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Physics For Scientists & Engineers
Found in: Page 284
Physics For Scientists & Engineers

Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

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Short Answer

Review: After a 0.300−kg rubber ball is dropped from a height of 1.75 m, it bounces off a concrete floor and rebounds to a height of 1.50 m.

(a) Determine the magnitude and direction of the impulse delivered to the ball by the floor.

(b) Estimate the time the ball is in contact with the floor and use this estimate to calculate the average force the floor exerts on the ball.

(a) The magnitude and direction of the impulse delivered to the ball by the floor is3.38j^ kg⋅m/s

.

(b) The average force the floor exerts on the ball is 7×102j^  N .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Concept

The impulse imparted to a particle by a net forcecan be calculated as the time integral of the force

I→=∫titfF→dt

where

I→= Impulse delivered to object

F→= Average friction force

Δt= Time interval in seconds.

The impulse-momentum theorem can help us to find the connection between the change in the horizontal momentum of the object to the horizontal force acting on it.

ΔPx=FxΔt

where

ΔPx= Change in the horizontal momentum

Fx= Average friction force

Δt= Time interval in seconds.

Step 2:Stating given data

The data given are listed below as

  • Mass of the ball, m=0.300 kg
  • The ball dropped from a height h1=1.75  m and rebounded to a height h2=1.50 m .

Step 3: Magnitude and direction of the impulse delivered to the ball by the floor

Part (a):

The impulse the floor exerts on the ball is equal to the change in momentum of the ball.

The magnitude and direction of the impulse delivered to the ball by the floor can be calculated as

ΔP→=m(v→f−v→i)=m(vf−vi)j^

We know if the initial speed is zero, then the relationship between the final velocity and the height is

v=2gh

Then the final and initial velocity can be calculated as

vi=2gh1=2×9.8×1.75=5.856≈5.86 m/svf=2gh2=2×9.8×1.50=5.422≈5.42 m/s

Substitute the values in the above expression, we get

ΔP→=(0.300 kg)[(5.42 m/s)−(−5.86 m/s)]j^=3.38j^  kg⋅m/s

Thus, the magnitude and direction of the impulse delivered to the ball by the floor is 3.38j^ kg⋅m/s

Step 4: Estimation of the time and finding the average force

Part (b):

Estimating the contact time interval to be0.05 s .

So from the impulse-momentum theorem, we can calculate the average force as

F→=ΔP→Δt

Substitute the values in the above expression, we get

F→=3.38  kg⋅m/s0.05 sF→=7×102j^  N

Thus, the average force the floor exerts on the ball is7×102j^  N .

Most popular questions for Physics Textbooks

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The front 1.20 mof a 1400−kgcar is designed as a “crumple zone” that collapses to absorb the shock of a collision. If a car traveling 25.0 m/sstops uniformly in, 1.20 m

(a) how long does the collision last,

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A 10.0 −gbullet is fired into a 200 −gblock of wood at rest on a horizontal surface. After impact, the block slides8.0 m before coming to rest. If the coefficient of friction between the block and the surface is,0.400 what is the speed of the bullet before impact?

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