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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.

PS-10-1-PT3-101-a.gif

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.

PS-10-1-PT3-101-b.gif

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.

PS-10-1-PT3-101-c.gif

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.

PS-10-1-PT3-101-d.gif

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

 

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