AP Physics C: Mechanics Linear Momentum — Worked Answer Explanations

Unit 4 · 12 questions explained

Below is a complete answer key for our AP Physics C: Mechanics 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 0.5 kg ball moving at $4$ m/s collides with a wall and bounces back at $4$ m/s. What is the magnitude of the impulse on the ball?
- A
  $0$ N·s
  Why not A: Added momenta algebraically without accounting for direction reversal.
- B
  $2$ N·s
  Why not B: Used only $m ∣ v ∣ = 0.5 \times 4 = 2$ as the change.
- C
  $4$ N·sCorrect
- D
  $8$ N·s
  Why not D: Doubled the final momentum without subtracting the initial.
Explanation
Taking toward the wall as positive: $p_{i} = + 2$ kg·m/s, $p_{f} = - 2$ kg·m/s. Impulse $J = Δ p = p_{f} - p_{i} = - 2 - 2 = - 4$ N·s. Magnitude $= 4$ N·s.

Key takeaway
Impulse equals change in momentum. When a ball reverses direction, the speeds add: $|J| = m(v_i + v_f)$.
Question 2 · Easy
Two ice skaters (60 kg and 40 kg) push off each other from rest. If the 40 kg skater moves at $3$ m/s to the right, what is the velocity of the 60 kg skater?
- A
  $3$ m/s to the left
  Why not A: Assumed equal speeds, ignoring different masses.
- B
  $2$ m/s to the leftCorrect
- C
  $4.5$ m/s to the left
  Why not C: Inverted the mass ratio.
- D
  $2$ m/s to the right
  Why not D: Correct magnitude but wrong direction (conservation requires opposite directions).
Explanation
System starts at rest: $p_{t o t a l} = 0$ . After push: $0 = 60 v + 40 (3) \Rightarrow v = - 2$ m/s, i.e., $2$ m/s to the left.

Key takeaway
Conservation of momentum: if the system starts at rest, $m_1 v_1 = -m_2 v_2$. The heavier skater moves slower.
Question 3 · Easy
A constant force $F = 8$ N acts on a 2 kg object for 3 s. What is the change in velocity?
- A
  $4$ m/s
  Why not A: Computed $F / m = 4$ but forgot to multiply by $Δ t$ .
- B
  $12$ m/sCorrect
- C
  $24$ m/s
  Why not C: Multiplied $F \times Δ t \times m$ instead of dividing by $m$ .
- D
  $3$ m/s
  Why not D: Set $Δ v = Δ t$ in m/s without using force or mass.
Explanation
Impulse $J = F Δ t = 8 \times 3 = 24$ N·s $= Δ p = m Δ v$ . So $Δ v = J / m = 24/2 = 12$ m/s.

Key takeaway
Impulse–momentum theorem: $J = F\Delta t = m\Delta v$. Divide total impulse by mass to get velocity change.
Question 4 · Easy
A 1 kg ball moving at $6$ m/s to the right undergoes a perfectly inelastic collision with a stationary 2 kg ball. What is their combined velocity after the collision?
- A
  $6$ m/s
  Why not A: Ignored the mass of the second ball.
- B
  $3$ m/s
  Why not B: Divided initial momentum by initial mass only.
- C
  $2$ m/sCorrect
- D
  $1$ m/s
  Why not D: Used total mass $= 3$ but initial momentum $= 1 \times 6/2$ (wrong).
Explanation
Conservation of momentum: $p_{i} = 1 \times 6 = 6$ kg·m/s. After collision: $(1 + 2) v = 6 \Rightarrow v = 2$ m/s.

Key takeaway
Perfectly inelastic: objects stick together. Apply $m_1 v_1 = (m_1+m_2)v_f$.
Question 5 · Medium
The impulse–momentum theorem in integral form states $\int_{t_{1}}^{t_{2}} F (t) d t = Δ p$ . A force $F (t) = 6 t^{2}$ N acts on a 2 kg object from $t = 0$ to $t = 2$ s starting from rest. Find the final speed.
- A
  $4$ m/s
  Why not A: Evaluated $F (2) / m = 24/2$ without integrating over time.
- B
  $6$ m/s
  Why not B: Used $F (2) \times t / m$ as a shortcut instead of the time integral.
- C
  $8$ m/sCorrect
- D
  $16$ m/s
  Why not D: Computed the impulse correctly ( $16$ N·s) but forgot to divide by $m$ .
Explanation
Impulse: $J = \int_{0}^{2} 6 t^{2} d t = [2 t^{3}]_{0}^{2} = 16 - 0 = 16$ N·s. By the impulse–momentum theorem: $J = m Δ v$ , so $Δ v = 16/2 = 8$ m/s. Starting from rest, $v_{f} = 8$ m/s.

Key takeaway
The integral impulse $\int F(t)\,dt$ equals the change in momentum. Divide by mass to find $\Delta v$.
Question 6 · Medium
A 2 kg ball moving at $5$ m/s to the right collides elastically with a 2 kg ball at rest. What are their velocities after the collision?
- A
  Both move at $2.5$ m/s
  Why not A: Average speed result — applies to inelastic, not elastic.
- B
  First stops; second moves at $5$ m/sCorrect
- C
  First moves at $5$ m/s; second stops
  Why not C: Reversed the two balls' outcomes.
- D
  First moves at $- 2.5$ m/s; second at $7.5$ m/s
  Why not D: Result for unequal masses; equal masses exchange velocities.
Explanation
For an elastic collision between equal masses: the moving ball stops and the stationary one takes its velocity. This follows from simultaneous conservation of momentum ( $m v_{1} = m v_{1}^{'} + m v_{2}^{'}$ ) and kinetic energy ( $v_{1}^{2} = v_{1}^{'2} + v_{2}^{'2}$ ). With $m_{1} = m_{2}$ : $v_{1}^{'} = 0$ , $v_{2}^{'} = v_{1} = 5$ m/s.

Key takeaway
Equal-mass elastic collision: balls exchange velocities. The moving ball stops; the stationary one moves at the original speed.
Question 7 · Medium
Three particles have masses and positions: $m_{1} = 2$ kg at $x_{1} = 0$ m, $m_{2} = 3$ kg at $x_{2} = 4$ m, and $m_{3} = 1$ kg at $x_{3} = 6$ m. Find the center of mass.
- A
  $\frac{10}{3} \approx 3.33$ m
  Why not A: Took the unweighted average of positions: $(0 + 4 + 6) /3 = 10/3$ .
- B
  $3$ mCorrect
- C
  $2$ m
  Why not C: Weighted by particle number rather than mass.
- D
  $4$ m
  Why not D: Used only $m_{2} x_{2} / m_{t o t a l}$ and neglected other contributions.
Explanation
$x_{c m} = \frac{m _{1} x _{1} + m _{2} x _{2} + m _{3} x _{3}}{m _{1} + m _{2} + m _{3}} = \frac{2 ( 0 ) + 3 ( 4 ) + 1 ( 6 )}{2 + 3 + 1} = \frac{0 + 12 + 6}{6} = \frac{18}{6} = 3$ m.

Key takeaway
$x_{cm} = \dfrac{\sum m_i x_i}{\sum m_i}$ — mass-weighted average of positions.
Question 8 · Medium
Two particles with masses $m_{1} = 4$ kg at $x = 0$ and $m_{2} = 2$ kg at $x = 6$ m. Where is the center of mass?
- A
  $3$ m
  Why not A: Took the midpoint without weighting by mass.
- B
  $2$ mCorrect
- C
  $4$ m
  Why not C: Weighted toward the lighter mass instead of the heavier one.
- D
  $1$ m
  Why not D: Used $(m_{2} / m_{t o t a l}) \times x_{2} = (2/6) \times 6 = 2$ but placed it at 1 m due to arithmetic error.
Explanation
$x_{c m} = \frac{4 ( 0 ) + 2 ( 6 )}{4 + 2} = \frac{12}{6} = 2$ m. The center of mass is closer to the heavier mass at $x = 0$ .

Key takeaway
The center of mass lies closer to the more massive object. $x_{cm} = \frac{\sum m_i x_i}{M}$.
Question 9 · Medium
A 3 kg object moving at $8$ m/s to the right undergoes a perfectly inelastic collision with a 5 kg object moving at $4$ m/s to the left. Find their common velocity after the collision.
- A
  $6$ m/s to the right
  Why not A: Added speeds without accounting for opposite directions.
- B
  $2$ m/s to the right
  Why not B: Divided net momentum by only $m_{1} = 3$ kg.
- C
  $0.5$ m/s to the rightCorrect
- D
  $1$ m/s to the left
  Why not D: Sign error; net momentum is to the right.
Explanation
Take right as positive. $p_{i} = 3 (8) + 5 (- 4) = 24 - 20 = 4$ kg·m/s. After collision: $(3 + 5) v = 4 \Rightarrow v = 4/8 = 0.5$ m/s to the right.

Key takeaway
In a perfectly inelastic collision, $\sum m_i v_i = (\sum m_i) v_f$. Assign consistent signs; the net momentum determines both magnitude and direction.
Question 10 · Hard
A 0.1 kg bullet traveling at $400$ m/s embeds in a 4.9 kg stationary block on a frictionless surface. A spring ( $k = 1000$ N/m) is in front of the block. What is the maximum spring compression?
- A
  $0.2$ m
  Why not A: Used the bullet's initial $K E$ instead of the post-collision $K E$ .
- B
  $0.4$ m
  Why not B: Forgot to square the velocity before applying $\frac{1}{2} m v^{2}$ .
- C
  $0.8$ m
  Why not C: Used total mass but bullet's original speed in energy.
- D
  $0.566$ mCorrect
Explanation
Step 1 — Collision (momentum conserved, KE not): $v_{f} = \frac{0.1 ( 400 )}{0.1 + 4.9} = \frac{40}{5} = 8$ m/s.

Step 2 — Spring compression (energy conserved): $\frac{1}{2} (5) (64) = \frac{1}{2} (1000) x^{2} \Rightarrow x^{2} = \frac{320}{1000} = 0.32 \Rightarrow x = 0.32 = 0.566$ m.

Key takeaway
Bullet-block + spring: two separate conservation laws. Use momentum for the collision (step 1), then energy for the spring compression (step 2).
Question 11 · Hard
Using the integral form of the impulse–momentum theorem, a force $F (t) = F_{0} e^{- t / τ}$ acts on mass $m$ from $t = 0$ to $t = \infty$ starting from rest. Find $v (\infty)$ .
- A
  $\frac{F _{0}}{m τ}$
  Why not A: Divided $F_{0}$ by $m τ$ instead of $F_{0} τ / m$ .
- B
  $\frac{F _{0} τ}{m}$ Correct
- C
  $\frac{F _{0}}{m}$
  Why not C: Forgot the factor $τ$ from the integral of $e^{- t / τ}$ .
- D
  $F_{0} m τ$
  Why not D: Multiplied rather than divided by $m$ .
Explanation
$J = \int_{0}^{\infty} F_{0} e^{- t / τ} d t = F_{0} [- τ e^{- t / τ}]_{0}^{\infty} = F_{0} (0 - (- τ)) = F_{0} τ$ .

By impulse–momentum theorem: $v_{\infty} = J / m = \frac{F _{0} τ}{m}$ .

Key takeaway
$\int_0^\infty e^{-t/\tau}\,dt = \tau$. The exponentially decaying force imparts finite momentum $J = F_0\tau$, giving terminal velocity $F_0\tau/m$.
Question 12 · Hard
In a one-dimensional elastic collision between mass $m_{1}$ (initial speed $v_{0}$ ) and stationary mass $m_{2}$ , the final velocities are $v_{1}^{'} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} v_{0}$ and $v_{2}^{'} = \frac{2 m _{1}}{m _{1} + m _{2}} v_{0}$ . Verify these satisfy energy conservation symbolically.
- A
  Only $v_{2}^{'}$ conserves energy; $v_{1}^{'}$ does not.
  Why not A: Confused momentum with energy conservation conditions.
- B
  Neither velocity satisfies energy conservation.
  Why not B: Error in algebraic verification.
- C
  Both velocities together conserve momentum but not energy.
  Why not C: Elastic collision by definition conserves both — checked only momentum.
- D
  Both velocities together conserve both momentum and kinetic energy.Correct
Explanation
Momentum: $m_{1} v_{1}^{'} + m_{2} v_{2}^{'} = m_{1} \frac{m _{1} - m _{2}}{m _{1} + m _{2}} v_{0} + m_{2} \frac{2 m _{1}}{m _{1} + m _{2}} v_{0} = \frac{m _{1} v _{0} [( m _{1} - m _{2} ) + 2 m _{2} ]}{m _{1} + m _{2}} = \frac{m _{1} v _{0} ( m _{1} + m _{2} )}{m _{1} + m _{2}} = m_{1} v_{0}$ . ✓

Energy: $\frac{1}{2} m_{1} v_{1}^{'2} + \frac{1}{2} m_{2} v_{2}^{'2} = \frac{m _{1} ( m _{1} - m _{2} ) ^{2} v _{0}^{2}}{2 ( m _{1} + m _{2} ) ^{2}} + \frac{4 m _{1}^{2} m _{2} v _{0}^{2}}{2 ( m _{1} + m _{2} ) ^{2}}$
$= \frac{m _{1} v _{0}^{2}}{2 ( m _{1} + m _{2} ) ^{2}} [(m_{1} - m_{2})^{2} + 4 m_{1} m_{2}] = \frac{m _{1} v _{0}^{2} ( m _{1} + m _{2} ) ^{2}}{2 ( m _{1} + m _{2} ) ^{2}} = \frac{1}{2} m_{1} v_{0}^{2}$ . ✓

Key takeaway
The elastic collision formulas are derived by simultaneously solving $p = p'$ and $KE = KE'$. Verifying both algebraically is a key Physics C skill.