Impulse and Momentum Question: Dont get what Im doing wrong

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A 64.9 g ball moving to the right at 11.1 m/s is struck by a bat causing it to move to the left with a speed of [y] m/s. If the bat and ball are in contact for 0.01683 s, what is the average force in N exerted by the ball on the bat?

Ok for this question Im getting -42.8, but thats wrong. First I find the impulse by using the formula p=mV and then I find the force by using the formula F=P/t.
 

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  • #2
quasar987
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A speed of [y] ? Do you have the real numerical value for that?

The answer is much simpler actually. It follows from the second law only.
 
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  • #3
quasar987
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Sorry for my first post, I hadnt' realized what you meant.

Then I'll agree with you that it's weird your way doesn't work. But try mine, does it give the same answer?

[tex]a_{av} = \frac{\Delta v}{\Delta t}[/tex]

and then

[tex]F_{av} = ma_{av}[/tex]
 
  • #4
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Yea, there are two ways to go about this problem. Your idea would work. However, you are making a small mistake. The impulse is not simply p, it is delta p. That is,
Impulse = p(final) - p(initial) = m (vf - vi).

Than you can go back to the idea of using F = I/t (I = impulse). I think you should review impulse, because that is a key idea.


As quasar said, you could also simply use newtons second law. Remember, the force the bat exerts on the ball is the same as the force the ball exerts on the bat (newtons third law).
 
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Parth Dave said:
Yea, there are two ways to go about this problem. Your idea would work. However, you are making a small mistake. The impulse is not simply p, it is delta p. That is,
Impulse = p(final) - p(initial) = m (vf - vi).

Than you can go back to the idea of using F = I/t (I = impulse). I think you should review impulse, because that is a key idea.


As quasar said, you could also simply use newtons second law. Remember, the force the bat exerts on the ball is the same as the force the ball exerts on the bat (newtons third law).
Ok I think I understand that, but how do I find the final velocity of the ball?

EDIT: Nevermind I put in -11 as the final velocity and ended up with the right answer. Thx for the help guys.
 
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quasar987
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Why is the final velocity -11 ?! :grumpy:
 
  • #7
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quasar987 said:
Why is the final velocity -11 ?! :grumpy:
Well I figured since the ball was travelling @ 11m/s and then got hit by the bat in the opposite direction it would be -11. I entered -11 into the formula and it worked :confused:
 
  • #8
Gokul43201
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The question, as stated in the OP, is clearly incomplete. You have to specify [y] to determine the applied force.

There is no reason for the speed of the rebounding ball to be fixed unless some additional constraint is imposed. If I swung the bat harder, surely I can make the ball go faster.

The easiest way to fix to question would be to say that the bat is stationary, and the collision is elastic. But none of this is sated in the question itself.

So, the questions just wrong - insufficient data.
 

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