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Practice Exam 1

42 multiple-choice and 4 free-response questions, in the format of the AP Physics 1: Algebra-Based Exam. Each section runs on its own clock and ends when time runs out.

Section	Questions	Time
Section I: Multiple Choice	42	85 minutes
Section II: Free Response	4	95 minutes

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Section I: Multiple Choice

42 questions · 85 minutes

Take north as positive. A car on a straight road has a velocity of $- 12$ m/s and a constant acceleration of $+ 2.0$ m/s². What is its velocity 4.0 s later, and what is it doing at that moment?
Answer: B.
1. A
  This adds 8.0 m/s to the speed, as if every acceleration speeds an object up. A positive acceleration on a car moving the negative way points against its motion, so the car slows down.
2. B
  Here $v_{x} = v_{x 0} + a_{x} t = - 12 + (2.0) (4.0) = - 4.0$ m/s. The sign is still negative, so the car still moves south, and its speed has fallen from 12 m/s to 4.0 m/s.
3. C
  This drops the minus sign on the starting velocity, so 12 minus 8.0 reads as a northward velocity. At 4.0 s the car has not stopped yet, so it is still moving south.
4. D
  This adds the sizes 12 and 8.0 and takes the direction from the acceleration. The sign of the velocity, not of the acceleration, tells you which way the car moves.
Take east as positive. A runner starts at the 0 m mark of a straight track and runs east to the 300 m mark in 60 s. She then turns around and walks west to the 180 m mark, which takes her 90 s. What is her average velocity for the whole 150 s?
Answer: D.
1. A
  This divides the total distance, 300 m plus 120 m, by the time. That is her average speed; average velocity uses the displacement.
2. B
  This averages her two velocities, 5.0 m/s and $- 1.3$ m/s. She spent longer walking than running, so a plain average of the two overweights the run.
3. C
  This divides her 180 m displacement by the 60 s of running only. The average is over the whole interval, 150 s.
4. D
  Average velocity is displacement over elapsed time. She ends 180 m east of her start, and 180 m divided by 150 s is 1.2 m/s east.
Take east as positive. A cart's velocity-time graph is a straight line that starts at $+ 4.0$ m/s at $t = 0$ and falls steadily to $- 2.0$ m/s at $t = 6.0$ s, crossing zero at $t = 4.0$ s. What is the cart's displacement from $t = 0$ to $t = 6.0$ s?
Answer: A.
1. A
  Displacement is the area between the graph and the time axis: $\frac{1}{2} (4.0) (4.0) = + 8.0$ m above the axis and $\frac{1}{2} (2.0) (- 2.0) = - 2.0$ m below it. The net is $+ 6.0$ m.
2. B
  This counts the area below the axis as positive. That gives the distance traveled; the 2.0 m the cart covers while moving west counts against its displacement.
3. C
  This treats the region as one triangle with a 6.0 s base and a 4.0 m/s height. The line crosses the axis at 4.0 s, so the area splits into two triangles on opposite sides of it.
4. D
  This stops at 4.0 s, where the cart turns around. The cart keeps moving for 2.0 s more, back toward the west, and that motion belongs to the displacement too.
Take east as positive. A car drives east along a straight road at 20 m/s. A small drone flies along the road just above it. A passenger in the car measures the drone's velocity as 5.0 m/s west. What is the drone's velocity measured by a person standing still beside the road?
Answer: C.
1. A
  This adds the two speeds without their signs. The drone moves west relative to the car, so its velocity relative to the car is $- 5.0$ m/s.
2. B
  This is the passenger's measurement. The person beside the road is not moving with the car, so the car's 20 m/s has to be combined in.
3. C
  The drone's velocity relative to the road is its velocity relative to the car plus the car's velocity relative to the road: $- 5.0 + 20 = + 15$ m/s.
4. D
  This gets the size right but keeps the direction the passenger saw. The drone drifts back past the car only because it moves east more slowly than the car does.
A ball is launched from level ground. Just after launch, its velocity has a horizontal component of 6.0 m/s and a vertical component of 20 m/s upward. Air resistance is negligible; use $g = 10$ m/s². At the highest point of its flight, what are the ball's speed and acceleration?
Answer: D.
1. A
  This treats the top as a moment when the ball stops. Only the vertical part of the velocity is zero there, and gravity does not switch off when it is.
2. B
  The acceleration is right, but this loses the horizontal motion. The horizontal component stays 6.0 m/s for the whole flight.
3. C
  The speed is right, but zero vertical velocity does not mean zero acceleration. Gravity still pulls down at the top, so the vertical velocity is changing there.
4. D
  At the top only the vertical velocity is zero, and the horizontal 6.0 m/s never changes because nothing pushes the ball sideways. Gravity is the only force the whole time, so the acceleration stays 10 m/s² downward.
A rod 1.2 m long has negligible mass. A 3.0 kg ball is fixed to its left end and a 1.0 kg ball to its right end. How far from the left end is the center of mass of the rod and balls?
Answer: A.
1. A
  Here $x_{cm} = \frac{(3.0) (0) + (1.0) (1.2)}{3.0 + 1.0} = 0.30$ m. The center of mass sits nearer the heavier ball.
2. B
  This splits the 1.2 m in the ratio 1 to 3 by dividing by 3. The masses add to 4.0 kg, so the rod divides into quarters, not thirds.
3. C
  This is the middle of the rod. That is the center of mass only when the two balls have equal mass.
4. D
  This swaps the masses, so the center of mass lands near the lighter ball. It belongs near the 3.0 kg ball on the left.
A skier is pulled up a straight, snowy slope at a steady speed by a tow rope that runs parallel to the slope. Friction from the snow on her skis is not negligible, and air resistance is negligible. Which list gives every force exerted on the skier, with its direction?
Answer: B.
1. A
  A normal force points perpendicular to the surface that exerts it, away from that surface. On a slope that direction is tilted away from vertical.
2. B
  Earth, the rope and the snow each exert a force on her, and the snow exerts two: one perpendicular to its surface and kinetic friction opposite to her motion. She slides up the slope, so friction points down it.
3. C
  Kinetic friction opposes the skier's motion relative to the snow. She slides up the slope, so friction points down it.
4. D
  A component of gravity is part of the gravitational force already listed, not a second force, and it is not drawn on a free-body diagram. This list also leaves out friction from the snow.
An astronaut and her suit have a mass of 100 kg. She floats at rest in deep space, 10 m from a 25 kg supply crate that is also at rest, with nothing else nearby. She pulls hand over hand on a light rope tied to the crate until she and the crate meet. How far from her starting point do they meet?
Answer: C.
1. A
  This assumes they meet halfway. The rope pulls on the two with forces of equal size, but the 25 kg crate gains speed four times as fast as she does, so it covers more of the gap.
2. B
  This swaps the masses and puts the meeting point near the crate. The heavier astronaut moves less, so they meet nearer her start.
3. C
  The pulls between her and the crate are internal to the astronaut-crate system, so its center of mass stays at rest, and they meet there. Measured from her start, it is $\frac{(25) (10)}{100 + 25} = 2.0$ m.
4. D
  This divides by her mass alone. The center of mass divides by the total mass of the system, 125 kg.
A block of mass $m$ rests on a frictionless ramp that makes an angle $θ$ with the horizontal. An ideal spring with spring constant $k$ runs parallel to the ramp, from the block up to a post fixed at the top of the ramp, and holds the block at rest. How far is the spring stretched from its relaxed length?
Answer: D.
1. A
  This has the spring hold up the block's whole weight. The ramp's normal force carries part of it, so along the ramp the spring has to balance only the part of gravity that points down the slope.
2. B
  This uses the part of gravity perpendicular to the ramp, which the normal force balances. The spring pulls along the ramp, so it balances the part along the ramp, $m g \sin θ$ .
3. C
  This makes the spring's vertical pull hold up the whole weight, as if the ramp pushed on nothing. Balance the forces along the ramp, where the normal force has no part, and the spring's pull comes out as $m g \sin θ$ .
4. D
  Along the ramp only two forces act: the spring's pull up the slope, $k Δ x$ , and the part of gravity down it, $m g \sin θ$ . The block is at rest, so $k Δ x = m g \sin θ$ and $Δ x = \frac{m g \sin θ}{k}$ .
A 3.0 kg block rests on a level, frictionless table. A light string tied to it runs over an ideal pulley at the table's edge to a 1.0 kg block hanging below. The blocks are released from rest. Use $g = 10$ m/s². What is the size of each block's acceleration?
Answer: B.
1. A
  This divides the hanging block's 10 N weight by the table block's mass alone. The same pull has to speed up both blocks, 4.0 kg in all.
2. B
  Take both blocks and the string as one system. The only unbalanced force on it is the hanging block's 10 N weight, so $a = \frac{10}{3.0 + 1.0} = 2.5$ m/s².
3. C
  This uses the weight of the block on the table as the driving force. The table's normal force balances that weight; only the hanging block's weight drives the system.
4. D
  This is free fall. The string pulls up on the hanging block, so it speeds up at less than $g$ .
A probe lands on planet Q and measures the gravitational field strength at its surface as 10 N/kg, the same as at Earth's surface. Planet Q's radius is twice Earth's radius. How does planet Q's mass compare with Earth's mass?
Answer: A.
1. A
  The surface field strength is $g = \frac{G M}{R^{2}}$ . Doubling $R$ divides $g$ by 4, so the mass must be 4 times as large to keep $g$ the same.
2. B
  This treats the field strength as falling off with $R$ . It falls off with $R^{2}$ , so doubling the radius needs 4 times the mass.
3. C
  This scales the mass with volume, $R^{3}$ , as if Q had Earth's density. Q's surface field would then be twice Earth's, not equal to it.
4. D
  This reads equal field strength as equal mass. The field also depends on the distance to the planet's center, and Q's surface is twice as far from it.
You press a 0.60 kg book flat against a vertical wall with a steady horizontal push, and nothing else holds it up. The coefficient of static friction between the book and the wall is 0.40. Use $g = 10$ m/s². What is the smallest push that keeps the book from sliding down the wall?
Answer: C.
1. A
  This multiplies the coefficient by the weight, $(0.40) (6.0)$ , as if the weight pressed the book into the wall. Here your push sets the normal force, and the weight is what friction has to hold up.
2. B
  This makes the push equal to the weight. A horizontal push cannot hold the book up by itself; it presses the book into the wall, and friction, at most 0.40 times that push, has to carry the 6.0 N.
3. C
  Friction has to hold up the 6.0 N weight, and it can be at most $μ_{s} F_{n}$ . The wall's normal force equals your push $F$ , so $0.40 F \geq 6.0$ N, and the smallest push is $\frac{6.0}{0.40} = 15$ N.
4. D
  This divides the mass by the coefficient and leaves out $g$ . Friction has to hold up the book's weight, $(0.60) (10) = 6.0$ N, not 0.60.
A car takes a flat, unbanked curve at the greatest speed it can without sliding. Static friction from the road is the only horizontal force on it. Later the same car takes a second flat, unbanked curve with the same road surface and twice the radius of the first. What is the greatest speed at which it can take the second curve?
Answer: D.
1. A
  The maximum friction is unchanged, but a wider curve needs less centripetal force at any given speed, so the same friction allows a higher speed.
2. B
  Setting $μ_{s} m g = \frac{m v^{2}}{r}$ makes $v^{2}$ , not $v$ , proportional to $r$ . Doubling the radius multiplies the speed by $\sqrt{2}$ , not 2.
3. C
  This turns the relationship over. The friction sets the largest centripetal acceleration, $μ_{s} g = \frac{v^{2}}{r}$ , so a larger radius allows a larger $v^{2}$ .
4. D
  The largest centripetal acceleration friction can give is $μ_{s} g = \frac{v^{2}}{r}$ , so $v = \sqrt{μ_{s} g r}$ . Doubling $r$ multiplies the greatest speed by $\sqrt{2}$ .
A student launches a lab cart from rest along a level track several times. For each launch she finds the work done on the cart from force-sensor and motion-sensor data, and that work equals the cart's translational kinetic energy $K$ afterward. The motion sensor also records the cart's speed $v$ . She graphs $K$ on the vertical axis against $v^{2}$ on the horizontal axis and gets a straight line through the origin with a slope of 0.30 J s²/m². What is the cart's mass?
Answer: A.
1. A
  Since $K = \frac{1}{2} m v^{2}$ , the slope of $K$ against $v^{2}$ is $\frac{1}{2} m$ . So $m = 2 (0.30) = 0.60$ kg.
2. B
  This takes the slope as the mass and forgets the one-half in $K = \frac{1}{2} m v^{2}$ .
3. C
  This halves the slope instead of doubling it. The slope is half the mass, so the mass is twice the slope.
4. D
  This takes the mass as 1 divided by the slope. The slope of $K$ against $v^{2}$ is $\frac{1}{2} m$ , so the mass is twice the slope.
A 0.50 kg puck on a level, frictionless table is tied to a string whose other end is fixed to a peg in the middle of the table. The puck moves in a horizontal circle of radius 0.80 m at a steady 4.0 m/s. How much work does the string do on the puck during one full revolution?
Answer: C.
1. A
  This is the puck's kinetic energy, $\frac{1}{2} (0.50) (4.0)^{2}$ . The string does not supply it; the puck already has it and keeps it.
2. B
  This multiplies the 10 N tension by the diameter of the circle. Work uses the displacement along the force, and the puck never moves along the string.
3. C
  The string pulls toward the center, perpendicular to the puck's velocity at every instant, so it does no work. That fits the puck's kinetic energy staying the same.
4. D
  This multiplies the 10 N tension by the 5.0 m circumference. The force is perpendicular to the path at every point, so none of that distance counts.
A 4.0 kg wooden block slides across a level floor, starting at 6.0 m/s. The coefficient of kinetic friction between the block and the floor is 0.20. Use $g = 10$ m/s². How far does the block slide before it stops?
Answer: B.
1. A
  This finds the kinetic energy as $\frac{1}{2} m v$ , leaving off the square on the speed, which gives 12 J instead of 72 J.
2. B
  Friction is $(0.20) (4.0) (10) = 8.0$ N, and its negative work must remove all $\frac{1}{2} (4.0) (6.0)^{2} = 72$ J of kinetic energy. So $d = \frac{72}{8.0} = 9.0$ m.
3. C
  This drops the one-half from the kinetic energy and uses 144 J instead of 72 J.
4. D
  This takes the friction force as $μ_{k} m = 0.80$ , leaving out $g$ . The normal force is the 40 N weight, so friction is 8.0 N.
Three small spheres, each of mass $m$ , sit at the corners of an equilateral triangle with sides of length $d$ , far from any other object. The gravitational potential energy of a pair is taken as zero when the two are infinitely far apart. What is the total gravitational potential energy of the three-sphere system?
Answer: A.
1. A
  The system has three pairs, each a distance $d$ apart and each with $- \frac{G m^{2}}{d}$ . The total is the sum over the pairs.
2. B
  This counts the two pairs that include one chosen sphere and misses the pair formed by the other two.
3. C
  This counts each pair twice, once from each sphere's point of view. Each pair adds its energy once.
4. D
  This counts only one pair. Potential energy belongs to every pair in the system, and three spheres make three pairs.
Questions 18 to 20 refer to the following.
A motor-driven winch lifts a 40 kg crate straight up a shaft on a steel cable. Air resistance is negligible; use $g = 10$ m/s². A motion sensor records the crate's height above the bottom of the shaft and its speed:
- $t = 0$ : height 0 m, speed 0
- $t = 2.0$ s: height 1.0 m, speed 1.0 m/s
- $t = 6.0$ s: height 5.0 m, speed 1.0 m/s
- $t = 8.0$ s: height 6.0 m, speed 0
The crate speeds up steadily from 0 to 2.0 s, rises at a constant 1.0 m/s from 2.0 s to 6.0 s, and slows steadily from 6.0 s to 8.0 s. It moves upward the whole time.
Take the crate and Earth as the system. By how much does the mechanical energy of the system change from $t = 0$ to $t = 8.0$ s?
Answer: D.
1. A
  The kinetic energy is zero at both ends, but the potential energy is not the same. The cable is outside the system and does positive work on it, so the system gains energy.
2. B
  This uses $m Δ y$ and leaves out $g$ . The change in gravitational potential energy is $m g Δ y$ .
3. C
  This adds the 20 J of kinetic energy the crate had while it moved at 1.0 m/s. At 8.0 s the crate is at rest again, so its kinetic energy is zero at both ends.
4. D
  The crate starts and ends at rest, so only the potential energy changes: $m g Δ y = (40) (10) (6.0) = 2400$ J. The cable, outside the system, does that much work on it.
Uses the material above question 18.

Again take the crate and Earth as the system. From $t = 6.0$ s to $t = 8.0$ s the crate slows to a stop while still rising. Which statement about the mechanical energy of the system during that interval is correct?
Answer: C.
1. A
  The crate loses only $\frac{1}{2} (40) (1.0)^{2} = 20$ J of kinetic energy but gains $(40) (10) (1.0) = 400$ J of potential energy. The rest has to come from outside the system.
2. B
  Slowing down lowers the kinetic energy by 20 J, but the 1.0 m rise raises the potential energy by 400 J, so the total goes up.
3. C
  The cable pulls up while the crate moves up, so it does positive work and moves energy into the system. Its tension, $(40) (10 - 0.50) = 380$ N, over 1.0 m adds 380 J, which matches $- 20 + 400$ J.
4. D
  The pull is less than the weight, which is why the crate slows. But the pull still points along the crate's upward motion, so its work on the system is positive.
Uses the material above question 18.

What is the average power the cable delivers to the crate from $t = 2.0$ s to $t = 6.0$ s?
Answer: B.
1. A
  This reads constant speed as no energy transfer. The kinetic energy is constant, but the cable raises the crate 4.0 m and does 1600 J of work in those 4.0 s.
2. B
  At constant speed the cable's pull equals the crate's 400 N weight, so $P = F v = (400) (1.0) = 400$ W. The same comes from 1600 J of work over 4.0 s.
3. C
  This multiplies the mass by the speed. The cable's force is the weight, $m g = 400$ N, not 40.
4. D
  This takes the height at 6.0 s, 5.0 m, as the distance. During the interval the crate rose only 4.0 m, from 1.0 m to 5.0 m.
A car drives along a straight, level road, and its engine keeps the power delivered to the car constant. At 10 m/s, the forward force the road exerts on the car's drive wheels is 3000 N. What is that forward force when the car is moving at 20 m/s?
Answer: D.
1. A
  This assumes the force stays fixed. Then the power would double as the speed doubles, and the power is held constant.
2. B
  This makes the force grow with the speed. With $P = F v$ fixed, a larger speed goes with a smaller force.
3. C
  This divides by the square of the speed ratio, as if the force went as $\frac{1}{v^{2}}$ . Power is force times speed to the first power.
4. D
  With constant power, $P = F v$ makes the force inversely proportional to the speed. Doubling the speed halves the force: $3000 (\frac{10}{20}) = 1500$ N.
A 0.045 kg golf ball flies through the air at 40 m/s. A 6.0 kg bowling ball rolls slowly down a lane at 0.30 m/s. Which statement compares the two balls correctly?
Answer: B.
1. A
  Speed alone does not decide momentum. The golf ball is fast but light, and both products mv come to 1.8 kg m/s.
2. B
  Each ball has mv = 1.8 kg m/s, so the momenta match. Kinetic energy grows with the square of speed, so the golf ball has about 36 J of translational kinetic energy against the bowling ball's 0.27 J.
3. C
  Equal momentum does not mean equal kinetic energy. K = p²/(2m), so for the same p the lighter golf ball carries far more kinetic energy.
4. D
  Mass alone does not decide momentum. The bowling ball is heavy but slow, and its mv is the same 1.8 kg m/s as the golf ball's.
A 0.40 kg soccer ball sits at rest, and a player kicks it horizontally. During the kick, the force the foot exerts on the ball rises steadily from 0 to 800 N over 0.005 s, then falls steadily back to 0 at 0.010 s. The foot's force is far larger than any other horizontal force on the ball while they touch. What is the ball's speed just after the kick?
Answer: C.
1. A
  4.0 is the impulse, in N s. It still has to be divided by the ball's mass to give a change in velocity.
2. B
  This uses only the area under the rising half of the force graph. The force keeps pushing while it falls back to zero, and that half adds the same impulse again.
3. C
  The impulse is the triangular area under the force graph, (1/2)(800 N)(0.010 s) = 4.0 N s. Starting from rest, the ball's speed is 4.0 N s divided by 0.40 kg, or 10 m/s.
4. D
  This treats the peak force as if it lasted the whole 0.010 s, a rectangle instead of a triangle. The average force is only half the peak.
Two vehicles collide at an icy intersection. A 1200 kg car was heading east at 20 m/s, and a 2000 kg pickup was heading north at 12 m/s. They lock together, and during the collision the forces between them are far larger than any friction from the ice. In which direction do the locked vehicles move just after the collision?
Answer: A.
1. A
  The car's momentum is $(1200) (20) = 24,000$ kg m/s east and the pickup's is $(2000) (12) = 24,000$ kg m/s north. The two-vehicle system keeps both, so its momentum, and the locked vehicles' velocity, points midway between east and north.
2. B
  Speed alone does not set the direction; the system's momentum does. The pickup's greater mass makes up for its lower speed, and both vehicles carry 24,000 kg m/s.
3. C
  Mass alone does not set the direction either. The car is lighter but faster, and its momentum is the same size as the pickup's.
4. D
  Kinetic energy has no direction, and this collision does not keep it anyway. The direction comes from the momentum, which is the same size for both vehicles.
You drop a 0.60 kg basketball from rest, and air resistance is negligible. While it falls, which statement about momentum is correct?
Answer: D.
1. A
  Momentum changes whenever a net force acts, collision or not. Gravity is a net force on the ball alone, so its downward momentum grows the whole way down.
2. B
  Work changes energy, not momentum. Gravity here is internal to the ball-Earth system, and internal forces cannot change that system's total momentum.
3. C
  Gravity is external only if you pick the ball by itself as the system. Include Earth and the pull becomes an internal interaction.
4. D
  Gravity is an interaction inside the ball-Earth system, so that system feels no net external force. The ball and Earth pull on each other equally, so their momentum changes are equal and opposite.
On smooth ice, a 20 kg curling stone slides at 2.0 m/s straight at an identical stone at rest and hits it head-on. Nothing on either stone can release stored energy, such as a compressed spring. Take the first stone's starting direction as forward. From a video, a student reports that just after the collision the first stone moves backward at 0.50 m/s and the second moves forward at 2.5 m/s. Which statement about the report is correct?
Answer: A.
1. A
  Momentum checks out: 40 kg m/s before and $(20) (- 0.50) + (20) (2.5) = 40$ kg m/s after. But the kinetic energy would rise from 40 J to 2.5 J + 62.5 J = 65 J, and with nothing to release stored energy, a collision cannot add kinetic energy to the system.
2. B
  This drops the minus sign on the first stone's backward velocity and gets 10 + 50 = 60 kg m/s. Moving backward, that stone has $- 10$ kg m/s, so the total after is 40 kg m/s, the same as before.
3. C
  Matching momentum is needed but is not enough. An elastic collision keeps the total kinetic energy the same, and here it would rise from 40 J to 65 J.
4. D
  One stone losing kinetic energy says nothing about the system. Here the second stone would gain more than the first loses, so the total would rise, which no collision can do without a source of stored energy.
A power drill's bit starts from rest and speeds up with constant angular acceleration, reaching an angular speed of 300 rad/s after 0.50 s. Through what angle does the bit turn during that 0.50 s?
Answer: A.
1. A
  With constant angular acceleration from rest, the average angular velocity is half the final, 150 rad/s. Over 0.50 s that gives 150 rad/s times 0.50 s = 75 rad.
2. B
  This multiplies the final angular speed by the time, as if the bit spun at 300 rad/s the whole time. It started from rest, so its average was only half that.
3. C
  This reads the angular speed, 300 rad/s, as if it were an angle. An angular speed has to be multiplied by a time to give an angle.
4. D
  This is the angular acceleration, 300 rad/s divided by 0.50 s, in rad/s². It tells how fast the bit speeds up, not how far it turns.
A dead tree trunk stands straight up. To pull it over, a worker ties a rope to the trunk 4.0 m above the ground and pulls on it with a force of 300 N. The rope makes a 30 degree angle with the trunk. What is the magnitude of the torque the rope exerts on the trunk about its base?
Answer: B.
1. A
  This applies the angle twice, once to shorten the distance and again to shrink the force. Only one factor of sin 30 degrees belongs, either on r or on F.
2. B
  The angle between the rope and the line from the base to the knot is 30 degrees, so the torque is rF sin 30 degrees = (4.0 m)(300 N)(0.50) = 600 N m. Only the part of the pull perpendicular to the trunk turns it.
3. C
  This uses cos 30 degrees, which gives the part of the pull along the trunk. That part points through the base and exerts no torque about it.
4. D
  This multiplies r by the whole force, as if the rope pulled at right angles to the trunk. At 30 degrees, only half the pull is perpendicular.
Four small balls, each of mass 0.20 kg, are fixed at the corners of a light, flat square frame 0.40 m on a side. The frame turns about an axis that lies along one side of the square, so two of the balls sit on the axis. Treating the balls as point objects, what is the rotational inertia of the system about this axis?
Answer: A.
1. A
  The two balls on the axis are 0 m from it and add nothing. The other two are 0.40 m away, so I = 2(0.20 kg)(0.40 m)² = 0.064 kg m².
2. B
  This counts all four balls at 0.40 m. The two balls on the axis have r = 0, so they add no rotational inertia.
3. C
  This uses r instead of r² for the two far balls. Rotational inertia depends on the square of the distance from the axis.
4. D
  This counts all four balls and also forgets to square r. Only the two balls off the axis count, each with mr².
A uniform plank 2.0 m long with a mass of 4.0 kg rests on a single pivot 0.50 m from its left end. A student wants to place a 5.0 kg block on the plank so that the plank stays level and at rest. Use $g = 10$ m/s². Where must the block go?
Answer: D.
1. A
  0.10 m is where the block sits measured from the plank's left end. The lever arm in the torque balance is measured from the pivot, so the block belongs 0.40 m from the pivot.
2. B
  The plank's own weight acts at its center, 0.50 m to the right of the pivot, and turns it clockwise. A block on the right would add to that torque rather than balance it.
3. C
  This takes the plank's lever arm as 1.0 m, the distance from the left end to the center. The lever arm is measured from the pivot, which is 0.50 m from the center.
4. D
  The plank's weight, 40 N at 0.50 m right of the pivot, exerts 20 N m one way. The block's 50 N weight must exert 20 N m the other way, so it sits 20/50 = 0.40 m left of the pivot.
A display turntable in a shop window has a rotational inertia of 0.15 kg m² about its axle. While it speeds up, its motor exerts a torque of 1.2 N m on it and friction in the bearing exerts a torque of 0.30 N m in the opposite sense. What is the magnitude of the turntable's angular acceleration?
Answer: B.
1. A
  This uses only the friction torque. The angular acceleration depends on the net torque, the motor's minus the bearing's.
2. B
  The net torque is 1.2 N m minus 0.30 N m, or 0.90 N m. Dividing by the rotational inertia gives 0.90/0.15 = 6.0 rad/s².
3. C
  This ignores the bearing friction and divides the motor's torque alone by the rotational inertia. Friction opposes the motor, so the net torque is smaller.
4. D
  This adds the two torques. They act in opposite senses, so the friction torque subtracts from the motor's.
A 6.0 kg bucket of water hangs from a light rope wound in a single layer around a drum of radius 0.10 m at the top of a well. A person turns a crank so that the drum raises the bucket at a steady speed while the drum turns through 8.0 revolutions. Friction at the drum's axle is negligible. Use g = 10 m/s². How much work does the crank's torque do on the drum?
Answer: D.
1. A
  This leaves out g, using 6.0 kg times 0.10 m as the torque. The rope pulls on the drum with the bucket's weight, 60 N, not with its mass.
2. B
  This multiplies the torque by 8.0, the number of revolutions. W = tau times the angle in radians, and 8.0 revolutions is 16 pi, about 50 rad.
3. C
  This converts each revolution to pi radians. One full revolution is 2 pi radians.
4. D
  At steady speed the crank's torque matches the rope's, (60 N)(0.10 m) = 6.0 N m. Eight revolutions is 16 pi rad, so W = (6.0 N m)(16 pi), about 300 J, which matches the bucket's gain in mgh.
A rotating platform at a science museum has a rotational inertia of 240 kg m² about its vertical axle, and friction at the axle is negligible. It turns at 1.0 rad/s with a 30 kg child standing on it 2.0 m from the axle. The child walks inward and stops 0.50 m from the axle. Treating the child as a point object, what is the platform's angular speed now?
Answer: D.
1. A
  This sets up the right ratio upside down. Moving the child inward lowers the rotational inertia, so the angular speed must go up, not down.
2. B
  The child pushes on the platform while walking, so the platform alone does not keep a constant angular velocity. Only the platform-child system keeps a constant angular momentum.
3. C
  This uses mr instead of mr² for the child. Rotational inertia depends on the square of the distance, so the child's share drops from 120 kg m² to 7.5 kg m².
4. D
  No external torque acts on the platform-child system, so its angular momentum stays (240 + 30 x 2.0²)(1.0) = 360 kg m²/s. The new rotational inertia is 240 + 30 x 0.50² = 247.5 kg m², so the angular speed is 360/247.5 = 1.5 rad/s.
A comet moves in an elliptical orbit around the Sun, and only the gravitational forces between the two act. The Sun's mass is so much larger that its own motion is negligible. As the comet travels from its farthest point from the Sun toward its closest point, which statement is correct?
Answer: A.
1. A
  The gravitational potential energy becomes more negative as the comet falls inward, so its kinetic energy rises. Gravity points straight at the Sun, so it exerts no torque about the Sun and the angular momentum stays constant.
2. B
  The comet's speed rises, but its distance from the Sun falls. With no torque about the Sun, the angular momentum stays the same even as the speed changes.
3. C
  A constant total does not mean each part is constant. In an elliptical orbit the system's energy shifts between potential and kinetic while the total stays fixed.
4. D
  Closer to the Sun, U = -GMm/r is more negative, so the potential energy decreases. Both rising at once would also make the total energy grow, which nothing here can do.
Questions 35 to 37 refer to the following.
A 0.40 kg glider on a level air track is attached to a horizontal ideal spring, and friction is negligible. Right is positive, and the glider's position is measured from its equilibrium position. A motion sensor records the glider's velocity against time. The record is not drawn here; in words, it is a smooth wave. The velocity is 0 at t = 0, reaches its greatest value, +0.60 m/s, at t = 0.25 s, is 0 again at t = 0.50 s, reaches -0.60 m/s at t = 0.75 s, and returns to 0 at t = 1.00 s. The pattern then repeats.

Using the glider's velocity record, what is the spring constant of the spring?
Answer: C.
1. A
  This drops the 4 pi² when solving T = 2 pi sqrt(m/k) for k. Squaring both sides gives k = 4 pi² m/T².
2. B
  This uses 2 pi instead of 4 pi², forgetting to square the 2 pi when squaring both sides of the period equation.
3. C
  The velocity first returns to zero while increasing at 1.00 s, the same point in the cycle as at t = 0, so T = 1.00 s. Then k = 4 pi² m/T² = 4 pi² (0.40 kg)/(1.00 s)² = 15.8 N/m.
4. D
  This takes the period as 0.50 s, the time between two zeros of velocity. That is only half a cycle; the glider is at the opposite end of its path at 0.50 s.
Uses the material above question 35.

Using the glider's velocity record, where is the glider at t = 0.50 s, and what is its acceleration then?
Answer: B.
1. A
  Zero velocity does not mean zero acceleration. At the end of its path the spring is stretched the most, so the force on the glider, and its acceleration, are at their largest.
2. B
  The glider moved right from 0 s to 0.50 s and stops at 0.50 s, so it is at its farthest right point. The spring pulls it back toward equilibrium, in the negative direction, with the largest force of the cycle.
3. C
  This reads the zero on a velocity graph as zero displacement. The glider passes equilibrium at its greatest speed, at 0.25 s and 0.75 s, not where it stops.
4. D
  A restoring force points opposite the displacement. At its greatest positive displacement, the glider's acceleration points in the negative direction.
Uses the material above question 35.

Using the glider's velocity record, how much energy is stored in the spring at t = 0.50 s?
Answer: C.
1. A
  The glider has no kinetic energy at 0.50 s, but the energy has not vanished. It is all stored in the stretched spring at that moment.
2. B
  This splits the energy evenly between kinetic and potential. At 0.50 s the glider is stopped, so none of the energy is kinetic.
3. C
  The total energy equals the greatest kinetic energy, (1/2)(0.40 kg)(0.60 m/s)² = 0.072 J. At 0.50 s the glider is stopped, so all of that energy is in the spring.
4. D
  This leaves out the 1/2 in K = (1/2)mv², doubling the energy.
A student lowers a small metal figurine with a mass of 0.60 kg into a graduated cylinder of water. The figurine sinks completely under the surface, and the water level rises from 120.0 mL to 170.0 mL. One milliliter is 1 cm³. What is the density of the metal?
Answer: D.
1. A
  This converts 50 cm³ to cubic meters by dividing by 100. A cubic meter holds (100)³ = 1,000,000 cubic centimeters.
2. B
  This converts 50 cm³ as if it were 50 liters, dividing by 1000. A cubic meter holds 1,000,000 cubic centimeters but only 1000 liters.
3. C
  This divides by 100² = 10,000, the factor for area. Volume needs the cube, 100³.
4. D
  The figurine pushes aside 50 cm³ of water, which is 5.0 x 10⁻⁵ m³. Its density is 0.60 kg divided by that volume, 12,000 kg/m³.
A U-shaped glass tube, open to the air at both ends, holds water (density 1000 kg/m³). Oil with a density of 800 kg/m³, which does not mix with water, is poured into the right arm until the oil column is 10.0 cm tall, measured from the oil-water boundary up to the oil's surface. Both liquids are at rest. How does the height of the oil's surface compare with the height of the water's surface in the left arm?
Answer: B.
1. A
  Equal surface heights would need equal densities. The oil is less dense, so it needs a taller column to make the same pressure at the boundary level.
2. B
  At the level of the oil-water boundary, the pressure is the same in both arms, so (800)(10.0 cm) = (1000)h and the water stands h = 8.0 cm above that level. The oil surface, 10.0 cm up, is 2.0 cm higher.
3. C
  This flips the density ratio and has the water stand 12.5 cm above the boundary. The denser liquid needs the shorter column to make the same pressure.
4. D
  8.0 cm is how far the water surface stands above the boundary level, not the difference between the two surfaces.
A wooden block floats at rest in fresh water (density 1000 kg/m³) with 60 percent of its volume below the surface. The block is moved to a tank of oil with a density of 800 kg/m³. Once it floats at rest in the oil, what percent of its volume is below the oil's surface?
Answer: C.
1. A
  This multiplies 60 percent by 0.80, as if a less dense liquid let the block ride higher. A less dense liquid gives less buoyant force per cubic meter, so the block sinks deeper.
2. B
  The submerged fraction depends on the liquid's density. The buoyant force must still equal the block's weight, and oil provides less force per cubic meter displaced.
3. C
  The block's density is 0.60 x 1000 = 600 kg/m³. Floating, its weight equals the weight of oil displaced, so the fraction under is 600/800 = 75 percent.
4. D
  This uses the ratio of the two liquids' densities, 800/1000. The fraction depends on the block's density compared with the oil's.
A beaker of water sits on a scale that reads 12.0 N. A student holds a string tied to a 0.40 kg stone with a volume of 1.5 x 10⁻⁴ m³ and lowers it until the stone hangs at rest, completely under water and not touching the beaker. No water spills. Use g = 10 m/s². What does the scale read now?
Answer: C.
1. A
  The water pushes up on the stone with the buoyant force, and by Newton's third law the stone pushes down on the water just as hard. That adds to the scale reading rather than taking away from it.
2. B
  The string holds up only part of the stone's weight. The water supports the rest, and that push on the water passes down to the scale.
3. C
  The water pushes up on the stone with a buoyant force of (1000)(1.5 x 10⁻⁴)(10) = 1.5 N, so the stone pushes down on the water with 1.5 N. The scale reads 12.0 N + 1.5 N = 13.5 N.
4. D
  This adds the stone's whole 4.0 N weight. The string still holds up 2.5 N of it, so only the 1.5 N buoyant force reaches the scale.
A garden hose with an inside cross-sectional area of 2.0 cm² fills a 20 L bucket in 40 s. The water fills the hose completely. One liter is 1.0 x 10⁻³ m³. What is the average speed of the water in the hose?
Answer: A.
1. A
  The volume flow rate is 0.020 m³ / 40 s = 5.0 x 10⁻⁴ m³/s. Dividing by the area, 2.0 x 10⁻⁴ m², gives v = 2.5 m/s.
2. B
  This converts 20 L to 0.20 m³. One liter is 10⁻³ m³, so 20 L is 0.020 m³.
3. C
  This divides the volume by the area and stops, which gives a length in meters, not a speed. The flow rate needs the 40 s first.
4. D
  This converts 2.0 cm² by dividing by 10⁶, the factor for volume. For area the factor is 100² = 10⁴, so 2.0 cm² is 2.0 x 10⁻⁴ m².

Section II: Free Response

4 questions · 95 minutes

Question 1: Mathematical Routines 10 points

A crate of mass 40 kg sits on the flat, level bed of a pickup truck. The back face of the crate is 1.2 m in front of the closed tailgate. Between the crate and the bed, the coefficient of static friction is $μ_{s} = 0.30$ and the coefficient of kinetic friction is $μ_{k} = 0.25$ . The truck starts from rest on a straight, level road and accelerates forward. Take forward as positive, and use $g = 10$ m/s².

Respond to parts (a), (b), (c) and (d). Draw any diagram on your own paper.
1. (a)
  
  The truck accelerates forward at 2.0 m/s², and the crate moves with the truck without sliding on the bed. Draw a free-body diagram of the crate. Represent each force with a labeled arrow that starts on a dot standing for the crate, and draw the arrows so their lengths show the relative sizes of the forces.
  Question 1, part (a): your response. It stays in this browser tab and is never sent anywhere.
  Score part (a)
  - 1 point Draws the gravitational force straight down and the normal force from the bed straight up, with arrows of equal length.Earns it: " $F_{g}$ down and $F_{N}$ up, drawn the same length"; "weight, 400 N, down; normal force from the truck bed, 400 N, up"Does not: a normal force drawn longer than gravity (the crate does not accelerate up or down, so the two balance); a normal force tilted forward or backward (the bed is level, so the force it exerts at right angles to its surface is vertical)
  - 1 point Draws static friction from the bed pointing forward, shorter than the gravity arrow, as the only horizontal force.Earns it: " $F_{f}$ forward, about a fifth as long as $F_{g}$ "; "friction from the bed pointing the way the truck accelerates, and nothing else sideways"Does not: friction drawn backward (without friction the crate would slide backward relative to the bed, so friction on it points forward); an extra backward arrow labeled inertia or force of motion (no object exerts it); a forward arrow for the truck pushing the crate in addition to friction (the bed touches the crate only from below, and its push along the surface is the friction)
  Model response
  Three arrows start on the dot. The gravitational force $F_{g}$ points straight down. The normal force $F_{N}$ from the bed points straight up and is the same length, because the crate does not speed up vertically. Static friction $F_{f}$ from the bed points forward and is about one fifth as long as $F_{g}$ : the crate needs $F_{f} = m a = (40) (2.0) = 80$ N, and its weight is $(40) (10) = 400$ N. Friction is the only horizontal force, and it points forward because it is what speeds the crate up along with the truck.
2. (b)
  
  Derive an expression for the greatest acceleration $a_{max}$ the truck can have without the crate sliding on the bed. Express your answer in terms of $μ_{s}$ and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
  Question 1, part (b): your response. It stays in this browser tab and is never sent anywhere.
  Score part (b)
  - 1 point Applies Newton's second law to the crate along the horizontal, with static friction as the only horizontal force, and sets that friction at its maximum value $μ_{s} F_{N}$ .Earns it: " $Σ F_{x} = m a$ , so $F_{f} = m a$ , and $F_{f}$ can be at most $μ_{s} F_{N}$ "Does not: uses $μ_{k}$ (the crate is not sliding, so the limit comes from static friction); writes $F_{f} = m g$ (friction is not the weight); writes $Σ F = 0$ horizontally (the crate accelerates with the truck)
  - 1 point Uses the vertical balance $F_{N} = m g$ and arrives at $a_{max} = μ_{s} g$ .Earns it: " $m a_{max} = μ_{s} m g$ , so $a_{max} = μ_{s} g$ "Does not: $a_{max} = μ_{s} m g$ (the mass was not divided out; this has units of force); $a_{max} = \frac{μ_{s}}{g}$ (an algebra slip)
  Model response
  Horizontally, Newton's second law gives $Σ F_{x} = m a$ . Static friction is the only horizontal force, so $F_{f} = m a$ .
  
  Vertically the crate does not accelerate: $F_{N} - m g = 0$ , so $F_{N} = m g$ .
  
  Static friction can be at most $μ_{s} F_{N} = μ_{s} m g$ , so the crate stays put only while $m a \leq μ_{s} m g$ . The greatest acceleration is
  $a_{max} = μ_{s} g$
  For this crate that is $(0.30) (10) = 3.0$ m/s².
3. (c)
  
  On another day the crate starts in the same place, and the driver accelerates forward from rest at a steady 4.0 m/s².
  
  i. Calculate the acceleration of the crate relative to the ground.
  
  ii. Calculate the time from the moment the truck starts until the tailgate reaches the crate.
  Question 1, part (c): your response. It stays in this browser tab and is never sent anywhere.
  Score part (c)
  - 1 point Recognizes that the crate slides, because 4.0 m/s² is more than the 3.0 m/s² static friction can give it, and uses kinetic friction to find $a_{c} = μ_{k} g = 2.5$ m/s², forward. A crate acceleration of 2.45 m/s² worked with g = 9.8 m/s², where the static limit is 2.94 m/s² and still less than 4.0 m/s², is also accepted.Earns it: "4.0 > 3.0, so it slides; $μ_{k} m g = m a_{c}$ , $a_{c} = 2.5$ m/s² forward"; "with $g = 9.8$ , also accepted: $4.0 > 2.94$ , so it slides; $a_{c} = (0.25) (9.8) = 2.45$ m/s² forward"Does not: 4.0 m/s² (assumes the crate keeps up with the truck); 3.0 m/s² (uses the static coefficient for a crate that is sliding); 2.5 m/s² backward (kinetic friction from the bed still points forward on the crate, because the crate slides backward relative to the bed)
  - 1 point Sets up the motion of both the truck and the crate from rest so that the truck gains 1.2 m on the crate: $\frac{1}{2} (4.0) t^{2} - \frac{1}{2} (2.5) t^{2} = 1.2$ m, or the same idea with a relative acceleration of 1.5 m/s².Earns it: " $\frac{1}{2} (1.5 {m/s}^{2}) t^{2} = 1.2$ m"Does not: $\frac{1}{2} (2.5) t^{2} = 1.2$ m, which gives about 0.98 s (that is how far the crate moves over the road, not how far the tailgate gains on it); $\frac{1}{2} (4.0) t^{2} = 1.2$ m, which gives about 0.77 s (treats the crate as staying where it started)
  - 1 point Calculates $t \approx 1.3$ s, with units. A time of about 1.2 s worked with g = 9.8 m/s² is also accepted.Earns it: " $t = \sqrt{\frac{2 (1.2)}{1.5}} = 1.26$ s"Does not: about 1.5 s (uses 3.0 m/s² for the crate, so the relative acceleration is 1.0 m/s²); about 0.89 s (drops the $\frac{1}{2}$ from the displacement equation)
  Model response
  i. To move with the truck the crate would need 4.0 m/s², but static friction can give it at most 3.0 m/s² (part b), so the crate slides on the bed. Kinetic friction is then the only horizontal force on it: $μ_{k} m g = m a_{c}$ , so $a_{c} = μ_{k} g = (0.25) (10) = 2.5$ m/s², forward.
  
  ii. Both start from rest at $t = 0$ . In a time $t$ the truck moves $\frac{1}{2} (4.0) t^{2}$ and the crate moves $\frac{1}{2} (2.5) t^{2}$ , so the tailgate closes in on the crate by $\frac{1}{2} (1.5) t^{2}$ . It reaches the crate when that equals the 1.2 m gap:
  $\frac{1}{2} (1.5 {m/s}^{2}) t^{2} = 1.2 m$ $t = \sqrt{\frac{2 (1.2)}{1.5}} \approx 1.3 s$
4. (d)
  
  The driver straps a second crate, identical to the first, on top of it, so the two crates move as one object. Indicate whether the greatest acceleration the truck can have without the stack sliding on the bed is greater than, less than, or equal to $a_{max}$ from part (b). Justify your answer. In your justification, include qualitative reasoning beyond mathematical derivations or expressions.
  Question 1, part (d): your response. It stays in this browser tab and is never sent anywhere.
  Score part (d)
  - 1 point Indicates that the greatest acceleration is equal to $a_{max}$ , with an attempt at a justification that uses forces.Earns it: "Equal, because the friction and the mass both double."Does not: greater, because the heavier stack presses harder on the bed (true, but the stack also needs more force to speed up); less, because the heavier stack is harder to speed up (true, but the bed can also exert more friction on it)
  - 1 point Indicates that doubling the mass doubles the normal force from the bed, so the largest static friction force the bed can exert doubles.Earns it: "The stack weighs twice as much, so the bed pushes up twice as hard and the maximum friction is twice as big."Does not: says friction depends on the area of contact (it does not, and the area is unchanged anyway); says the maximum friction stays the same
  - 1 point Indicates that the stack also needs twice the forward force for any given acceleration, so the larger friction exactly keeps pace and the greatest acceleration does not change.Earns it: "Twice the mass needs twice the force to get the same acceleration, and twice the friction is available, so the limit is the same 3.0 m/s²."Does not: " $a_{max} = μ_{s} g$ has no mass in it" alone (correct, but it gives no reasoning beyond the expression)
  Model response
  Equal. The stack has twice the mass, so Earth pulls on it twice as hard and the bed has to push up on it twice as hard. The most static friction the bed can exert is proportional to that normal force, so it doubles too. But the stack also needs twice the forward force to speed up at any given rate. The friction available and the friction needed grow in the same proportion, so the truck can still accelerate at up to 3.0 m/s² before the stack slides. That agrees with part (b), where the mass dropped out of $a_{max} = μ_{s} g$ .
Your score: 0 of 10, from the boxes you ticked.
Question 2: Translation Between Representations 12 points

A homemade yo-yo is two wooden disks of radius $R$ glued to the ends of a short axle of radius $\frac{R}{2}$ . The axle's mass is negligible compared with the disks'. The yo-yo has total mass $M$ , and its rotational inertia about its central axis is $\frac{1}{2} M R^{2}$ . A light string is wound around the axle, and its free end is tied to a fixed hook above. A student holds the yo-yo at rest with the string vertical and lets go. The yo-yo falls straight down and spins as the string unwinds from the axle without slipping. Air resistance is negligible.

Seen from the front, the string leaves the left side of the axle, so the yo-yo turns clockwise as it falls. Take down as positive for straight-line motion and clockwise as positive for rotation.

Respond to parts (a), (b), (c) and (d). Draw any diagram or graph on your own paper.
1. (a)
  
  Draw a force diagram of the yo-yo as seen from the front. Draw a circle to stand for the yo-yo and a smaller circle inside it for the axle. Represent each force exerted on the yo-yo with a labeled arrow that starts at the point where that force is exerted, and draw the arrows so their lengths show the relative sizes of the forces.
  Question 2, part (a): your response. It stays in this browser tab and is never sent anywhere.
  Score part (a)
  - 1 point Draws the gravitational force pointing down from the center of the yo-yo, and the tension pointing up from the left edge of the axle, a distance $\frac{R}{2}$ to the left of the center. No other forces are drawn.Earns it: " $F_{g}$ down from the center; $F_{T}$ up from the left side of the axle"Does not: tension drawn from the center (a force through the center exerts no torque about it, so the yo-yo could not start spinning); tension drawn from the outer rim, a distance $R$ from the center (the string is wound on the axle); an extra arrow for the spin or for a torque (torque is not a force)
  - 1 point Draws the tension arrow shorter than the gravity arrow, because the yo-yo's center speeds up downward, so the net force on it points down.Earns it: " $F_{T}$ clearly shorter than $F_{g}$ "Does not: equal arrows (then the net force would be zero and the center would not speed up); tension longer than gravity
  Model response
  Two arrows. The gravitational force $F_{g}$ starts at the center of the yo-yo and points straight down. The tension $F_{T}$ starts on the left edge of the axle, $\frac{R}{2}$ to the left of the center, where the string leaves it, and points straight up. The tension arrow is shorter, because the yo-yo's center speeds up as it falls, so the net force on it is downward.
2. (b)
  
  Derive an expression for the magnitude $a$ of the acceleration of the yo-yo's center. Express your answer in terms of $g$ . Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
  Question 2, part (b): your response. It stays in this browser tab and is never sent anywhere.
  Score part (b)
  - 1 point Applies Newton's second law to the translation of the yo-yo's center: $M g - F_{T} = M a$ .Earns it: " $Σ F = M a$ : $M g - F_{T} = M a$ "Does not: $M g = M a$ (leaves out the tension); $F_{T} - M g = M a$ with down taken as positive (a sign error)
  - 1 point Applies Newton's second law in rotational form about the center, with the tension's torque $F_{T} \frac{R}{2}$ and the given rotational inertia: $F_{T} \frac{R}{2} = \frac{1}{2} M R^{2} α$ .Earns it: " $Σ τ = I α$ : $F_{T} (\frac{R}{2}) = \frac{1}{2} M R^{2} α$ "Does not: uses $F_{T} R$ for the torque (the string pulls at the axle, a distance $\frac{R}{2}$ from the center); includes a torque from gravity about the center (gravity acts at the center, so its lever arm is zero)
  - 1 point Connects the straight-line and rotational accelerations at the axle's radius: $a = α \frac{R}{2}$ .Earns it: "no slipping on the axle, so $a = \frac{R}{2} α$ "Does not: $a = α R$ (uses the outer radius; this leads to $a = \frac{g}{2}$ )
  - 1 point Arrives at a correct expression: $a = \frac{g}{3}$ .Earns it: " $a = \frac{g}{3}$ "Does not: $a = \frac{g}{2}$ (from using $a = α R$ ); $a = g$ (treats the yo-yo as falling freely)
  Model response
  Translation, with down positive: $Σ F = M a$ , so
  $M g - F_{T} = M a$
  Rotation about the center, with clockwise positive: gravity acts at the center and exerts no torque about it, so $Σ τ = I α$ gives
  $F_{T} (\frac{R}{2}) = \frac{1}{2} M R^{2} α$
  The string comes off the axle without slipping, so the center moves down exactly as fast as string unwinds from a circle of radius $\frac{R}{2}$ : $a = α \frac{R}{2}$ , or $α = \frac{2 a}{R}$ . Then
  $F_{T} \frac{R}{2} = \frac{1}{2} M R^{2} \cdot \frac{2 a}{R} = M R a$
  so $F_{T} = 2 M a$ . Putting that into the first equation: $M g - 2 M a = M a$ , so
  $a = \frac{g}{3}$
3. (c)
  
  The yo-yo falls a total distance $h$ before the string runs out. Take the system to be the yo-yo and Earth, with gravitational potential energy $U_{g} = 0$ when the yo-yo has fallen the distance $h$ .
  
  On one set of axes, sketch three graphs as functions of the distance $y$ the yo-yo has fallen, from $y = 0$ to $y = h$ : the system's gravitational potential energy $U_{g}$ , the yo-yo's translational kinetic energy $K_{trans}$ , and its rotational kinetic energy $K_{rot}$ . Label each graph, and mark its value at $y = 0$ and at $y = h$ in terms of $M$ , $g$ and $h$ .
  Question 2, part (c): your response. It stays in this browser tab and is never sent anywhere.
  Score part (c)
  - 1 point Sketches $U_{g}$ as a straight line falling from $M g h$ at $y = 0$ to zero at $y = h$ .Earns it: " $U_{g}$ : a straight line from $M g h$ down to 0"Does not: a curved $U_{g}$ graph (gravitational potential energy changes by the same amount for each meter fallen); $U_{g}$ starting at zero (the zero is set at $y = h$ )
  - 1 point Sketches $K_{trans}$ and $K_{rot}$ each as a straight line rising from zero at $y = 0$ .Earns it: "both kinetic energies are straight lines through the origin"Does not: curves bending upward (that would be the shape against time, not against distance; with a constant net force, each meter fallen adds the same energy); a kinetic energy that does not start at zero (the yo-yo starts from rest)
  - 1 point Shows $K_{rot}$ ending at $\frac{2}{3} M g h$ and $K_{trans}$ ending at $\frac{1}{3} M g h$ , so the rotational line is twice as high as the translational line everywhere and the three energies always add to $M g h$ .Earns it: "at $y = h$ : $U_{g} = 0$ , $K_{trans} = \frac{1}{3} M g h$ , $K_{rot} = \frac{2}{3} M g h$ "Does not: $K_{trans} = K_{rot} = \frac{1}{2} M g h$ (assumes an even split without working it out); $K_{trans} = M g h$ with $K_{rot}$ extra (the energies would then add to more than the system started with)
  Model response
  There are no nonconservative forces doing work: the string does not slip on the axle, and the point where it pulls is momentarily at rest, so the tension does no work. The yo-yo-Earth system's mechanical energy stays at $M g h$ .
  
  When the center moves at speed $v$ , the yo-yo spins at $ω = \frac{v}{R / 2} = \frac{2 v}{R}$ , so
  $K_{rot} = \frac{1}{2} (\frac{1}{2} M R^{2}) {(\frac{2 v}{R})}^{2} = M v^{2} = 2 K_{trans}$
  The rotational kinetic energy is always twice the translational, so the energy given up, $M g y$ , splits one third and two thirds.
  
  The graph: $U_{g}$ is a straight line from $M g h$ at $y = 0$ down to 0 at $y = h$ . $K_{trans}$ is a straight line from 0 up to $\frac{1}{3} M g h$ , and $K_{rot}$ is a straight line from 0 up to $\frac{2}{3} M g h$ , twice as steep.
4. (d)
  
  A student says the yo-yo reaches $y = h$ with the same speed as a small ball dropped from rest through the same height $h$ , because mechanical energy is conserved for both. Using your graph from part (c), derive an expression for the yo-yo's speed at $y = h$ in terms of $g$ and $h$ . Justify why your answer agrees with your answer to part (b). Then indicate whether the student's claim is correct and justify your answer.
  Question 2, part (d): your response. It stays in this browser tab and is never sent anywhere.
  Score part (d)
  - 1 point Uses the energies from the graph to find the speed, either from the translational kinetic energy alone, $\frac{1}{2} M v^{2} = \frac{1}{3} M g h$ , or from all of $M g h$ becoming kinetic energy with the rotational part written in terms of $v$ , $M g h = \frac{1}{2} M v^{2} + M v^{2}$ . Either way $v = \sqrt{\frac{2 g h}{3}}$ .Earns it: " $v = \sqrt{\frac{2}{3} g h}$ from $K_{trans} = \frac{1}{3} M g h$ "; "At $y = h$ all of $M g h$ is kinetic: $M g h = \frac{1}{2} M v^{2} + M v^{2} = \frac{3}{2} M v^{2}$ , so $v = \sqrt{\frac{2 g h}{3}}$ "Does not: $\frac{1}{2} M v^{2} = M g h$ , $v = \sqrt{2 g h}$ (puts all of the energy into translation and ignores the graph)
  - 1 point Justifies the agreement with part (b) by showing that part (b) gives the same speed: constant acceleration $\frac{g}{3}$ from rest over a distance $h$ gives $v^{2} = 2 (\frac{g}{3}) h$ .Earns it: " $v^{2} = 2 a h = \frac{2 g h}{3}$ , the same as from the energies"Does not: states that the parts agree without showing the speed from part (b)
  - 1 point Justifies that the claim is wrong: both systems conserve mechanical energy, but the ball turns all of its lost potential energy into translational kinetic energy and reaches $\sqrt{2 g h}$ , while the yo-yo puts two thirds of it into spinning, so it reaches the bottom more slowly.Earns it: "Energy is conserved for both, but the yo-yo's goes mostly into spin, so less is left for falling. The ball reaches $\sqrt{2 g h}$ , the yo-yo only $\sqrt{2 g h / 3}$ ."Does not: "The yo-yo is slower because the string takes energy away." (the tension does no work on the yo-yo here, so mechanical energy is conserved); agrees with the student
  Model response
  From the graph, at $y = h$ the translational kinetic energy is $\frac{1}{3} M g h$ :
  $\frac{1}{2} M v^{2} = \frac{1}{3} M g h \Rightarrow v = \sqrt{\frac{2 g h}{3}}$
  From part (b), the center speeds up from rest at a constant $\frac{g}{3}$ over a distance $h$ , so $v^{2} = v_{0}^{2} + 2 a h = 0 + 2 (\frac{g}{3}) h$ , which is the same speed. The energy graph and the force derivation agree.
  
  The student is wrong. Mechanical energy is conserved for both, but that only fixes the total. A dropped ball has no spin to speak of, so all of its lost potential energy becomes translational kinetic energy and it reaches $\sqrt{2 g h}$ . The yo-yo has to spin faster and faster as it falls, and two thirds of its energy goes into that spin, so only a third is left for falling. It reaches the bottom at $\sqrt{\frac{2 g h}{3}}$ , slower than the ball by a factor of $\sqrt{3}$ .
Your score: 0 of 12, from the boxes you ticked.
Question 3: Experimental Design and Analysis 10 points

A physics class has a clamp stand with a fixed pivot, light string, a meterstick, a stopwatch, a protractor, and a solid steel sphere 4.0 cm in diameter. The sphere has a small hook, so it can hang from the string as a pendulum bob.

Steel sphere, released from about 10 degrees
Length $ℓ$ (m) Time for 10 swings (s) Period $T$ (s)

0.30 11.1 1.11
0.50 14.1 1.41
0.70 16.9 1.69
0.90 19.0 1.90
1.10 21.1 2.11

Respond to parts (a), (b), (c) and (d). Draw any graph on your own graph paper.
1. (a)
  
  The students want to determine experimentally whether the period of a pendulum depends on the angle from which it is released, for release angles from 5 degrees up to 60 degrees.
  
  i. Describe a procedure the students could use, using only the equipment listed. Say which quantity they vary, which quantities they keep the same, and what they measure.
  
  ii. Briefly describe a method to reduce experimental uncertainty in the measured period.
  Question 3, part (a): your response. It stays in this browser tab and is never sent anywhere.
  Score part (a)
  - 1 point Varies only the release angle, set with the protractor, at several angles spread across the range in the question, while keeping the length from the pivot to the center of the sphere the same (there is only one sphere, so the answer need not say the sphere stays the same), and measures the period with the stopwatch at each angle. Any fixed length and any set of angles that spans roughly 5 to 60 degrees earns the point, for example 5, 15, 30, 45 and 60 degrees, or every 10 degrees from 10 to 60.Earns it: "Hang the steel sphere so the length from the pivot to its center is 0.80 m. Pull it back to 5, 15, 30, 45 and 60 degrees in turn, measured with the protractor, release it from rest each time, and time its period. Keep the same length every time."; "Use one string length, 0.50 m, and the same sphere throughout. Release from rest at 5, 10, 20, 30, 40, 50 and 60 degrees, set with the protractor, and time 10 swings at each angle to get the period."Does not: changes the string length between releases as well as the angle (two variables change at once); tries only angles close together, such as 5 and 10 degrees (that cannot show whether the period changes across the range asked about); pushes the bob at the start instead of releasing it from rest (the swing then goes past the angle that was set)
  - 1 point Reduces uncertainty in the period by timing many complete swings, such as 10, and dividing by the number of swings, or by repeating the timing several times at each angle and averaging. Any number of swings well above one, and any number of repeated trials, earns the point.Earns it: "Time 10 full back-and-forth swings and divide by 10. Do it three times at each angle and average."; "Time 20 swings instead of one, and repeat each angle five times and take the mean."Does not: times a single swing (reaction time is then a large fraction of the period); times from one side to the other and calls that the period (that is half a cycle); "repeat the experiment" with no detail of what is repeated or how the results are combined
  Model response
  i. Hang the steel sphere from the pivot and use the meterstick to set the length from the pivot to the sphere's center to 0.80 m. Pull the sphere aside to 5 degrees, measured with the protractor, release it from rest, and measure the period with the stopwatch. Repeat at 15, 30, 45 and 60 degrees. Keep the same length for every release, so the only thing that changes is the release angle. The quantity measured is the period.
  
  ii. Instead of timing one swing, time 10 complete back-and-forth swings and divide by 10, which shrinks the effect of reaction time at the start and stop. Do this three times at each angle and average.
2. (b)
  
  Describe how the students could analyze their data from part (a) to determine whether the period depends on the release angle.
  Question 3, part (b): your response. It stays in this browser tab and is never sent anywhere.
  Score part (b)
  - 1 point Describes a graph of the period $T$ against the release angle (or an equivalent comparison of the average periods at the different angles).Earns it: "Plot T on the vertical axis against the release angle on the horizontal axis."Does not: plots period against length (length was held fixed in this experiment); plots period against time
  - 1 point States how the result answers the question: points along a horizontal line, with the periods at different angles differing by no more than the spread of repeated trials at one angle, mean the period does not depend on the release angle over that range; periods that change with angle by more than that spread, even only at the larger angles, mean it does.Earns it: "If the points lie on a flat line, with the differences between angles no bigger than the differences between repeated trials, the angle does not matter. If the periods at the big angles are longer by more than that spread, it does."Does not: "if the line goes through the origin, the period is proportional to the angle" with no criterion for no dependence; decides from a comparison of just two single trials with no allowance for uncertainty
  Model response
  Plot the average period $T$ on the vertical axis against the release angle on the horizontal axis. If the points lie along a horizontal line, so that the periods at different angles differ by no more than the spread between repeated trials at the same angle, the period does not depend on the release angle over this range. If the periods change with the angle by more than that spread, even if only at the larger angles, the period does depend on the release angle.
3. (c)
  
  In a second experiment, a group uses the steel sphere and changes the length $ℓ$ of the pendulum, measured from the pivot to the center of the sphere. For each length they release the bob from about 10 degrees and time 10 complete swings. Their data are in the table. They want to use the data to determine an experimental value for the acceleration due to gravity, $g$ .
  
  i. Indicate the quantities the students could plot on the vertical and horizontal axes to produce a linear graph whose slope can be used to determine $g$ . Include units.
  
  ii. On graph paper, plot the quantities you indicated in part (c)(i). Label each axis with its quantity and units, choose a numerical scale for each, and plot the five data points.
  
  iii. Draw a best-fit line for the data you plotted in part (c)(ii).
  Question 3, part (c): your response. It stays in this browser tab and is never sent anywhere.
  Score part (c)
  - 1 point Indicates a pair of quantities that gives a linear graph whose slope can be used to find $g$ : $T^{2}$ (in s²) on the vertical axis against $ℓ$ (in m) on the horizontal axis. $ℓ$ against $T^{2}$ , or $T$ against $\sqrt{ℓ}$ , also earns the point.Earns it: "Vertical: period squared, $T^{2}$ , in s². Horizontal: length $ℓ$ in m."Does not: $T$ against $ℓ$ (that graph curves, because $T$ grows as the square root of $ℓ$ ); time for 10 swings against length, unsquared (also curved)
  - 1 point Labels both axes with quantities and units consistent with part (c)(i), and uses linear numerical scales that spread the data over most of the graph.Earns it: " $T^{2}$ (s²) from 0 to 5.0 in steps of 0.5; $ℓ$ (m) from 0 to 1.2 in steps of 0.1"Does not: unevenly spaced scale marks, such as the data values themselves written at equal spacing; axes without units
  - 1 point Plots all five points correctly on the chosen scale. For $T^{2}$ against $ℓ$ the points are (0.30, 1.23), (0.50, 1.99), (0.70, 2.86), (0.90, 3.61) and (1.10, 4.45). For other axes accepted in part (c)(i), the same five data points placed on those axes earn the point.Earns it: five points within half a grid division of the values aboveDoes not: plots $T$ instead of $T^{2}$ after labeling the axis $T^{2}$ ; plots the time for 10 swings, squared, instead of the period squared
  - 1 point Draws a single straight line that follows the trend of the points, with points on both sides of it, rather than joining the points dot to dot.Earns it: a straight line passing close to all five points and close to the originDoes not: a line joining the first and last points only, when it leaves the middle points all on one side; a zigzag connecting each point to the next
  Model response
  i. Plot the period squared, $T^{2}$ , in s², on the vertical axis, and the length $ℓ$ , in m, on the horizontal axis. Squaring $T_{p} = 2 π \sqrt{\frac{ℓ}{g}}$ gives $T^{2} = \frac{4 π^{2}}{g} ℓ$ , a straight line through the origin with slope $\frac{4 π^{2}}{g}$ .
  
  ii. Squaring the periods gives 1.23, 1.99, 2.86, 3.61 and 4.45 s². The vertical axis runs from 0 to 5.0 s² in steps of 0.5, and the horizontal axis from 0 to 1.2 m in steps of 0.1. The five points are plotted at (0.30 m, 1.23 s²), (0.50 m, 1.99 s²), (0.70 m, 2.86 s²), (0.90 m, 3.61 s²) and (1.10 m, 4.45 s²).
  
  iii. A straight line through the middle of the points passes very close to the origin and close to every point, with some points just above it and some just below.
4. (d)
  
  Using your best-fit line from part (c)(iii), calculate an experimental value for $g$ .
  Question 3, part (d): your response. It stays in this browser tab and is never sent anywhere.
  Score part (d)
  - 1 point Calculates the slope of the best-fit line from two points read off the line (not two data points, unless they lie on the line), with units. For $T^{2}$ against $ℓ$ the slope is about 4.0 s²/m. For $ℓ$ against $T^{2}$ the slope is about 0.248 m/s², and for $T$ against $\sqrt{ℓ}$ it is about $2.0 {s/m}^{1 / 2}$ .Earns it: "slope $= \frac{4.44 - 0.81}{1.10 - 0.20} = 4.03$ s²/m"Does not: divides a single $T^{2}$ by its $ℓ$ instead of finding the slope of the line; a slope with no units, or with units that do not match the axes, such as m/s² for $T^{2}$ against $ℓ$
  - 1 point Uses the slope to calculate $g$ in the way the student's axes require: $g = \frac{4 π^{2}}{slope}$ for $T^{2}$ against $ℓ$ , $g = 4 π^{2} \times slope$ for $ℓ$ against $T^{2}$ , or $g = \frac{4 π^{2}}{{slope}^{2}}$ for $T$ against $\sqrt{ℓ}$ . Each gives about 9.79 m/s². Any value from about 9.4 to 10.2 m/s² that follows correctly from the student's own line earns the point.Earns it: " $g = \frac{4 π^{2}}{4.03 s^{2} /m} = 9.79$ m/s²"Does not: $g = slope$ , about 4 m/s² (forgets the $4 π^{2}$ ); $g = \frac{slope}{4 π^{2}}$ , about 0.10 m/s² (for $T^{2}$ against $ℓ$ , the relationship is inverted)
  Model response
  Two points on my best-fit line, not data points, are (0.20 m, 0.81 s²) and (1.10 m, 4.44 s²):
  $slope = \frac{4.44 - 0.81}{1.10 - 0.20} = 4.03 s^{2} /m$
  The line is $T^{2} = \frac{4 π^{2}}{g} ℓ$ , so the slope is $\frac{4 π^{2}}{g}$ :
  $g = \frac{4 π^{2}}{4.03 s^{2} /m} \approx 9.79 {m/s}^{2}$
Your score: 0 of 10, from the boxes you ticked.
Question 4: Qualitative/Quantitative Translation 8 points

Two sealed plastic storage boxes have the same outside dimensions. Each is rectangular, with vertical sides and a flat bottom of area $A$ , so every horizontal slice through either box has area $A$ . Box 1 is empty and has mass $M_{1}$ . Box 2 has sand inside, so its mass $M_{2}$ is greater than $M_{1}$ . Each box floats upright at rest in a tank of fresh water of density $ρ_{w}$ , with its bottom level and part of the box above the surface. The tank is wide, so the water level in the tank does not change noticeably when a box sinks farther.

A student sets a stone of mass $m$ on the center of each box's lid. Each box sinks a little farther and comes to rest again, still upright, with its bottom level and part of the box above the surface. Let $Δ d_{1}$ and $Δ d_{2}$ be the extra distances that Box 1 and Box 2 sink when the stone is added.

Respond to parts (a), (b) and (c).
1. (a)
  
  Indicate whether $Δ d_{2}$ is greater than, less than, or equal to $Δ d_{1}$ . Justify your answer using qualitative reasoning beyond references to equations.
  Question 4, part (a): your response. It stays in this browser tab and is never sent anywhere.
  Score part (a)
  - 1 point Indicates that $Δ d_{2}$ is equal to $Δ d_{1}$ , with an attempt at a justification that uses forces or displaced water.Earns it: "They sink the same extra distance, because the same stone needs the same extra push from the water."Does not: greater, because Box 2 is heavier (its extra weight already set its starting depth); less, because the stone is a smaller fraction of Box 2's mass
  - 1 point Connects the claim to the balance of forces on a floating box: each box is at rest before and after, so the buoyant force on it must grow by exactly the stone's weight, which is the same for both boxes.Earns it: "Both boxes float at rest, so the buoyant force equals the weight each time. Adding the stone adds the same weight to each, so each needs the same extra buoyant force."Does not: says the buoyant force increases without saying by how much or why; says the buoyant force is the same on both boxes before the stone is added (Box 2 is heavier, so it needs more)
  - 1 point Justifies the equal depths from the buoyant force: the buoyant force is the weight of the water a box pushes aside, so the same extra force means the same extra volume of the same water, and with the same bottom area that volume is the same extra depth.Earns it: "The extra push has to come from pushing aside extra water whose weight equals the stone's weight. Same water, same weight, same volume, and the boxes have the same bottom, so the same extra depth."Does not: "the stone is the same" with no link to displaced water; says the denser box displaces more water when the stone is added
  Model response
  Equal. Each box floats at rest both before and after the stone goes on, so each time the water's upward push balances the box's weight. Adding the stone adds the same weight, $m g$ , to each box, so the water has to push up on each one harder by that same amount. The water pushes up harder only by being pushed aside more: the buoyant force is the weight of the water displaced. The same extra force therefore means the same extra volume of the same water. Both boxes have the same bottom area and vertical sides, and they stay upright, so that extra volume is the same extra depth for each. How heavy a box was to begin with only decided where it started floating.
2. (b)
  
  Derive an expression for $Δ d$ , the extra distance a floating box of mass $M$ sinks when the stone is added. Express your answer in terms of $M$ , $m$ , $A$ , $ρ_{w}$ and physical constants, as appropriate. Your answer need not contain every one of these symbols. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
  Question 4, part (b): your response. It stays in this browser tab and is never sent anywhere.
  Score part (b)
  - 1 point Begins with the zero net force on the floating box, before and after the stone is added: the buoyant force equals the total weight each time.Earns it: " $Σ F = 0$ , so $F_{b} = M g$ before and $F_{b}^{'} = (M + m) g$ after"Does not: uses only the stone's weight, with no statement of the force balance it comes from; sets the buoyant force equal to the stone's weight alone before the stone is added
  - 1 point Writes the buoyant force with the submerged volume as the bottom area times the submerged depth: $F_{b} = ρ_{w} A d g$ , before and after.Earns it: " $ρ_{w} A d_{1} g = M g$ and $ρ_{w} A d_{2} g = (M + m) g$ "Does not: uses the whole volume of the box (only the part below the surface displaces water); uses the density of the box or the stone in place of the water's
  - 1 point Arrives at a correct expression: $Δ d = \frac{m}{ρ_{w} A}$ .Earns it: " $Δ d = d_{2} - d_{1} = \frac{m}{ρ_{w} A}$ "Does not: $Δ d = \frac{M + m}{ρ_{w} A}$ (that is the whole depth after the stone is added, not the extra depth); $Δ d = \frac{m g}{ρ_{w} A}$ (the $g$ on each side should cancel)
  Model response
  The box floats at rest, so $Σ F = 0$ and the buoyant force equals the weight. The buoyant force is $F_{b} = ρ_{w} V g$ , where $V$ is the volume below the surface. With the bottom level and the sides vertical, that is the bottom area times the submerged depth $d$ , so $V = A d$ .
  
  Before the stone: $ρ_{w} A d_{1} g = M g$
  
  After the stone: $ρ_{w} A d_{2} g = (M + m) g$
  
  Subtracting the first equation from the second:
  $ρ_{w} A (d_{2} - d_{1}) g = m g$ $Δ d = d_{2} - d_{1} = \frac{m}{ρ_{w} A}$
3. (c)
  
  Does your expression from part (b) agree with your reasoning in part (a)? Justify why or why not.
  Question 4, part (c): your response. It stays in this browser tab and is never sent anywhere.
  Score part (c)
  - 1 point Makes a statement about how $Δ d$ depends, or does not depend, on the box's mass $M$ according to the expression from part (b). The point is earned for addressing the dependence, even if it is described incorrectly.Earns it: "The expression has no $M$ in it."Does not: "yes, they agree" with no reference to what the expression depends on
  - 1 point Correctly links the expression to the claim in part (a): because $Δ d$ does not depend on $M$ , and the boxes differ only in mass ( $m$ , $A$ and $ρ_{w}$ are the same for both), the two boxes sink the same extra distance, as claimed.Earns it: " $Δ d = \frac{m}{ρ_{w} A}$ has the stone's mass, the water's density and the bottom area in it, all the same for both boxes, and not the box's mass, so $Δ d_{1} = Δ d_{2}$ , which is what I said in part (a)."Does not: says the expression shows the heavier box sinks farther; says they agree because both boxes float
  Model response
  Yes. The expression $Δ d = \frac{m}{ρ_{w} A}$ contains the stone's mass, the water's density and the box's bottom area, but not the box's own mass $M$ . The two boxes differ only in $M$ , so the expression gives the same extra depth for both, $Δ d_{1} = Δ d_{2}$ . That is what I argued in part (a): each box needs the same extra buoyant force, so it pushes aside the same extra water, and the same bottom area turns that into the same extra depth.
Your score: 0 of 8, from the boxes you ticked.

Steel sphere, released from about 10 degrees
Length $ℓ$ (m)	Time for 10 swings (s)	Period $T$ (s)
0.30	11.1	1.11
0.50	14.1	1.41
0.70	16.9	1.69
0.90	19.0	1.90
1.10	21.1	2.11

Practice Exam 2

42 multiple-choice and 4 free-response questions, in the format of the AP Physics 1: Algebra-Based Exam. Each section runs on its own clock and ends when time runs out.

Section	Questions	Time
Section I: Multiple Choice	42	85 minutes
Section II: Free Response	4	95 minutes

A four-function, scientific, or graphing calculator is allowed on both sections of the exam.

One section at a time. When a section's clock reaches zero, the section ends and is scored.
Multiple choice is scored when its section ends, with an explanation for every choice.
Free response is never scored by the page. When the section ends, you see the scoring guide and a model response for each part and tick the points your answer earns.
At the end you get your score and the time you used for each section.
Want the answers as you go? Pick Study mode below. Each multiple-choice answer is checked as soon as you give it, with its explanation, and each free-response question has a button that opens its scoring guide.
Your answers are kept in this browser tab if the page reloads, and nowhere else.

Time for each section Standard time Time and a half Double time Untimed Study mode

Section I: Multiple Choice

42 questions · 85 minutes

Take east as positive. Two friends jog along a straight beach path from the same bench to a lifeguard tower 200 m east of it. Ali jogs straight to the tower. Bea jogs 300 m east to a pier, then turns and jogs 100 m west to the tower. Which choice gives each jogger's displacement and the distance each one jogs?
Answer: B.
1. A
  This treats distance as the size of the displacement. That only works when the path never turns back; Bea's 100 m back west still counts toward her distance.
2. B
  Both end 200 m east of the bench, so both displacements are +200 m. Distance is a scalar that counts every meter of path, so Bea's is 300 m + 100 m = 400 m.
3. C
  This adds the sizes of Bea's two legs as if both pointed east. Her legs are +300 m and -100 m, and their vector sum is +200 m.
4. D
  This takes Bea's last leg as her whole displacement. Displacement runs from where she started to where she ended, which is 200 m east.
Take east as positive. A commuter train runs along a straight track. At $t = 0$ it is at $x = + 500$ m, moving east at 20 m/s. At $t = 40$ s it is at $x = + 1100$ m, again moving east at 20 m/s. In between, it slowed down through a work zone and then sped back up. What are the train's average velocity and average acceleration over the 40 s?
Answer: D.
1. A
  This averages the starting and ending velocities. That shortcut only works for constant acceleration, and this train slowed down and sped up in between.
2. B
  This divides the final position, 1100 m, by the time. The train started at +500 m, so its displacement is only 600 m.
3. C
  The slowing down in the work zone does not enter the average. Average acceleration is the change in velocity over the time interval, and the velocity ended where it began.
4. D
  Average velocity is displacement over time, (1100 m - 500 m)/(40 s) = 15 m/s east. Average acceleration uses only the starting and ending velocities, which are equal, so it is 0.
Take east as positive. A toy car moves along a straight track. Its position-time graph from $t = 0$ to $t = 4.0$ s is a smooth curve: the position increases the whole time, steeply at first, and the curve flattens until it is level at $t = 4.0$ s. During this interval, which way do the car's velocity and acceleration point?
Answer: C.
1. A
  A flattening curve does not mean the car turns back. The position still increases, so the car is still moving east.
2. B
  This reads a rising position graph as a positive acceleration. The acceleration follows how the slope changes, and here the slope is getting smaller.
3. C
  The slope of the position graph is the velocity, and it is positive but shrinking, so the car moves east while slowing down. Slowing down means the acceleration points opposite the velocity, which is west.
4. D
  This treats the curve as if it were a straight line. A straight position graph means constant velocity, but this one bends, so the velocity is changing.
Take east as positive. A train moves east along a straight, level track at a steady 15 m/s. A passenger sets a toy remote-control car on the aisle floor, and it drives east along the aisle, speeding up steadily from 1.0 m/s to 3.0 m/s relative to the train in 2.0 s. As measured by a person standing still on the platform, what are the toy car's velocity at the end of the 2.0 s and its acceleration?
Answer: C.
1. A
  3.0 m/s is the velocity a passenger on the train measures. The platform observer also sees the toy carried along at the train's 15 m/s.
2. B
  This subtracts the toy's velocity from the train's. The toy drives the same way the train moves, so the velocities add.
3. C
  The platform observer adds the train's velocity to the toy's: 15 m/s + 3.0 m/s = 18 m/s east. The train moves at a steady velocity, so both observers measure the same acceleration, (3.0 - 1.0)/2.0 = 1.0 m/s² east.
4. D
  This mixes reference frames, taking the final velocity from the platform (18 m/s) and the initial one from the train (1.0 m/s). From the platform the toy went from 16 m/s to 18 m/s, so the acceleration is 1.0 m/s².
A small launcher on level ground fires a ball at 15 m/s, aimed 25 degrees above the horizontal, and the ball lands back on the ground. Then it fires an identical ball at the same speed, aimed 65 degrees above the horizontal. Air resistance is negligible. Compared with the first ball, how do the second ball's time in the air and horizontal distance compare?
Answer: B.
1. A
  Time in the air is set by the vertical component of the launch velocity, not by the launch speed alone. The 65 degree launch has the larger vertical component.
2. B
  The steeper launch has a larger vertical component, so the ball stays up longer, but its horizontal component is smaller by just the right factor. The horizontal distance depends on $\sin θ \cos θ$ , which is the same for 25 and 65 degrees.
3. C
  A longer flight does not by itself mean a longer distance. The steeper ball also moves more slowly sideways, and the two effects cancel here.
4. D
  This looks only at the smaller horizontal component and forgets the extra time it has to act. Distance is horizontal velocity times time, and the product comes out equal.
A sign stand is shaped like a T. A thin, uniform post 1.8 m long with a mass of 3.0 kg stands straight up, and a thin, uniform crossbar 0.80 m long with a mass of 1.5 kg is fixed across its top, centered on it. How far above the bottom of the post is the center of mass of the stand?
Answer: A.
1. A
  Each uniform piece acts as if its mass sat at its own middle: the post's at 0.90 m and the crossbar's at the top, 1.8 m. Weighting each by its mass, $\frac{(3.0) (0.90) + (1.5) (1.8)}{4.5} = 1.2$ m.
2. B
  This averages the two middles, 0.90 m and 1.8 m, as if the pieces had equal masses. The post has twice the crossbar's mass, so the center of mass sits closer to the post's middle.
3. C
  This swaps the masses, giving the crossbar 3.0 kg and the post 1.5 kg. Each mass goes with the middle of its own piece.
4. D
  This divides by the number of pieces, 2, instead of the total mass, 4.5 kg. A center of mass above the top of the stand is a sign the weighting went wrong.
A small box B rests on top of a larger box A, and box A rests on the floor. Both boxes are at rest. Which list gives every force exerted on box A, with its direction?
Answer: D.
1. A
  This leaves out box B. B touches A, and every object in contact with A exerts a force on it.
2. B
  B's weight is the force Earth exerts on B, so it does not belong on A's diagram. What A feels is B's contact push, a normal force.
3. C
  A normal force is not always up. It points away from the surface that exerts it, and B sits on top of A, so B pushes A down.
4. D
  Box A touches the floor and box B, and Earth pulls on it, so three forces act on it. Box B pushes down on A's top surface, because a normal force points away from the surface that exerts it.
A pickup truck tows a small car forward along a level road with a light rope that does not stretch, and both are speeding up. How do the force the rope exerts on the car and the force the rope exerts on the truck compare?
Answer: C.
1. A
  A rope can only pull along itself, toward its own middle. At the truck's end that pull points backward, toward the car.
2. B
  This assumes the end doing the dragging must pull harder. With a rope of negligible mass, the tension is the same at both ends.
3. C
  A light rope that does not stretch has the same tension all along it. It pulls the car forward and pulls back on the truck with that same tension.
4. D
  This mixes up the rope with the truck's drive. The road pushes the truck forward harder than the rope pulls it back, but the rope's own pulls at its two ends are equal.
A hot-air balloon drifts east at a steady 3.0 m/s, a few meters above a field. A rider lets go of a small sandbag over the side. Air resistance on the sandbag is negligible. Seen from the ground, the sandbag keeps moving east at 3.0 m/s while it falls. Which statement explains why its eastward velocity stays the same?
Answer: A.
1. A
  Gravity is the only force on the sandbag, and it points straight down. Forces are balanced in the east-west direction, so the velocity changes only in the vertical direction.
2. B
  Gravity is unbalanced, so the net force is not zero. The sandbag speeds up downward while its eastward velocity holds steady.
3. C
  No force from the balloon travels with the sandbag after it lets go. An object needs no force to keep moving at a constant velocity.
4. D
  Its speed does change, because its downward velocity grows. Only the eastward part of its velocity stays the same.
A roofer lowers a bag of tools of mass $m$ to the ground on a rope. Near the ground the bag is still moving down, but it is slowing down with an acceleration of magnitude $a$ . Air resistance is negligible. Which expression gives the magnitude of the force the rope exerts on the bag?
Answer: B.
1. A
  This takes the acceleration as downward because the bag moves down. When an object slows, its acceleration points opposite its velocity, so here it points up.
2. B
  Slowing down while moving down means the bag's acceleration points up. Taking up as positive, $T - m g = m a$ , so $T = m (g + a)$ , more than the bag's weight.
3. C
  This sets the rope's pull equal to the bag's weight. That holds only at a constant velocity, and this bag is slowing.
4. D
  This is the net force alone. The rope must also hold up the bag's weight, $m g$ .
An astronaut on the Moon, where the gravitational field is about one sixth as strong as on Earth, pushes a 120 kg equipment cart horizontally across level ground. The cart's wheels turn freely, so friction is negligible. Compared with doing the same thing on Earth, how large a horizontal push gives the cart the same acceleration?
Answer: D.
1. A
  This confuses weight with inertia. The cart weighs one sixth as much on the Moon, but it is just as hard to speed up horizontally.
2. B
  This turns the ratio upside down, and weight does not enter a horizontal push at all. The cart's inertial mass is the same on the Moon as on Earth.
3. C
  This squares the field ratio, as if an inverse-square rule applied here. The cart's mass does not depend on where it is, so the push is unchanged.
4. D
  How much a push changes the cart's motion depends on its inertial mass, which is 120 kg everywhere. The weaker gravity changes the cart's weight, not its inertia.
A worker drags a heavy crate across a level floor at a steady speed. First she pulls with a horizontal rope. Then she pulls with the rope tilted 30 degrees above the horizontal, and the crate still slides along the floor without lifting. How does the kinetic friction force the floor exerts on the crate compare in the two cases?
Answer: A.
1. A
  Kinetic friction is $μ_{k} F_{N}$ . The tilted rope pulls partly upward, so the floor no longer has to hold up the crate's whole weight, and the normal force drops. With the same coefficient, the friction drops too.
2. B
  The coefficient is the same, but the friction is the coefficient times the normal force. Pulling partly upward lowers the normal force, so it lowers the friction.
3. C
  How hard the worker pulls does not set kinetic friction; the normal force does. The upward part of her pull makes the normal force smaller, so the friction is smaller.
4. D
  Kinetic friction does not depend on the speed. It depends on the normal force, which is smaller when the rope pulls partly upward.
A roller coaster car carries a rider over the top of a hump, where the track curves like a vertical circle. On one run, at the very top of the hump, the seat pushes up on the rider with a force equal to half her weight. On a later run, at the very top, the seat exerts no force on her at all. How does her speed at the top on the later run compare with her speed there on the first run?
Answer: B.
1. A
  This makes the speed proportional to the net force. The net force needed for circular motion is $\frac{m v^{2}}{r}$ , so doubling it doubles the speed squared, not the speed.
2. B
  At the top, $m g - F_{N} = \frac{m v^{2}}{r}$ . On the first run $F_{N} = \frac{1}{2} m g$ , so $v_{1}^{2} = \frac{1}{2} g r$ . On the later run $F_{N} = 0$ , so $v_{2}^{2} = g r$ . The speed squared doubles, so the speed is $\sqrt{2}$ , about 1.4, times as large.
3. C
  This has the seat pushing harder at a higher speed, which is true at the bottom of a dip. At the top of a hump the center of the circle is below the rider, and the faster she goes, the less the seat pushes.
4. D
  This squares the factor of 2 when it should take its square root. The speed squared doubles, so the speed grows by a factor of $\sqrt{2}$ .
Delivery drone P has a mass of 3.0 kg and flies at 4.0 m/s. Drone Q has half the mass of drone P and flies three times as fast. Which claim about their translational kinetic energies is correct?
Answer: A.
1. A
  Since $K = \frac{1}{2} m v^{2}$ , half the mass halves the kinetic energy, but three times the speed multiplies it by 9. Q has $\frac{1}{2} \times 9 = 4.5$ times as much as P.
2. B
  Mass does count, but speed counts more, because it is squared. Q's tripled speed more than makes up for its halved mass.
3. C
  Q does have more, but not for this reason. Kinetic energy is proportional to mass, so Q's halved mass halves what its speed alone would give.
4. D
  Halving the mass halves the kinetic energy, but tripling the speed multiplies it by 9, not 3, because speed is squared. The two effects do not cancel.
A 2.0 kg cart starts at rest on a level track where friction is negligible. A motorized arm moves it along the track, in the $+ x$ direction. The arm's force on the cart against the cart's position, described in words: a steady $+ 6.0$ N from $x = 0$ to $x = 2.0$ m, then falling in a straight line from $+ 6.0$ N to 0 between $x = 2.0$ m and $x = 3.0$ m, then a steady $- 4.0$ N from $x = 3.0$ m to $x = 4.0$ m. What is the cart's speed at $x = 4.0$ m?
Answer: C.
1. A
  This counts the last meter as +4.0 J. The force there points opposite the motion, so that work is negative and takes energy out.
2. B
  This sets 11 J equal to $m v^{2}$ and drops the $\frac{1}{2}$ in the kinetic energy.
3. C
  The work is the signed area under the force-position graph: 12 J + 3.0 J - 4.0 J = 11 J. All of it goes into kinetic energy, so $11 = \frac{1}{2} (2.0) v^{2}$ and $v = 3.3$ m/s.
4. D
  This takes the area from 2.0 m to 3.0 m as a 6.0 J rectangle. The force falls to zero there, so the area is a triangle worth 3.0 J.
A 1500 kg car speeds up along a level, straight on-ramp from 12 m/s to 18 m/s over a distance of 75 m. Assuming the net force on the car is constant, how large is it?
Answer: D.
1. A
  This squares the change in speed, $(6.0)^{2}$ , instead of taking the change in the squared speed. Kinetic energy depends on $v^{2}$ , so $18^{2} - 12^{2}$ is not $6^{2}$ .
2. B
  This leaves out the $\frac{1}{2}$ in the kinetic energy.
3. C
  This uses only the final kinetic energy, as if the car started from rest. It was already moving at 12 m/s.
4. D
  The net work equals the change in kinetic energy: $\frac{1}{2} (1500) (18^{2} - 12^{2}) = 135,000$ J. Dividing by 75 m gives 1800 N.
A pinball plunger is pulled back against an ideal spring. Compressing the spring 2.0 cm from its relaxed length stores elastic potential energy $U$ . How much more energy is stored when the compression is increased from 2.0 cm to 5.0 cm?
Answer: A.
1. A
  Elastic potential energy grows with the square of the compression. At 5.0 cm, 2.5 times the compression, the spring stores $(2.5)^{2} U = 6.25 U$ in all, which is $5.25 U$ more than the $U$ already stored.
2. B
  This treats the energy as proportional to the compression, so 2.5 times the compression would hold $2.5 U$ . The energy goes with the square of the compression.
3. C
  $6.25 U$ is the total stored at 5.0 cm. The question asks how much more than the $U$ already there at 2.0 cm.
4. D
  This squares only the extra 3.0 cm of squeeze, $(1.5)^{2} U$ . The energy depends on the whole compression measured from the relaxed length, not on the added part alone.
Questions 18 to 19 refer to the following.
A 60 kg bungee jumper steps off a bridge platform from rest. Her cord is tied to the platform and has a relaxed length of 15 m. While it is slack the cord exerts no force on her; once she falls past 15 m, treat it as an ideal spring. She falls until she stops for an instant 35 m below the platform, where the cord is stretched 20 m past its relaxed length. Treat her as a single object, and ignore the cord's mass and air resistance. Use $g = 10$ m/s².

Take the gravitational potential energy of the jumper-Earth system to be zero with the jumper at her lowest point. How does the gravitational potential energy of the system while she stands on the platform compare with its value at the moment the cord first becomes taut?
Answer: B.
1. A
  This takes the taut point to be 15 m above the zero. The cord goes taut after she has fallen 15 m, which leaves her 20 m above her lowest point.
2. B
  Gravitational potential energy is proportional to the height above the zero. The platform is 35 m above her lowest point and the taut point is 35 m - 15 m = 20 m above it, so the ratio is 35/20 = 1.75. Both values are positive, because both points are above the chosen zero.
3. C
  This divides the taut point's height by the platform's. The platform is the higher point, so the system has more gravitational potential energy there, not less.
4. D
  This squares the ratio of the heights. Near Earth's surface the gravitational potential energy is $m g y$ , proportional to the height itself.
Uses the material above question 18.

Call the jumper's mass $m$ , the cord's relaxed length $L$ , and its stretch at her lowest point $x$ . Which expression gives the spring constant of the cord?
Answer: D.
1. A
  This sets the cord's pull at the lowest point equal to her weight, $k x = m g$ . She is not in equilibrium there; she is at rest for an instant with the cord pulling much harder than her weight.
2. B
  This uses her whole drop, $L + x$ , as the stretch. The cord's energy depends on how far it is stretched past its relaxed length, which is only $x$ .
3. C
  This counts only the energy she loses while the cord stretches, $m g x$ . Her free fall through $L$ before the cord goes taut also adds energy that ends up in the cord.
4. D
  She is at rest at the top and at the bottom, so all the gravitational potential energy lost over her whole drop, $m g (L + x)$ , ends up in the cord: $\frac{1}{2} k x^{2} = m g (L + x)$ , so $k = \frac{2 m g (L + x)}{x^{2}}$ . With her numbers that is 105 N/m.
A box slides down a rough ramp at a steady speed. Take the box and Earth as the system; the ramp is outside it. What happens to the mechanical energy of the system as the box slides down?
Answer: C.
1. A
  A steady speed keeps the kinetic energy the same, but the box is also losing height. Mechanical energy includes the gravitational potential energy, which is falling.
2. B
  Earth is inside this system, so gravity's effect is already counted as the change in potential energy. Counting it again as outside work double counts it.
3. C
  Kinetic energy stays the same while gravitational potential energy falls, so the mechanical energy drops. The ramp is outside the system, and its friction does negative work that carries energy out.
4. D
  Energy is conserved overall, but this system is not closed. Friction from the ramp moves energy out of it, into thermal energy of the box and ramp.
A cyclist and her bike have a combined mass $m$ . She rides at a steady speed $v$ up a straight road that rises a height $h$ for every distance $L$ traveled along it. At what rate does the gravitational potential energy of the cyclist-bike-Earth system increase?
Answer: B.
1. A
  This treats her as rising straight up at speed $v$ . Only her vertical rise changes the gravitational potential energy, and she rises $h$ for every $L$ she rides.
2. B
  In a time $Δ t$ she rides $v Δ t$ along the road and rises $\frac{h}{L} v Δ t$ . The potential energy grows by $m g$ times that rise, so the rate is $\frac{m g v h}{L}$ .
3. C
  This turns the slope upside down. The road rises less than it runs, so her vertical speed is smaller than $v$ , by the factor $\frac{h}{L}$ .
4. D
  This is the part of her weight along the road, a force. A rate of energy change also needs her speed.
Two remote-control boats move along the same straight line on a pond, heading toward each other. A 0.80 kg boat moves east at 5.0 m/s, and a 1.0 kg boat moves west at 3.0 m/s. Take east as positive. What is the total momentum of the two-boat system?
Answer: A.
1. A
  Each momentum carries its velocity's sign: (0.80)(+5.0) = +4.0 kg m/s and (1.0)(-3.0) = -3.0 kg m/s. Their sum is +1.0 kg m/s, which points east.
2. B
  This multiplies the total mass, 1.8 kg, by the difference of the speeds, 2.0 m/s. A system's momentum is the sum of each boat's own mv, each with its sign.
3. C
  Adding 4.0 and 3.0 treats momentum as a scalar. The boats move in opposite directions, so their momenta partly cancel.
4. D
  This multiplies the total mass, 1.8 kg, by the average of the two speeds, 4.0 m/s. A system's momentum is the sum of each part's mv with its sign, not its total mass times an averaged speed.
Two gliders start at rest on a level air track, and friction is negligible. Glider 2 has twice the mass of glider 1. A student pushes each one with the same constant net force for the same 0.50 s. At the end of the push, how do glider 2's momentum and speed compare with glider 1's?
Answer: B.
1. A
  A larger mass does not take a larger impulse from the same push. The impulse is the force times the time, and both are the same.
2. B
  Each glider gets the same impulse, $F Δ t$ , so each gains the same momentum from rest. Momentum is $m v$ , so glider 2, with twice the mass, ends with half the speed.
3. C
  The change in momentum is set by the impulse, which is the same for both. Only the speed depends on the mass.
4. D
  Equal momenta do not mean equal speeds. With twice the mass, glider 2 needs only half the speed to carry the same momentum.
You toss a ball of mass $m$ straight up at speed $v$ and catch it when it comes back down to the height where it left your hand. Air resistance is negligible. From release to catch, what impulse does Earth's gravity deliver to the ball?
Answer: D.
1. A
  The ball comes back at the same speed, but moving the other way. Momentum is a vector, so its change is not zero.
2. B
  $m v$ is the change in momentum on the way up only, from $m v$ upward to zero. Gravity keeps pulling on the way down and delivers the same again.
3. C
  The size is right but the direction is not. Impulse points along the force, and gravity pulls down for the whole flight.
4. D
  The momentum goes from $m v$ upward to $m v$ downward, a change of $2 m v$ downward. Gravity is the only force, and $m g$ acting over the flight time, $\frac{2 v}{g}$ , gives the same $2 m v$ .
A 600 kg space probe coasts in a straight line at 50 m/s, so far from any planet or star that no outside forces act on it. Explosive bolts then split it into a 200 kg capsule and a 400 kg engine stage, which fly apart. What is the speed of the center of mass of the two pieces after the split?
Answer: A.
1. A
  The bolts exert forces inside the system, and those cannot change its total momentum. With no external force, the center of mass keeps moving at 50 m/s.
2. B
  The explosion does add kinetic energy, but internal forces cannot change the system's momentum. The pieces gain speed relative to each other, not as a whole.
3. C
  The bolts push the capsule forward just as hard as they push the stage backward. Those equal and opposite impulses leave the total momentum unchanged.
4. D
  The individual speeds are not needed. The total momentum, 30,000 kg m/s, is unchanged, so the center of mass still moves at 30,000 divided by 600 kg.
A golf ball is dropped onto a concrete floor and bounces straight up. Air resistance is negligible. The ball rises to only 64 percent of the height it was dropped from. Which claim about the collision between the ball and the floor is correct?
Answer: A.
1. A
  Rising to 64 percent of the height means the ball left the floor with 64 percent of the kinetic energy it hit with. Earth's share of the kinetic energy is negligible, so the system's kinetic energy fell, and that is what makes a collision inelastic. The rest became thermal energy and sound.
2. B
  Momentum is conserved in every collision of an isolated system, elastic or not, so it cannot tell the two apart. Only the kinetic energy can.
3. C
  Sticking together makes a collision perfectly inelastic, but objects that bounce apart can still lose kinetic energy. This ball does, since it does not rise back to its starting height.
4. D
  A reversed direction says nothing about kinetic energy. A ball that bounced straight back at the speed it arrived with would reverse its momentum and keep all of its kinetic energy.
On a bicycle, the chain runs without slipping over a front sprocket of radius 0.10 m and a rear sprocket of radius 0.040 m. How do the rear sprocket's angular speed and the chain's speed where it wraps around the rear sprocket compare with the same two quantities at the front sprocket?
Answer: D.
1. A
  This multiplies by 0.040/0.10. The chain moves at one speed over both sprockets, so $ω = \frac{v}{r}$ makes the smaller sprocket turn faster, not slower.
2. B
  Points on one rigid object share an angular speed, but the two sprockets are separate objects joined by a chain. What they share is the chain's speed.
3. C
  The chain does not stretch or slip, so it moves at one speed all the way around. Only the angular speed differs between the sprockets.
4. D
  The chain does not stretch or slip, so it moves at the same speed over both sprockets, and each sprocket's rim moves with it: $v = r ω$ . With the same $v$ , the rear sprocket's angular speed is $\frac{0.10}{0.040} = 2.5$ times the front's.
A traffic signal hangs from the far end of a straight, light arm attached to a vertical pole. The arm slants upward at 30 degrees above the horizontal, and the signal pulls straight down on the end of the arm. A crew remounts the arm so that it slants at 60 degrees above the horizontal, with the same signal hanging the same distance along the arm from the joint. By what factor does the magnitude of the torque the signal's pull exerts about the joint change?
Answer: C.
1. A
  This scales the torque with the angle itself. The torque depends on the lever arm, which does not grow in proportion to the angle.
2. B
  This uses the sines of 30 and 60 degrees, the angles between the arm and the horizontal. The angle that matters is the one between the arm and the downward pull, and the lever arm, the horizontal distance to the signal, shrinks as the arm tilts up.
3. C
  The lever arm is the horizontal distance from the joint to the line of the downward pull, $d \cos θ$ for a signal a distance $d$ along the arm. Tilting the arm from 30 to 60 degrees multiplies the torque by $\frac{\cos 60^{\circ}}{\cos 30^{\circ}}$ , about 0.58.
4. D
  The force and the distance along the arm are unchanged, but the torque depends on the lever arm, the perpendicular distance to the line of the pull. A steeper arm brings the signal closer to the joint, measured horizontally.
Two light rods each turn about an axle through one end. A 0.20 kg clip is fastened to rod 1, 0.30 m from its axle. A 0.10 kg clip is fastened to rod 2, 0.60 m from its axle. Treat each clip as a point object. How does the rotational inertia of rod 2 and its clip about its axle compare with that of rod 1 and its clip?
Answer: B.
1. A
  This lets the mass decide and leaves out the distances. Rotational inertia is $m r^{2}$ , so rod 2's doubled distance counts twice over.
2. B
  Rotational inertia is $m r^{2}$ : (0.20)(0.30)² = 0.018 kg m² for rod 1 and (0.10)(0.60)² = 0.036 kg m² for rod 2. Half the mass at twice the distance doubles $m r^{2}$ .
3. C
  This multiplies mass by distance, which does come out equal. Rotational inertia depends on the square of the distance.
4. D
  This squares the distance ratio and leaves out the mass. Rod 2's clip has half the mass, which halves that factor of 4.
Two workers carry a uniform beam of mass $M$ and length $L$ , holding it level and at rest. Worker A lifts straight up at the left end. Worker B lifts straight up at a point $\frac{L}{4}$ from the right end. How large an upward force does worker B exert?
Answer: D.
1. A
  This puts the beam's weight $\frac{L}{4}$ from worker A and worker B a full $L$ away. Measured from A, the weight acts at the center, $\frac{L}{2}$ away, and B lifts $\frac{3 L}{4}$ away.
2. B
  $\frac{1}{3} M g$ is worker A's force. Worker B holds the beam closer to its center, so B carries the larger share.
3. C
  An even split would be right only if both workers held the beam equally far from its center. B is $\frac{L}{4}$ from the center and A is $\frac{L}{2}$ from it.
4. D
  Taking torques about worker A's hands, $F_{B} (\frac{3 L}{4}) = M g (\frac{L}{2})$ , so $F_{B} = \frac{2}{3} M g$ . Worker A lifts the remaining $\frac{1}{3} M g$ .
Two identical hockey pucks rest on frictionless ice. A string is wrapped several times around the rim of puck 1. The string on puck 2 is tied to a small hook at its center. At the same moment, each string is pulled horizontally with the same constant force; the string on puck 1 comes off the rim along a line tangent to it. Air resistance is negligible. Just after the pulls begin, how do the motions of the pucks compare?
Answer: A.
1. A
  The net force on each puck is the same pull, so both centers of mass accelerate at F/m. Only puck 1's string has a lever arm about the puck's center, so only puck 1 feels a torque and starts to spin.
2. B
  The pull does not split into a share for moving and a share for turning. The net force alone sets the acceleration of the center of mass, and it is the same pull on both pucks.
3. C
  The pulls are equal in size, but their lines of action differ. Puck 1's string pulls along a line one radius from its center, so it exerts a torque.
4. D
  A net force always accelerates the center of mass, wherever on the object it acts. Puck 1 both speeds up and starts to spin.
A grindstone has a rotational inertia of 0.25 kg m² about its axle. A motor speeds it up from 20 rad/s to 40 rad/s. By how much does the grindstone's rotational kinetic energy increase?
Answer: C.
1. A
  50 J is the energy the stone had at 20 rad/s, and also what you get by squaring the 20 rad/s change in angular speed. Kinetic energy goes with the square of each angular speed, so find both energies and subtract.
2. B
  Doubling the angular speed does not double the kinetic energy. Since $K = \frac{1}{2} I ω^{2}$ , it grows four times, from 50 J to 200 J.
3. C
  The energy is $\frac{1}{2} (0.25) (20)^{2} = 50$ J at first and $\frac{1}{2} (0.25) (40)^{2} = 200$ J after, an increase of 150 J.
4. D
  200 J is the final rotational kinetic energy. The stone already had 50 J at 20 rad/s.
The rotor of a lab centrifuge has a rotational inertia $I$ about its axle and spins at angular speed $ω$ . When the motor shuts off, a brake exerts a constant torque of magnitude $τ$ that opposes the spin, and no other torque acts. How long does the rotor take to stop?
Answer: B.
1. A
  This multiplies the angular momentum by the torque. The angular impulse, torque times time, must equal the change in angular momentum, so divide instead.
2. B
  The brake's angular impulse must remove all of the angular momentum, $I ω$ : $τ Δ t = I ω$ , so $Δ t = \frac{I ω}{τ}$ .
3. C
  This leaves out the rotational inertia. The torque has to remove angular momentum, not angular speed, and a rotor with more rotational inertia takes longer to stop.
4. D
  This is the kinetic energy divided by the torque. That gives the angle, in radians, that the rotor turns while it stops, not the time.
A large circular platform turns on a vertical axle with negligible friction, and both it and a student standing at its edge are at rest. Seen from above, the student walks counterclockwise along the edge for a few seconds, and the platform turns clockwise under her. Then she stops walking and stands still on the platform. Which claim about the platform after she stops is correct?
Answer: B.
1. A
  Her feet are on the platform, so as she stops they push on it and exert a torque. That torque is internal to the student-platform system, which is why the system's total angular momentum stays at zero.
2. B
  Take the student and platform together. The axle exerts no torque, and gravity and the vertical pushes exert none about the axle, so the system's angular momentum stays at its starting value, zero. Once she stands still on the platform they share one angular velocity, and $(I_{platform} + I_{student}) ω = 0$ means $ω = 0$ .
3. C
  Her stopping steps do push the platform counterclockwise, but only until it stops along with her. Turning counterclockwise while she stands still would give the system angular momentum it never had.
4. D
  If the platform kept turning clockwise with her on it, the student and platform together would have clockwise angular momentum. They started with none, and no external torque acts to supply any.
Friction and air resistance are negligible in each case. Which of these motions is simple harmonic?
Answer: D.
1. A
  On either side, the net force on the block, gravity plus the normal force, points down the slope toward the middle and has the same size wherever the block is on that side. A restoring force that does not grow with displacement gives periodic motion, but not simple harmonic.
2. B
  The bounce repeats, but between bounces the only force is the ball's constant weight. Simple harmonic motion needs a restoring force proportional to displacement.
3. C
  At 70 degrees the restoring torque is far from proportional to the angle. A pendulum is close to simple harmonic only at small angles.
4. D
  The spring force changes in proportion to the stretch, so the net force on the block is proportional to its displacement from its resting position and points back toward it. That is the condition for simple harmonic motion.
A block hangs at rest from an ideal spring and stretches it a distance $d$ beyond its relaxed length. The block is pulled down a little farther and released, and it bobs up and down. Which expression gives the period of the oscillation?
Answer: B.
1. A
  This is $\sqrt{\frac{m}{k}}$ without the factor of $2 π$ that the period needs.
2. B
  At rest the spring force balances the weight, $k d = m g$ , so $\frac{m}{k} = \frac{d}{g}$ . Then $T = 2 π \sqrt{\frac{m}{k}} = 2 π \sqrt{\frac{d}{g}}$ .
3. C
  This is the frequency, the reciprocal of the period. The period is the time for one full cycle.
4. D
  This turns $\frac{m}{k}$ upside down. From $k d = m g$ , $\frac{m}{k}$ is $\frac{d}{g}$ , not $\frac{g}{d}$ .
A glider on a level air track is attached to a horizontal ideal spring and oscillates with an amplitude of 0.20 m. Friction is negligible. At what distance from equilibrium is the energy stored in the spring equal to the glider's kinetic energy?
Answer: C.
1. A
  This takes a quarter of the amplitude. At 0.050 m the spring holds only one sixteenth of the total energy.
2. B
  Halfway out, the spring holds only a quarter of the total energy, because its energy goes with the square of the stretch. The energies split evenly farther out than that.
3. C
  Equal shares means the spring holds half the total: $\frac{1}{2} k x^{2} = \frac{1}{2} (\frac{1}{2} k A^{2})$ , so $x = A / \sqrt{2}$ , about 0.14 m.
4. D
  At 0.17 m the glider moves at half its greatest speed. Kinetic energy goes with the square of speed, so the glider holds only a quarter of the energy there.
A student mixes 100 g of water, with a density of 1.00 g/cm³, and 100 g of rubbing alcohol, with a density of 0.80 g/cm³. Assume the volume of the mixture is the sum of the two volumes. What is the density of the mixture?
Answer: A.
1. A
  The volumes are 100 cm³ and 125 cm³, so 200 g fills 225 cm³. That is 0.89 g/cm³, or about 890 kg/m³.
2. B
  Averaging the two densities works only for equal volumes. Equal masses means more alcohol than water by volume, which pulls the density lower.
3. C
  Densities do not add. A mixture's density has to lie between the densities of the two liquids in it.
4. D
  This divides the total mass by the water's 100 cm³ alone. The alcohol's 125 cm³ is part of the mixture's volume too.
A 1200 kg car is parked on a level road. Its four tires share its weight equally, and each tire touches the road over an area of 150 cm². Use g = 10 m/s². What pressure does each tire exert on the road?
Answer: C.
1. A
  This converts 150 cm² to 1.5 m². A square meter is 10,000 cm², so 150 cm² is 0.015 m².
2. B
  This comes from converting 150 cm² to 0.15 m², or from using the car's mass where its weight belongs. Either way the result is ten times too small.
3. C
  Each tire supports a quarter of the 12,000 N weight, 3000 N, spread over 0.015 m². That is 3000 N divided by 0.015 m², or $2.0 \times 10^{5}$ Pa.
4. D
  This puts the whole car's weight on one tire's contact area. Each tire supports only a quarter of it.
A 0.20 kg wooden ball with a volume of $5.0 \times 10^{- 4}$ m³ is held completely under the water in a tank, and then it is released. The water's density is 1000 kg/m³; use g = 10 m/s². Treat the buoyant force and gravity as the only forces on the ball just after release. Which claim about the ball just after release is correct?
Answer: D.
1. A
  Water pressure grows with depth, so the water below the ball pushes up harder than the water above pushes down. That difference is the buoyant force, and it points up.
2. B
  The pressure is not the same on every side, because it grows with depth. The bottom of the ball is deeper than its top, so the water pushes up on it harder than it pushes down.
3. C
  The buoyant force equals the weight of the displaced water, not the object's own weight. The two match only for an object floating at rest, and this ball is held fully under.
4. D
  The buoyant force is the weight of the displaced water, (1000)(5.0 x 10⁻⁴)(10) = 5.0 N upward, and the ball's weight is only 2.0 N. The net force is 3.0 N upward, so the ball speeds up upward.
Questions 41 to 42 refer to the following.
A gardener empties a wide rain barrel with a siphon: a flexible hose with the same inside cross-sectional area all along its length, completely filled with water. The hose runs from below the water surface in the barrel, up over the barrel's rim, and down to a flower bed, where its open end points down into the air. The highest point of the hose is a height $h$ above the water surface in the barrel, and the open end is a distance $d$ below that surface. The barrel is wide enough that its water level drops negligibly during the questions below. Treat the water as an ideal fluid with density $ρ$ , and take the air pressure everywhere to be $P_{0}$ .

How fast does water leave the open end of the hose?
Answer: C.
1. A
  This uses the rise $h$ to the top of the hose. The speed at the open end depends on how far that end is below the water surface in the barrel.
2. B
  This drops the factor of 2. Bernoulli's equation between the surface and the open end gives $\frac{1}{2} v^{2} = g d$ .
3. C
  The barrel's surface and the open end are both at air pressure, and the surface is nearly at rest. Bernoulli's equation then gives $ρ g d = \frac{1}{2} ρ v^{2}$ , so $v = \sqrt{2 g d}$ .
4. D
  This uses the whole drop from the top of the hose to the open end. The water starts nearly at rest at the barrel's surface, not at the top of the hose, so only the depth $d$ counts.
Uses the material above question 41.

Point X is in the hose at its highest point, point Y is in the hose at the same height as the water surface in the barrel, and point Z is just inside the open end. How do the water pressures $P_{X}$ , $P_{Y}$ and $P_{Z}$ at these points compare?
Answer: A.
1. A
  Water leaves the open end into the air, so $P_{Z} = P_{0}$ . Y is level with the barrel's surface, where the pressure is $P_{0}$ and the water is nearly still, but at Y the water moves, so $P_{Y} = P_{0} - \frac{1}{2} ρ v^{2}$ . X is a height $h$ higher, at the same speed, so $P_{X}$ is lower than $P_{Y}$ by $ρ g h$ .
2. B
  The speed is the same everywhere in the hose, since its area is uniform, but the height is not. With equal speeds, Bernoulli's equation makes the pressure lower wherever the water is higher.
3. C
  Y is level with the barrel's surface, but the water at Y is moving and the water at the surface is not. The kinetic energy per volume the moving water carries, $\frac{1}{2} ρ v^{2}$ , comes out of its pressure, so $P_{Y}$ is below $P_{0}$ , which is $P_{Z}$ .
4. D
  This has the pressure rising with height. In the hose the speed is the same everywhere, so Bernoulli's equation makes the pressure fall as the water rises, and the top of the hose has the lowest pressure.

Section II: Free Response

4 questions · 95 minutes

Question 1: Mathematical Routines 10 points

A class wants to measure how fast a spring launcher fires a steel ball, without a photogate. The launcher fires a ball of mass $m = 0.020$ kg horizontally into a hole drilled in the side of a wooden block of mass $M = 0.100$ kg. The block rests on a level tabletop, and the ball lodges in the hole. The block, with the ball inside it, slides a distance $d = 0.20$ m along the tabletop before it stops. The coefficient of kinetic friction between the block and the tabletop is $μ_{k} = 0.25$ . The collision lasts so short a time that the block does not move noticeably during it. Air resistance is negligible, and use $g = 10$ m/s².

Respond to parts (a), (b), (c) and (d). Draw any diagram on your own paper.
1. (a)
  
  Draw a free-body diagram of the block with the ball inside it while it slides along the tabletop after the collision. Represent each force with a labeled arrow that starts on a dot standing for the block, and draw the arrows so their lengths show the relative sizes of the forces.
  Question 1, part (a): your response. It stays in this browser tab and is never sent anywhere.
  Score part (a)
  - 1 point Draws exactly three forces with correct directions: the gravitational force downward, the normal force from the tabletop upward, and kinetic friction from the tabletop horizontal and opposite the direction the block slides. No forward force is drawn.Earns it: " $F_{g}$ down, $F_{N}$ up, $F_{f}$ pointing back, opposite the motion"Does not: an arrow forward labeled as the push of the ball or "force of motion" (the ball is part of the sliding object now, and nothing pushes it forward once the collision is over); friction drawn in the direction of motion
  - 1 point Draws the normal force and the gravitational force with equal lengths, and the friction arrow clearly shorter than both, since friction is $0.25$ of the normal force.Earns it: vertical arrows the same length, friction about a quarter of their lengthDoes not: normal force shorter or longer than gravity on a level table with no vertical acceleration; friction drawn as long as the normal force
  Model response
  A dot for the block with three arrows starting on it: $F_{g}$ , the gravitational force exerted by Earth, straight down; $F_{N}$ , the normal force exerted by the tabletop, straight up and the same length as $F_{g}$ ; and $F_{f}$ , kinetic friction exerted by the tabletop, horizontal and pointing opposite the direction the block slides, about a quarter as long as $F_{N}$ .
2. (b)
  
  Derive an expression for the speed $v_{0}$ of the ball just before it enters the block. Express your answer in terms of $m$ , $M$ , $μ_{k}$ , $d$ and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
  Question 1, part (b): your response. It stays in this browser tab and is never sent anywhere.
  Score part (b)
  - 1 point Applies conservation of momentum to the collision, with the ball and block moving together afterward: $m v_{0} = (m + M) v_{1}$ , where $v_{1}$ is their speed just after the collision.Earns it: " $p_{i} = p_{f}$ , so $m v_{0} + M (0) = (m + M) v_{1}$ "Does not: conserves kinetic energy through the collision, $\frac{1}{2} m v_{0}^{2} = \frac{1}{2} (m + M) v_{1}^{2}$ (the ball lodges in the block, so the collision is perfectly inelastic); writes $m v_{0} = M v_{1}$ , leaving the ball's mass out of the moving object after the collision
  - 1 point Relates $v_{1}$ to the slide using either energy or kinematics with the friction force $μ_{k} (m + M) g$ : $\frac{1}{2} (m + M) v_{1}^{2} = μ_{k} (m + M) g d$ , or $a = μ_{k} g$ with $v_{1}^{2} = 2 a d$ .Earns it: " $Δ K = W_{f}$ : $0 - \frac{1}{2} (m + M) v_{1}^{2} = - μ_{k} (m + M) g d$ "; " $a = \frac{F_{f}}{m + M} = μ_{k} g$ , and $0 = v_{1}^{2} - 2 μ_{k} g d$ "Does not: uses $μ_{k} M g$ for the friction force but $(m + M)$ for the kinetic energy, so the masses no longer cancel; uses $v_{1} = \sqrt{2 g d}$ , a free-fall relation with no friction in it
  - 1 point Combines the two relations correctly, substituting $v_{1} = \sqrt{2 μ_{k} g d}$ into the momentum equation.Earns it: " $v_{1} = \sqrt{2 μ_{k} g d}$ , so $m v_{0} = (m + M) \sqrt{2 μ_{k} g d}$ "Does not: substitutes $v_{0}$ for $v_{1}$ in the sliding equation, treating the ball's speed before the collision as the block's speed after it
  - 1 point Arrives at a correct expression: $v_{0} = \frac{m + M}{m} \sqrt{2 μ_{k} g d}$ .Earns it: " $v_{0} = (1 + \frac{M}{m}) \sqrt{2 μ_{k} g d}$ "Does not: $v_{0} = \frac{m}{m + M} \sqrt{2 μ_{k} g d}$ (the mass ratio is upside down, which would make the ball slower than the block it pushed); $v_{0} = \frac{M}{m} \sqrt{2 μ_{k} g d}$ (drops the ball's own mass from the moving pair)
  Model response
  The collision is short, so the friction impulse during it is negligible and momentum is conserved:
  $m v_{0} + M (0) = (m + M) v_{1}$
  After the collision the pair slides to a stop. The normal force equals $(m + M) g$ , so friction is $μ_{k} (m + M) g$ . By the work-energy theorem,
  $0 - \frac{1}{2} (m + M) v_{1}^{2} = - μ_{k} (m + M) g d \Rightarrow v_{1} = \sqrt{2 μ_{k} g d}$
  Substituting into the momentum equation,
  $v_{0} = \frac{m + M}{m} \sqrt{2 μ_{k} g d}$
3. (c)
  
  Calculate the speed $v_{0}$ of the ball just before it enters the block.
  Question 1, part (c): your response. It stays in this browser tab and is never sent anywhere.
  Score part (c)
  - 1 point Calculates $v_{0} = \frac{0.120}{0.020} \sqrt{2 (0.25) (10) (0.20)} = 6.0$ m/s, with units. An answer of about 5.9 m/s, from correctly using g = 9.8 m/s², is also accepted, as is a correct substitution into an incorrect expression from part (b).Earns it: " $v_{0} = 6 \sqrt{1.0} = 6.0$ m/s"Does not: 1.0 m/s (that is the speed of the block and ball just after the collision, not the ball's speed before it); a number with no units
  Model response
  $v_{0} = \frac{0.020 + 0.100}{0.020} \sqrt{2 (0.25) (10 {m/s}^{2}) (0.20 m)} = 6 \sqrt{1.0} m/s = 6.0 m/s$
  The block and ball leave the collision at 1.0 m/s, a sixth of the ball's speed.
4. (d)
  
  A student suggests a shortcut that skips the collision. She sets the ball's kinetic energy just before it enters the block equal to the energy dissipated by friction while the block slides, $\frac{1}{2} m v_{0}^{2} = μ_{k} (m + M) g d$ , and solves for $v_{0}$ . Indicate whether the speed she calculates is greater than, less than, or equal to the actual speed of the ball. Justify your answer. In your justification, include qualitative reasoning beyond mathematical derivations or expressions.
  Question 1, part (d): your response. It stays in this browser tab and is never sent anywhere.
  Score part (d)
  - 1 point Indicates that her speed is less than the actual speed, with an attempt at a justification that uses energy in the collision.Earns it: "Less, because energy is lost when the ball hits the block."Does not: greater, because her equation leaves out the collision (leaving it out ignores the kinetic energy lost there, which makes her estimate too low, not too high); equal, because energy is conserved (total energy is, but kinetic energy of the ball and block is not)
  - 1 point States that the collision is perfectly inelastic: the ball lodges in the block, so kinetic energy of the ball and block is not conserved during the collision and part of it becomes thermal energy and sound (or goes into deforming the wood). Momentum, not kinetic energy, carries through the collision.Earns it: "The ball sticks in the block, so the collision is inelastic. Some of the ball's kinetic energy warms the wood and makes a sound before the block even starts sliding."Does not: says friction during the collision removes the energy (the collision is too short for friction to matter; the energy goes in the ball and wood squashing together)
  - 1 point Connects the two: friction dissipates only the kinetic energy left after the collision, which is less than the ball's kinetic energy before it, so setting the ball's kinetic energy equal to friction's share undercounts it and gives a speed that is too small. A numerical comparison (her method gives about 2.4 m/s against the actual 6.0 m/s) supports the point but cannot replace the reasoning.Earns it: "Friction only takes away what is left after the hit. The ball started with more than that, so a speed worked out from friction's share is too low."Does not: "her answer is $\sqrt{6}$ = 2.4 m/s, which is less" with no reason why the energy accounting falls short
  Model response
  Less than the actual speed. The ball lodges in the block, so the collision is perfectly inelastic. Momentum is conserved through it, but kinetic energy is not: as the ball and wood squash together, much of the ball's kinetic energy becomes thermal energy and sound before the block starts to slide. Here five sixths of it goes that way. Friction then removes only the kinetic energy that is left after the collision. The student treats friction's share as if it were all of the ball's kinetic energy, which is too little energy, so the speed she gets is too small. Her equation gives about 2.4 m/s, well under the actual 6.0 m/s.
Your score: 0 of 10, from the boxes you ticked.
Question 2: Translation Between Representations 12 points

A carnival ride is a large hollow cylinder that turns about its vertical central axis. Riders stand with their backs against the inside wall, a distance $r$ from the axis. The cylinder speeds up, and once it turns fast enough, the floor drops away. A rider of mass $m$ then stays in place against the wall, moving in a horizontal circle at a constant speed $v$ , with nothing under her feet. The coefficient of static friction between the rider's clothing and the wall is $μ_{s}$ , which is less than 1. Air resistance is negligible. Treat the rider as a single object.

Respond to parts (a), (b), (c) and (d). Draw any diagram or graph on your own paper.
1. (a)
  
  Draw a free-body diagram of the rider while she is held in place against the wall with the floor gone. Represent each force with a labeled arrow that starts on a dot standing for the rider, and draw the arrows so their lengths show the relative sizes of the forces. Indicate which direction is toward the axis of the cylinder.
  Question 2, part (a): your response. It stays in this browser tab and is never sent anywhere.
  Score part (a)
  - 1 point Draws the normal force from the wall horizontal and pointing toward the axis, with no other horizontal force: no separate "centripetal force" arrow and no outward arrow.Earns it: " $F_{N}$ , from the wall, pointing toward the axis; no other sideways arrow"Does not: an outward "centrifugal force" arrow (no object exerts it); a normal force pointing away from the axis, as if the rider pushed the wall; a "centripetal force" drawn in addition to the normal force (the normal force is what makes the centripetal acceleration)
  - 1 point Draws static friction from the wall pointing straight up and the gravitational force pointing straight down, with equal lengths.Earns it: friction up and gravity down, same lengthDoes not: friction drawn downward or along the direction of motion; friction drawn shorter than gravity while the rider is at rest vertically
  - 1 point Draws the normal force longer than the friction and gravitational forces. Since friction equals $m g$ and can be at most $μ_{s} F_{N}$ with $μ_{s}$ less than 1, the normal force must be greater than $m g$ .Earns it: normal arrow clearly the longest of the threeDoes not: normal force drawn equal to or shorter than gravity (then the wall could not supply enough friction to hold her up)
  Model response
  A dot for the rider with three arrows starting on it. $F_{N}$ , the normal force exerted by the wall, points horizontally toward the axis and is the longest arrow. $F_{f}$ , static friction exerted by the wall, points straight up. $F_{g}$ , the gravitational force exerted by Earth, points straight down and is the same length as $F_{f}$ . The normal force is longer than both because friction must equal $m g$ while never exceeding $μ_{s} F_{N}$ , and $μ_{s}$ is less than 1.
2. (b)
  
  Derive an expression for the lowest speed $v_{min}$ at which the rider can stay in place against the wall without sliding down. Express your answer in terms of $r$ , $μ_{s}$ and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
  Question 2, part (b): your response. It stays in this browser tab and is never sent anywhere.
  Score part (b)
  - 1 point Applies Newton's second law toward the axis, with the normal force as the only force in that direction: $F_{N} = \frac{m v^{2}}{r}$ .Earns it: " $Σ F_{c} = m a_{c}$ : $F_{N} = m \frac{v^{2}}{r}$ "Does not: sets the normal force equal to $m g$ , as on a floor; writes $F_{N} - \frac{m v^{2}}{r} = 0$ with an outward force balancing the normal force
  - 1 point Applies Newton's second law vertically, with no vertical acceleration: $F_{f} = m g$ .Earns it: "vertically $a = 0$ , so $F_{f} - m g = 0$ "Does not: sets $F_{f} = μ_{s} m g$ , using gravity as the normal force
  - 1 point Uses the maximum static friction at the lowest speed: $m g = μ_{s} F_{N} = μ_{s} \frac{m v_{min}^{2}}{r}$ (or the inequality $m g \leq μ_{s} \frac{m v^{2}}{r}$ ).Earns it: "At the slowest speed friction is at its limit, $F_{f, max} = μ_{s} F_{N}$ , so $m g = μ_{s} m v^{2} / r$ "Does not: uses $μ_{k}$ , though the rider does not slide relative to the wall
  - 1 point Arrives at a correct expression: $v_{min} = \sqrt{\frac{g r}{μ_{s}}}$ .Earns it: " $v_{min} = \sqrt{g r / μ_{s}}$ "Does not: $v_{min} = \sqrt{μ_{s} g r}$ (that would make a rougher wall need a faster ride); an answer that still contains $m$ without cancelling it
  Model response
  Toward the axis, the normal force alone gives the centripetal acceleration:
  $Σ F = m a \Rightarrow F_{N} = \frac{m v^{2}}{r}$
  Vertically the rider does not accelerate, so $F_{f} = m g$ . Static friction can be at most $μ_{s} F_{N}$ , and at the lowest speed it is exactly at that limit:
  $m g = μ_{s} \frac{m v_{min}^{2}}{r} \Rightarrow v_{min} = \sqrt{\frac{g r}{μ_{s}}}$
3. (c)
  
  On one set of axes, sketch two graphs as functions of the rider's speed $v$ , from $v = 0$ to a speed above $v_{min}$ :
  
  i. the greatest static friction force the wall can exert on the rider at that speed, and
  
  ii. the size of the gravitational force exerted on the rider.
  
  Label each graph, and mark the speed $v_{min}$ on the horizontal axis.
  Question 2, part (c): your response. It stays in this browser tab and is never sent anywhere.
  Score part (c)
  - 1 point Sketches the greatest static friction force as a curve that starts at zero at $v = 0$ and curves upward, getting steeper as $v$ grows ( $μ_{s} m v^{2} / r$ , a parabola).Earns it: an upward-opening half parabola from the originDoes not: a straight line through the origin (the normal force grows with $v^{2}$ , not $v$ ); a horizontal line at $μ_{s} m g$
  - 1 point Sketches the gravitational force as a horizontal line at $m g$ , above zero, the same at every speed.Earns it: a flat line labeled $m g$ Does not: a line that rises or falls with speed
  - 1 point Marks $v_{min}$ at the speed where the friction curve crosses the gravity line, consistent with the student's own curves.Earns it: "the two graphs cross at $v_{min}$ "Does not: $v_{min}$ marked where the friction curve leaves zero, or at a point unrelated to the crossing
  Model response
  Speed $v$ runs along the horizontal axis and force along the vertical axis. Graph i, labeled $F_{f, max} = μ_{s} m v^{2} / r$ , starts at the origin and curves upward like half a parabola, rising more steeply as the speed increases. Graph ii, labeled $F_{g} = m g$ , is a horizontal line above zero. The curve starts below the line, crosses it, and stays above it. The crossing is at $v_{min}$ , marked on the horizontal axis: at slower speeds the wall cannot supply as much friction as gravity needs, and at faster speeds it can.
4. (d)
  
  The operator covers the wall with a rougher liner, which has a larger coefficient of static friction with the riders' clothing. Describe how each of your graphs from part (c) would change, if at all, and determine whether $v_{min}$ increases, decreases, or stays the same. Justify your answer, and show that it agrees with your expression from part (b).
  Question 2, part (d): your response. It stays in this browser tab and is never sent anywhere.
  Score part (d)
  - 1 point Describes the friction curve as rising more steeply (higher at every speed above zero) while the gravity line is unchanged, so the crossing moves to a lower speed and $v_{min}$ decreases.Earns it: "The friction curve gets steeper, the $m g$ line stays put, so they cross sooner: $v_{min}$ goes down."Does not: says the gravity line moves (the rider's weight does not depend on the wall); says $v_{min}$ decreases with no reference to how the graphs change
  - 1 point Shows agreement with part (b): in $v_{min} = \sqrt{g r / μ_{s}}$ , $μ_{s}$ is in the denominator, so a larger $μ_{s}$ gives a smaller $v_{min}$ , the same conclusion the graphs give.Earns it: " $v_{min}$ is proportional to $1 / \sqrt{μ_{s}}$ , so a larger $μ_{s}$ lowers it, matching the earlier crossing."Does not: states the expression with no comment on how $v_{min}$ depends on $μ_{s}$
  Model response
  The greatest static friction at each speed is $μ_{s} m v^{2} / r$ , so a larger $μ_{s}$ makes that curve rise more steeply: at every speed the wall can now supply more friction. The gravity line stays where it is, because the rider's weight has nothing to do with the wall. A steeper curve reaches the flat line at a lower speed, so $v_{min}$ decreases. That agrees with part (b): $v_{min} = \sqrt{g r / μ_{s}}$ has $μ_{s}$ in the denominator, so increasing $μ_{s}$ lowers the lowest safe speed.
Your score: 0 of 12, from the boxes you ticked.
Question 3: Experimental Design and Analysis 10 points

A physics class has a straight wooden ramp 1.2 m long that can be propped up at one end, a meterstick, a caliper, an electronic balance, a stopwatch, and a photogate connected to a timer. The photogate measures how long an object blocks its light beam. With the beam set level with the center of a rolling cylinder, the cylinder's speed as it passes through is its diameter divided by that blocking time. The class also has a uniform solid steel cylinder. Each object in this question rolls down the ramp without slipping.

Steel pipe released from rest at different heights
Height $h$ (m) Speed $v$ at the photogate (m/s)

0.050 0.76
0.100 1.04
0.150 1.30
0.200 1.48
0.250 1.68

Respond to parts (a), (b), (c) and (d). Draw any graph on your own graph paper.
1. (a)
  
  The students want to determine experimentally whether the time the solid cylinder takes to roll from rest down the full length of the ramp depends on the angle of the ramp. Describe a procedure the students could use, using only the equipment listed. Say which quantity they vary, which quantities they keep the same, and what they measure and how.
  Question 3, part (a): your response. It stays in this browser tab and is never sent anywhere.
  Score part (a)
  - 1 point Varies the ramp's angle, by changing the height of the raised end and measuring it with the meterstick (or stating the angle found from that height and the ramp's length), while keeping the same cylinder, the same release point and distance rolled, and the release from rest for every trial.Earns it: "Raise the end to several heights measured with the meterstick. Each time, release the same cylinder from rest at the top end, so it always rolls the full 1.2 m."; "Change the angle; sin(angle) = height of the raised end / 1.2 m. Same cylinder, same starting point, released from rest."Does not: changes the cylinder or the release point along with the angle (two variables change at once); rolls the cylinder at a single angle several times (the angle never changes, so the test cannot show its effect)
  - 1 point Measures the time to roll the full length of the ramp in a workable way: the stopwatch started at release and stopped when the cylinder reaches the bottom end, at each angle. Repeating the timing and averaging is good practice but is not required for this point.Earns it: "Start the stopwatch when I let go and stop it when the cylinder reaches the bottom. Do three rolls at each height and average them."Does not: measures only the speed at the bottom with the photogate (the question asks about the time to roll down); "time how long it takes" with no start and stop points for the timing
  Model response
  Prop up one end of the ramp and use the meterstick to measure how high the raised end sits above the table. Set it in turn to 0.10, 0.20, 0.30, 0.40 and 0.50 m; since the ramp is 1.2 m long, the sine of its angle is that height divided by 1.2 m. At each height, hold the cylinder at rest against a mark at the top of the ramp, let go without pushing, and start the stopwatch at the same moment. Stop it when the cylinder reaches the bottom end of the ramp. Time three rolls at each height. The only thing that changes is the ramp's angle; the cylinder, the release point at the top, the distance rolled and the release from rest stay the same.
2. (b)
  
  Describe how the students could analyze their data from part (a) to determine whether the time to roll down the ramp depends on the ramp's angle.
  Question 3, part (b): your response. It stays in this browser tab and is never sent anywhere.
  Score part (b)
  - 1 point Describes a graph of the time (or its average) against the ramp's angle, the height of the raised end, or the sine of the angle, or an equivalent comparison of the average times at the different angles.Earns it: "Plot average time on the vertical axis against the height of the raised end on the horizontal axis."Does not: plots time against the cylinder's mass (the mass was not changed); plots the speed at the bottom against the angle when the question is about the time
  - 1 point States the criterion: times that change with the angle by more than the spread of repeated trials at one angle mean the time depends on the angle; points that lie along a horizontal line within that spread mean it does not. A best-fit line whose slope is clearly different from zero, given that spread, also earns the point.Earns it: "If the times get shorter as the ramp gets steeper, by more than the scatter in the repeat rolls, the angle matters. If the points are flat within that scatter, it does not."; "Fit a line to time against height. A slope clearly different from zero, compared with the scatter of the repeat trials, means the time depends on the angle."Does not: decides from a single roll at each angle with no allowance for uncertainty; "if the graph is a straight line, the time depends on the angle" (a horizontal straight line means it does not)
  Model response
  Plot the average time on the vertical axis against the height of the raised end, or against the sine of the angle, on the horizontal axis. If the average times change from one angle to the next by more than the spread among the three rolls at a single angle, the time depends on the angle. If all the points lie along a horizontal line within that spread, it does not. A steeper ramp should give shorter times.
3. (c)
  
  In a second experiment, a group uses a thick-walled steel pipe with mass $M = 0.80$ kg and outer radius $R = 0.025$ m. They release it from rest at several points on the ramp. For each release, $h$ is the vertical distance the pipe's center drops between the release point and the photogate, and the photogate, with its beam level with the pipe's center, gives its speed $v$ there as its outer diameter divided by the blocking time. Their data are in the table. The students correctly determine that an object of mass $M$ , radius $R$ and rotational inertia $I$ about its axis, rolling without slipping from rest, reaches a speed given by
  $v^{2} = \frac{2 g h}{1 + \frac{I}{M R^{2}}}$
  They want to use the data to determine an experimental value for the pipe's rotational inertia $I$ .
  
  i. Indicate the quantities the students could plot on the vertical and horizontal axes to produce a linear graph whose slope can be used to determine $I$ . Include units.
  
  ii. On graph paper, plot the quantities you indicated in part (c)(i). Label each axis with its quantity and units, choose a numerical scale for each, and plot the five data points.
  
  iii. Draw a best-fit line for the data you plotted in part (c)(ii).
  Question 3, part (c): your response. It stays in this browser tab and is never sent anywhere.
  Score part (c)
  - 1 point Indicates $v^{2}$ (in m²/s²) on the vertical axis against $h$ (in m) on the horizontal axis. $h$ against $v^{2}$ , or $v$ against $\sqrt{h}$ , also earns the point.Earns it: "Vertical: $v^{2}$ in m²/s². Horizontal: $h$ in m."Does not: $v$ against $h$ (that graph curves, because $v$ grows as the square root of $h$ ); $v^{2}$ against $I$ ( $I$ is the unknown, not a measured quantity)
  - 1 point Labels both axes with quantities and units consistent with part (c)(i), and uses linear numerical scales that spread the data over most of the graph.Earns it: " $v^{2}$ (m²/s²) from 0 to 3.0 in steps of 0.5; $h$ (m) from 0 to 0.30 in steps of 0.05"Does not: the data values themselves written at equal spacing as the scale; axes without units
  - 1 point Plots all five points correctly on the chosen scale. For $v^{2}$ against $h$ the points are (0.050, 0.58), (0.100, 1.08), (0.150, 1.69), (0.200, 2.19) and (0.250, 2.82). For other axes accepted in part (c)(i), the same data placed on those axes earn the point.Earns it: five points within half a grid division of the values aboveDoes not: plots $v$ instead of $v^{2}$ after labeling the axis $v^{2}$
  - 1 point Draws a single straight line that follows the trend of the points, with points on both sides of it, rather than joining the points dot to dot.Earns it: a straight line passing close to all five points and close to the originDoes not: a zigzag connecting each point to the next; a line forced through the first and last points that leaves the middle points all on one side
  Model response
  i. Plot $v^{2}$ , in m²/s², on the vertical axis against $h$ , in m, on the horizontal axis. The equation is $v^{2} = (\frac{2 g}{1 + I / (M R^{2})}) h$ , a straight line through the origin whose slope contains $I$ .
  
  ii. Squaring the speeds gives 0.58, 1.08, 1.69, 2.19 and 2.82 m²/s². The vertical axis runs from 0 to 3.0 m²/s² in steps of 0.5, and the horizontal axis from 0 to 0.30 m in steps of 0.05. The points sit at (0.050 m, 0.58 m²/s²), (0.100 m, 1.08 m²/s²), (0.150 m, 1.69 m²/s²), (0.200 m, 2.19 m²/s²) and (0.250 m, 2.82 m²/s²).
  
  iii. A straight line through the middle of the points passes very close to the origin, with some points just above it and some just below.
4. (d)
  
  Using your best-fit line from part (c)(iii), calculate an experimental value for the rotational inertia $I$ of the pipe about its axis.
  Question 3, part (d): your response. It stays in this browser tab and is never sent anywhere.
  Score part (d)
  - 1 point Calculates the slope of the best-fit line from two points read off the line (not two data points, unless they lie on the line), with units. For $v^{2}$ against $h$ the slope is about 11.2 m/s². For $h$ against $v^{2}$ it is about 0.089 s²/m, and for $v$ against $\sqrt{h}$ about $3.3 m^{1 / 2} /s$ .Earns it: "slope $= \frac{2.80 - 0.00}{0.250 - 0.000} = 11.2$ m/s²"Does not: divides a single $v^{2}$ by its $h$ instead of finding the slope of the line; a slope with no units
  - 1 point Uses the slope $S$ to calculate $I$ in the way the student's axes require. For $v^{2}$ against $h$ , $S = \frac{2 g}{1 + I / (M R^{2})}$ , so $I = M R^{2} (\frac{2 g}{S} - 1) = (0.80) (0.025)^{2} (\frac{20}{11.2} - 1) \approx 3.9 \times 10^{- 4}$ kg m². Any value from about $3.4 \times 10^{- 4}$ to $4.5 \times 10^{- 4}$ kg m² that follows correctly from the student's own line earns the point. With g = 9.8 m/s², which is accepted, the same slope gives about $3.8 \times 10^{- 4}$ kg m².Earns it: " $I = (0.80) (0.025)^{2} (\frac{20}{11.2} - 1) = 3.9 \times 10^{- 4}$ kg m²"Does not: $I = M R^{2} (\frac{S}{2 g} - 1)$ , which comes out negative (the ratio is inverted); $I = \frac{2 g}{S} - 1$ , about 0.79, with no $M R^{2}$ and no units (that is only the ratio $I / (M R^{2})$ )
  Model response
  Two points on my best-fit line, not data points, are (0 m, 0 m²/s²) and (0.250 m, 2.80 m²/s²):
  $S = \frac{2.80 - 0}{0.250 - 0} = 11.2 {m/s}^{2}$
  The slope is $S = \frac{2 g}{1 + I / (M R^{2})}$ , so
  $\frac{I}{M R^{2}} = \frac{2 g}{S} - 1 = \frac{20}{11.2} - 1 = 0.79$ $I = 0.79 (0.80 kg) (0.025 m)^{2} \approx 3.9 \times 10^{- 4} {kg m}^{2}$
  That sits between a thin hoop of the same mass and radius, $M R^{2} = 5.0 \times 10^{- 4}$ kg m², and a solid cylinder, $\frac{1}{2} M R^{2} = 2.5 \times 10^{- 4}$ kg m², which is what a thick-walled pipe should give.
Your score: 0 of 10, from the boxes you ticked.
Question 4: Qualitative/Quantitative Translation 8 points

A cart rolls forward along a straight, level track, and its wheels roll freely. A light string tied to the front of the cart runs over a pulley at the end of the track to a hanging weight, so the cart speeds up with a constant forward acceleration of magnitude $a$ . A small spring launcher is fixed upright on the cart. At the instant the cart's speed is $v_{0}$ , the launcher fires a small ball. Just after the ball leaves the launcher, its velocity as measured by a student standing beside the track has a horizontal component $v_{0}$ , forward, and a vertical component $u$ , upward. The ball's mass is so small that firing it does not change the cart's motion. Air resistance is negligible. The ball rises and falls back to the height from which it was launched, and the cart is still speeding up with acceleration $a$ when it does.

Respond to parts (a), (b) and (c).
1. (a)
  
  When the ball falls back to its launch height, is it behind the launcher, directly at the launcher, or ahead of the launcher? Justify your answer using qualitative reasoning beyond references to equations.
  Question 4, part (a): your response. It stays in this browser tab and is never sent anywhere.
  Score part (a)
  - 1 point Indicates that the ball comes down behind the launcher, with an attempt at a justification that compares the horizontal motions of the ball and the cart.Earns it: "Behind, because the cart keeps speeding up and the ball does not."Does not: at the launcher, because the ball was moving with the cart when it left (true at launch, but the cart changes speed afterward); ahead, because the ball keeps its forward speed while the cart is held back by the string (the string pulls the cart forward, not back)
  - 1 point States that once the ball leaves the launcher, the only force on it is gravity, which is vertical, so nothing changes its horizontal velocity: it moves forward at a steady $v_{0}$ for the whole flight.Earns it: "In the air only gravity pulls on the ball, straight down, so its forward speed stays $v_{0}$ ."Does not: says the ball keeps the cart's acceleration because it was launched from the cart; says the ball slows down horizontally because nothing is pushing it forward
  - 1 point States that the string keeps pulling the cart forward, so the cart's speed grows above $v_{0}$ during the flight; in the same time the cart therefore covers more horizontal distance than the ball, which leaves the ball behind the launcher. Reasoning relative to the launcher also earns the point: the ball starts with no horizontal velocity relative to the launcher, and the launcher keeps gaining speed while nothing speeds the ball up horizontally, so the ball drifts back relative to it.Earns it: "The cart goes faster than $v_{0}$ the whole time the ball is up, so it gets farther in the same time and the ball lands behind it."; "Relative to the cart, the ball starts with no sideways speed. The cart keeps speeding up and nothing pushes the ball forward, so the ball falls back relative to the launcher."Does not: says the cart moves farther because it is heavier than the ball (the mass is not the reason; the extra speed from the string is)
  Model response
  Behind the launcher. When the ball leaves, it is moving forward at $v_{0}$ , the same as the cart. In the air the only force on the ball is gravity, which points straight down, so nothing speeds the ball up or slows it down horizontally; it keeps moving forward at $v_{0}$ . The cart is different. The string keeps pulling it forward, so its speed rises above $v_{0}$ for the whole flight. Both are moving for the same time, and the cart is going faster than the ball at every instant after launch, so the cart covers more ground. When the ball comes back down, the launcher has moved ahead of it.
2. (b)
  
  Derive an expression for the horizontal distance $D$ between the ball and the launcher at the moment the ball falls back to its launch height. Express your answer in terms of $a$ , $u$ , $v_{0}$ and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
  Question 4, part (b): your response. It stays in this browser tab and is never sent anywhere.
  Score part (b)
  - 1 point Finds the time of flight from the vertical motion: $0 = u t - \frac{1}{2} g t^{2}$ , so $t = \frac{2 u}{g}$ .Earns it: " $Δ y = u t - \frac{1}{2} g t^{2} = 0$ gives $t = 2 u / g$ "Does not: $t = \frac{u}{g}$ (the time to reach the top, only half the flight); uses $v_{0}$ in the vertical motion
  - 1 point Writes the horizontal displacements over that time: the ball moves $v_{0} t$ and the cart moves $v_{0} t + \frac{1}{2} a t^{2}$ , so $D = \frac{1}{2} a t^{2}$ . Finding the ball's horizontal displacement relative to the launcher directly, with zero initial relative velocity and a relative acceleration of size $a$ backward, so $D = \frac{1}{2} a t^{2}$ , also earns the point.Earns it: " $x_{ball} = v_{0} t$ , $x_{cart} = v_{0} t + \frac{1}{2} a t^{2}$ , so $D = x_{cart} - x_{ball} = \frac{1}{2} a t^{2}$ "; "Relative to the launcher the ball starts with zero horizontal velocity and has acceleration $a$ backward, so $D = \frac{1}{2} a t^{2}$ ."Does not: gives the ball the acceleration $a$ as well, so the distance comes out zero; leaves out the cart's starting velocity $v_{0}$ but keeps it for the ball, so $v_{0}$ does not cancel
  - 1 point Arrives at a correct expression: $D = \frac{1}{2} a {(\frac{2 u}{g})}^{2} = \frac{2 a u^{2}}{g^{2}}$ .Earns it: " $D = 2 a u^{2} / g^{2}$ "Does not: $D = \frac{a u^{2}}{2 g^{2}}$ (uses the time to the top instead of the whole flight)
  Model response
  Vertically, the ball starts with velocity $u$ upward and has acceleration $g$ downward. It is back at its launch height when $Δ y = 0$ :
  $Δ y = u t - \frac{1}{2} g t^{2} = 0 \Rightarrow t = \frac{2 u}{g}$
  Horizontally, using $Δ x = v_{0} t + \frac{1}{2} a_{x} t^{2}$ , the ball has $a_{x} = 0$ and the cart has $a_{x} = a$ :
  $x_{ball} = v_{0} t x_{cart} = v_{0} t + \frac{1}{2} a t^{2}$ $D = x_{cart} - x_{ball} = \frac{1}{2} a {(\frac{2 u}{g})}^{2} = \frac{2 a u^{2}}{g^{2}}$
3. (c)
  
  A student predicts that if the ball were fired at a moment when the cart was moving faster, with the same acceleration $a$ and the same upward component $u$ , the ball would land farther behind the launcher. Does your expression from part (b) support the student's prediction? Justify your answer, and connect it to your reasoning in part (a).
  Question 4, part (c): your response. It stays in this browser tab and is never sent anywhere.
  Score part (c)
  - 1 point Addresses how $D$ depends on $v_{0}$ in the expression from part (b). For a correct expression, $D = \frac{2 a u^{2}}{g^{2}}$ does not contain $v_{0}$ , so the distance is the same whatever the cart's speed at launch and the expression does not support the prediction. A conclusion that follows from an incorrect expression in part (b) also earns this point.Earns it: "No. $v_{0}$ is not in my expression, so $D$ stays the same."Does not: "Yes, a faster cart goes farther" with no reference to the expression
  - 1 point Justifies the result in terms of part (a): the ball and the cart both start with the same horizontal velocity $v_{0}$ , so that shared velocity carries them forward equally and cancels; the gap comes only from the speed the cart gains after launch, which depends on $a$ and the flight time ( $2 u / g$ ), not on $v_{0}$ .Earns it: "Both start at $v_{0}$ , so that part moves them the same distance. Only the extra speed the cart picks up from the string opens the gap, and that depends on $a$ and how long the ball is in the air."Does not: "a faster cart covers more distance, but so does the ball" with no statement of what does open the gap
  Model response
  No. My expression $D = \frac{2 a u^{2}}{g^{2}}$ has no $v_{0}$ in it, so firing the ball when the cart is moving faster leaves the distance unchanged. That fits my reasoning in part (a). At launch the ball and cart move forward at the same $v_{0}$ , and that shared speed carries both of them forward by the same $v_{0} t$ , whatever $v_{0}$ is. The ball ends up behind only because the cart keeps gaining speed after the launch, and how much it gains depends on its acceleration $a$ and on how long the ball is in the air, $2 u / g$ . Neither of those depends on $v_{0}$ .
Your score: 0 of 8, from the boxes you ticked.

Full practice exams

Practice Exam 1

Section I: Multiple Choice

Section II: Free Response

Question 1: Mathematical Routines 10 points

Question 2: Translation Between Representations 12 points

Question 3: Experimental Design and Analysis 10 points

Question 4: Qualitative/Quantitative Translation 8 points

Your results

Practice Exam 2

Section I: Multiple Choice

Section II: Free Response

Question 1: Mathematical Routines 10 points

Question 2: Translation Between Representations 12 points

Question 3: Experimental Design and Analysis 10 points

Question 4: Qualitative/Quantitative Translation 8 points

Your results

Steel pipe released from rest at different heights
Height $h$ (m)	Speed $v$ at the photogate (m/s)
0.050	0.76
0.100	1.04
0.150	1.30
0.200	1.48
0.250	1.68