AP Physics Practice Questions: Momentum & Collisions

AP Physics multiple choice questions on momentum, impulse, and collisions. Each question includes detailed solutions at the bottom of the page.

1. A 2.0 kg ball moving at 3.0 m/s to the right collides head-on with a 4.0 kg ball moving at 2.0 m/s to the left. If the collision is perfectly elastic, what is the velocity of the 2.0 kg ball after collision?

   A) 4.33 m/s left
   B) 4.33 m/s right
   C) 3.67 m/s left
   D) 3.67 m/s right
   E) 5.00 m/s left

2. A 0.10 kg baseball is pitched at 40 m/s. The batter hits it straight back at 50 m/s. If the bat was in contact with the ball for 0.005 s, what was the average force exerted by the bat?

   A) 180 N
   B) 360 N
   C) 900 N
   D) 1800 N
   E) 3600 N

3. A 5.0 kg block moving at 6.0 m/s collides and sticks to a stationary 3.0 kg block. What percentage of the initial kinetic energy is lost in the collision?

   A) 25%
   B) 37.5%
   C) 50%
   D) 62.5%
   E) 75%

4. A 0.02 kg bullet is fired into a 3.0 kg ballistic pendulum block. The block rises 0.15 m. What was the bullet's initial speed?

   A) 120 m/s
   B) 150 m/s
   C) 180 m/s
   D) 259 m/s
   E) 240 m/s

5. Two ice skaters push off from each other. Skater A (mass 60 kg) moves at 2.0 m/s. If skater B (mass 90 kg) moves in the opposite direction, what is skater B's speed?

   A) 0.67 m/s
   B) 1.0 m/s
   C) 1.33 m/s
   D) 1.5 m/s
   E) 3.0 m/s

Solutions

1. C) 3.67 m/s left
   Let right be positive. For elastic collisions:
   \( v_{1f} = \dfrac{(m_1 - m_2)v_{1i} + 2m_2 v_{2i}}{m_1 + m_2} \)
   \( v_{1f} = \dfrac{(2 - 4) \times 3 + 2 \times 4 \times (-2)}{2 + 4} = \dfrac{(-2) \times 3 + 8 \times (-2)}{6} = \dfrac{-6 - 16}{6} = \dfrac{-22}{6} = -3.67 \, \text{m/s} \)
   Negative means left

2. D) 1800 N
   Impulse = change in momentum = \( F_{avg} \Delta t \)
   \( \Delta p = m(v_f - v_i) = 0.10(50 - (-40)) = 0.10 \times 90 = 9.0 \, \mathrm{kg·m/s} \)
   \( F_{avg} = \dfrac{\Delta p}{\Delta t} = \dfrac{9.0}{0.005} = 1800 \, \text{N} \)

3. B) 37.5%
   Using conservation of momentum for perfectly inelastic collision:
   \( m_1 v_1 = (m_1 + m_2) v_f \)
   \( 5.0 \times 6.0 = 8.0 \times v_f \) ⇒ \( v_f = 30/8 = 3.75 \, \text{m/s} \)
   Initial KE: \( \dfrac{1}{2} \times 5.0 \times 6.0^2 = 90 \, \text{J} \)
   Final KE: \( \dfrac{1}{2} \times 8.0 \times 3.75^2 = 56.25 \, \text{J} \)
   Loss: \( 90 - 56.25 = 33.75 \, \text{J} \)
   Percentage: \( (33.75/90) \times 100\% = 37.5\% \)
   

4. D) 259 m/s
   First, velocity of block+bullet after collision from energy conservation:
   \( \dfrac{1}{2}(M+m)v^2 = (M+m)gh \) ⇒ \( v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.15} = \sqrt{2.94} = 1.715 \, \text{m/s} \)
   Now, conservation of momentum for collision:
   \( m_b v_b = (M+m)v \) ⇒ \( 0.02 v_b = 3.02 \times 1.715 \)
   \( v_b = \dfrac{3.02 \times 1.715}{0.02} = \dfrac{5.179}{0.02} = 258.95 \, \approx 259 \text{m/s} \)
   

5. C) 1.33 m/s
   Conservation of momentum: \( 0 = m_A v_A + m_B v_B \)
   \( 0 = 60 \times 2.0 + 90 \times v_B \)
   \( v_B = -120/90 = -1.33 \, \text{m/s} \) (opposite direction)
   Magnitude: 1.33 m/s