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Chapter 7: Problem 4

A cue stick hits a cue ball with an average force of \(24 \mathrm{N}\) for a duration of \(0.028 \mathrm{s}\). If the mass of the ball is \(0.16 \mathrm{kg}\) how fast is it moving after being struck?

Short Answer

Expert verified
Answer: The final velocity of the cue ball is 4.2 m/s.

Step by step solution

01

Identify the given information and the formula we will use.

We are given the mass of the cue ball (m = 0.16 kg), the force exerted by the cue stick (F = 24 N), and the duration of the force (t = 0.028 s). We will use Newton's second law of motion, given by F = ma, to find the acceleration of the ball.
02

Calculate the acceleration.

To find the acceleration (a), we will rearrange the formula F = ma to solve for a: a = F / m Now, we can plug in the values for the force and mass: a = (24 N) / (0.16 kg) = 150 \(\frac{m}{s^2}\) So, the acceleration of the cue ball is 150 m/s².
03

Calculate the final velocity using the formula v = u + at.

To find the final velocity (v) of the cue ball, we will use the formula v = u + at, where u is the initial velocity, a is the acceleration, and t is the time duration. Since the cue ball is initially at rest, its initial velocity (u) is 0, and our formula becomes: v = 0 + (150 m/s²)(0.028 s) Now, we can multiply the acceleration by the time duration: v = (150 m/s²)(0.028 s) = 4.2 m/s So, the final velocity of the cue ball after being struck is 4.2 m/s.

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