Answer:
7.39 m/s is the velocity of the cue ball after impact.
Explanation:
Given that,
[tex]\text { Mass of the pool cue is } 0.48 \mathrm{kg} .\left(\mathrm{m}_{\mathrm{i}}\right)[/tex]
[tex]\text { Moving at velocity } 3.39 \mathrm{m} / \mathrm{s} .(\mathrm{v_i})[/tex]
[tex]\text { Mass of the cue ball that } 0.48 \text { mass ball hits is } 0.22 \mathrm{kg} .\left(\mathrm{m}_{\mathrm{f}}\right)[/tex]
[tex]\text { We need to find the velocity ( } \mathrm{v}_{\mathrm{f}} \text { ) of the cue ball of mass } 0.22 \mathrm{kg} \text { . }[/tex]
We know that,
If two objects collide then the “total momentum” before is equal to the “total momentum” after collision.
[tex]\mathrm{m}_{\mathrm{i}} \mathrm{v}_{\mathrm{i}}=\mathrm{m}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}[/tex]
Substitute the given values in the formula.
[tex]0.48 \times 3.39=0.22 \times \mathrm{V}_{\mathrm{f}}[/tex]
[tex]\frac{1.6272}{0.22}=\mathrm{V}_{\mathrm{f}}[/tex]
[tex]\mathrm{v}_{\mathrm{f}}=7.39 \mathrm{m} / \mathrm{s}[/tex]
