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: ^^^^^^^^^

Step-by-step explanation:

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Answer 2

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

Step-by-step 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 * 3.39=0.22 * \mathrm{V}_{\mathrm{f}}`

`(1.6272)/(0.22)=\mathrm{V}_{\mathrm{f}}`

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