COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
2nd Edition
ISBN: 9781319414597
Author: Freedman
Publisher: MAC HIGHER
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Chapter 7, Problem 66QAP
To determine

(a)

The magnitude and direction of the impulse given to the ball by the bat, if a baseball bat strikes a ball when both are moving at 31.3 m/s (relative to the ground) toward each other. The bat and ball are in contact for 1.20 ms, after which the ball is traveling at a speed of 42.5 m/s, the mass of the bat and the ball are 850 g and 145 g, respectively?

Expert Solution
Check Mark

Answer to Problem 66QAP

The magnitude and direction of the impulse given to the ball by the bat = 10.7 kg m/s along positive x axis.

Explanation of Solution

Given info:

  mbat=850g=0.85kgmball=145g=0.145kgvball,fx=42.5m/svball,ix=31.3m/s

Formula used:

Calculation:

Impulse given to the ball,

  mballΔvball,x=mball(vball,fxvball,ix)=0.145(42.5(31.3))=10.7kgm/s

Conclusion:

The magnitude and direction of the impulse given to the ball by the bat = 10.7 kg m/s along positive x axis.

To determine

(b)

The magnitude and direction of the impulse given to the bat by the ball?

Expert Solution
Check Mark

Answer to Problem 66QAP

The magnitude and direction of the impulse given to the bat by the ball = 10.7 kg m/s along negative x axis.

Explanation of Solution

Given info:

  mbat=850g=0.85kgmball=145g=0.145kgvball,fx=42.5m/svball,ix=31.3m/s

Formula used:

Calculation:

By conservation of momentum equation, we have,

  mbatvbat,ix+mballvball,ix=mbatvbat,fx+mballvball,fxvbat,fx=mbatvbat,ix+mballvball,ixmballvball,fxmbatvbat,fx=0.85×31.3+0.145×(31.3)0.145×42.50.85=18.7m/s

Impulse given to the bat,

  mbatΔvbat,x=mbat(vbat,fxvbat,ix)=0.85(18.7(31.3))=10.7kgm/s

Conclusion:

The magnitude and direction of the impulse given to the bat by the ball = 10.7 kg m/s along negative x axis.

To determine

(c)

What average force does the bat exert on the ball?

Expert Solution
Check Mark

Answer to Problem 66QAP

The average force of the bat on the ball is 8.92×103N in the positive x direction.

Explanation of Solution

Given info:

  mbat=850g=0.85kgmball=145g=0.145kgvball,fx=42.5m/svball,ix=31.3m/sΔt=1.20ms=1.2×103s

Formula used:

Force,

F = Impulse/Time

Calculation:

Average force,

  Favg,x=ΔpΔt=10.71.2×103=8.92×103N

Conclusion:

The average force of the bat on the ball is 8.92×103N in the positive x direction.

To determine

(d)

Why doesn't the force shatter the bat?

Expert Solution
Check Mark

Answer to Problem 66QAP

The force value is 8.92×103N, which is very high. But the contact time is 1.2 ms which is very low. Because of that the bat does not shatter.

Explanation of Solution

Given info:

The average force of the bat on the ball is 8.92×103N in the positive x direction.

Calculation:

The force value is 8.92×103N, which is very high. But the contact time is 1.2 ms which is very low. Because of that the bat does not shatter.

Conclusion:

The force value is 8.92×103N, which is very high. But the contact time is 1.2 ms which is very low. Because of that the bat does not shatter.

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COLLEGE PHYSICS LL W/ 6 MONTH ACCESS

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Impulse Derivation and Demonstration; Author: Flipping Physics;https://www.youtube.com/watch?v=9rwkTnTOB0s;License: Standard YouTube License, CC-BY