Question

What is the cue ball's momentum after the collision? What is the 5-ball's kinetic energy after the collision? What was the cue ball's momentum before the collision? What was the speed of the cue ball before the collision?

A cue ball of mass 23 g hits the 5-ball, which has a mass of 16 g. After the collision, the 5-ball has a speed of 10.03 m/s while the cue ball moves at 1.53 m/s.

Answer 1

1. Cue ball momentum after = `0.035 kg.m.s^(-1)`
2. 5-Ball’s KE after = `0.80 m.s^(-1)`
3. Cue ball momentum before = `0.20 kg.m.s^(-1)`
4. Cue ball speed before = `8.5 m.s^(-1)`

Explanation:

Terms:
`u` is an initial velocity
`v` is a final velocity
`m` is a mass
`p` is a momentum
Subscript `c` denotes a property of the cue ball
(e.g. `m_c` is the mass of the cue ball.)
Subscript `5` denotes a property of the 5-ball

Convert mass values to kg:
Cue ball: `m_c= 23/1000 = 0.023 g`
5 ball: `m_5=16/1000 = 0.016 g`

1. Cue ball’s momentum after the collision
   `p_(c2) = m_c × v_c = 0.023 × 1.53 = 0.03519`
   `⇒ p_(c2) = 0.035 kg.m.s^(-1)`

2. 5-Ball’s kinetic energy after the collision
   `E_k=½ m v^2 = ½ × 0.016 × 10.03^2 = 0.8048072`
   `⇒ E_k = 0.80 J`

3. Use conservation of momentum to calculate the momentum of the cue ball before the collision
   Total momentum before = total momentum after
   Note that I am assuming that the 5-ball was stationary before the collision (the question does not state this). I am also assuming that the two balls move off in the same direction.
   `p_(c1) + 0 = p_(c2) + m_5 × v_5`
   `p_(c1)` is the cue ball’s momentum before the collision.
   `p_(c1) = 0.03519 + 0.016 × 10.03 = 0.19567`
   `⇒ p_(c1) = 0.20 kg.m.s^(-1)`

4. Calculate the speed of the cue ball before the collision
   `p_(c1) = m_c × u_c ⇒ u_c = p_(c1) / m_c = 0.19567 / 0.023 = 8.5073913`
   `⇒ u_(c) = 8.5 m.s^(-1)`