AP Physics C Mechanics Quiz: Conservation Of Linear Momentum
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Conservation Of Linear MomentumQuestion 1 of 8

A firework of mass 3.0 kg3.0\,\text{kg} moves horizontally at +4.0 m/s+4.0\,\text{m/s} (no air resistance). It explodes into two pieces: piece 1 has mass 1.0 kg1.0\,\text{kg} and piece 2 has mass 2.0 kg2.0\,\text{kg}. Immediately after, piece 2 moves at +1.0 m/s+1.0\,\text{m/s}. With external impulse negligible, momentum is conserved: p⃗i=(3.0)(+4.0)ı^\vec p_i=(3.0)(+4.0)\hat\imath and p⃗f=(1.0 v⃗1+2.0⋅+1.0ı^)\vec p_f=(1.0\,\vec v_1+2.0\cdot +1.0\hat\imath). Calculate the velocity of piece 1 after the event given the initial conditions.

v⃗1=+10 ı^ m/s\vec v_1=+10\,\hat\imath\,\text{m/s}
v⃗1=+14 ı^ m/s\vec v_1=+14\,\hat\imath\,\text{m/s}
v⃗1=−10 ı^ m/s\vec v_1=-10\,\hat\imath\,\text{m/s}
v⃗1=+6 ı^ m/s\vec v_1=+6\,\hat\imath\,\text{m/s}
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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Conservation Of Linear Momentum

Practice Conservation Of Linear 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 Conservation Of Linear 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

A firework of mass 3.0 kg3.0\,\text{kg} moves horizontally at +4.0 m/s+4.0\,\text{m/s} (no air resistance). It explodes into two pieces: piece 1 has mass 1.0 kg1.0\,\text{kg} and piece 2 has mass 2.0 kg2.0\,\text{kg}. Immediately after, piece 2 moves at +1.0 m/s+1.0\,\text{m/s}. With external impulse negligible, momentum is conserved: p⃗i=(3.0)(+4.0)ı^\vec p_i=(3.0)(+4.0)\hat\imath and p⃗f=(1.0 v⃗1+2.0⋅+1.0ı^)\vec p_f=(1.0\,\vec v_1+2.0\cdot +1.0\hat\imath). Calculate the velocity of piece 1 after the event given the initial conditions.

  1. v⃗1=+10 ı^ m/s\vec v_1=+10\,\hat\imath\,\text{m/s} (correct answer)
  2. v⃗1=+14 ı^ m/s\vec v_1=+14\,\hat\imath\,\text{m/s}
  3. v⃗1=−10 ı^ m/s\vec v_1=-10\,\hat\imath\,\text{m/s}
  4. v⃗1=+6 ı^ m/s\vec v_1=+6\,\hat\imath\,\text{m/s}
Explanation: This question tests AP Physics C: Mechanics understanding of linear momentum conservation in isolated systems. Linear momentum, defined as the product of mass and velocity, is conserved in isolated systems where no external forces act. In this scenario, a firework explodes into two pieces with no air resistance, making it an isolated system where momentum conservation applies. Choice A correctly calculates piece 1's velocity: initial momentum = (3.0 kg)(+4.0 m/s) = 12.0 kg·m/s, final momentum = (1.0 kg)(v1) + (2.0 kg)(+1.0 m/s), so 12.0 = v1 + 2.0, giving v1 = +10.0 m/s. Choice B incorrectly adds rather than conserves momentum. To help students: Emphasize that explosions are internal forces that don't violate momentum conservation. Practice problems with fragmenting objects helps students recognize that pieces can have different velocities while total momentum remains constant.

Question 2

A 3.0 kg3.0\,\text{kg} cart moving at v⃗i=+2.0 m/s ı^\vec v_i=+2.0\,\text{m/s}\,\hat\imath explodes into two pieces of 1.0 kg1.0\,\text{kg} and 2.0 kg2.0\,\text{kg}. The 1.0 kg1.0\,\text{kg} piece leaves at v⃗1=+8.0 m/s ı^\vec v_1=+8.0\,\text{m/s}\,\hat\imath. With external impulse negligible, calculate the 2.0 kg2.0\,\text{kg} piece's velocity.

  1. v⃗2=−1.0 m/s ı^\vec v_2=-1.0\,\text{m/s}\,\hat\imath (correct answer)
  2. v⃗2=+1.0 m/s ı^\vec v_2=+1.0\,\text{m/s}\,\hat\imath
  3. v⃗2=−2.0 m/s ı^\vec v_2=-2.0\,\text{m/s}\,\hat\imath
  4. v⃗2=+4.0 m/s ı^\vec v_2=+4.0\,\text{m/s}\,\hat\imath
Explanation: This question tests AP Physics C: Mechanics understanding of linear momentum conservation in isolated systems. Linear momentum, defined as the product of mass and velocity, is conserved in isolated systems where no external forces act. In this scenario, a cart explodes into two pieces with negligible external impulse, making this an internal force problem where total momentum before equals total momentum after. Choice A correctly applies conservation: initial momentum = (3.0 kg)(+2.0 m/s) = 6.0 kg·m/s; after explosion, (1.0 kg)(+8.0 m/s) + (2.0 kg)(v2) = 6.0, giving 8.0 + 2.0v2 = 6.0, so v2 = -2.0/2.0 = -1.0 m/s. Choice B incorrectly assumes both pieces must move in the same direction. To help students: Emphasize that in explosions, pieces can move in opposite directions while conserving total momentum. Practice setting up equations systematically and checking that calculated velocities make physical sense.

Question 3

Two skaters on frictionless ice push off from rest: skater 1 (75 kg75\,\text{kg}) moves at v⃗1=+1.6 m/s ı^\vec v_1=+1.6\,\text{m/s}\,\hat\imath after the push. With negligible external forces so ∑p⃗=0⃗\sum \vec p=\vec 0, calculate skater 2's velocity if m2=50 kgm_2=50\,\text{kg}.

  1. v⃗2=−2.4 m/s ı^\vec v_2=-2.4\,\text{m/s}\,\hat\imath (correct answer)
  2. v⃗2=+2.4 m/s ı^\vec v_2=+2.4\,\text{m/s}\,\hat\imath
  3. v⃗2=−1.1 m/s ı^\vec v_2=-1.1\,\text{m/s}\,\hat\imath
  4. v⃗2=−3.0 m/s ı^\vec v_2=-3.0\,\text{m/s}\,\hat\imath
Explanation: This question tests AP Physics C: Mechanics understanding of linear momentum conservation in isolated systems. Linear momentum, defined as the product of mass and velocity, is conserved in isolated systems where no external forces act. In this scenario, two skaters push off from rest on frictionless ice, creating an isolated system where initial momentum is zero and must remain zero. Choice A correctly applies conservation: initial momentum = 0; final momentum = (75 kg)(+1.6 m/s) + (50 kg)(v2) = 120 + 50v2 = 0, giving v2 = -120/50 = -2.4 m/s. Choice C incorrectly calculates the velocity magnitude without proper consideration of the mass ratio. To help students: Emphasize that action-reaction pairs during push-offs create equal and opposite momentum changes. Practice problems with different mass ratios to reinforce the inverse relationship between mass and velocity.

Question 4

A 0.60 kg0.60\,\text{kg} firework traveling at v⃗i=+5.0 m/s ı^\vec v_i=+5.0\,\text{m/s}\,\hat\imath explodes into 0.20 kg0.20\,\text{kg} and 0.40 kg0.40\,\text{kg} pieces. The 0.40 kg0.40\,\text{kg} piece has v⃗=+2.0 m/s ı^\vec v=+2.0\,\text{m/s}\,\hat\imath. With ∑p⃗\sum \vec p conserved, calculate the 0.20 kg0.20\,\text{kg} piece's velocity.

  1. v⃗=+11 m/s ı^\vec v=+11\,\text{m/s}\,\hat\imath (correct answer)
  2. v⃗=−1.0 m/s ı^\vec v=-1.0\,\text{m/s}\,\hat\imath
  3. v⃗=+7.0 m/s ı^\vec v=+7.0\,\text{m/s}\,\hat\imath
  4. v⃗=+9.0 m/s ı^\vec v=+9.0\,\text{m/s}\,\hat\imath
Explanation: This question tests AP Physics C: Mechanics understanding of linear momentum conservation in isolated systems. Linear momentum, defined as the product of mass and velocity, is conserved in isolated systems where no external forces act. In this scenario, a firework explodes into two pieces with total momentum conserved throughout the explosion process. Choice A correctly applies conservation: initial momentum = (0.60 kg)(+5.0 m/s) = 3.0 kg·m/s; after explosion, (0.20 kg)(v) + (0.40 kg)(+2.0 m/s) = 3.0, giving 0.20v + 0.80 = 3.0, so v = 2.2/0.20 = +11 m/s. Choice C incorrectly assumes the lighter piece moves at an intermediate velocity. To help students: Emphasize that in explosions, lighter pieces often achieve higher velocities to conserve momentum. Practice problems with varying mass ratios to build intuition about velocity distributions.

Question 5

Two ice skaters initially at rest on frictionless ice push off each other: skater A (60 kg60\,\text{kg}) and skater B (40 kg40\,\text{kg}). Afterward, A moves at v⃗A=−2.0 m/s ı^\vec v_A=-2.0\,\text{m/s}\,\hat\imath. With negligible external forces so total momentum remains 0⃗\vec 0, calculate skater B's velocity.

  1. v⃗B=+3.0 m/s ı^\vec v_B=+3.0\,\text{m/s}\,\hat\imath (correct answer)
  2. v⃗B=+1.3 m/s ı^\vec v_B=+1.3\,\text{m/s}\,\hat\imath
  3. v⃗B=−3.0 m/s ı^\vec v_B=-3.0\,\text{m/s}\,\hat\imath
  4. v⃗B=+2.0 m/s ı^\vec v_B=+2.0\,\text{m/s}\,\hat\imath
Explanation: This question tests AP Physics C: Mechanics understanding of linear momentum conservation in isolated systems. Linear momentum, defined as the product of mass and velocity, is conserved in isolated systems where no external forces act. In this scenario, two skaters initially at rest push off each other on frictionless ice, creating an isolated system with zero initial momentum that must remain zero. Choice A correctly applies conservation: initial momentum = 0; final momentum = (60 kg)(-2.0 m/s) + (40 kg)(vB) = -120 + 40vB = 0, giving vB = +3.0 m/s. Choice B incorrectly calculates the velocity ratio without considering the mass difference. To help students: Emphasize that when a system starts at rest, the total momentum must remain zero. Practice problems with different mass ratios to reinforce that lighter objects gain proportionally higher velocities.

Question 6

Two ice skaters initially at rest on level ice (friction negligible) push off each other. Skater 1 has mass 50 kg50\,\text{kg} and skater 2 has mass 75 kg75\,\text{kg}. After the push, skater 1 moves at +3.0 m/s+3.0\,\text{m/s} along +ı^+\hat\imath. The system is isolated horizontally, so p⃗i=0⃗\vec p_i=\vec 0 and p⃗f=50(+3.0)ı^+75 v⃗2\vec p_f=50(+3.0)\hat\imath+75\,\vec v_2. Calculate the velocity of skater 2 after the event given the initial conditions.

  1. v⃗2=−2.0 ı^ m/s\vec v_2=-2.0\,\hat\imath\,\text{m/s} (correct answer)
  2. v⃗2=+2.0 ı^ m/s\vec v_2=+2.0\,\hat\imath\,\text{m/s}
  3. v⃗2=−1.5 ı^ m/s\vec v_2=-1.5\,\hat\imath\,\text{m/s}
  4. v⃗2=−3.0 ı^ m/s\vec v_2=-3.0\,\hat\imath\,\text{m/s}
Explanation: This question tests AP Physics C: Mechanics understanding of linear momentum conservation in isolated systems. Linear momentum, defined as the product of mass and velocity, is conserved in isolated systems where no external forces act. In this scenario, two ice skaters initially at rest push off each other on frictionless ice, creating an isolated system with zero initial momentum. Choice A correctly calculates skater 2's velocity: initial momentum = 0, final momentum = (50 kg)(+3.0 m/s) + (75 kg)(v2) = 0, so 150 + 75v2 = 0, giving v2 = -150/75 = -2.0 m/s. Choice B has the wrong sign, failing to recognize that the skaters must move in opposite directions. To help students: Emphasize that when starting from rest, objects must move in opposite directions to conserve zero momentum. Practice problems with different mass ratios helps students see the inverse relationship between mass and velocity.

Question 7

A 4.0 kg block sliding at 6.0 m/s on a frictionless surface collides elastically with a 2.0 kg block moving at 3.0 m/s in the opposite direction. After the collision, the 2.0 kg block moves at 7.0 m/s in its original direction. A student calculates that the 4.0 kg block's final velocity should be 1.0 m/s in its original direction. Which statement best describes this result?

  1. The calculation is correct because momentum is conserved and the collision satisfies elastic collision conditions for these masses and velocities
  2. The calculation violates conservation of momentum; the 4.0 kg block should have final velocity 2.0 m/s in its original direction
  3. The calculation violates conservation of kinetic energy; this collision cannot be elastic with the given final velocity of the 2.0 kg block (correct answer)
  4. The calculation is incorrect because both momentum and energy conservation are violated; the collision scenario is physically impossible
Explanation: Check both conservation laws. Taking the 4.0 kg block's initial direction as positive: Initial momentum = (4.0)(6.0)+(2.0)(−3.0)=24−6=18 kg\cdotpm/s(4.0)(6.0) + (2.0)(-3.0) = 24 - 6 = 18\text{ kg·m/s}. Final momentum = (4.0)(1.0)+(2.0)(7.0)=4+14=18 kg\cdotpm/s(4.0)(1.0) + (2.0)(7.0) = 4 + 14 = 18\text{ kg·m/s}. Momentum is conserved. Initial KE = 12(4.0)(6.0)2+12(2.0)(−3.0)2=72+9=81 J\frac{1}{2}(4.0)(6.0)^2 + \frac{1}{2}(2.0)(-3.0)^2 = 72 + 9 = 81\text{ J}. Final KE = 12(4.0)(1.0)2+12(2.0)(7.0)2=2+49=51 J\frac{1}{2}(4.0)(1.0)^2 + \frac{1}{2}(2.0)(7.0)^2 = 2 + 49 = 51\text{ J}. Kinetic energy is not conserved (81 J ≠ 51 J), so this cannot be an elastic collision. The student's calculation satisfies momentum conservation but violates energy conservation for an elastic collision.

Question 8

A railroad car of mass MM moves at constant speed v0v_0 on a horizontal track. Rain falls vertically downward at a rate of λ\lambda kg/s into the car. Assuming the rainwater accumulates in the car and moves with it, what is the speed of the car as a function of time tt, considering that the horizontal component of momentum must be conserved?

  1. v(t)=v0(1−λtM)v(t) = v_0\left(1 - \frac{\lambda t}{M}\right) for small tt
  2. v(t)=Mv0M+λtv(t) = \frac{Mv_0}{M + \lambda t} for all t≥0t \geq 0 (correct answer)
  3. v(t)=v0−λtv0Mv(t) = v_0 - \frac{\lambda t v_0}{M} for small tt
  4. v(t)=v0MM+λtv(t) = v_0\sqrt{\frac{M}{M + \lambda t}} for all t≥0t \geq 0
Explanation: The rain falls vertically, so it has zero horizontal momentum. The horizontal momentum of the system (car + accumulated rain) must remain constant at Mv0Mv_0. At time tt, the total mass is M+λtM + \lambda t (original car mass plus accumulated rainwater). By conservation of horizontal momentum: (M+λt)v(t)=Mv0(M + \lambda t)v(t) = Mv_0, which gives v(t)=Mv0M+λtv(t) = \frac{Mv_0}{M + \lambda t}. Choice A is incorrect as it's only a linear approximation valid for small times. Choice C incorrectly treats this as uniform acceleration. Choice D incorrectly applies energy considerations instead of momentum conservation.