument. We can also reference if the passage mentions anything specifically about the wood used in the violins and how it affects violins. 

We have a small excerpt from the passage above, and our fundamental frequency equation below. First thing we want to note is the excerpt says minor changes in the thickness and density of the wood can produce significant differences in an instrument’s sound. Here’s the thing, we’re not told what these changes are, nor what differences are ultimately produced. We know these changes can affect the sound of a violin, but we have no way to tell how. Looking at our equation, we can tell we can adjust fundamental frequency by altering any of the three variables given. This is a tangible change. If we increase tension in the string, for example, we can guarantee that we have a higher tone instrument. Fundamental frequency would be higher. If we increase the mass per unit length, or rho, then fundamental frequency would be lower. So, let’s reference this equation as the only way we’re told in the passage to make a lower-toned instrument.

  1. heavier wood in the violin. We said changes in the wood will produce differences in the sound of the instrument, but we don’t know what this difference will be. Any changes in wood will contradict this statement from the passage.
  2. thicker wood in the violin. Another change in the wood in the violin. We know there will be changes, but we don’t know what changes would occur.
  3. heavier strings on the violin. This answer choice deals with the strings of the violin, so we can look at our formula up top. If we increase mass, fundamental frequency is lower. This matches our breakdown, so let’s keep answer choice C. We can now eliminate answer choices A and B because those contradict our passage.
  4. denser wood in the violin. Yet another answer choice that discusses changes in the wood. We don’t know if this change would make a lower-toned instrument. We can eliminate this answer choice. We’re left with our correct answer, answer choice C-heavier strings on the violin.

98) This is another passage-based question. We need to reference what the passage says about the fundamental frequency of the A string on a cello.

We have an excerpt from the passage here. We’re focused on the cello and the passage says the cello is tuned one octave below the viola, which means the frequencies of the cello strings are half that of the viola strings. This means we’re going to reference the frequency of the A string on the viola, and the frequency of the A string on a cello should be half of that. The fundamental frequency of the A string on the viola 440 Hz, it’s only 220 Hz on a cello.

This was a math problem, and we did minimal rounding or approximating. It doesn’t make as sense to compare our answer choices one at a time here. We can pick our correct answer from the choices listed here, that’s answer choice B: 220 hertz.

99) This is another passage-based question. We need to reference what the passage says about the fundamental notes of a violin’s strings. Note, all of our answers are given in hertz and whole numbers.

We have an excerpt from the passage here. It says its strings are tuned with decreasing frequency to the notes E, A, D, G, where A has a frequency of 440 Hz. We also know the fundamental tones of the strings of a violin are separated by a perfect fifth. A perfect fifth corresponds to a pair of pitches with a frequency ratio of 2:3. The fundamental frequency of each string is 2/3 that of the next, higher frequency string. We can write out the notes E-A-D-G. We know A is 440 Hz. E is going to be 3/2 multiplied by 440 hertz. D is going to be 440 hertz multiplied by 2/3. And G is going to be the frequency of D multiplied by 2/3. Solving for these values, we get frequencies of 660, 440, 293, and 196 hertz.

This was essentially a math problem, and we did minimal rounding or approximating. We can pick our correct answer from the choices listed here, that’s answer choice D.

 

Physics Question Pack: Passage 17

100) We’re actually told in our question stem that we want to answer this question using the passage. We can go back to our passage and reference where it talks about the energy released following an atom splitting. Is that always the case when the test maker mentions the passage? Not always, but for these solutions specifically, I want to be as thorough as possible. As you’re practicing on your own, I don’t recommend going back to the passage often, unless it’s for specific details. The reason I reference the passage fairly often in these solutions is so you can clearly see my thinking and what I’m referencing from the passage.  

We have an excerpt from our passage here. It says There is a large electrical repulsion between these two fragments that causes them to move apart and gain kinetic energy. We’re also told the force between the two nuclei depends on the distance between the centers of the nuclei and the charges of the fragments. That’s shown in the form of an equation as well. So we ultimately want an answer choice that explains that the atom splits, and the like-charges of the fragments repel one another. That’s ultimately the source of the energy. 

  1. fast-moving electrons. Does this answer choice sound like our breakdown, first of all? It does not. In fact, there’s no mention of fast-moving electrons in the entire passage, or the question stem. This answer choice isn’t answering the specific question being asked. When an atom splits, the energy released doesn’t have to do with fast-moving electrons.
  2. the short-range attraction of the nucleons. This answer choice is the opposite of our breakdown of the question. Remember in the passage, the repulsive force is actually balanced by short-range attractive forces-this is the opposite of our breakdown.
  3. mutual attraction of the fragments. Another answer choice that contradicts what we were told in the passage and in our breakdown. The fragments repel one another, there’s not mutual attraction.
  4. mutual repulsion of the fragments. The passage does mention the large electrical repulsion between fragments that causes them to move apart and gain kinetic energy. In our breakdown we said we want an answer choice that mentions the fragments repelling one another because of the charges. And only one answer choice matches that requirement: answer choice D.

101) In other words, which of the following quantities aren’t influenced by mass, but are influenced by charge. We can go back to the passage to reference the specific equations given to us, but we’re going to use general knowledge to determine how mass and charge influence our answer choices.

We have an excerpt from our passage here, and we have our four answer choices below. Normally you’ll see me really break down the question before jumping into the possible answers, but this question is very open-ended, so we’ll reference our answer choices concurrently. We said we want to know which of the following quantities aren’t influenced by mass, but are influenced by charge? We also know Newton’s 3rd law, both fragments will experience the same equal and opposite force, and move away with the same momentum. 

  1. Magnitude of the force of one fragment on the other. We’re given this equation in the first sentence of this excerpt. We notice charge will influence this force, but mass is not found in this equation. Let’s keep comparing
  2. Magnitude of acceleration. We said in our breakdown that momentum is conserved, so the fragments move away with the same momentum. Formula for momentum is m1v1=m2v2. If we have different values for M1 and M2, we have to have different values for V1 and V2 as well. Answer choice A is going to remain superior.
  3. Speed. Reasoning here is going to be the same as answer choice B. The fragments move away with the same momentum. Formula for momentum is m1v1=m2v2. If we have different values for M1 and M2, we have to have different values for V1 and V2 as well
  4. Kinetic energy. Kinetic energy is ½ mv squared. Because momentum is conserved, kinetic energy will not be the same value if we have different masses. If you were unsure, you could also test this with various masses and velocities and find that kinetic energy is not equal. Answer choice A is going to remain our best option here.

102) We want to know why there cannot be any naturally-occurring elements with more than 92 protons. We can flip back to our passage to see where this is discussed. We’re going to touch on the forces that hold together the nucleus and explain why any nucleus with more than 92 protons is not going to occur naturally. We’re ultimately going to explain the repulsion that takes place in the nucleus.

We have an excerpt from our passage here. It says there is a large repulsive force between the positive charges in the nucleus. This force is balance by a short-range attractive force, the strong nuclear force. Protons are positively charged and neutrons are uncharged. So, we have nothing but positive charges in the same vicinity in the nucleus. We know particles of like charges will repel one another, but that should also mean there should be no reason for the concentrated atomic nucleus to exist. That’s where the strong nuclear force comes in. We’re told all of the repulsive forces between the positively charged protons are balanced. The strong nuclear force is what binds the protons and neutrons together to form that nucleus. The strong force ends up balancing the positive, repulsive forces, but it can ultimately only do that to a certain extent. Adding more protons will make the Coulomb force stronger than this short-range nuclear force.

  1. All of the heavier elements have radioactively decayed. We just spent an entire passage relating the balance between the strong nuclear force, and the repulsive Coulomb force. And does this answer choice explain the reason why there are no naturally occurring elements with more protons in their nucleus than uranium’s nucleus? At best, this is a side effect of what’s described in the passage, but it’s not a reason why. Let’s try and find a superior answer.
  2. All of the heavier elements are stable. Does this answer choice sound like it answers the question being asked? If anything, it sounds like a good reason for elements with additional protons to be naturally occurring. If heavier elements are more stable, they would be more likely to occur.
  3. The range of the strong nuclear force is too short to hold them together. This answer choice matches our breakdown. The strong nuclear force is what binds the protons and neutrons together to form that nucleus. But it can ultimately only do that to a certain extent. The maximum number of protons in a naturally occurring element is therefore maxed out at 92 protons, or the number in the nucleus of uranium. Probably the best answer choice we’ll have for this question.
  4. The heavier elements can be made only in nuclear reactors. This answer choice tells us what happens because there are no naturally occurring elements with more protons in their nucleus that uranium. We want to know why. This answer isn’t answering the specific question being asked. There can be elements that are both naturally occurring and made in nuclear reactors. But this choice doesn’t answer the why that’s posed in the question stem. Answer choice C remains our best answer here.

103) Looking at the answer choices, we want to know which of the graphs properly shows force between fragments vs separation of fragments. We can flip back to our passage to see where this is discussed. We’re going to note the equation for the repulsive force between the two nuclei and relate it to separation of the fragments.

We have an excerpt from our passage here, and first sentence gives us the formula for the repulsive force between the two nuclei. We’re relating force and separation in our question stem, so we can relate the two as follows: Force is proportional to 1 over distance squared. So, our graph should be one of an inverse square. Force decreases as separation increases exponentially. Or said differently, as separation increases exponentially, the force between fragments decreases.

a.

Answer choice A shows a linear increase. Recall what we said: As separation increases, force is going to decrease. And just think of that logically, visualize the two fragments getting further away from one another. The force between the fragments is going to get smaller as that distance increases. We’ll eliminate this answer choice.

b.

Answer choice B shows an exponential increase, but reasoning here is going to be similar to answer choice A. As separation increases, force is going to decrease. And just think of that logically, visualize the two fragments getting further away from one another. The force between the fragments is going to get smaller as that distance increases

c.

Answer choice C shows a linear decrease. This is better than answer choices A and B, but we said we’re expecting an exponential decrease in separation, and an inverse square graph.

d.

Answer choice D matches our breakdown. We have force decreasing as separation increases. We can confidently eliminate answer choices A-C. We’re left with our correct answer, answer choice D.

 

Physics Question Pack: Passage 18

104) We’re going to have to identify what’s going on in this beta decay, and also use our knowledge of beta decay to pick from one of our four answer choices. We can also use the following visual before going through our specific example:

Carbon 14 has 6 protons and an atomic mass of 14 atomic mass units. We know carbon has 6 protons, because that’s carbon’s atomic number. The 14 we said is the atomic mass, or the number of nucleons-protons and neutrons in the nucleus. Carbon 14 has 8 neutrons and 6 protons. Nitrogen 14 has 7 protons (That’s nitrogen’s atomic number). An atomic mass of 14 means it has 7 neutrons as well. So, we go from 6 protons and 8 neutrons to 7 protons and 7 neutrons.

  1. Electron. This matches our breakdown exactly. We said we were expecting a negative charge emitted by the carbon, or an electron. Let’s still go through the additional answer choices to confirm this is the best answer.
  2. Alpha. This answer doesn’t make sense in this case. We’re told we have beta decay, we’re not expecting an alpha particle to be emitted. We’re expecting an electron. Answer choice A remains the best option.
  3. Neutron. We need a negative charge that comes from the electron, not the neutral neutron. We can eliminate this answer choice as well. 
  4. Positron. That was our other potential answer in this situation with our beta decay. But like we’ve mentioned, the electron and the negative charge were what we were looking for in terms of an answer. We can eliminate this answer choice as well. We’re left with our correct answer, answer choice A.

105) We’re going to need to reference the passage for the half-life of the carbon-14 isotope. We’re going to use general knowledge to actually explain what this half-life means and solve for the atoms of carbon-14 present when the object was created 18,000 years ago.

We have part of our passage here. It says This isotope is unstable and experiences spontaneous beta decay with a half-life of approximately 6,000 years. Half-life is the time required for half of the nuclei in a sample of a specific isotope to undergo radioactive decay. We’re ultimately trying to solve for the atoms of carbon-14 present when the object was created 18,000 years ago. The object currently has 1000 atoms present. 18,000 years ago, said differently, is 3 half-lives ago. So, the number of atoms is doubled every 6000 years we go back. The first 6000 years takes us to 2000 atoms present, the next 6000 years takes us to 4000 atoms present, and the third and final 6000 years takes us to 8000 atoms present.

This was a math problem, and we did minimal rounding or approximating. That means we can compare all of our answer choices at once. We can pick our predicted answer, answer choice D: 8,000 atoms.

106) We’re going to need to reference the passage for the mass of the beta particle. We’re going to use our general knowledge to solve for average kinetic energy.

We have part of our passage above. We’re told the mass of the beta particle is 9 x 10^-31 kg, and we were told in our question stem that average velocity is 3 x 10^7 m/s. The equation for kinetic energy is ½ mv squared, so let’s plug in our known values for mass and velocity.

First thing we want to do is square our velocity. Let’s separate this into 3 squared and 10^7 squared. If we’re raising an exponent to a different power, multiply the exponents. V squared is now 9.0 x 10^14. We can now multiply this by ½ M

How do we multiply two numbers in scientific notation? What we’re going to do is multiply our coefficients first, then we add exponents while keeping the same base. So, in this case we multiply 9 times 9 times 1/2, which gives us 40.5. We add the exponents on our base 10 and get 10^-17.

We want to keep our number in scientific notation. We use LARS (left add, right subtract) and move our decimal point on 40.5 one place to the left, that means we add one to the exponent. Final answer here for kinetic energy is going to be 4.05 x 10^-16 J.

Just like our last question, this was a math problem, and we did minimal rounding or approximating. That means we can compare all of our answer choices at once. We can pick our calculated answer: answer choice C.

107) We’re given a specific wavelength for maximum light emission of a scintillator, and want to know the energy of the light photon. We’re going to need to reference the passage for Planck’s constant and the velocity of electromagnetic radiation. These values should be general knowledge, but we can take a quick look at our passage to make sure we have the correct values.

We have part of our passage here. We’re told the velocity of electromagnetic radiation in vacuum is 3 x 10^8 m/s. Planck’s constant is 6.6 x 10^-34 JxS. The equation for photon energy is E=hc/lambda. We can plug in our given values for h, c, and lambda. We’re told lambda is 450 nanometers. We can convert nanometers to meters by multiplying by a factor. 1 nanometer is 10^-9 meters. So, in scientific notation wavelength is 4.5 x 10^-7 meters. 

Next, let’s multiply our numerator. How do we multiply two numbers in scientific notation? What we’re going to do is multiply our coefficients first, then we add exponents while keeping the same base. So, in this case we can round. 6.6 is approximately 6 2/3 and multiplied by 3 gives us 20. Adding the exponents together gives us 10^-26. We have approximately 2.0 x 10^-25 meters in our numerator. We divide by our denominator 4.5 x 10^-7 meters. 

How do we divide two numbers in scientific notation? We’re going to turn this into two fractions instead of trying to solve the entire problem at once. So instead this becomes:

(2.0/4.5) and (10^-25 / 10^-7). After simplifying and converting our number to scientific notation again, this gives us a value of 4.4 x 10^-17 Joules

Just like our last few questions, this was a math problem, but we did some rounding. We want to be very careful when comparing answer choices. First thing we notice, is the exponents in our answer choices are several hundredfold away from one another. This means we can look for an exponent that matches our calculation. Answer choice A not only matches our calculated exponent, but also the 4.4 we calculated. We can stick with answer choice A as our best option here.

 

Physics Question Pack: Passage 19

108) We can reference the passage to get details about the child on the toboggan. We’ll use our knowledge of different types of energy as well to explain the conversion taking place between points A and B.

Energy cannot be created nor destroyed in an isolated system, it can be internally converted to any other form of energy. Now, the conservation of mechanical energy says the mechanical energy of an isolated system remains constant without friction. But in this situation, we do have frictional force. We’re told the toboggan and rider are opposed by a constant 60-N frictional force. The child slides down the hill from point A to point B in the figure below. We have potential energy, which is the energy difference between the energy of an object in a given position, and its energy at a reference position. At the top of the hill we have potential energy, and when the child begins to slide down, this is converted to kinetic energy. But we also have thermal energy because the frictional force acts on the toboggan.

  1. Kinetic to potential and thermal. This sounds like the opposite of our breakdown. We expect potential energy at the top of the hill at point A. That is converted to kinetic and thermal as the toboggan heads down the hill.
  2. Kinetic and potential to thermal. This doesn’t match our breakdown exactly. As the toboggan is sliding, the potential energy is converted to kinetic and thermal. As far as being compared to answer choice A, this answer is slightly better in my opinion, because it doesn’t switch the order of the kinetic to potential energy conversion. 
  3. Potential to kinetic and thermal. This matches our breakdown better than answer choice B. We started with potential energy, and that’s converted to kinetic energy as the toboggan slides. Because of the frictional force, there’s also thermal energy. Answer choice C is our best option of the three we’ve gone through.
  4. Potential and kinetic to thermal. This is the same as answer choice B, just stated differently. We can use the same reasoning here to eliminate this answer choice. Answer choice C is going to be our best, correct answer.

109) We can reference the passage to get details about the child on the toboggan. We’ll also have to know how to solve for the energy lost to friction using our physics general knowledge.

We’re told in the passage the toboggan and rider are opposed by a constant 60-N frictional force. We’re also told the child is 54 kg, the toboggan is 6 kg, air resistance is negligible, and g is 10 m/s^2, and the distance from point A to point B is given as “l”. The key to answering this question is knowing which of this information we actually need to solve for the amount of energy lost to friction.

Energy lost due to frictional force is the product of that force multiplied by the distance across which the force acts. We’re told the frictional force is 60-N. And this force is constant along the distance “l.” Meaning the amount of energy lost due to friction is the 60 Newtons frictional force times “l”

  1. (60 N)(20 m) 20 meters is the vertical distance traveled from point A to point B, not the total distance apart along the slope. 
  2. (60 N)(l) This matches our calculated value. We have the frictional force multiplied by the distance across which the force is acting. Let’s keep this answer choice and we can eliminate answer choice A.
  3. (60 kg)(g)(20 m) This would be a way to solve for gravitational potential energy, or mgh. That’s not what we’re looking for in this case. We can stick with answer choice B as our best answer. 
  4. (60 kg)(g)(l) This answer corresponds to the correct units, but it’s essentially the same as answer choice C that incorporates the mass of the child and toboggan and “g.” We can eliminate answer choice D. We’re left with our correct answer, answer choice B.

110) We can reference the passage to get details about the child on the sled. Then we’ll be calculating the coefficient of friction using our general knowledge. 

We’re told in the passage we have a 47-kg child with a 3-kg sled. And later in the passage we’re told to the right of point B the sled and rider are opposed by a 50-N frictional force. 

The kinetic friction force (fk) between systems moving relative to one another is given by fk = μkN where μk is the coefficient of kinetic friction, which depends on both of the materials. Ultimately, we’re setting that frictional force equal to the coefficient of friction multiplied by normal force. Normal force is the mass of the child and sled multiplied by “g.” That’s 50 total kg multiplied by g, or 500 newtons. So 500 newtons multiplied by the coefficient of friction is equal to 50 newton frictional force. Solving for the coefficient of friction gives us 0.1.

This was a math problem, and we did minimal rounding or approximating. We can pick our correct answer from the choices listed here, that’s answer choice A: 0.1.

111) In other words, we want to compare the speed of the sled and rider at point B if the starting point of the rider is 10 m lower instead.

We’ll have to reference the height of point A on the hill, which was 20 meters. And we’re told the new starting point is 10 m lower than point A. We’ll focus on conservation of energy, and specifically the conversion of potential energy to kinetic energy. And we also want to note the sled and its rider slid free of friction along the hill.

We’re converting potential energy to kinetic energy. If potential energy at the top of the hill at 20 meters is mgh, then that same potential energy is mg x ½ h at half height. That’s going to be equal to kinetic energy at the bottom of the hill which is ½ mv^2. The mass on both sides cancels out in both equations.

gh=1/2 v^2 and g 1/2h = 1/2V^2. 

Isolating v^2 gives us 2gh for point A. And gh for the lower location. Taking the square root of both sides gives us root 2gh for point A. Root gh for the lower location. The difference between the two ends up being the root 2. When the sled starts at point A, it goes faster by a factor of root 2. When the sled starts lower, it moves at a speed slower by a factor of root 2.

This was another math problem, and we did minimal rounding or approximating. We came up with an exact answer, which is that starting from the new location will make the speed of the rider slower by a factor of root 2. We can pick our correct answer, answer choice D.

 

Physics Question Pack: Passage 20

112) So, if we’re looking at a series of several nodes, we want to know the spacing between two nodes next to one another. This answer is going to come from our general knowledge. We’re going to need to understand nodes and antinodes. And quick glance at our answer choices shows we’ll want to have an answer in terms of wavelength.

Nodes are points along a standing wave where the wave amplitude is zero. Essentially, what’s happening here is waves add with opposite phases and cancel each other out. The nodes occur in intervals of half wavelengths. You actually see this when looking at the figure in the passage.

Essentially, we see what looks like a sine curve going through the axis. Below, I’ve outlined in red the full wavelength. We have a node when the curves crossed, or in ½ lamda intervals.

Not much subjectivity required. We can pick the answer that matches our prediction. Answer choice D, lambda over 2

113) We have a laser-cavity mode, and a standing-wave pattern viewed as a superposition of two waves. What’s a superposition? It’s essentially saying the sum of the two traveling waves shown as one. We want to know the properties of the individual waves that make up that superposition. We can look at the standing wave pattern illustrated in the passage to get a sense of the properties of the waves. We’ll also have to know how superposition of two waves works; quick glance at our answer choices shows we want an answer in terms of amplitude and direction.

We have a short excerpt from the passage shown here. It says A standing-wave pattern in a laser cavity is illustrated in the figure below. What do the two waves here look like? We have a standing wave made up of the solid wave and the dashed wave-each look equal in magnitude. Also, at any point along the waveforms, the two waves are in opposite directions.

We said we’re expecting our waves are equal in magnitude from our figure. That means we can automatically eliminate answer choices C and D and we’re left with answer choices A and B. We also said the waves are in opposite directions, meaning we can also eliminate answer choice B. We’re left with our correct answer, answer choice A.

114) We want to know the cavity mode number of a gas laser with a specific cavity length, and we’re given single oscillation frequency. We can solve this question using the equation given in the passage and solving for the cavity mode number. We’re given frequency, and we know speed of light, so we’ll solve for wavelength.

We have a short excerpt from the passage shown here. We have our equation here as well that we’ll use to solve for our cavity mode number. The question stem said the gas laser has a cavity length of 1/3 m. As we mentioned, we have frequency, so we can solve for wavelength first. Speed=frequency x wavelength. Plug in our values for speed and frequency and use our math strategies to solve for wavelength. Isolating wavelength gives us 3.3 x 10^-7 m.

We can use that value to plug into our equal, along with our length of 1/3 meters. Solving for the cavity-mode number gives us 2.0 x 10^6

This was a math problem. We did minimal rounding or approximating. We can select the answer choice that matches our prediction here. Answer choice B says 2.0 x 10^6.

115) We can solve this question using general knowledge. We’re given beat wavelength here, and we just went over how to relate speed (c), frequency, and wavelength in our previous question. We know the relationship from our previous question: Speed (c)=frequency x wavelength. Isolating wavelength gives us (c / frequency). We’re given the expression for beat frequency in the question stem, so our final expression is:

Speed / (fm+1 – fm) OR c / (fm+1 – fm)

This was another math problem. We did minimal rounding or approximating, and in fact, we didn’t even deal with any numbers. That shows how important knowing the relationship between variables and being able to manipulate equations can be. We weren’t given any specific values, but we were still able to get to the correct answer. We can pick the answer that matches our calculated value: answer choice D.

 

Physics Question Pack: Questions 116-120

116) We’re considering net forces here, meaning the forces can be in the same direction, opposite directions, or some combination of the two. That means the net force can range anywhere from 4-N (opposite directions, we take the difference) to 18-N (same direction, we add the forces together). Only answer choice that doesn’t fall within this range is answer choice A: 2N.

117) Resistivity is the property of a material that quantifies how strongly it resists or conducts electric current. A low resistivity indicates a material that readily allows electric current and vice versa. It is calculated as:

ρ=R•A/L

R is the electrical resistance of a uniform specimen of the material

l is the length of the specimen

A is the cross-sectional area of the specimen

For this specific question, we want to know about resistance. We can isolate resistance and get:

R=ρ•L/A

The resistance depends on resistivity, length of the wire, and the cross-sectional area of the wire. The only answer choice that matches our breakdown and our resistance equation is answer choice B.

118) Key here is going to be that p>f, which means the focal length is greater than the distance of the object in front of the lens. We also know we have a converging lens which can form a real or virtual image of an object. When the object is a distance from the lens greater than the focal length, like we have in our question stem, a real and inverted image will be formed.  

  1. Real and pointing down. This answer choice matches our breakdown.
  2. Real and pointing up. While we do expect the image to be real, we expect an inverted image. Answer choice A remains superior.
  3. Virtual and pointing down. Although we do expect the image to be pointing down, we’re expecting a real image, not a virtual image.
  4. Virtual and pointing up. This answer choice contradicts everything we said in our breakdown. We can eliminate answer choice D. We’re sticking with answer choice A as our best answer.

119) In this case, we have electron capture which we’re told is a form of β+ decay. In electron capture an electron will be “captured” by the nucleus, and a proton is changed to a neutron. The net change is a decrease in protons (and atomic number), but no net change in mass (protons and neutrons have the same mass). We expect 47Be to therefore change to 37Li. The 7 remains the same because we said there’s no net change in mass. The 4 protons decrease to 3 protons, which also means we change the identity of Be to Li. This matches answer choice B. 

120) The law of reflection for the case of our question, states that when a ray of light reflects off a surface, the angle of incidence θ is equal to the angle of reflection θ’. Right away we can eliminate answer choices A and D. 

Next, we have to consider the relationship between the angles of incidence and refraction. We can consider Snell’s law: n1sinθ1 = n2sinθ2. In this case θ2 is α in our figure. 

The question stem tells us the index of refraction of the medium (n2) is 1.5, and we should know for air, n1, is 1.0. Because n2 is greater than n1, in order for Snell’s law to work, sinθ1 has to be greater than sin α. That means θ1>α.

Combine the two parts of our breakdown: 

θ is equal to θ’, and θ1>α. This matches answer choice B.



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