AP Physics 1 Linear Momentum — Worked Answer Explanations

Unit 4 · 12% of the AP exam · 8 questions explained

Below is a complete answer key for our AP Physics 1 Linear Momentum practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Linear Momentum practice test and come back here to review, or head back to the Linear Momentum unit overview.

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Question 1 · Easy
A force-vs-time graph shows a constant force of $8 N$ acting on an object for $0.5 s$ , then $0 N$ for the remaining time. What is the impulse delivered during the entire force application?
- A
  $0.5 N\cdots$
  Why not A: Used time only.
- B
  $4.0 N\cdots$ Correct
- C
  $8.0 N\cdots$
  Why not C: Used force only without integrating over time.
- D
  $16 N\cdots$
  Why not D: Doubled the area under the curve.
Explanation
Impulse equals the area under the $F$ -vs- $t$ curve: $J = (8 N) (0.5 s) = 4.0 N\cdots$ .

Key takeaway
Impulse is the area under the force-vs-time graph.
Question 2 · Easy
A $0.20 kg$ baseball moving at $40 m/s$ is caught by a fielder. What is the magnitude of the impulse delivered to the ball during the catch?
- A
  $0.20 N\cdots$
  Why not A: Used mass without multiplying by velocity.
- B
  $2.0 N\cdots$
  Why not B: Off by a factor of 4.
- C
  $8.0 N\cdots$ Correct
- D
  $160 N\cdots$
  Why not D: Used $\frac{1}{2} m v^{2}$ instead of $m v$ .
Explanation
Impulse magnitude equals change in momentum: $∣ J ∣ = ∣ m Δ v ∣ = 0.20 \times 40 = 8.0 N\cdots$ .

Key takeaway
Impulse equals change in momentum: $J = \Delta p$.
Question 3 · Easy
A $1.0 kg$ cart moving at $4.0 m/s$ collides and sticks to a $3.0 kg$ cart at rest. What is the speed of the combined carts after the collision?
- A
  $0.5 m/s$
  Why not A: Off by a factor of 2.
- B
  $1.0 m/s$ Correct
- C
  $2.0 m/s$
  Why not C: Averaged the speeds without weighting by mass.
- D
  $4.0 m/s$
  Why not D: Forgot to add the second mass to the denominator.
Explanation
Conservation of momentum: $(1.0) (4.0) + (3.0) (0) = (1.0 + 3.0) v_{f}$ , so $v_{f} = 1.0 m/s$ .

Key takeaway
In a perfectly inelastic collision, $v_f = \dfrac{p_{total,initial}}{m_{total}}$.
Question 4 · Easy
Which of the following collisions conserves both momentum AND kinetic energy?
- A
  A car crashing into a wall and crumpling.
  Why not A: Crumple zones absorb energy — inelastic.
- B
  Two billiard balls bouncing off each other.Correct
- C
  A bullet embedding in a block of wood.
  Why not C: Perfectly inelastic — KE lost to deformation/heat.
- D
  A meteor sticking to a planet on impact.
  Why not D: Perfectly inelastic.
Explanation
Billiard ball collisions are nearly elastic — both momentum and KE are conserved. The other choices are all inelastic (they conserve momentum but not KE).

Key takeaway
All isolated collisions conserve momentum; only elastic collisions also conserve kinetic energy.
Question 5 · Medium
An astronaut of mass $80 kg$ floating at rest in space throws a $2.0 kg$ wrench at $10 m/s$ . What is the recoil speed of the astronaut?
- A
  $0.10 m/s$
  Why not A: Off by a factor of 2.5.
- B
  $0.25 m/s$ Correct
- C
  $2.0 m/s$
  Why not C: Used wrench mass instead of astronaut mass.
- D
  $10 m/s$
  Why not D: Same as wrench speed; missed momentum conservation.
Explanation
Total initial momentum is zero. After the throw: $m_{a} v_{a} = m_{w} v_{w}$ , so $v_{a} = (2.0) (10) /80 = 0.25 m/s$ in the opposite direction.

Key takeaway
From rest, recoil speed scales as $v_{recoil} = m v / M$.
Question 6 · Medium
A $0.10 kg$ ball hits a wall at $20 m/s$ and rebounds at $15 m/s$ (opposite direction). The collision lasts $0.020 s$ . What is the average force the wall exerts on the ball?
- A
  $25 N$
  Why not A: Subtracted speeds instead of adding them.
- B
  $100 N$
  Why not B: Off by a factor of 1.75.
- C
  $175 N$ Correct
- D
  $350 N$
  Why not D: Doubled the answer.
Explanation
$Δ p = m (v_{f} - v_{i}) = 0.10 (- 15 - 20) = - 3.5 N\cdots$ . Force magnitude: $∣ F ∣ = ∣Δ p ∣/Δ t = 3.5/0.020 = 175 N$ .

Key takeaway
When direction reverses, $|\Delta v| = |v_i| + |v_f|$, not $|v_f| - |v_i|$.
Question 7 · Medium
Cart A ( $2 kg$ , $3 m/s$ east) collides with Cart B ( $4 kg$ , $1 m/s$ west) on a frictionless track. After they stick together, what is the velocity of the combined carts?
- A
  $0.33 m/s$ eastCorrect
- B
  $0.33 m/s$ west
  Why not B: Sign error in choosing the east direction as positive.
- C
  $1.0 m/s$ east
  Why not C: Used only Cart A's momentum.
- D
  $2.0 m/s$ east
  Why not D: Forgot to divide by total mass.
Explanation
Take east positive: $p_{i} = (2) (3) + (4) (- 1) = 2 kg\cdotm/s$ east. After: $v_{f} = 2/ (2 + 4) \approx 0.33 m/s$ east.

Key takeaway
Pick a positive direction first; momentum is a vector with sign.
Question 8 · Hard
Two balls of equal mass collide elastically head-on. Ball A moves at $+ 5 m/s$ , Ball B at $- 3 m/s$ . What are their velocities after the collision?
- A
  Both balls have velocity $+ 1 m/s$ .
  Why not A: Inelastic-collision answer.
- B
  $v_{A} = - 3 m/s$ and $v_{B} = + 5 m/s$ .Correct
- C
  $v_{A} = + 3 m/s$ and $v_{B} = + 5 m/s$ .
  Why not C: Sign-error on Ball A's velocity.
- D
  $v_{A} = - 5 m/s$ and $v_{B} = + 3 m/s$ .
  Why not D: Negated both velocities; only the swap is correct.
Explanation
For an elastic 1D collision between equal masses, the velocities are exchanged: Ball A takes Ball B's prior velocity and vice versa.

Key takeaway
Equal-mass elastic collisions swap velocities.