   Chapter 6, Problem 11P

Chapter
Section
Textbook Problem

A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.960 m. What impulse was given to the ball by the floor?

To determine
The impulse given to the ball by the floor.

Explanation

Given Info:

The height from where the ball is dropped is 1.25m ., the height that the ball rebounded is 0.960m

Explanation:

Formula to calculate the velocity of the ball just before the impact is,

Vb=vi2+2aΔh

• Δh is the change in height of the ball
• vi is the initial velocity of the ball
• a is the acceleration of the ball

Since, the ball is in a free fall; the magnitude of the acceleration of the ball will be equal to the acceleration due to gravity. The ball is at rest initially. Thus, the initial velocity of the ball is zero.

Substitute zero for vi , 9.8m/s2 for a and 1.25m for Δh to find the velocity of the ball just before the impact,

vi=0+2(9.8m/s2)(1.25m)=4.95m/s

Thus, the velocity of the ball just before the impact of the ball with the floor is 4.95m/s .

Formula to calculate the rebound velocity of the ball is,

Vr=vf22aΔh

• Δh is the change in height of the ball
• vf is the final velocity of the ball after the rebound
• a is the acceleration of the ball

Since, the ball is rebounding; the magnitude of the acceleration of the ball will be equal to the acceleration due to gravity. The ball is at rest finally. Thus, the final velocity of the ball is zero.

Substitute 0 for vf , 9.8m/s2 for a and 0.960m for Δh to find the rebound velocity of the ball,

vi=0+2(9

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