AP Physics C Mechanics Quiz: Momentum

20 questions · exam conditions

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MomentumQuestion 1 of 20

Two skaters collide elastically: $m_1=50\,\text{kg}$ with $v_{1i}=+4.0\,\text{m/s}$ and $m_2=70\,\text{kg}$ with $v_{2i}=0$ . Afterward $v_{2f}=+3.0\,\text{m/s}$ . What is the final velocity of skater 1?

-0.20\,\text{m/s}

+0.20\,\text{m/s}

+1.0\,\text{m/s}

-1.0\,\text{m/s}

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Momentum

Practice Momentum in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Momentum, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$-0.20\,\text{m/s}$ (correct answer)
$+0.20\,\text{m/s}$
$+1.0\,\text{m/s}$
$-1.0\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in elastic collisions where both momentum and kinetic energy are conserved. In elastic collisions, knowing the masses, initial velocities, and one final velocity allows us to calculate the other final velocity using momentum conservation. In this problem, a 50 kg skater moving at +4.0 m/s collides elastically with a stationary 70 kg skater, and after collision the second skater moves at +3.0 m/s. Choice A (-0.20 m/s) is correct because applying momentum conservation: m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f gives us (50)(4.0) + (70)(0) = (50)v₁f + (70)(3.0), which simplifies to 200 = 50v₁f + 210, yielding v₁f = -10/50 = -0.20 m/s. Choice C (+1.0 m/s) is incorrect because it would not conserve momentum, as the total final momentum would exceed the initial momentum. To help students: In elastic collisions, always verify both momentum and energy conservation. The negative final velocity indicates the first skater rebounds backward, which is physically reasonable when a lighter object collides with a heavier stationary object.

Question 2

A stationary $10\,\text{kg}$ device explodes into $6.0\,\text{kg}$ and $4.0\,\text{kg}$ fragments; the $6.0\,\text{kg}$ fragment has $-5.0\,\text{m/s}$ . Determine the velocity of the $4.0\,\text{kg}$ fragment after the explosion.

$+7.5\,\text{m/s}$ (correct answer)
$-7.5\,\text{m/s}$
$+3.0\,\text{m/s}$
$+5.0\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in explosion problems where fragments move in opposite directions. When a stationary object explodes, the total momentum must remain zero, requiring the momentum vectors of all fragments to sum to zero. In this problem, a stationary 10 kg device explodes into 6.0 kg and 4.0 kg fragments, with the 6.0 kg fragment moving at -5.0 m/s. Choice A (+7.5 m/s) is correct because applying momentum conservation: 0 = m₁v₁ + m₂v₂ gives us 0 = (6.0)(-5.0) + (4.0)v₂, which simplifies to 0 = -30 + 4v₂, yielding v₂ = +7.5 m/s. Choice B (-7.5 m/s) is incorrect because it has the wrong sign, which would result in both fragments moving in the same direction, creating net momentum from an initially stationary system. To help students: In explosion problems from rest, fragments must move in opposite directions. The lighter fragment moves faster to compensate for its smaller mass, maintaining zero total momentum.

Question 3

A stationary $8.0\ \text{kg}$ object explodes into $5.0\ \text{kg}$ and $3.0\ \text{kg}$ fragments; the $5.0\ \text{kg}$ fragment moves at $-4.0\ \text{m/s}$ . Determine the $3.0\ \text{kg}$ fragment velocity.

$+6.7\ \text{m/s}$ (correct answer)
$-6.7\ \text{m/s}$
$+2.4\ \text{m/s}$
$+4.0\ \text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in explosion problems where fragments move in opposite directions. When an object at rest explodes, the total momentum remains zero, requiring the momentum vectors of all fragments to sum to zero. In this problem, a stationary 8.0 kg object explodes into 5.0 kg and 3.0 kg fragments, with the 5.0 kg fragment moving at -4.0 m/s. Choice A (+6.7 m/s) is correct because initial momentum = 0, so (5.0 kg)(-4.0 m/s) + (3.0 kg)(v) = 0, giving -20.0 + 3.0v = 0, thus v = +20.0/3.0 = +6.67 m/s ≈ +6.7 m/s. Choice B (-6.7 m/s) is incorrect because it has the wrong sign, failing to recognize that the fragments must move in opposite directions. To help students: Use vector diagrams to visualize how explosion fragments must have opposite momentum vectors. Practice with different mass splits to show how the lighter fragment always moves faster to conserve momentum.

Question 4

A stationary $5.0\,\text{kg}$ object explodes into $1.0\,\text{kg}$ and $4.0\,\text{kg}$ fragments; the $1.0\,\text{kg}$ fragment moves $+20\,\text{m/s}$ . Determine the $4.0\,\text{kg}$ fragment velocity.

$-5.0\,\text{m/s}$ (correct answer)
$+5.0\,\text{m/s}$
$-20\,\text{m/s}$
$-4.0\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in explosion scenarios where a stationary object breaks into fragments. When an object at rest explodes, the total momentum must remain zero, meaning the momentum vectors of all fragments must sum to zero, resulting in fragments moving in opposite directions. In this problem, a 5.0 kg object at rest explodes into 1.0 kg and 4.0 kg fragments, with the 1.0 kg fragment moving at +20 m/s, and we need the velocity of the 4.0 kg fragment. Choice A (-5.0 m/s) is correct because the initial momentum is zero, so: 0 = (1.0 kg)(+20 m/s) + (4.0 kg)(

v_2

), giving 0 = 20 + 4v_2, thus v_2 = -20/4 = -5.0 m/s. Choice C (-20 m/s) is incorrect because it simply uses the opposite of the first fragment's velocity without considering the mass ratio. To help students: Emphasize that in explosions from rest, the momentum magnitudes are inversely proportional to the masses. Use momentum vector diagrams to show how fragments must have opposite directions to maintain zero total momentum.

Question 5

A stationary $5.0\ \text{kg}$ object explodes into $2.0\ \text{kg}$ and $3.0\ \text{kg}$ fragments; the $2.0\ \text{kg}$ fragment moves at $+9.0\ \text{m/s}$ . Determine the $3.0\ \text{kg}$ fragment velocity.

$-6.0\ \text{m/s}$ (correct answer)
$+6.0\ \text{m/s}$
$-3.0\ \text{m/s}$
$-13.5\ \text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in explosion problems where an initially stationary object breaks into fragments. In explosions, the total momentum must remain zero if the system starts at rest, meaning the momentum vectors of all fragments must sum to zero. In this problem, a stationary 5.0 kg object explodes into 2.0 kg and 3.0 kg fragments, with the 2.0 kg fragment moving at +9.0 m/s. Choice A (-6.0 m/s) is correct because initial momentum = 0, so (2.0 kg)(+9.0 m/s) + (3.0 kg)(v) = 0, giving v = -18.0/3.0 = -6.0 m/s. Choice D (-13.5 m/s) is incorrect because it might result from incorrectly applying mass ratios or calculation errors. To help students: Emphasize that explosions from rest always have zero total momentum, making the fragments move in opposite directions. Use vector diagrams to show how momentum vectors must cancel, and practice with different mass ratios to build intuition.

Question 6

A stationary $6.0\,\text{kg}$ object explodes into $2.0\,\text{kg}$ and $4.0\,\text{kg}$ fragments; the $2.0\,\text{kg}$ fragment moves $+12\,\text{m/s}$ . Determine the $4.0\,\text{kg}$ fragment velocity.

$-6.0\,\text{m/s}$ (correct answer)
$+6.0\,\text{m/s}$
$-3.0\,\text{m/s}$
$-24\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in explosion scenarios where an initially stationary object breaks apart. Linear momentum conservation states that in the absence of external forces, the total momentum of a system remains constant, which for an initially stationary object means the total final momentum must be zero. In this problem, a 6.0 kg object at rest explodes into two fragments (2.0 kg and 4.0 kg), with the 2.0 kg fragment moving at +12 m/s, and we need to find the velocity of the 4.0 kg fragment. Choice A (-6.0 m/s) is correct because the initial momentum is zero, so: 0 = (2.0 kg)(+12 m/s) + (4.0 kg)(

v_2

), which gives 0 = 24 + 4v_2, so v_2 = -24/4 = -6.0 m/s. Choice D (-24 m/s) is incorrect because it represents the momentum of the first fragment divided by 1 kg instead of the mass of the second fragment. To help students: Emphasize that explosions from rest require the final momenta to sum to zero, creating equal and opposite momentum components. Use vector diagrams to show how the fragments must move in opposite directions to conserve momentum.

Question 7

A $6.0\,\text{kg}$ object at rest explodes into $2.0\,\text{kg}$ and $4.0\,\text{kg}$ fragments; the $2.0\,\text{kg}$ fragment moves at $+9.0\,\text{m/s}$ . Determine the velocity of the $4.0\,\text{kg}$ fragment.

$-4.5\,\text{m/s}$ (correct answer)
$+4.5\,\text{m/s}$
$-18\,\text{m/s}$
$-3.0\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in explosion problems where an initially stationary object breaks into fragments. In explosions, the total momentum of the system must remain zero if the object was initially at rest, meaning the momentum vectors of all fragments must sum to zero. In this problem, a 6.0 kg object at rest explodes into 2.0 kg and 4.0 kg fragments, with the 2.0 kg fragment moving at +9.0 m/s. Choice A (-4.5 m/s) is correct because applying momentum conservation: 0 = m₁v₁ + m₂v₂ gives us 0 = (2.0)(9.0) + (4.0)v₂, which simplifies to 0 = 18 + 4v₂, yielding v₂ = -4.5 m/s. Choice B (+4.5 m/s) is incorrect because it has the wrong sign, which would create net momentum in the positive direction, violating conservation for an initially stationary system. To help students: Remember that explosions from rest must have zero total momentum after the event. Use the fact that momentum vectors must cancel out, and always check that the heavier fragment moves slower than the lighter fragment to conserve momentum.

Question 8

A $0.20\,\text{kg}$ ball moving at $+8.0\,\text{m/s}$ experiences a constant $-24\,\text{N}$ force for $0.050\,\text{s}$ on a frictionless line. Calculate the velocity of the ball resulting from the interaction.

$+2.0\,\text{m/s}$ (correct answer)
$+14\,\text{m/s}$
$-2.0\,\text{m/s}$
$+6.0\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically using the impulse-momentum theorem with forces acting over time. The impulse-momentum theorem relates the impulse (force × time) to the change in momentum, allowing calculation of final velocity from initial conditions and applied force. In this problem, a 0.20 kg ball moving at +8.0 m/s experiences a -24 N force for 0.050 s on a frictionless surface. Choice A (+2.0 m/s) is correct because applying the impulse-momentum theorem: FΔt = m(vf - vi) gives us (-24)(0.050) = (0.20)(vf - 8.0), which simplifies to -1.2 = 0.20vf - 1.6, yielding vf = 0.4/0.20 = +2.0 m/s. Choice B (+14 m/s) is incorrect because it might result from adding the impulse effect instead of properly accounting for the negative force direction. To help students: Always pay attention to force direction - negative forces oppose motion. The ball slows down but continues in the positive direction because the impulse isn't large enough to reverse its motion completely.

Question 9

A system consists of three particles with masses $m_1 = 2.0$ kg, $m_2 = 3.0$ kg, and $m_3 = 1.0$ kg. At time $t = 0$ , their momentum vectors are $\vec{p_1} = (4.0, 6.0)$ kg⋅m/s, $\vec{p_2} = (-2.0, 8.0)$ kg⋅m/s, and $\vec{p_3} = (6.0, -2.0)$ kg⋅m/s. If no external forces act on the system, what is the velocity of the center of mass?

$(1.33, 2.00)$ m/s with total system momentum magnitude of 14.4 kg⋅m/s (correct answer)
$(1.33, 2.00)$ m/s with total system momentum magnitude of 8.0 kg⋅m/s
$(2.67, 4.00)$ m/s with total system momentum magnitude of 14.4 kg⋅m/s
$(2.67, 4.00)$ m/s with total system momentum magnitude of 8.0 kg⋅m/s

Explanation: The total momentum of the system is

\vec{P}_{total} = \vec{p_1} + \vec{p_2} + \vec{p_3} = (4.0, 6.0) + (-2.0, 8.0) + (6.0, -2.0) = (8.0, 12.0)

kg⋅m/s. The magnitude is

|\vec{P}_{total}| = \sqrt{8.0^2 + 12.0^2} = \sqrt{64 + 144} = \sqrt{208} = 14.4

kg⋅m/s. The total mass is

M = 2.0 + 3.0 + 1.0 = 6.0

kg. The velocity of the center of mass is

\vec{v}_{cm} = \frac{\vec{P}_{total}}{M} = \frac{(8.0, 12.0)}{6.0} = (1.33, 2.00)

m/s. Choice B has the wrong momentum magnitude. Choice C has the wrong velocity (this would be if you forgot to divide by total mass). Choice D has both the wrong velocity and wrong momentum magnitude.

Question 10

A rocket of initial mass $M_0$ ejects fuel at a constant rate $\frac{dm}{dt} = -k$ with exhaust velocity $v_e$ relative to the rocket. If the rocket starts from rest in space (no external forces), what is the rocket's velocity when its mass has decreased to $\frac{M_0}{2}$ ?

$v_e \ln(2)$ , derived from the rocket equation with mass ratio considerations (correct answer)
$\frac{v_e M_0}{2k}$ , using the impulse-momentum theorem for continuous mass ejection
$v_e \ln(\frac{M_0}{2})$ , from integrating the thrust equation over the mass change interval
$\frac{v_e}{2}$ , since half the mass is ejected at velocity $v_e$ relative to the rocket

Explanation: The rocket equation (Tsiolkovsky equation) is

\Delta v = v_e \ln\left(\frac{M_0}{M_f}\right)

, where

M_0

is initial mass and

M_f

is final mass. When the mass decreases to

\frac{M_0}{2}

, we have

M_f = \frac{M_0}{2}

. Therefore,

\Delta v = v_e \ln\left(\frac{M_0}{M_0/2}\right) = v_e \ln(2)

. Since the rocket starts from rest, this is also the final velocity. Choice B incorrectly uses impulse-momentum without considering the changing mass of the rocket. Choice C has the wrong argument in the logarithm (should be the mass ratio, not the final mass). Choice D incorrectly assumes a simple proportional relationship and ignores the exponential nature of rocket propulsion.

Question 11

A cannon ( $m_c=250\,\text{kg}$ ) fires a projectile ( $m_p=5.0\,\text{kg}$ ) at $+180\,\text{m/s}$ relative to the ground. Find the recoil velocity of the cannon after the event.

$-3.6\,\text{m/s}$ (correct answer)
$+3.6\,\text{m/s}$
$-0.90\,\text{m/s}$
$-9.0\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in recoil problems where an initially stationary system separates into moving parts. In recoil situations, the total momentum must remain zero if the system starts at rest, meaning the momentum of the projectile and the recoiling object must be equal in magnitude but opposite in direction. In this problem, a 250 kg cannon fires a 5.0 kg projectile at +180 m/s relative to the ground. Choice A (-3.6 m/s) is correct because applying momentum conservation: 0 = mcvc + mpvp gives us 0 = (250)vc + (5.0)(180), which simplifies to 0 = 250vc + 900, yielding vc = -900/250 = -3.6 m/s. Choice B (+3.6 m/s) is incorrect because it has the wrong sign, which would create net momentum in the positive direction instead of maintaining zero total momentum. To help students: Emphasize that in recoil problems from rest, the objects must move in opposite directions. The lighter object moves faster and the heavier object moves slower, with their momenta canceling exactly.

Question 12

A $0.20\,\text{kg}$ tennis ball initially at $-15\,\text{m/s}$ experiences $+120\,\text{N}$ for $0.050\,\text{s}$ . Calculate the final velocity of the ball.

$+15\,\text{m/s}$ (correct answer)
$-15\,\text{m/s}$
$+30\,\text{m/s}$
$+7.5\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically using the impulse-momentum theorem to find velocity changes when forces act over time. The impulse-momentum theorem states that the impulse (force × time) equals the change in momentum, allowing us to calculate final velocities when initial conditions and applied forces are known. In this problem, a 0.20 kg tennis ball initially moving at -15 m/s experiences a +120 N force for 0.050 s, and we must find the final velocity. Choice A (+15 m/s) is correct because impulse = FΔt = (120 N)(0.050 s) = 6.0 N·s = Δp = m(vf - vi), so 6.0 = 0.20(vf - (-15)), giving 6.0 = 0.20vf + 3.0, thus vf = 3.0/0.20 = +15 m/s. Choice B (-15 m/s) is incorrect because it suggests no change in speed, ignoring the substantial impulse applied to the ball. To help students: Emphasize the vector nature of impulse and momentum, ensuring proper sign conventions are maintained. Practice problems involving direction changes and use impulse-momentum bar charts to visualize the momentum change process.

Question 13

Two skaters collide elastically: $m_1=50\,\text{kg}$ at $+4.0\,\text{m/s}$ , $m_2=70\,\text{kg}$ at rest. What is the final velocity of $m_2$ ?

$+3.33\,\text{m/s}$ (correct answer)
$+2.86\,\text{m/s}$
$-3.33\,\text{m/s}$
$+1.67\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically finding the second object's velocity in an elastic collision where the first object hits a stationary target. In elastic collisions between two objects where one is initially at rest, specific formulas relate the final velocities to the initial conditions and mass ratio. In this problem, a 50 kg skater at +4.0 m/s collides elastically with a 70 kg skater at rest, and we need the final velocity of the second skater. Choice A (+3.33 m/s) is correct because for elastic collisions with the second object initially at rest, v2f = (2m1/(m1+m2))v1i = (2(50)/(50+70))(4.0) = (100/120)(4.0) = 3.33 m/s. Choice B (+2.86 m/s) is incorrect because it might result from using the wrong formula or making calculation errors with the mass ratio. To help students: Emphasize memorizing or deriving the elastic collision formulas for the special case of one object at rest. Practice problems with various mass ratios and use energy bar charts alongside momentum diagrams to verify both conservation laws are satisfied.

Question 14

A $2.0\,\text{kg}$ cart at $+3.0\,\text{m/s}$ sticks to a $1.0\,\text{kg}$ cart at rest. What is the final velocity?

$+2.0\,\text{m/s}$ (correct answer)
$+1.0\,\text{m/s}$
$+3.0\,\text{m/s}$
$+0.67\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in perfectly inelastic collisions where objects stick together. Linear momentum, the product of mass and velocity, remains constant in isolated systems, meaning the total momentum before collision equals the total momentum after. In this problem, a 2.0 kg cart moving at +3.0 m/s collides with and sticks to a 1.0 kg cart initially at rest, and we must find their common final velocity. Choice A (+2.0 m/s) is correct because applying conservation of momentum: initial momentum = (2.0 kg)(+3.0 m/s) + (1.0 kg)(0 m/s) = 6.0 kg·m/s, and final momentum = (2.0 + 1.0 kg)(v_f) = 3.0v_f, so v_f = 6.0/3.0 = +2.0 m/s. Choice B (+1.0 m/s) is incorrect because it might result from dividing the initial velocity by the total mass without considering the initial momentum. To help students: Use the systematic approach of calculating total initial momentum, then dividing by total final mass. Visual representations like momentum bar charts can reinforce the conservation concept and help students avoid common calculation errors.

Question 15

A $10\,\text{kg}$ cannon fires a $0.50\,\text{kg}$ projectile at $+80\,\text{m/s}$ from rest. Find the recoil velocity of the cannon.

$-4.0\,\text{m/s}$ (correct answer)
$+4.0\,\text{m/s}$
$-0.25\,\text{m/s}$
$-8.0\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in recoil situations where a projectile is fired from an initially stationary system. Linear momentum conservation requires that the total momentum before firing (zero for a system at rest) equals the total momentum after firing, resulting in the cannon recoiling in the opposite direction to the projectile. In this problem, a 10 kg cannon fires a 0.50 kg projectile at +80 m/s, and we must find the cannon's recoil velocity. Choice A (-4.0 m/s) is correct because applying conservation of momentum: initial momentum = 0, final momentum = (0.50 kg)(+80 m/s) + (10 kg)(v_c) = 0, which gives 40 + 10v_c = 0, so v_c = -40/10 = -4.0 m/s. Choice B (+4.0 m/s) is incorrect because it has the wrong sign, suggesting the cannon moves in the same direction as the projectile, which violates momentum conservation. To help students: Stress that recoil always occurs in the opposite direction to the projectile motion. Practice with various mass ratios and use free-body diagrams to reinforce the action-reaction principle underlying momentum conservation.

Question 16

A $6.0\ \text{kg}$ cart at $+2.0\ \text{m/s}$ collides elastically with a $2.0\ \text{kg}$ cart at rest. What is the final velocity of the $2.0\ \text{kg}$ cart?

$+1.0\ \text{m/s}$
$+3.0\ \text{m/s}$ (correct answer)
$+1.5\ \text{m/s}$
$-3.0\ \text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in elastic collisions where both momentum and kinetic energy are conserved. In elastic collisions, objects bounce off each other without losing mechanical energy, requiring simultaneous solution of momentum and energy conservation equations. In this problem, a 6.0 kg cart at +2.0 m/s collides elastically with a 2.0 kg cart at rest. Choice B (+3.0 m/s) is correct because for elastic collisions with one object initially at rest, the final velocity of the initially stationary object is v_2f = (2m_1/(

m_1+m_2

))v_1i = (2×6.0/(6.0+2.0))×2.0 = (12.0/8.0)×2.0 = 3.0 m/s. Choice C (+1.5 m/s) is incorrect because it might result from treating this as an inelastic collision or using incorrect elastic collision formulas. To help students: Teach the specific formulas for elastic collisions, especially the special case where one object is initially at rest. Use energy bar charts alongside momentum calculations to verify that kinetic energy is conserved in elastic collisions.

Question 17

A $40\ \text{kg}$ cannon fires a $2.0\ \text{kg}$ projectile at $+120\ \text{m/s}$ from rest. Find the recoil velocity of the cannon.

$-6.0\ \text{m/s}$ (correct answer)
$+6.0\ \text{m/s}$
$-2.4\ \text{m/s}$
$-60\ \text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in recoil problems where momentum is conserved in firing projectiles. When a cannon fires a projectile, the system's total momentum remains zero if initially at rest, causing the cannon to recoil in the opposite direction. In this problem, a 40 kg cannon fires a 2.0 kg projectile at +120 m/s from rest. Choice A (-6.0 m/s) is correct because initial momentum = 0, so (2.0 kg)(+120 m/s) + (40 kg)(v) = 0, giving v = -240/40 = -6.0 m/s. Choice D (-60 m/s) is incorrect because it might result from using the wrong mass ratio or confusing the projectile and cannon masses. To help students: Draw clear before/after diagrams showing the recoil effect, emphasize that the more massive object moves slower to conserve momentum. Practice with various mass ratios to show how recoil velocity depends inversely on mass.

Question 18

Two skaters collide elastically: $m_1=50\,\text{kg}$ at $+4.0\,\text{m/s}$ , $m_2=70\,\text{kg}$ at rest. What is the final velocity of $m_1$ ?

$-0.67\,\text{m/s}$ (correct answer)
$+0.67\,\text{m/s}$
$+2.33\,\text{m/s}$
$-2.33\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in elastic collisions between two objects where both momentum and kinetic energy are conserved. In elastic collisions, we must apply both conservation laws to find the final velocities, which for a moving object colliding with a stationary object of different mass results in specific velocity relationships. In this problem, a 50 kg skater at +4.0 m/s collides elastically with a 70 kg skater at rest, and we need to find the final velocity of the first skater. Choice A (-0.67 m/s) is correct because for elastic collisions with one object initially at rest, v1f = ((m1-m2)/(m1+m2))v1i = ((50-70)/(50+70))(4.0) = (-20/120)(4.0) = -0.67 m/s. Choice B (+0.67 m/s) is incorrect because it has the wrong sign, failing to recognize that the lighter object bounces backward when colliding with a heavier stationary object. To help students: Teach the specific formulas for elastic collisions with one object at rest, emphasizing how mass ratios determine whether objects bounce back or continue forward. Use simulations to visualize how different mass ratios affect post-collision motion.

Question 19

Two cars stick together: $m_1=1000\,\text{kg}$ at $+25\,\text{m/s}$ , $m_2=1500\,\text{kg}$ at $+5\,\text{m/s}$ . What is the final velocity?

$+13\,\text{m/s}$ (correct answer)
$+12\,\text{m/s}$
$+15\,\text{m/s}$
$+20\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in perfectly inelastic collisions where objects moving in the same direction stick together. Linear momentum conservation requires that the vector sum of momenta before collision equals the total momentum after, which for same-direction motion simplifies to algebraic addition. In this problem, two cars moving in the positive direction (1000 kg at +25 m/s and 1500 kg at +5 m/s) stick together, and we need their final velocity. Choice A (+13 m/s) is correct because the total initial momentum = (1000 kg)(+25 m/s) + (1500 kg)(+5 m/s) = 25000 + 7500 = 32500 kg·m/s, and with total mass = 2500 kg, the final velocity = 32500/2500 = +13 m/s. Choice B (+12 m/s) is incorrect because it might result from calculation errors or rounding too early in the problem. To help students: Stress the importance of calculating total momentum first before dividing by total mass. Use tables to organize given information and ensure all quantities are properly accounted for in the conservation equation.

Question 20

A $0.058\,\text{kg}$ tennis ball changes from $-22\,\text{m/s}$ to $+v_f$ when a $310\,\text{N}$ force acts for $0.012\,\text{s}$ . Calculate the velocity of the ball resulting from the interaction.

$+42\,\text{m/s}$ (correct answer)
$+12\,\text{m/s}$
$-42\,\text{m/s}$
$+64\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically using the impulse-momentum theorem to find velocity changes. The impulse-momentum theorem states that the impulse (force × time) equals the change in momentum, or FΔt = mΔv = m(vf - vi). In this problem, a 0.058 kg tennis ball initially moving at -22 m/s experiences a 310 N force for 0.012 s. Choice A (+42 m/s) is correct because applying the impulse-momentum theorem: FΔt = m(vf - vi) gives us (310)(0.012) = (0.058)(vf - (-22)), which simplifies to 3.72 = 0.058(vf + 22), yielding vf = 3.72/0.058 - 22 = 64.1 - 22 = +42.1 m/s. Choice D (+64 m/s) is incorrect because it represents the total velocity change without accounting for the initial velocity, a common error when students forget to subtract the initial velocity. To help students: Always write out the impulse-momentum equation completely, including initial velocity with its proper sign. Practice problems involving direction changes to reinforce the importance of sign conventions in momentum calculations.