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Chapter 7: Problem 82

A 3.0 -kg ball of clay with a speed of \(21 \mathrm{~m} / \mathrm{s}\) is thrown against a wall and sticks to the wall. What is the magnitude of the impulse exerted on the ball?

Short Answer

Expert verified
Answer: The magnitude of the impulse exerted on the ball is 63 kg m/s.

Step by step solution

01

Identify the given quantities

In this problem, we are given the mass (m) of the ball of clay as 3.0 kg and its initial velocity (v_initial) as 21 m/s. The ball sticks to the wall, which means that its final velocity (v_final) will be 0 m/s.
02

Calculate the initial momentum of the ball

The momentum is given by mass times velocity: \(momentum = m \times v\). To find the initial momentum, we plug in the values for mass and initial velocity: \(momentum_{initial} = m \times v_{initial}\) \(momentum_{initial} = 3.0 \thinspace kg \times 21 \thinspace m/s\) \(momentum_{initial} = 63 \thinspace kg \thinspace m/s\)
03

Calculate the final momentum of the ball

As the ball sticks to the wall, its final velocity is 0 m/s. Therefore, the final momentum will also be 0: \(momentum_{final} = m \times v_{final}\) \(momentum_{final} = 3.0 \thinspace kg \times 0 \thinspace m/s\) \(momentum_{final} = 0 \thinspace kg \thinspace m/s\)
04

Calculate the change in momentum

The change in momentum (or impulse) is the difference between the final and initial momentums: \(I = momentum_{final} - momentum_{initial}\) \(I = 0 \thinspace kg \thinspace m/s - 63 \thinspace kg \thinspace m/s\) \(I = -63 \thinspace kg \thinspace m/s\)
05

Find the magnitude of the impulse

The magnitude of the impulse is the absolute value of the impulse: \(|I| = |-63 \thinspace kg \thinspace m/s|\) \(|I| = 63 \thinspace kg \thinspace m/s\) The magnitude of the impulse exerted on the ball is 63 kg m/s.

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