Ap1
Question
A mass is suspended from the ceiling by one rope at an angle as indicated. The mass is held in place by a second, horizontal rope. Without knowing the exact angle from the
ceiling, except that it is greater than 45 degrees, which statement best represents the possible value for tension in the horizontal line?
(A) T > mg
(B) T = mg
(C) T < mg
(D) T= \(mg/\sqrt{2}\)
Answer/Explanation
Ans:C
T < mg. Since the mass is stationary, we conclude that the net force is zero. The upward tension of the angled rope must be mg in order to cancel out gravity. The free-body diagram on the hanging mass is:
Since the angle θ is greater than 45 degrees, the vertical component of tension \(T_2\) must be bigger than the horizontal component (To see this quickly, draw an extreme case like
85 degrees). Therefore, the horizontal component of tension must be less than mg. The horizontal rope T must cancel this horizontal component of \(T_2\)
Question
Pushing on a mass of 15 kg with a sideways force of 120 N is not enough to get the mass to start sliding across the floor. Which of the following best describes the coefficient of static friction between the floor and the mass?
(A) μ = 0.8
(B) 0.8 < μ < 1.25
(C) μ < 0.8
(D) μ = 1.25
Answer/Explanation
Ans:(B)
μ > 0.8. The normal force between the mass and the floor is 150 N. The needed sideways push to overcome static friction is greater than 120 N. Dividing these two forces would give us:
\( \mu_s = \frac{120}{150} = 0.8\)
Since a greater force is needed, the coefficient of static friction must be greater than this value.
Question
While standing on an elevator and experiencing a downward acceleration of \(4.9 m/s^2\), what is the force between the floor and your feet?
(A) mg
(B) 0.5 mg
(C) 1.5 mg
(D) 4.9 mg
Answer/Explanation
Ans:(B) 0.5 mg. The free-body diagram consists of only two forces.
Newton’s second law gives \( \mu_s +N − mg = ma\). Let a = −0.5 g, which is half the value of g and downward: N − mg = −0.5 mg ,N = 0.5 mg
Question
After throwing a ball at an upward angle, the reason the ball continues to go horizontally while falling due to gravity is that
(A) the ball’s inertia keeps it going
(B) the force of the throw keeps it going
(C) the energy from gravitation potential energy keeps replenishing the lost energy
(D) air pressure keeps the ball from falling too quickly
Answer/Explanation
Ans:(A)
The ball’s inertia keeps it going. Newton’s first law says that a body’s inertia maintains its current motion once in motion. After the force of the throw, the only force on the ball is the vertical force of gravity. This causes the ball to accelerate in the vertical direction while the ball’s sideways velocity in maintained due to its inertia.
Question
Consider the situation in which your hand is pushing a book across a rough desktop with lots of friction. Of the many forces involved, consider only these two individual
forces: the force on your hand from the book and the force the book is experiencing from your hand. While the book is accelerating from rest to its final velocity, which
statement best compares the force experienced by your hand compared with that
experienced by the book?
(A) Force on book > force on hand
(B) Force on hand > force on book
(C) Force on hand = force on book
(D) The relationship between these two forces depends on the frictional force
Answer/Explanation
Ans:(C)
Force on hand = force on book. Newton’s third law of action-reaction states that the force pair of interaction between two objects is always equal and opposite. The acceleration of either object has no role in the third law. The acceleration comes about from the net force on a single object.
Question
A puck comes to a stop in 20 meters after an initial speed of 10 m/s. What is the coefficient of kinetic friction between the puck and the floor?
(A) The mass is required to calculate this answer.
(B) 0.25
(C) 0.5
(D) 0.025
Answer/Explanation
Ans:(B)
To find force, we must first find acceleration:
\(v_f^2 = v_i^2 + 2ad\)
(0)2 = (10 m/s)2 + (2a) (20 m)
\(a = -2.5 m/s^2\)
The net force horizontally is the friction. There are no other horizontal forces:
Newton’s second law gives us:
f = ma = −m(2.5)
Our model for friction gives us:
f = −μN
This value is negative since it is to the left. Since the vertical forces cancel out, N = mg.
Substituting for friction from above and mg for N:
−m(2.5) = −μmg
Note that mass cancels:
μ = (2.5/g) = 0.25
Question
A rock is dropped from an extremely high cliff and experiences free-fall conditions. Which of the following statements is NOT true about the rock’s velocity, acceleration, or displacement between the 3rd and 4th seconds of falling?
(A) The rock will fall an additional 4.9 meters.
(B) The rock will speed up by 9.8 m/s.
(C) The rock’s acceleration will remain 9.8 m/s2.
(D) The rock’s displacement, velocity, and acceleration vectors are all directed downward.
Answer/Explanation
Ans:(A)
The rock will fall an additional 4.9 meters is NOT true. An object in free fall will accelerate by a constant 9.8 m/s2. Since the rock is already moving downward, this leads to an increase in speed of 9.8 m/s in that 1-second interval. However, the rock will fall much farther than 4.9 m as it is already moving close to 30 m/s downward by the start of the 3rd second. It will fall an additional 35 meters in that 4th second. The rock does indeed fall 4.9 meters in the 1st second of the drop but falls an increasingly greater distance each second thereafter.
Question
When using the kinematics equation \(x = x_0 + v_{x0}t + ½ a_xt^2\), which of the following is always assumed to be true?
(A) \(a_x\) = zero
(B) \(x > x_o\)
(C) \(V_{x0}\) is positive.
(D) \(a_x\) is constant.
Answer/Explanation
Ans: (D)
\(a_x\) is constant. All the equations of motion are predicated on the assumption of constant acceleration. Any permutation of signs or values will work in any of the equations of motion. If the acceleration is not constant, one must use graphical methods or approximate the acceleration as constant over a narrow interval of time.
Question
Consider a projectile launched at 30 degrees above a flat surface. The projectile experiences no air resistance and lands at the same height from which it was launched.
Compare the initial and final conditions of the following vector components:
\(V_x\) \(V_y\) \(a_x\) \(a_y\)
(A) They are all the same at both instances.
(B) They are all different.
(C) \(V_x\) and \(V_y\) have changed, but \(a_x\) , \(a_y\) are constant.
(D) They are all the same except for \(V_y\).
Answer/Explanation
Ans:(D)
They are all the same except for \(V_y\). Projectile motion involves a constant vertical acceleration of −g and no horizontal acceleration at all. Therefore, ax and ay are both
constant throughout the flight. Since ax is zero, there is no change in \(V_x\). So \(V_x\) is also constant. \(V_y\), however, is always changing. For a symmetric problem such as this one, the projectile lands with the same vertical speed as it had initially but in the opposite direction so that \(V_{yf} = −V_{yi}\).
Question
A physics teacher is swirling a bucket around in a vertical circle. If the bucket slips out of his hand when it is directly overhead, the bucket will
(A) fly upward initially
(B) drop straight down
(C) fly out horizontally initially
(D) fly out up and at an angle initially
Answer/Explanation
Ans:(C) The bucket will fly out horizontally initially. Objects executing circulation motion have velocity vectors tangent to their motion while experiencing an inward acceleration. If the inward force is released, Newton’s first law dictates the object follow its velocity vector. Note that gravity will cause the object to fall after it is released, but the initial motion is horizontal.
Question
If the Moon was twice as far from the center of Earth as it is currently, how would its orbital period change?
(A) The Moon’s orbital period would double.
(B) The Moon’s orbital period would increase by a factor of 4.
(C) The Moon’s orbital period would increase by a factor of \(2\sqrt{2}\).
(D) The Moon’s orbital period would decrease by a factor of 2.
Answer/Explanation
Ans:(C)
The Moon’s orbital period would increase by a factor of \(2\sqrt{2}\). Kepler’s third law of orbital bodies states that their orbital periods squared are proportional to their orbital
radius cubed:
Note that you can also obtain this result by examining the equation for orbital speed from Newton (v = (Gm/r)½). This shows that the orbital speed of the Moon must decrease by the square root of 2. Coupling this with the knowledge the new orbit is now twice as large, it will take 2(2½) or 2 2 times as long to orbit once around Earth as it used to.
Question
If you went to another planet, which of the following would be true?
(A) Your mass would be the same, but your weight would change.
(B) Your mass would change, but your weight would be the same.
(C) Your mass would be the same, and your weight would be the same.
(D) Your mass would change, and your weight would change.
Answer/Explanation
Ans:(A)
Your mass would be the same, but your weight would change. Mass is a property of an object and measures the object’s inertia regardless of location. Weight is a force due
to gravity and depends both on the object’s mass and on the planetary gravity the object is currently experiencing.
Question
You lift a heavy suitcase twice. Each time you lift it to the same height, but the second time you do it in half the time. Compare the work done to and power delivered to the suitcase.
(A) Same work, same power
(B) Twice the work, twice the power
(C) Twice the work, same power
(D) Same work, twice the power
Answer/Explanation
Ans:(D)
Same work, twice the power.
Work = force × distance
For lifting operations, the force is assumed to be equal to the weight of the object for most of the lift. (Briefly, at the beginning of the lift, the lifting force must be greater than
the weight in order to get the object moving. This brief force must be different in the second case in order to get the suitcase moving faster. However, this initial force is generally ignored in these calculations as it does not happen over an appreciable distance, just as the decrease in lifting force at the end of the lift is likewise ignored.) The same force
over the same distance yields the same work. If the details of the forces contributing to the work are troublesome, you may prefer to think about the energy gained by the suit-
case (W = ΔE). Since the suitcase has gained the same amount of energy in both cases by being lifted, the work done must likewise be the same:
Power = work / time
If the same work is done in half the time, twice the power must be supplied.
Question
A constant horizontal force of 12 N is applied for 5 meters to move an initially stationary mass of 5 kg. The friction between the floor and the mass does a total of −40
joules of work to the mass. The final speed (in m/s) of the object is
(A) 2(6) ½
(B) 2(2) ½
(C) 2(10)½
(D) 4
Answer/Explanation
Ans:(B)
From the work-energy theorem:
\(W_{net}\) = ΔKE
Work done by the horizontal force = (12 N)(5 m) = 60 J
Word done by friction = −40 J
\(W_{net}\) = +60 −40 = 20 J
So the kinetic energy of the mass has gone up by 20 J. Initially, it had no KE since it was stationary. Therefore:
Question
A skier begins a downhill run at a height of 9 meters and a speed of 2 m/s. If you ignore friction, what will be her speed when she is 3 meters from the bottom of the slope?
(A) 6.0 m/s
(B) 12.8 m/s
(C) 11.1 m/s
(D) 7.7 m/s
Answer/Explanation
Ans:(C)
Mechanical energy is conserved since there are no nonconservative forces involved:
Question
A rubber ball (m = 0.250 kg) strikes a wall horizontally at 3.50 m/s and rebounds elastically. What is the magnitude of impulse delivered to the wall in N · s?
(A) 0.875
(B) 1.75
(C) 2.50
(D) 8.75
Answer/Explanation
Ans:(B)
Impulse = Δmomentum
\(= mv_f − mv_i\)
Since the collision is elastic, no kinetic energy is lost. Assuming the wall remains station-
ary, the rubber ball must rebound with the same speed but in the opposite direction:
\(v_f = −v_i\)
Substituting for \(v_f\) above:
Impulse \(= m(−v_i) − mv_i = −2 mv_i = −2 (0.25 kg)(3.5 m/s) = −1.75 kg m/s\)
Here the negative sign indicates that the impulse delivered to the ball is in the opposite direction from the ball’s initial velocity. Note the ball and the wall receive equal and
opposite impulses (from Newton’s third law) and the problem asks for magnitude (absolute value).
Since FΔt is also impulse, units of N · s must be the same as kg · m/s and is quickly shown: N · s = (\(kg · m/s^2\))s = Kg · m/s
Question
A running back (m = 85 kg) running at 1.5 m/s is tackled from the side by another player (m = 75 kg) running perpendicularly to the running back’s original heading at (1.75 m/s). What is the resulting speed of the two entangled players just after the
tackle?
(A) 2.2 m/s
(B) 0.25 m/s
(C) 1.6 m/s
(D) 1.1 m/s
Answer/Explanation
Ans:(D)
Total momentum must be conserved through the collision. So we need to find only the magnitude of the initial momentum of the system. Each player is originally
running perpendicularly to the other. So their individual momentums are different components. Imagine one as an x-component and the other as a y-component. Therefore,
the resulting magnitude is found using the Pythagorean theorem. Dividing by total mass yields the speed of the combined players after impact:
Question
Which of the following ranks the torques applied by the three equal forces A, B, and C to the long thin rod about its fixed axis as indicated?
(A) \(τ_C > τ_B > τ_A\)
(B) \(τ_A = τ_B > τ_C\)
(C) \(τ_B = τ_C > τ_A\)
(D) \(τ_B > τ_A > τ_C\)
Answer/Explanation
Ans:(D)
Torque = (force) (distance from axis of rotation) sin θ
Force C is applied along the lever arm. So force C has an angle of 180°, resulting in no torque.
Force A has a much shorter lever arm than force B. Therefore the torque supplied by A is much less than the torque supplied by B.
Question
A large spherical cloud of dust in deep space that is spinning slowly collapses under its own gravitational force into a spherical cloud 10 times smaller in diameter. How does its rotational rate change?
(A) The rotational rate does not change.
(B) The rotational rate increases by a factor of 10.
(C) The rotational rate decreases by a factor of 10.
(D) The rotational rate increases by a factor of 100.
Answer/Explanation
Ans:(D)
The rotational rate increases by a factor of 100. The moment of inertia (I) for a sphere is proportional to \(R^2\). Since the shape of the cloud is not changing, we can determine that
the moment of inertia has decreased by a factor of \(10^2\). Since angular momentum is conserved:
\(I_1ω_1= I_2ω_2\)
The new angular velocity (ω2) must be 100 times bigger:
\(I_1ω_1= I_2ω_2= (I_1/100) (100 ω_1)\)
Question
A simple circuit consisting of a battery, switch, and a lightbulb in series is constructed. Which is the best explanation of what happens when the switch is closed for the first time?
(A) The lightbulb goes out immediately.
(B) Electrons are allowed to flow through the switch from the battery. When they reach the bulb, it lights.
(C) Electrons are allowed to flow through the switch. When those electrons already within the bulb move, the lightbulb lights.
(D) Electrons on the negative side of the switch are allowed to first flow. When they reach the bulb, it lights.
Answer/Explanation
Ans:(C)
When those electrons already within the bulb move, the lightbulb lights. Closing a switch completes the path, allowing electrons to move around the circuit. All components within the circuit have electrons that respond to the voltage difference and begin to flow. This resulting movement of electrons already within the bulb causes the lightbulb to glow. Recall that the drift velocities of electrons in simple circuits that make up the current are only about 0.03 cm/s.
Question
A single resistor attached to a constant voltage source has an additional, identical resistor inserted between one end of the resistor and the voltage supply. Which of these
is the correct description of what will happen to the voltage drop across (V) and the current running through the original resistor (I) as a result?
(A) I and V will remain unchanged.
(B) I and V will both be halved.
(C) I will be halved; V will remain unchanged.
(D) V will be halved; I will remain unchanged.
Answer/Explanation
Ans:(B)
The effective resistance of the entire circuit is doubled, meaning the voltage supply will send out only half the current:
\(V = IR\)
\(V_{same} / R_{doubled} = I\)
The voltage used for the entire circuit must be shared between the two resistors. Since the two resistors are identical, the split will be 50/50.
Question
A single resistor attached to a constant voltage source has an additional, identical resistor inserted beside it, forming its own connection across the voltage supply. Which
of these is the correct description of what will happen to the voltage drop across (V) and the current running through the original resistor (I) as a result?
(A) I and V will remain unchanged.
(B) I and V will both be halved.
(C) I will be halved; V will remain unchanged.
(D) V will be halved; I will remain unchanged.
Answer/Explanation
Ans:(A)
The new resistor forms its own parallel path and therefore does not affect the voltage for original path. Since the original resistor has the same voltage and resistance, it will draw the same current. Note, however, that the voltage supply must now furnish twice the current with the second resistor in place.
Question
As two sine waves run into each other on a single rope under tension, as shown above, what is the resulting sequence of events?
(A) Initial constructive interference following by unchanged waves emerging on
opposite sides
(B) Initial constructive interference following by waves bouncing backward
(C) Initial partial destructive interference followed by constructive interference followed by unchanged waves emerging on opposite sides
(D) Initial partial destructive interference followed by constructive interference followed by waves bouncing backward
Answer/Explanation
Ans:(C)
After the waves superpose, they emerge as they were before as if they had never met. As they begin to overlap, the leading trough of the wave on the left will superpose with leading peak of the wave on the right to interfere destructively. However, since the other halves of the waves are unaffected, this is only partial destruction. When the waves are completely superposed, trough will be on
trough and peak on peak, leading to a constructive interference.
Question
As you stand on the side of the road, an ambulance approaches your position at high speed. Which option best describes the sound of the siren as observed by you as compared with the driver’s perception?
Answer/Explanation
Ans:(A)
Wave speed is determined by the medium, which is the air. So wave speed will not be observed to be different by either observer. Wavelength will appear to be shorter to the stationary observer as the
crests are compressed by the ambulance moving toward you between peaks:
Wavelength × frequency = wave speed
Since wavelength is shorter, the frequency must be higher in order to give the same speed. This higher frequency is the directly observable higher pitch that is known as the
Doppler shift.
Question
Two cars approach an intersection at right angles, as pictured above. What is their relative speed to each other?
(A) 10 mph
(B) 45 mph
(C) 57 mph
(D) 80 mph
Answer/Explanation
Ans: (C)
Relative velocity is the difference in the velocity vectors. When the vectors are subtracted, the resultant is the hypotenuse of a right triangle:
Question
Which of the following statements are true regarding collisions? Select two answers.
(A) An elastic collision conserves both net momentum and net kinetic energy.
(B) An inelastic collision conserves mechanical energy but not net momentum.
(C) An elastic collision between two objects has equal but opposite impulses delivered.
(D) An inelastic collision between two objects is one in which both objects come to rest during the collision.
Answer/Explanation
Ans: (A) and (C)
All collisions conserve net momentum. Elastic collisions also conserve kinetic energy. All collisions involve equal but opposite exchanges of momentums. Inelastic collisions simply indicate that kinetic energy is not conserved.
Section ii frq ap1
Question
A static, nonuniform wedge-shaped mass, M, (with center of mass x as indicated) with a base length of L (shown below) is supported at two points. The first support point is at the bottom left corner of the wedge. The second support point is located at a point ¾ of the length of the wedge away from that corner.
(a) Using as givens M, L, and g (the gravitational field strength), solve for the two unknown contact forces at the points of support (F1 and F2).
(b) Support #2 is removed. The wedge begins to fall by rotating around support #1 with an angular acceleration of 2 rad/s2. If the mass of the wedge is 3.8 kg and L = 75 cm, determine the moment of inertia of the nonuniform wedge about the top of support #1.
(c) As the wedge continues to rotate clockwise but remains in contact with support #1:
(i) Does the angular acceleration increase, decrease, or remain the same? Why?
(ii) Does the moment of inertia of the wedge increase, decrease, or remain the same? Why?
Answer/Explanation
Ans:
(a) Both net force and net torque must add up to zero. Make a free-body diagram:
\(F_{net} = +F_1 + F_2 − Mg = ma = 0\)
\(F_1 + F_2 = Mg\)
We now have two unknowns but only one equation! For torque, we are free to choose any axis of rotation we want since the wedge is not actually moving. We should place
the axis of rotation at either contact point in order to eliminate one variable. If we place our axis of rotation at the F1 contact point and use minus for clockwise rotations and plus for counterclockwise rotations:
Net torque \(= F_1(0) – Mg (L/3) + F_2 (3L/4) = Iα = 0\)
\(F_2(3L/4) = Mg (L/3)\)
\(F_2 = 4 Mg/9\)
Combine this equation with the one found above.
\(F_1 = Mg – 4 Mg/9 = 5 Mg/9\)
(b)
Torque = Iα
Since \(F_2\) is gone, the torque is supplied only by the mass of the object:
Torque = FR sin θ
\(Torque = Mg (L/3) (1)\)
Substituting into the above equation:
\(MgL/3 = Iα = I (2)\)
\(I = MgL/6 = (3.8)(9.8)(0.75)/6 = 4.7 kg · m^2\)
Note that \(I = mr^2\) cannot be used as the object has mass spread out over many different r values.
(c) (i) As the object rotates, neither the lever arm nor the force (Mg) changes. However, the angle between the lever arm and force goes from 90° initially toward 0° at the lowest point of the downward swing. Since torque is proportional to sin θ, the torque (and thus the angular acceleration) will decrease.
(ii) Since neither the distribution of mass nor the axis of rotation changes during the swing, the object’s moment of inertia remains constant.
Question
An object of unknown mass is fired at an unknown angle with an initial speed of 25 m/s. Ignore frictional effects. If only 45 percent of its initial kinetic energy is still present as kinetic energy at the projectile’s highest point, determine the time in flight for the projectile.
Answer/Explanation
Ans:
This is a projectile motion (Chapter 3) and energy conservation (Chapter 5) problem. Energy conservation tells us that the other 55 percent of the initial energy is in gravitational potential energy.
Determine the time to fall from that height. Remember there is no vertical velocity at the highest point!
Double this value to obtain time in flight: 3.8 seconds
Question
Two solid spheres of radius R made of the same type of steel are placed in contact, as shown in the figures above. The magnitude of the gravitational force that they exert on each other is \(F_1\).
When two other solid spheres of radius 3R made of this steel are placed in contact, what is the magnitude of the gravitational force that they exert on each other?
(A) \(F_1\)
(B) 3\(F_1\)
(C) 9\(F_1\)
(D) 81\(F_1\)
Answer/Explanation
Answer: D
Question
The figure above shows three resistors connected in a circuit with a battery. Which of the following correctly ranks the energy E dissipated in the three resistors during a given time
interval?
(A) \(E_{300}Ω > E_{200}Ω > E_{100}Ω\)
(B) \(E_{300}Ω > E_{100}Ω > E_{200}Ω\)
(C) \(E_{200}Ω > E_{300}Ω > E_{100}Ω\)
(D) \(E_{200}Ω > E_{100}Ω > E_{300}Ω\)
Answer/Explanation
Answer: C
Question
A person driving a car suddenly applies the brakes. The car takes 4 s to come to rest while traveling 20 m at constant acceleration. Can the speed of the car immediately before the brakes were applied be determined without first determining the car’s acceleration?
(A) Yes, by dividing the distance (20 m) by the time (4 s).
(B) Yes, by determining the average speed while braking and doubling it.
(C) No, because the acceleration is needed to use standard equations such as \(∆x = v_ot +1/2 at^2\).
(D) No, because the fundamental relationship that defines velocity contains acceleration.
Answer/Explanation
Answer: B
Question
While traveling in its elliptical orbit around the Sun, Mars gains speed during the part of the orbit where it is getting closer to the Sun. Which of the following can be used to explain this gain in speed?
(A) As Mars gets closer to the Sun, the Mars–Sun system loses potential energy and Mars gains kinetic energy.
(B) A component of the gravitational force exerted on Mars is perpendicular to the direction of motion, causing an acceleration and hence a gain in speed along that direction.
(C) The torque exerted on Mars by the Sun during this segment of the orbit increases the Mars–Sun system’s angular momentum.
(D) The centripetal force exerted on Mars is greater than the gravitational force during this segment of the orbit, causing Mars to gain speed as it gets closer to the Sun.
Answer/Explanation
Answer: A
Question
The graphs above represent the position x, velocity v, and acceleration a as a function of time t for a marble moving in one dimension. Which of the following could describe the motion of the marble?
(A) Rolling along the floor and then bouncing off a wall
(B) Rolling down one side of a bowl and then rolling up the other side
(C) Rolling up a ramp and then rolling back down
(D) Falling and then bouncing elastically off a hard floor
Answer/Explanation
Answer: C
Question
Multi-Correct: Students need to select all the correct answers to the question below in order to
earn credit.
A race car going around a flat, unbanked circular track gradually increases speed as it completes one full trip around the track. Which of the following can explain why the car gains speed?
(A) Energy stored in the fuel is converted to mechanical energy.
(B) A component of the frictional force exerted by the ground on the tires is directed toward the center of the circle.
(C) A component of the frictional force exerted by the ground on the tires is in the direction of motion.
(D) The car’s velocity and acceleration are perpendicular.
Answer/Explanation
Answer: A and C
Question
You are given a set of chimes that consists of eight hollow metal tubes open at both ends, as shown above. The chimes are played by striking them with a small hammer to produce musical sounds. Your task is to use the chimes to determine the speed of sound in air at room temperature. You have available a set of tuning forks and other common laboratory equipment
but are not allowed to use electronic equipment, such as a sound sensor. (A tuning fork vibrates when struck and produces sound at a particular frequency, which is printed on the tuning fork.)
(a) Describe your experimental procedure in enough detail so that another student could perform your experiment. Include what measurements you will take and how you will take them.
(b) Describe how you will use your measurements to determine the speed of sound, in enough detail that another student could duplicate your process.
(c) Describe one assumption you made about the design of your experiment, and explain how it might affect the value obtained for the speed of sound.
(d) A student doing a different experiment to determine the speed of sound in air obtained wavelength and period measurements and created the following plot of the data. Use the graph to calculate the speed of sound and include an explanation of your method.
Answer/Explanation
Ans:
Question
A student of mass 50.0 kg swings on a playground swing, which is very light compared to the student. A friend releases the seat of the swing from rest at a height of 1.00 m above the lowest
point of the motion. The student swings down and, at the lowest point of the motion, grabs a jug of water of mass 4.00 kg. The jug is initially at rest on a small table right next to the swing, so it does not move vertically as the student grabs it. The student keeps swinging forward while holding the jug, and the seat reaches a maximum height H1 above the lowest point. Air resistance and friction are negligible.
(a) Indicate whether H1 is greater than, less than, or equal to 1.00 m.
—————-Greater than 1.00 m
—————-Less than 1.00 m
—————-Equal to 1.00 m
Justify your answer qualitatively, with no equations or calculations.
(b) Explain how H1 can be calculated. You need not actually do the calculations, but provide complete instructions so that another student could use them to calculate H1.
(c) The student now swings backward toward the starting point. At the lowest point of the motion, the student drops the water jug. Indicate whether the new maximum height that the seat reaches is greater than, less than, or equal to H1 .
——————–Greater than H1
——————–Less than H1
——————–Equal to H1
Justify your answer.
Answer/Explanation
Ans:
Question
The figure above shows four resistors connected in a circuit with a battery. Which of the following correctly ranks the potential difference, ΔV, across the four resistors?
(A) \(∆V_{4R} > ∆V_{3R} > ∆V_{2R} > ∆V_R\)
(B) \(∆V_{4R} > ∆V_{3R} > ∆V_{2R} = ∆V_R\)
(C)\(∆V_{4R} = ∆V_{3R} > ∆V_{2R} > ∆V_R\)
(D) \(∆V_{2R} = ∆V_R > ∆V_{3R} > ∆V_{4R}\)
Answer/Explanation
Answer: B
Refer to the diagram below for questions (a)–(b)
The figure above represents an electric field created by charged objects that are not shown. The field vectors and the locations W and Z are in the plane of the page.
Question(a)
At which location is the electric potential greater?
(A) W
(B) Z
(C) Neither; the potential is the same at both locations.
(D) It cannot be determined without knowing the values of the charges on the objects creating
the electric field.
Answer/Explanation
Answer: B
Question(b)
A small charged bead held inside the electric field has an electric force exerted upon it. At which location does the electric force have a greater magnitude?
(A) W
(B) Z
(C) Neither; the magnitude of the force is the same at both locations.
(D) It cannot be determined without knowing the sign of the charge on the bead.
Answer/Explanation
Answer: A
Question
A student writes the following information for a process that involves a fixed quantity of ideal gas.
W = −P∆V
∆U = Q + W
P = \(2.0\times 10^5\) Pa
∆V = \(- 2.0\times 10^{-3} m^3\)
∆U = – 600J
Which of the following descriptions best represents the process?
(A) The gas expands at a constant pressure of 200 kPa.
(B) The gas is cooled at constant volume until its pressure falls to 200 kPa.
(C) The gas is compressed at a constant pressure of 200 kPa.
(D) The gas is heated and its pressure increases at constant volume.
Answer/Explanation
Answer: C
Question
An electron is at point P in a uniform magnetic field directed into the page, as depicted above. For which of the following states of motion of the electron is the magnetic force exerted on the electron equal to zero?
(A) The electron is not moving.
(B) The electron is moving perpendicularly into the page.
(C) The electron is moving perpendicularly out of the page.
(D) The electron is moving.
Answer/Explanation
Answer: A, B, C
Question
The figure above shows a pipe of height \(H_P\) and cross-sectional area \(A_P\) attached to the top of a tank of height \(H_T\) and cross-sectional area \(A_T\) . The pipe and tank are completely filled with water. The force exerted by the water on the bottom of the tank depends on which of the given quantities?
(A) \(A_P\)
(B) \(A_T\)
(C) \(H_P\)
(D) \(H_T\)
Answer/Explanation
Answer: B, C, D
Question
The figure above represents a glass lens that has one flat surface and one curved surface. After incoming parallel rays pass through the lens, the rays pass through a focal point. The focal length f is the distance from the center of the lens to the focal point.
(a) The rays undergo refraction and change direction at the right surface of the lens, as shown. Explain why the angle of refraction of ray 1 is greater than that of ray 2.
(b) The index of refraction of the glass is nglass, and the radius of curvature of the lens’s right edge is R. (The radius of curvature is the radius of the sphere of which that edge is a part.
A smaller R corresponds to a lens that curves more.) A teacher who wants to test a class’s understanding about lenses asks the students if the equation f = nglassR makes sense for
the focal length of the lens in air. Is the teacher’s equation reasonable for determination of the focal length? Qualitatively explain your reasoning, making sure you address the dependence of the focal length on both R and nglass.
(c) An object is placed a distance f / 2 (half of the focal length) to the left of the lens. On which side of the lens does the image form, and what is its distance from the lens in terms of f ? Justify your answer. (Assume this is a thin lens.)
(d) The lens is now placed in water, which has an index of refraction that is greater than air but less than the glass. Indicate below whether the new focal length is greater than, less than, or
equal to the focal length f in air.
——————Greater than in air
——————Less than in air
——————The same as in air
Justify your answer qualitatively, with no equations or calculations.
Answer/Explanation
Ans:
Question
Two small conducting spheres attached to very thin silk filaments hang from a very large metal beam. The silk attached to sphere 1 is dry and therefore nonconducting, while the silk attached
to sphere 2 is wet, making it a conductor. Initially the spheres are uncharged. A student takes a positively charged plastic rod and moves it close to both spheres, holding the rod the same distance from each sphere as shown above. Throughout the experiment, the rod never touches the spheres, and the spheres never touch each other.
(a) Which of the spheres moves toward the rod? If they both do, which one moves closer to the rod? Explain your reasoning, using words and diagrams. The wet silk filament attached to sphere 2 gradually dries out. After the silk is completely dry, the rod is moved far away from both spheres.
(b) Indicate below how the two spheres are now positioned, relative to their original vertical positions.
————–Closer to each other
————–Farther from each other
————–In their original vertical positions
Explain your reasoning.
(c) Which of the following describes how the forces exerted by the spheres on each other now compare?
—————Sphere 1 exerts a greater force on sphere 2 than sphere 2 exerts on sphere 1.
—————Sphere 2 exerts a greater force on sphere 1 than sphere 1 exerts on sphere 2.
—————Sphere 1 exerts the same force on sphere 2 that sphere 2 exerts on sphere 1.
Briefly explain your reasoning.
Answer/Explanation
Ans:
Ap2
Question
Compare the two columns of water pictured below. They have the same height. Which of the following statements compares the total weight and the pressure at the bottom of the columns?
(A) Weight and pressure are greater for the wider column.
(B) Weight and pressure are the same for both.
(C) Weight is greater for the wider column, while pressure is the same.
(D) Weight is the same, while pressure is greater for the wider column.
Answer/Explanation
Ans:(C)
Although the wider column has a greater volume of water and therefore more mass, the fact that the
height remains constant means any unit area on the bottom of the column will experience the same force/area (pressure).
Question
The exact same boat will ride higher (have more volume exposed to the air) while floating on a fluid if
(A) the fluid’s density is raised
(B) the fluid’s temperature is raised
(C) the boat’s density is raised
(D) the boat shifts its cargo to a lower deck
Answer/Explanation
Ans:(A)
The buoyancy force that lifts the boat upward is proportional to the weight of the fluid displaced. The boat will rise or sink until the buoyant force and the boat’s weight cancel. Raising the fluid’s temperature will usually make the fluid less dense, requiring a greater displacement of water (i.e., the boat will ride lower). Raising the boat’s density will increase its weight, requiring a greater displacement of water. Finally, shifting the cargo around will have no effect as the weight and shape of the boat remain unaffected.
Question
Fluid is poured into an open U-shaped tube as shown below. If a person blows across the top of the right hand side’s opening, what will happen to the fluid in the tube?
(A) It will remain the same.
(B) It will rise on the right-hand side and lower on the left-hand side.
(C) It will rise on both sides.
(D) It will rise on the left-hand side and lower on the right-hand side.
Answer/Explanation
Ans:(B)
Blowing over the right-hand side will lower the pressure at that height:
\(\frac{1}{2}ρv^2 +ρgh + P= constant = constant\)
Remember that h remains the same while v is raised, forcing P lower. This causes animbalance in pressure between the right-hand side (lower pressure) and left-hand side (higher pressure). This difference in pressure causes a net force pushing the fluid upward on the right and downward on the left.
Question
The Sun’s rays strike a black surface that is directly on top of a liquid. The liquid is then observed to swirl as it transfers the heat to the bottom layers of the liquid. The correct sequence of transfers in thermal energy in this story is
(A) Radiative → convective → conductive
(B) Radiative → conductive → convective
(C) Convective → conductive → convective
(D) Convective → conductive → radiative
Answer/Explanation
Ans:(B)
Radiative transfer is the transfer of energy via electromagnetic radiation (such as sunlight). Conductive transfer is the transfer of thermal energy via direct contact (such as a floating object in contact with the water). Convective transfer is the transfer of thermal energy due to the movement of a fluid (like currents in the water).
Question
Which of the following is the best explanation of the statement, “A roomful of neon gas has a lower average speed than 1 mole of helium gas at the same temperature and pressure.”
(A) This is not a correct statement.
(B) There is more than 1 mole of neon gas in an average room.
(C) Helium atoms have more degrees of freedom than do neon atoms.
(D) Helium has a lower mass per particle than neon.
Answer/Explanation
Ans:(D)
Average kinetic energy is directly related to temperature. Since both gases are at STP (standard temperature and pressure),
their average kinetic energies are the same. However, since helium has a lower mass, it must have a higher velocity to obtain the same kinetic energy \((½mv^2)\).
Question
Consider the following process for a gas. Which of the following statements is true concerning this process?
(A) Path D (from point 4 to point 1) is isobaric.
(B) No work was done to or by the gas during path A (from point 1 to point 2).
(C) No net work was done to or by the gas during the entire cycle (from point 1 back topoint 1 through A, B, C, D).
(D) No work was done to or by the gas during path B (from point 2 to point 3).
Answer/Explanation
Ans:(D)
The term isobaric indicates constant pressure, so choice (A) is false. During path B, the gas expands and therefore does work to its surroundings. So choice (B) is false. Over the entire process, the area of the enclosed area is the work done by the gas. Since this area does not equal zero, choice (C) is false. Process D involves an expansion or contraction (displacement) of the gas. Therefore no work is done during the part of the process.
Question
Two identical spheres originally in contact undergo induced charge separation. The two spheres are slowly separated until a gap of 12 cm is separating their surfaces. The net force of attraction between the two oppositely charged large spheres after they have been separated by 12 cm is measured. The entire experiment is then repeated with two smaller but otherwise similar spheres.
The force between the smaller spheres compared with the force between the larger
spheres is best described as
(A) the same since the charges and separations are the same
(B) smaller since the charges on each small sphere are closer together
(C) bigger since the center of the spheres are closer
(D) the same since the average force felt by any one particular charge is the same in
both cases
Answer/Explanation
Ans:(C)
Remember \(kq_1q_2/r^2\).
The charges \((q_1)\) and \((q_2)\) are the same,and k is a constant. So the only variation in the two situations is the average r between the spheres. In the case of the 2 smaller spheres, the
average distance (represented by the center of the spheres) is actually closer. Therefore the force is stronger in that case.
Question
One large conducting sphere with a net charge of +8 microcoulombs is put into contact with a slightly smaller conducting sphere that initially has a charge of –6
microcoulombs. Which statement best describes the charge distribution after the two spheres are separated?
(A) They are both neutral.
(B) The larger sphere has +8 microcoulombs of charge, while the smaller retains –6 microcoulombs of charge.
(C) Both spheres have +1 microcoulomb of charge.
(D) The larger sphere has slightly more than +1 microcoulomb of charge, while the smaller sphere has slightly less than +1 microcoulomb of charge.
Answer/Explanation
Ans:(D)
When the spheres contact each other, electrons move off of the smaller sphere to the larger sphere and attempt to neutralize both spheres. Since this is a net +2 microcoulombs of charge, the electrons distribute themselves over both surface areas of the spheres, keeping the positive charges as far from each other as possible. Since the larger sphere has more surface area, it captures a greater portion of the excess charge than does the smaller sphere.
Question
Which of the following pictures is an incorrect representation of the electric field near various charged conductors?
Answer/Explanation
Ans:(C) Field lines should point away from the positive and toward the negative, which all the pictures show. Field lines should never cross, and none of these diagrams has that. Field lines should also be perpendicular to the surface of conductors. Picture (C) violates this principle.
Question
A standard RC (resistor-capacitor) circuit takes approximately 2 seconds to charge the capacitor fully. If the resistor is 100 ohms and the voltage across the RC combination is 12 volts, what is the current through the capacitor?
(A) One must know the capacitance of the capacitor in order to answer this question.
(B) 0.12 A
(C) 0.06 A
(D) 0 A
Answer/Explanation
Ans:(D) A fully charged capacitor cannot accept any more charge; therefore, the current going into it is zero.
Question
If each of the batteries in the circuit below supplies a voltage V and if each resistor has resistance R, what will be the reading on the meter pictured?
(A) 0
(B) V/3
(C) 2V/3
(D) 2V/3R
Answer/Explanation
Ans:(B)
The meter pictured is a voltmeter. The voltage supplied by the battery will be split evenly among the three identical resistors. Having the additional battery in parallel
does not affect the potential difference. The current drawn from each battery will be half of the current flowing through each resistor.
Question
The primary difference between a permanent magnet (an object with identifiable north and south poles) and magnetic material that does not have identifiable north and south poles is
(A) electron configuration
(B) magnetic domain alignment
(C) magnetic materials have no poles at all
(D) magnetic materials are lacking one of the two poles
Answer/Explanation
Ans:(B)
Electron configuration is the difference between a non- magnetic and a magnetic material. Any magnetic material can be made to have overall north and south poles by lining up all of the individual magnetic domains within the material. (Each domain has its own north and south poles.) Magnetic poles do not exist in isolation as all magnetic field lines loop back on themselves.
Question
A very long wire carries a steady current. What is the magnetic field at a distance of R from the wire?
(A) The magnetic field is zero.
(B) The magnetic field loops around the wire and gets weaker proportional to \(1/R^2\) as one moves to larger R values.
(C) The magnetic field loops around the wire and gets weaker proportional to 1/R as one moves to larger R values.
(D) The magnetic field extends radially outward and gets weaker proportional to \(1/R^2\) as one moves to larger R values.
Answer/Explanation
Ans:(C)
The moving charges in the wire create loops of magnetic field around them. As R increases, the field strength decreases proportional to 1/R rather than \(1/R^2\) because the source of the field is aline rather than a point.
Question
Two parallel wires carrying current near each other exert magnetic forces on each other. If the direction and magnitude of the current in both wires is reversed and doubled, what will happen to the magnetic forces?
(A) They will reverse directions and double in magnitude.
(B) They will reverse directions and quadruple in magnitude.
(C) They will remain in the same directions and double in magnitude.
(D) They will remain in the same directions and quadruple in magnitude.
Answer/Explanation
Ans:(D)
Reversing the direction of one current will switch the direction of its magnetic field. However, when using the right-hand rule and since the direction of the moving charges within the
reversed field is also reversed, the same direction of force is found. Doubling the source current will double the field strength, which then is acting on twice as much current and creating 4 times the force.
Question
A conducting loop of wire is held steady parallel to the ground within a strong magnetic field directed upward. The setup is viewed from above. As the loop of wire is quickly pulled out of a magnetic field, what will the loop experience?
(A) A temporary clockwise current that dies out soon after the loop is out of the field
(B) A temporary counterclockwise current that dies out soon after the loop is out ofthe field
(C) A temporary clockwise current that dies out as soon as the loop is out of the field
(D) A temporary counterclockwise current that dies out as soon as the loop is out of
the field
Answer/Explanation
Ans:(D)
Only while the flux of the field through the loop is changing will there be an induced emf causing the current. The induced current will be such that its magnetic field will oppose the change. Since the magnetic field flux upward through the loop is getting smaller as the loop is pulled out, the induced current will create an additional upward magnetic field. The right-hand rule for current indicates a counterclockwise current is needed.
Question
A charge is accelerated by an electric field. While the charge is accelerating, which of the following is occurring?
(A) A static magnetic field is created.
(B) A new, static electric field is created.
(C) Only a varying magnetic field is created.
(D) An electromagnetic wave is created.
Answer/Explanation
Ans:(D)
When a charge is accelerating, the changing velocity of the charge creates a changing magnetic field. The changing magnetic field induces a changing electric field. This dual oscillation is self-perpetuating and is known as electromagnetic radiation (or light when the frequency of oscillation lies within the visible range)
Question
A light beam is traveling in a vacuum and then strikes the boundary of another optically dense material. If the light comes to the interface at an angle of incidence of 30 degrees on the vacuum side. It will emerge with an angle of refraction
(A) equal to 30 degrees
(B) less than 30 degrees
(C) greater than 30 degrees
(D) the relative indices of refraction must be known in order to find the solution
Answer/Explanation
Ans:(B)
Since the second medium is slower, it will have a larger n value, thereby requiring a smaller sin θ value. So a smaller angle is needed:
\(n_1 sin θ_1 = n_2 sin θ_2\)
Question
A beam of light incident at 50 degrees travels from a medium with an index of refraction of 2.5 into one with an index of refraction of 1.1. The refracted angle of light that emerges is
(A) 20 degrees
(B) 50 degrees
(C) 90 degrees
(D) there is no solution
Answer/Explanation
Ans:(D)
This is a situation of total internal reflection. Since 50 degrees is beyond the critical angle for this pair of materials, all of the light will be reflected and none will be refracted. Remember that a 90-degree “refraction” happens right at the critical angle.
Question
A radio telescope uses a curved reflecting material to focus radio signals from orbiting satellites or distance objects in space onto a detector. Which of the following would be the biggest problem with designing a radio telescope with a convex reflecting dish?
(A) The image produced is virtual.
(B) The image produced is inverted.
(C) The detection equipment would have to be placed in the line of sight.
(D) There is no real problem with designing a convex reflecting dish for radio waves.
Answer/Explanation
Ans:(A)
Convex mirrors are diverging. This means the real, reflected rays do not come to an actual focus. The focus is virtual and located behind the mirror. So any detection equipment placed at the focus would not actually receive a signal!
Question
When examining a two-slit diffraction pattern, which of the following would you notice as the slits were brought closer together?
(A) The interference pattern would contain more maximums. They would be more closely spaced.
(B) The location of the central interference peak would shift.
(C) The interference pattern would spread out.
(D) The interference peaks themselves would each become sharper (narrower).
Answer/Explanation
Ans:(C)
D sin θ = mλ
Wavelength does not change. So as D get smaller, sin θ must increase. Therefore, the location of the bright, constructive interference fringes (integer m’s) are found at greater angles.
Question
A convergent lens can produce which of the following from a single object in front of the lens? Select two answers.
(A) Real and upright image
(B) Virtual and upright image
(C) Real and inverted image
(D) Virtual and inverted image
Answer/Explanation
Ans:(B) and (C)
The image is virtual and upright, or it is inverted and real. If the distance of the object is less than 1 focal length from the convergent lens, the image is virtual and upright. If the distance of the object is more than 1 focal length from the convergent lens, the image is inverted and real.
Question
A radioactive isotope undergoes gamma decay. Afterward, the nucleus
(A) contains fewer nucleons
(B) has the same number of nucleons but a different ratio of protons/neutrons
(C) has the same mass number but a lower mass
(D) undergoes beta decay
Answer/Explanation
Ans:(C)
Since the nucleus has emitted a photon of electromagnetic energy, the net charge on the nucleus must remain the same. Since energy left the nucleus, the mass of the nucleus as a whole must be
decreased by \(E/c^2\).
Question
When examining the spectrum for a specific element, a bright spectral line with the highest frequency represents
(A) an electron excited to the atom’s highest energy state
(B) an electron falling from the highest energy state in the atom to the next to highest state
(C) an electron excited from its ground state to the next to lowest energy state
(D) an electron falling from a higher energy state to a much lower energy state
Answer/Explanation
Ans:(D)
Light is emitted (bright spectral line) when an electron falls from an excited state to a lower energy state. The greater the energy difference is (as opposed to the starting energy state),
the higher the frequency of light emitted (E = hf).
Question
Some objects can be made to exhibit an interference pattern with the two-slit experiment, and some objects cannot. What is the difference between those that can and those that cannot?
(A) Objects are either fundamentally a wave or fundamentally a particle.
(B) All objects can exhibit either wavelike properties or particle-like properties depending on the situation.
(C) Whether an interference pattern can be observed depends on the object’s de Broglie wavelength.
(D) All objects can be made to exhibit an interference pattern under the right circumstances.
Answer/Explanation
Ans:(C)
The de Broglie wavelength is the connection between classically particle-like objects and their wavelike properties. Interference patterns have been successfully shown for many particles. Macroscopic objects, however, have such large masses that their de Broglie wavelengths make observing a diffraction pattern impossible: λ = h/p
Question
For an RC circuit, which of the following is the correct analogy for the instantaneous current behavior in a pathway containing a capacitor? Choose the option that would give the correct instantaneous current values in the capacitor’s path.
Answer/Explanation
Ans:(A)
An uncharged capacitor will initially appear to be a simple wire as the first charge to hit one side of the capacitor draws its opposite on the other side. On the other hand, a fully charged capacitor accepts no additional charges and blocks any current from entering its pathway. These two states are connected by an exponential function:
\(I = (V/R) e^{−t/RC}\)
Remember to compare t = 0 to t >> RC.
Question
Which of the following statements are true concerning isolated fundamental particles? Select two answers.
(A) An isolated, fundamental particle cannot have charge.
(B) An isolated, fundamental particle cannot have potential energy.
(C) An isolated, fundamental particle cannot have internal energy.
(D) An isolated, fundamental particle cannot have mass.
Answer/Explanation
Ans:(B) and (C)
Potential energies are energies of relationship and therefore cannot belong to an isolated particle with no internal structure. Likewise, internal energy refers to energies (both kinetic and potential) carried by the various internal constituents of nonfundamental particles.
frq ap2
Question
Use the following circuit to answer the questions below.
(a) When the battery is first attached to the circuit and the capacitor is still uncharged, how much current is being supplied by the battery? Explain your reasoning.
(b) After the capacitor has been fully charged, how much current is being supplied by the battery? Explain your reasoning.
(c) After the capacitor has been fully charged, the battery is removed, leaving an open circuit where the battery was currently located. Determine the direction of current through the 10-ohm resistor and
(i) The total amount of charge that passes through the 10-ohm resistor
(ii) The initial instantaneous current after the battery has been removed
(iii) The approximate time for the capacitor to be half discharged
Answer/Explanation
Ans:(a)
Initially the top path acts as a simple 5-ohm resistor because the capacitor is uncharged and will act as if current is passing right through it. The 10-ohm resistor is in parallel with the capacitor, resulting in an equivalent resistance of 10/3 ohms:
1/5 + 1/10 = 1/R
2/10 + 1/10 = 1/R
3/10 = 1/R
R = 3 Ω
Ohm’s law gives us a 3-amp current coming from the battery:
V = IR
10 = I (10/3)
I = 3 A
(b)
After the capacitor has been fully charged, it will act as an open circuit, no longer allowing charge to flow through the upper path. Equivalent resistance for the circuit will now be 10 ohms
V = IR
10 V = I (10 Ω)
I = 1 A
(c) The direction of current is constant throughout the problem, from right to left, whether it is being supplied by the battery or by the capacitor.
(i) C=Q/V
5=Q/10
The entire 10 V is on the capacitor at the end because the 5-ohm resistor is not using any voltage (I = 0):
Q = 50 μC
(ii) Initially, the 10 volts of the capactor will be driving current through an equivalent resistance of 15 ohms (5 + 10 in series):
V = IR
10 volts = I (15 ohms)
I = 0.67 amps
Note that right away, the voltage will begin to decrease across the capacitor as it discharges, therefore driving increasingly smaller currents through the resistors.
(iii) As an exponential decreasing function, the time constant of RC provides an approximate time for losing about half the charge:
RC=(15 ohms )(5 microfarads)
=75 microseconds
Question
A student is provided with a solenoid attached to an ammeter and with a bar magnet. The student monitors the ammeter as she performs the following sequence of events. Describe qualitatively her observation of the ammeter (comparing current sign and magnitude) at each step of her procedure.
(a) The student slowly inserts the north pole of the bar magnet into the solenoid.
(b) She holds the magnet steady at this location.
(c) The student withdraws the magnet quickly from the solenoid.
(d) She repeats the steps but uses the south pole of the magnet.
What factors about this procedure and the equipment affect the magnitude of current observed?
Answer/Explanation
Ans:This is a magnetism and electromagnetism problem (see Chapter 13 for more
information).
(a) The student will see a modest current in one direction as the bar magnet enters the solenoid. Note that without knowing the actual orientation of the loops, the magnet, and the ammeter, the actual sign of the current cannot be determined.
(b) She will observe no current while the magnet is being held still.
(c) The student will observe a current in the opposite direction as it is withdrawn. The magnitude of this current will be larger as the change in magnetic flux is happening quicker.
(d) The same as described above but with opposite signs for the current. The number of coils, the total resistance of the coils, the strength of the magnet, the speed at which the magnet is moved, and possibly the orientation of the magnet relative to the cross- sectional area of the solenoid all affect the magnitude of the observed current.