Question

A 0.48 kg pool cue moving at 3.39 m/s hits the 0.22 kg cue ball that was a rest and the pool cue stops after the impact. What is the velocity of the cue ball after the impact (in m/s)?

Answer 1

Answer:

7.39 m/s is the velocity of the cue ball after impact.

Explanation:

Given that,

$\text { Mass of the pool cue is } 0.48 \mathrm{kg} .\left(\mathrm{m}_{\mathrm{i}}\right)$

$\text { Moving at velocity } 3.39 \mathrm{m} / \mathrm{s} .(\mathrm{v_i})$

$\text { Mass of the cue ball that } 0.48 \text { mass ball hits is } 0.22 \mathrm{kg} .\left(\mathrm{m}_{\mathrm{f}}\right)$

$\text { We need to find the velocity ( } \mathrm{v}_{\mathrm{f}} \text { ) of the cue ball of mass } 0.22 \mathrm{kg} \text { . }$

We know that,  

If two objects collide then the “total momentum” before is equal to the “total momentum” after collision.  

$\mathrm{m}_{\mathrm{i}} \mathrm{v}_{\mathrm{i}}=\mathrm{m}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}$

Substitute the given values in the formula.

$0.48 \times 3.39=0.22 \times \mathrm{V}_{\mathrm{f}}$

$\frac{1.6272}{0.22}=\mathrm{V}_{\mathrm{f}}$

$\mathrm{v}_{\mathrm{f}}=7.39 \mathrm{m} / \mathrm{s}$