# 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.

Dec 20, 2015
1. Cue ball momentum after = $0.035 k g . m . {s}^{- 1}$
2. 5-Ball’s KE after = $0.80 m . {s}^{- 1}$
3. Cue ball momentum before = $0.20 k g . m . {s}^{- 1}$
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: ${m}_{c} = \frac{23}{1000} = 0.023 g$
5 ball: ${m}_{5} = \frac{16}{1000} = 0.016 g$

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
${p}_{c 1}$ 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)