Question

While playing pool, the white cue ball strikes the 8 ball. Before the collision, the cue ball's velocity was 10 m/s and the 8 ball was at rest. If both ball's have a mass of 0.15 kg and the cue ball's velocity after the collision is 2 m/s, what is the 8 ball's velocity?

Answer 1

Answer is given in the pic.

Answer 2

Answer:

The final velocity of the 8-ball after the collision, $v_{8}$, measured is $8m/s$.

Explanation:

Given,

The mass of the cue ball, $m_{c}$ = $0.15kg$

The mass of the 8-ball, $m_{8}$ = $0.15kg$

Before collision,

The cue ball's initial velocity, $u_{c}$ = $10m/s$

The 8-ball's initial velocity, $u_{8}$ = $0m/s$ ( at rest )

After collision,

The cue ball's final velocity, $v_{c}$ = $2m/s$

The 8-ball's final velocity, $v_{8}$ =?

As we know, by the law of conservation of momentum:

• Total momentum before the collision = total momentum after the collision

Therefore,

• $m_{c} u_{c} + m_{8}u_{8} = m_{c} v_{c} + m_{8}v_{8}$

• $0.15\times 10 + 0.15\times 0 = 0.15\times 2 + 0.15\times v_{8}$
• $0.15v_{8} = 1.5 -0.3$
• $0.15v_{8} = 1.2$
• $v_{8}$ = $8m/s$

Hence, the final velocity of the 8-ball after the collision, $v_{8}$ = $8m/s$.