CAP 2015 Solutions

This post will cover the multiple choice section of the 2015 CAP exam.

CAP 2015 Question 1

As shown in the picture, a ball is attached to a ceiling and a wall with massless ropes A and B. Rope A is at angle \theta from the vertical direction, and rope B is horizontal. The system is static. If rope B is cut, what is the ratio of the tension in rope A immediately after it is cut to the tension in A before it is cut?

\displaystyle a)  \, \, 1

\displaystyle b) \, \, \cos{\theta}

\displaystyle c) \, \,  \frac{1}{\cos{\theta} }

\displaystyle d) \, \, \cos^{2}\theta

\displaystyle e) \, \, \cos{\theta}\sin{\theta}

Solution

The answer is d)

Let the ball have a mass M and the initial tension in A be T_A. Then balancing forces vertically, T_A \cos{\theta} = Mg.

Once B is cut, what happens to A? There’s one thing you know for sure. The length of the string remains constant. Therefore the acceleration in the radial direction is zero. Mathematically, if the new tension is T_A', then

T_A'  = Mg \cos{\theta}. Dividing our two equations, we get

T_A'/T_A = \cos^{2} \theta

CAP 2015 Question 2

Consider a circuit made of a wire with uniform resistance in a shape of a circle as shown in the picture. The circle is connected diagonally from point A to point B with the same type of wire. If the current passing through the circuit is i_0, what is the current passing through the wire AB as a function of angle \theta?

\displaystyle a) \, \, 0

\displaystyle b) \, \, \frac{\theta}{\pi - \theta} i_0

\displaystyle c) \, \, \frac{\pi - 2 \theta}{\pi + 2}i_0

\displaystyle d) \, \, \frac{\pi-2 \theta}{\pi+4}i_0

Solution

The answer is d)

Whenever you see a problem like this on a multiple choice test, the first thing you should do is look at a special case of \theta. What’s the most obvious choice? \theta = 0. The possibilities when \theta = 0 are

 \displaystyle a) \, \, 0

\displaystyle b) \, \, 0

\displaystyle c) \, \, \frac{\pi }{\pi + 2}i_0

\displaystyle d) \, \, \frac{\pi}{\pi+4}i_0

Notice that the answer is definitely not 0 because certainly some current will pass through AB! But the other two options are different! So if we solve the case when \theta = 0, we can unambiguously determine the answer. It turns out this is very easy. It’s just three resistors in parallel! But what are their resistances? Let the resistance of AB be R. From geometry, the circular arcs are \frac{\pi}{2} times longer. And since resistance is proportional to length (we are assuming a constant resistivity), their resistance is \frac{\pi}{2} times higher. Therefore the total resistance is

\displaystyle  \frac{1}{R_{tot}} = \frac{1}{R} + \frac{1}{\frac{\pi}{2} R} + \frac{1}{\frac{\pi}{2} R}

\displaystyle \Rightarrow \frac{R_{tot}}{R} = \frac{\pi}{4+\pi}

You may be able to immediately see that the current through AB is \frac{R_{tot}}{R} i_0. If not, here’s an explanation of why. The voltage drop over the resistors is just R_{tot}i_0. But this must be the voltage drop over AB as well! If the current in AB is i_{AB}, then

\displaystyle  i_{AB} R = R_{tot}i_0 \Rightarrow i_{AB} = \frac{R_{tot}}{R} i_0 = \frac{\pi}{4+\pi} i_0

This is the answer you would have gotten using d).

 

cap 2015 Question 3

Graph of speed vs time

The above graph represents the speed of a car as a function of time. We know that as the car speeds up there is a friction force with air that can be approximately considered to be proportional to the speed of the car. Which of the following graphs can be the force of the engine as a function of time?

 

a)

b)

c)

d)

e)

Solution

The answer is c).

We are given F \propto v. If you’re unfamiliar with this notation, it just means that F is proportional to v. Therefore since v is accelerating uniformly, v \propto t \Rightarrow F \propto t. So F must increase in a straight line in the beginning. This means the answer must be b) or c). What’s different about the two? In b), the final force is equal to the initial force. But this can’t possibly be true because there’s now a strong force from air resistance! Therefore c) must be correct.

 

cap 2015 Question 4

If electromagnetic radiation with intensity I (power per unit area) is absorbed by a surface, it exerts a pressure on the surface that is given by I \cos{\theta}, where c is the speed of light and \theta is the angle the light rays hit the surface relative to a direction perpendicular to the surface. The Sun has a radiation power of P = 3.9 \cdot 10^{26} \, W. The absorption of sunlight by Earth causes a force on Earth that pushes it away from the Sun. Which of the following is closest to the magnitude of this force?

\displaystyle a) \, \,10^7 \, N

\displaystyle b) \, \, 10^9 \, N

\displaystyle b) \, \, 10^{11} \, N

\displaystyle b) \, \, 10^{13} \, N

Solution

The answer is b).

Let R_E be the radius of the Earth and R_{ES} be the distance between the Earth and the Sun. From the data sheet, we know that  R_E = 6.371 \cdot 10^6 \, m and R_{ES} = 1.49598 \cdot 10^{11} \, m. We also know that c = 3.00 \cdot 10^8 \, \frac{m}{s}. What’s the intensity I absorbed by the earth? Well power is proportional to area, and the area the sunlight is spread over is the area of a sphere with radius R_{ES}. This is

A_S = 4 \pi R_{ES}^2

But the total power of the sun’s light on (not in) this sphere is just the power of the sun! So the intensity is

\displaystyle I = \frac{P}{A_S} = \frac{3.9 \cdot 10^{26} \, W}{4 \pi (1.49598 \cdot 10^{11} \, m)^2 } = 1.4 \cdot 10^{3} \, \frac{W}{m^2}

Now we need the area of the Earth perpendicular to the sun’s rays. Why? Because the sunlight only cares about the surface perpendicular to itself. Think about this a little bit.

Hopefully you got A_E = \pi R_E^2. Just imagine looking at the silhouette of the Earth against the backdrop of the sun! All you see is a circle of radius R_E! The force is now

\displaystyle F = \frac{I A_E}{c} = \frac{1.4 \cdot 10^{3} \, \frac{W}{m^2} \cdot \pi (6.371\cdot 10^6 \, m)^2}{3.00 \cdot 10^8 \, \frac{m}{s}} = 6.0 \cdot 10^8 \, N \sim 10^9 \, N

cap 2015 Question 5

As shown in the picture a cylinder with volumetric thermal expansion coefficient of \beta_{c} = 3 \cdot 10^{-6} \, / {}^{\circ} C is completely submerged and floating in a fluid at the temperature 50^{\circ} C. The fluid has volumetric thermal expansion coefficient of \beta_{f} = 8 \cdot 10^{-5} \, / {}^{\circ} C. If we cool down the cylinder and the fluid to 0^{\circ} C what percentage of the cylinder’s height will be out of the fluid?

\displaystyle a) \, \,  0.39 \%

\displaystyle b) \, \, 0.42 \%

\displaystyle c) \, \, 0.78 \%

\displaystyle d) \, \,  0.84 \%

Solution

The answer is a).

First notice that if you have an object with density \rho_{obj} in a liquid of density \rho_{l} such that \rho_{l} >= \rho_{obj}, the submerged fraction is \alpha = \frac{\rho_{obj}}{\rho_{l}}. This is a useful equation to remember. If you’re unfamiliar with it, here’s a quick derivation. Let the object’s volume be V. Then the buoyant force is

F_{B} = \alpha V \rho_{l} g

This must balance the total weight of the object. Therefore

\alpha V \rho_{l} g= V \rho_{obj} g \Rightarrow \alpha = \frac{\rho_{obj}}{\rho_{l}}

OK. Let’s use this equation. Initially, the submerged fraction is exactly 1. Therefore without even using the equation, we know that \rho_{obj} = \rho_{l} = \rho. However, after a change in temperature, the volumes change. The total volume of the object is multiplied by a factor 1-\beta_{c} \Delta T. Since the mass stays the same, the density is multiplied by a factor of \frac{1}{1-\beta_{c}  \Delta T } . It’s the same for the liquid except the coefficient of expansion is \beta_{f}. Therefore the ratio of densities is now

\displaystyle \alpha =  \frac{ \frac{\rho}{1-\beta_{c} \Delta T} }{\frac{\rho}{1-\beta_{f} \Delta T} } = \frac{1-\beta_{f} \Delta T}{1-\beta_{c} \Delta T}

Plugging in values,

\displaystyle \alpha =  \frac{1-8 \cdot 10^{-5} \, / {}^{\circ} C \cdot 50^{\circ} C }{1-3 \cdot 10^{-6} \, / {}^{\circ} C \cdot 50^{\circ} C} = 0.996

Therefore the fraction outside the fluid is 1-\alpha = 3.9 \cdot 10^{-3} = 0.39 \%

cap 2015 Question 6

The amount of sunlight absorbed by the Earth’s atmosphere is approximately proportional to the length of air through which it travels to reach to the Earth. Which of the following is closest to the ratio of the absorption of sunlight during sunset relative to the absorption when the Sun is exactly at the center of the sky? The effective height of the atmosphere is about 10 km.

Solution

The answer is b).

This is just a geometry question. Let the Earth’s radius be R and the atmosphere’s thickness be d. From the data table, R = 6371 \, km and from the question d = 10^4 \, m. Look at the diagram below.

When the sun is directly overhead, it simply travels a distance d from B to A. However, at sunset the light travels along path CA. Notice that OA \perp CA so we can apply Pythagorean’s Theorem on \triangle OAC.

OA = R,

OC = d+R, so

AC = \sqrt{(OC)^2-(OA)^2} = \sqrt{(R+d)^2-R^2}

Therefore since sunlight absorption is proportional to distance through the atmosphere, the ratio is

\displaystyle \gamma = \frac{\sqrt{(R+d)^2-R^2}}{d} =\frac{ \sqrt{(6371 \, km + 10 \, km)^2 - (6371 \, km)^2}} {10 \, km} = 36

cap 2015 Question 7

A beam of red light is made up of a stream of photons as shown above. The size of dots represents the photon energy, and the spacing represents the spatial distribution between photons.:

If we use photons of half the wavelength of red light, keeping the intensity constant, what should the stream look like?

a)

b)

c)

d)

e)

Solution

The answer is c).

First notice that halving the wavelength doubles the energy, since E = \frac{hc}{\lambda}. Because the size represents the energy, doubling the energy should increase the size. Therefore we can immediately rule out d) and e). Next notice that since the intensity is constant and the energy of each photon is doubled, the number of photons per unit time should be halved. In other words, there is twice as much space between each photon, which corresponds to c).

cap 2015 Question 8

To paint the surface of a solid metal sculpture we need 100 buckets of paint. If we melt the sculpture and create 1000 smaller but otherwise identical sculptures, how many buckets of paint are needed to paint the surface of all 1000 small sculptures? Assume that the original and the smaller sculptures are all solid with no cavity inside.

\displaystyle a) \, \, 10

\displaystyle b) \, \, 100

\displaystyle c) \, \, 1000

\displaystyle d) \, \, 10000

Solution

The answer is c).

The volume of each small sculpture is a thousandth of the original sculpture. Since volume goes down by the cube of the length, we know that the length of the small sculpture is a tenth of the original sculpture (10^3 = 1000). Because area goes down by the square of the length, the surface area of each small sculpture is a hundredth of the original. But since there are a thousand of these sculptures, there combined surface area is 1000 \cdot \frac{1}{100} = 10 times the original surface area. Since the number of buckets is proportional to the area, we need 10 \cdot 100 = 1000 buckets.

cap 2015 Question 9

A person is standing on a platform that is sliding down the hill as shown in the picture. The slope of the hill is 30^{\circ} with respect to the horizontal. The person is standing on a scale positioned on the surface of the platform and the scale is reading only 150 lbs while his real weight is $160 lbs$. Which of the following is the coefficient of kinetic friction between the platform and the slope?

\displaystyle a) \, \, \frac{\sqrt{3}}{2}

\displaystyle b) \, \, \frac{\sqrt{3}}{4}

\displaystyle c) \, \, \frac{\sqrt{3}}{3}

\displaystyle d) \, \, \frac{\sqrt{2}}{3}

\displaystyle e) \, \, there is not enough information.

Solution

The answer is b).

We can immediately find the acceleration of the person in the direction of gravity. If his mass is m_p, then the normal force from the scale is \frac{150}{160} mg = \frac{15}{16} mg. Therefore his net acceleration downwards is a_{y} = g - \frac{15}{16}g = \frac{1}{16}g. Since the acceleration of the block must be along the ramp, if the total acceleration is a, then a_{y} = a \sin{30^{\circ}} = \frac{a}{2} \Rightarrow a = \frac{g}{8}

We’re done with the person now. Let’s just treat the platform and the person as one composite object of mass M. This is just a standard block down the ramp! Call the normal force from the hill onto the platform N and the coefficient of kinetic friction \mu. By balancing forces perpendicular to the ramp, we see that the normal force is

N = Mg \cos{30^{\circ}} = Mg \frac{\sqrt{3}}{2}

Using Newton’s Law along the ramp, we get

Mg \sin{30^{\circ}} - \mu N = \frac{Mg}{2} - \mu Mg \frac{\sqrt{3}}{2}  = Ma

Simplifying,

\mu = \frac{\sqrt{3}}{4}

cap 2015 Question 10

A constant amount of ideal gas, at the temperature T_0, undergoes a process that changes its pressure from P_0 to 2P_0. Then its volume is increased from V_0 to 3V_0 at a constant pressure, as shown on the P-V diagram:

 

Which of the P-T diagrams below correctly reflects this process?

a)

b)

c)

d)

Solution

The answer is b).

Notice that in the branch 1-2, V is constant. Therefore PV = nRT \Rightarrow T \propto P. Since P doubles, T must double as well. So the answer is either a) or b). Next notice that in the branch 2-3, P is constant. Therefore PV = nRT \Rightarrow latex T \propto V$. Since V triples, T must triple. Therefore T goes from 2T_0 to 6T_0, which is reflected in b).

cap 2015 question 11

A sentence from a book by a famous bestselling author Dan Brown:

The pilot nodded. “Altitude sickness. We were at sixty thousand feet. You’re thirty percent lighter up there. Lucky we only did a puddle jump. If we’d gone to Tokyo I’d have taken her all the way up a hundred miles. Now that’ll get your insides rolling.”

Is the statement correct?

\displaystyle a)  Yes
\displaystyle b)  No, the distance from the surface of the Earth have to be about 1000 km for a person to feel 30% lighter.

\displaystyle c)  No, the person will not feel lighter at higher altitude since the upward force of airplanes engine compensates for the lack of gravity
\displaystyle a)  No both arguments 2 and 3 are true

Solution

The first thing to notice is that sixty thousand feet is not that much. Let’s call h the altitude above the Earth where the gravitational force is 30 \% weaker. Then by Newton’s gravitational law,

\frac{GM}{(R_E+h)^2} = 30 \% g

Notice we have divided out the mass of the person. We can rewrite this as

\frac{GM}{R_E^2}\frac{1}{(1+\frac{h}{R_E})^2} = 30 \% g%

But \frac{GM}{R_E^2} = g! It’s just the acceleration at the surface of the Earth! So

\frac{1}{(1+\frac{h}{R_E} )^2} =  30 \% \Rightarrow \frac{h}{R_E} = \frac{1}{\sqrt{30 \%}}-1 \approx 0.83

Using the data sheet, R_E = 6.371 \cdot 10^6 \, m \Rightarrow h = 5.3 \cdot 10^6 \, m which is much much greater than sixty thousand feet. It’s actually about ten million feet. But 5.3 \cdot 10^6 \, m \sim 1000 km so b) is correct.

Notice c) is not correct because suppose we really were ten million feet away. Then the normal force will be 30 \% lighter and we will feel 30% lighter as well. So the answer is b).

cap 2015 Question 12

The figure below shows some light rays coming from point P and passing through a lens.
Which of the distances shown in the diagram corresponds to the focal length of this lens?

a) \, \, 1

b) \, \, 2

c) \, \, 3

d) \, \, 4

Solution

The answer is c)

You just have to know that when you send a beam towards a converging lens parallel to the ground, the ray will pass through the focus. In addition, the focus lies on the ground. Therefore just by looking at the diagrams, c) as the answer.

cap 2015 question 13

A passenger of a train that is moving with the speed of 25 m/s sees from his seat far from a window that it takes 6 seconds for another train to pass the window completely. If the length of the second train is 300 \, m and it is going in the opposite direction, what is its velocity?

\displaystyle a) \, \, 15 \, \frac{m}{s}

\displaystyle b) \, \, 20 \, \frac{m}{s}

\displaystyle c) \, \, 25 \, \frac{m}{s}

\displaystyle d) \, \, 30 \, \frac{m}{s}

\displaystyle e) \, \, 35 \, \frac{m}{s}

solution

The answer is c).

Let’s go to the passenger’s frame. Here, he measures the passing train to have a speed of \frac{300 \, m}{6 \, s} = 50 \, \frac{m}{s}. However, the speed he measures is faster than the speed we measure because the passenger is racing towards the other train. In fact, it is faster by precisely 25 \frac{m}{2} because that is the passengers speed as measured by us. Therefore the other train has a speed of (50-25) \,  \frac{m}{2} = 25 \, \frac{m}{s}

cap 2015 Question 14

An observer is running toward a mirror with a speed of 15 \, km/h, and the mirror is moving toward the observer with a speed of 10  \, km/h. What is the relative speed with which the observer sees his image approaching him?

\displaystyle a) \, \, 10 \, \frac{km}{h}

\displaystyle b) \, \, 20 \, \frac{km}{h}

\displaystyle c) \, \, 25 \, \frac{km}{h}

\displaystyle d) \, \, 30 \, \frac{km}{h}

\displaystyle e) \, \, 50 \, \frac{km}{h}

Solution

The answer is e).

Consider the frame where the mirror is standing still. Then in that frame, the observer is running towards the mirror at a speed of 15 \, \frac{km}{h} + 10 \, \frac{km}{h} = 25 \, \frac{km}{h}. It’s obvious in this frame that his image is running towards him at 25 \, \frac{km}{h} as well. Therefore their relative speed is (25 + 25) \, \frac{km}{h} = 50 \, \frac{km}{h}.

cap 2015 question 15

Three identical closed containers are filled with gases at the same temperature. Container A is filled with 64 \, g of oxygen, container B is filled with 84 \, g of nitrogen, and container C is filled with 8 \, g of hydrogen. Which is the correct ranking of the pressures in the containers?

solution

The answer is d)

Since the gases are at the same temperature and volume, the container with the highest pressure is the one with the greatest number of moles. How do we find the number of moles? It’s just the mass divided by the mass per mole! And these are given in the data sheet. The molar mass of H_2 is 2.016 \, g/mol, the molar mass of O_2 is 31.998 \, g/mol, and the molar mass of N_2 is 28.013 \, g/mol.

Let’s do some division.

The number of moles in container A is \frac{64 \, g}{31.998 \, g/mol} = 2.0 \, mol.

The number of moles in container B is \frac{84 \, g}{28.013 \, g/mol} = 3.0 \, mol.

The number of moles in container C is \frac{8 \, g}{2.016 \, g/mol} = 4.0 \, mol.

Therefore P_A < P_B < P_C 

cap 2015 question 16

A ball is placed on a massless spring that is held at an angle of \theta with respect to the horizontal. The spring is then compressed a distance of x and released. When the ball reaches the maximum height of its trajectory, it is traveling at a speed v. Then a different ball, weighing four times as much as the first, is placed on the spring which is still at an angle \theta. The spring is again compressed a distance x and released. Compared to the first ball, the second ball reaches a maximum height that is

solution

This is a tricky question that many students have a hard time understanding. The reason it’s easy to mess up is partly because the question wasn’t very clear. When they say “A ball is placed on a massless spring,” they mean that that the ball is initially in equilibrium with the spring. I imagine that the spring is being launched up a ramp, and that initially the ball is perfectly still. Here’s the difficult part. As long as the ball is connected to the spring, we can completely forget about gravity and take the initial position to be the rest length of the spring. Here’s a helpful way to think about it. Imagine you have a mass hanging from a spring hanging from the ceiling. Than we know experimentally that the mass undergoes oscillations around the equilibrium point. But this equilibrium point is not the relaxed length of the spring. It is shifted by exactly \frac{mg}{k} downwards to provide a force of k \cdot \frac{mg}{k} = mg back upwards to compensate for gravity. Going back to our problem, the speed u of the ball as it passes it’s equilibrium point and is released from the spring satisfies

1/2 m u^2 = 1/2 kx^2 \Rightarrow u^2 \propto \frac{1}{m}

Where m is the ball’s mass.

But the final height of our ball is the final height of another ball thrown vertically with a speed u \sin{\theta} because horizontal components do not matter at all. Therefore by conservation of energy, the height h is

mgh = 1/2 m u^2 \sin^{2} \theta \Rightarrow h \propto u^2

Since u^2 \propto \frac{1}{m},

h \propto 1/m

Increasing the mass by a factor of 4 therefore decreases the height by a factor of 4. This corresponds to b).

 

cap 2015 question 17

A laser beam propagating in glass (shown in white) hits the rectangular gap filled with air (shown in gray). Which line shown on the figure below represents correct path of the beam?

 
solution

The answer is d).

You should be able to solve this problem without any math at all. After doing a lot of Snell’s law problems, you should know intuitively where the light will turn. But here’s the math for completeness. Consider the first refraction from glass to air.

By Snell’s Law, n_g \sin{\theta_g} = n_{air} \sin{\theta_{air}}

where n_g is the index of refraction in glass, n_{air} is the index of refraction in air, \theta_g is the angle of incidence and \theta_{air} is the angle of retraction. Since n_g > n_{air}, \theta_air > \theta_g. The only path that satisfies this equation is path 4. Therefore the answer is d).

cap 2015 question 18

A parallel beam of light of frequency 6.9 \cdot 10^{14} \, Hz enters a glass plate with an index of refraction n = 1.5. The frequency of light in the glass is:

\displaystyle a) \, \, 4.6 \cdot 10^{14} \, Hz

\displaystyle b) \, \, 6.9 \cdot 10^{14} \, Hz

\displaystyle c) \, \, 10.36 \cdot 10^{14} \, Hz

\displaystyle d) \, \, 13.8 \cdot 10^{14} \, Hz

solution

The answer is b).

The frequency of light does not change as light travels through different mediums. I like to think that if the frequency were not the same, there would be a “traffic jam” at the interface between the air and the glass. In some time \Delta t, the number of waves incident on the glass is f \Delta t and the number leaving is f' \Delta t. These two must be equal or else photons will clump up at the surface. Please keep in mind that this is by no means a derivation. It’s just an easy way to remember things. Anyways, this gives us the answer immediately.

cap 2015 question 19

Students are watching a science fiction movie where the blood spilled during a battle on the a spaceship with its engines turned off form spheres floating in air. The students think it is unphysical.

\displaystyle a)  The students are right, because the blood should not form spheres.
\displaystyle b)  The students are wrong, because in free fall the liquid should form into spheres due to atmospheric pressure.

\displaystyle c)  The students are wrong, because in free fall the liq- uid should form the spheres due to the surface tension.

\displaystyle d) The students are right, because the gravitational forces should immediately pull the blood toward the walls or other objects on the ship.

solution

The answer is c).

First of all, d) is definitely wrong because gravitational force is much much much too weak. b) is also wrong because in space there is no atmospheric pressure. a) is wrong because the blood has to form a sphere by symmetry. If we neglect any interaction with the ship, there is absolutely no preferred direction. The only perfectly symmetrical shape in math is a sphere. This leaves c). We know that liquids like blood have surface tension that resists an increase in area. In other words, you have to put in energy to increase the radius of the sphere. It’s easy to see the shape that minimizes area for a given volume is a sphere. For example, imagine the shape is a very very thin pizza. Then as you make it arbitrary thin, the area gets arbitrarily large! You want to make the shape super compact, and the most compact shape in math is a sphere. Therefore c) is correct.

cap 2015 question 20

Unstable elementary particles are produced traveling at v = 0.6c with respect to an observer in a labora- tory. These particles have a typical intrinsic lifetime of 100ns. In the frame of the observer, how long will the particles typically last before decaying?

\displaystyle a)  They won’t decay while moving

\displaystyle b) \, \, 100 \, ns

\displaystyle c) \, \, 125 \, ns

\displaystyle d) \, \, 175 \, ns

\displaystyle e) \, \, 80 \, ns

solution

The key observation is that in the particles frame, it decays in 100 \, ns. By time dilation, 100 \, ns in the particle’s frame will be longer in the lab frame. The gamma value is

\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = \frac{1}{\sqrt{1-0.6^2}} = 1.25

Therefore 100 \, s in the particle’s frame is

\gamma \cdot 100 \, ns = 1.25 \cdot 100 \, ns = 125 \, ns

cap 2015 question 21

A charged moving object enters a volume where a uniform field is present. After some time the object moves in a circular orbit. Which field was in this area?

a)  gravitational

b)  magnetic

c)  electrostatic

d)  both a or c are possible

e)  both b or c are possible

solution

It isn’t gravitational because in a constant gravitational field, the acceleration will always point in a constant direction. However, if the orbit is circular, the acceleration constantly changes directions. In the same way, it can’t be electrostatic. This leaves magnetic. And this is definitely possible! You’ve probably done a problem like this before, so you recognize the problem. If not, here’s a quick derivation. Let’s say that the particle is initially moving in the xy plane and that the magnetic field points along the negative z axis. Then you should verify yourself that at each point the force points perpendicular to the velocity. From there, it’s easy to see that the particle moves in a circle.

cap 2015 question 22

Students are watching a science fiction movie where the crew of one spaceship watches an explosion on the other spaceship. After a short time interval the crew hear the sound of the explosion. The students think it is unphysical.

solution

The answer is d).

It’s important to remember that sound requires air. Sound is just pressure oscillations in air, so in the vacuum of space, there’s no medium for it to move. Therefore d) is correct. a) is obviously wrong. If sound doesn’t propagate at all, it certainly can’t propagate at the speed of light. c) is also obviously wrong. Since the sounds never even reach the ship, the metal couldn’t have blocked anything. And of course c) is wrong.

cap 2015 question 23

A beam of electrons is sent through a small hole in a piece of foil. The places where the electrons hit on a distant screen are recorded. If we make the hole smaller, the region where the electrons are hitting the screen will be

a)  bigger

b)  the same

c)  smaller

solution

The answer is a).

This is another problem where you should know the answer intuitively. Treat the electrons as a wave. Then the setup is essentially that of a single slit diffraction pattern. Therefore you know that if you decrease the hole, the wave flares out more dramatically. This makes the region larger.

cap 2015 question 24

An astronaut takes a pendulum up to the International Space Station (ISS). The ISS orbits 330 \, km above the surface of the Earth. The radius of the Earth is 6371 \, km. Compared to when it is at ground level, the pendulum on the ISS swings with a period that is:

a) \, \, 49.3 × 10−3 times as long.

b) \, \, 0.95 times as long.
c) \, \, 1.05 times longer.
d) \, \, 20.3 times longer.

e)  The pendulum does not swing on the ISS.

solution

 

The answer is e)

This is a pretty fun problem. It’s very tempting to go through the calculation to find the free fall acceleration on the ISS, then use the formula for the period of a pendulum. However, you have to remember that there is no gravity on the ISS! The centrifugal force due to rotation around the Earth exactly cancels out the gravitational force from the Earth. That’s why astronauts always look like they’re floating! Now obviously if there’s no gravity, the pendulum will never swing!

cap 2015 question 25

Two objects, A and B, appear to be the same length in

a reference frame when A is stationary and B is mov-

ing with speed 3c along its length. In the frame of

reference where B is stationary and A is moving, what is the ratio of their lengths?

\displaystyle a) \, \, L_A/LB = 5/4

\displaystyle b) \, \, L_A/LB = 16/25

\displaystyle c) \, \, L_A/LB = 4/5

\displaystyle d) \, \, L_A/LB = 9/25

\displaystyle e) \, \, L_A/LB = 25/16

solution

The answer is b)

By length contraction, B is actually contracted from it’s rest frame by \gamma = \frac{1}{\sqrt{1-v^2/c^2}} = \frac{1}{\sqrt{1-(3/5)^2}} = 5/4. However in that frame latex A$ is also contracted by the same factor of 5/4. Therefore L_A/L_B = 4/5 \cdot 4/5 = 16/25.

 

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