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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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Most popular questions from this chapter

A 3.0-kg body is initially moving northward at \(15 \mathrm{m} / \mathrm{s}\) Then a force of \(15 \mathrm{N},\) toward the east, acts on it for 4.0 s. (a) At the end of the 4.0 s, what is the body's final velocity? (b) What is the change in momentum during the \(4.0 \mathrm{s} ?\)
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