Question

A 105 g cue ball is traveling to the right at 12.5 m/s. Meanwhile, a 104 g 8-ball is moving at 13.4 m/s to the left. After the collision, the cue ball is moving off to the left at a velocity of 8.9 m/s. What is the magnitude and direction of the final velocity of the 8-ball?

Answer 1

The velocity of the 8-ball after collision is
`v_2 = 8.12 \ m/s` and the direction is towards that right

Step-by-step explanation:

From the question we are told that

The mass of the cue ball is
`m_1  =  105 \  g`

The speed of the cue ball is
`u_1 =  12.5 \  m/s`

The mass of the 8-ball is
`m_2  =  104 \  g`

The velocity of the 8 ball is
`u_2 =  - 13.4 \  m/s`

The negative sign is because it is moving left

The velocity of the cue ball after collision is
`v_1 =  -8.9 \  m/s`

Generally from the law of momentum conservation

`m_1 *  u_1  +  m_2 *  u_2  =  m_1 * v_1  +  m_2 *  v_2`

`105 *  12.5  +  104 *  - 13.4 =   104* -8.9 +  104*  v_2`

=>
`-81.1 = -925.6 + 104v_2`

=>
`104v_2 = 844.5`

=>
`v_2 = 8.12 \ m/s`