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1.A cue ball (mass = 0.165 kg) is at rest on a frictionless pool table. The ball is hit dead center by a pool
stick, which applies an impulse of + 1.50 Ns to the ball. The ball then slides along the table and makes
an elastic head-on collision with a second ball of equal mass that is initially at rest. Find the velocity of
the second ball just after it is struck

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Answer:

Steps and solution in the attached pictureSteps and solution in the attached picture.

Explanation:

Steps and solution in the attached pictureSteps and solution in the attached picture.

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The velocity of the second ball just after it is struck will be 9.091 m/s.

What is an impulse?

To quantify the effect of force acting over time to change the momentum of an object is referred to as impulse.

To calculate the velocity in the given scenario,

Given that,

mV=1.5

[tex]V=\frac{impulse}{mass}[/tex]

[tex]V=\frac{1.5}{0.165}[/tex]

V=9.091 m/s

Thus, the velocity will be 9.091 m/s.

For more details regarding impulse, visit:

https://brainly.com/question/16980676

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