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More than just a momentum problem

  • Thread starter macgirl06
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  • #1
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When a bullet traveling at 721 m/s strikes a block of wood originally at rest on a frictionless surface, the bullet emerges from the other side of the block of wood traveling at 349 m/s. If the mass of the bullet is 5.38 g and the mass of the block is 744 g, what is the speed of the block after the collision?


I don't know what to do, I tried doing m1v1 + m2v2 = m1v1' + m2v2', and I couldn't get the answer. Any help would be greatly appreciated.

Homework Statement

 

Answers and Replies

  • #2
Doc Al
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I don't know what to do, I tried doing m1v1 + m2v2 = m1v1' + m2v2', and I couldn't get the answer.
But this is "just" a conservation of momentum problem, so that should work. Show exactly what you did. What values did you use?
 
  • #3
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I did the following:

(0.0058kg)(721 m/s) - (0.0058kg)(349m/s) \ 0.744 kg

and that didnt work

and thnks for the fast reply
 
  • #4
Doc Al
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44,892
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I did the following:

(0.0058kg)(721 m/s) - (0.0058kg)(349m/s) \ 0.744 kg
That looks perfectly OK to me:
Speed of block = [(0.0058kg)(721 m/s) - (0.0058kg)(349m/s)]/(0.744 kg)

Check your arithmetic; perhaps you made an error there.
 
  • #5
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I have tried that many many times, are you sure there arent any tricks to this question or something you have overlooked?
 
  • #6
Doc Al
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This is as straightforward a momentum conservation problem as you are likely to find. What answer did you get and why do you think it's wrong?
 
  • #7
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I got it, it was just a calculational error. Thanks for the clarification!
 

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