Force exerted by a wall

  • Thread starter Nirupt
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  • #1
Nirupt
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Homework Statement



Above you had a ball travelling horizontally at 25.0 m/s bouncing off a brick wall and rebounding at 20.0 m/s. The ball was in contact with the wall for 3.50 ms, and you found the magnitude of the average acceleration during the time the ball was in contact with the wall. Now, what was the average force that the wall exerted on the ball during this collision, if the mass of the ball is 180

Homework Equations



Where it says above all I solved a question asking for acceleration which was 1428.57 m/s^2

The Attempt at a Solution


Not sure.. my only guess is that it goes with a free fall equation or an x not known since the ball was throw horizontally
 

Answers and Replies

  • #2
Simon Bridge
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Hint: Newton's laws.
 
  • #3
Astrum
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[tex]\vec{F}=\frac{d\vec{p}}{dt}[/tex], this is related to impulse.

Impulse is the change in momentum.
 
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  • #4
haruspex
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Where it says above all I solved a question asking for acceleration which was 1428.57 m/s^2
But that's the wrong answer for the scenario in the OP.
Presumably you arrived at that by dividing the change in velocity by the time taken, Δv/Δt.
Average force is defined similarly: change in momentum divided by time taken.
 
  • #5
Nirupt
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I probably should have posted my optional answers:

2.32 * 10^3 N

1.61 * 10^3 N

952 N

4.06 * 10^4 N


I'll post my attempt later tonight when I am home.
 
  • #6
haruspex
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I'll post my attempt later tonight when I am home.
But you need to get the average acceleration right first. it's nearly ten times the figure you posted.
 
  • #7
Nirupt
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1.43 * 10^3 was the answer for acceleration
 
  • #8
haruspex
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1.43 * 10^3 was the answer for acceleration
Are you saying that's the answer you got or that it is the correct answer? And are you saying it is the average acceleration for this scenario?:
a ball travelling horizontally at 25.0 m/s bouncing off a brick wall and rebounding at 20.0 m/s. The ball was in contact with the wall for 3.50 ms
(20+25)/.0035 = 1.29*104.
You appear to have calculated (25*2)/0.035.
 
  • #9
Nirupt
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Ah thanks for the correction. So.. this makes F = Ma easier to use correct?

1.29*10^4*180 = 2.32*10^3 N since mass and acceleration are known
 
  • #10
haruspex
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That's it.
 

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