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Mv=mv conservation of momentum

  • Thread starter Kchu
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  • #1
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i dont know how to solve this problem is it a conservation of momentum problem?

because cant you just use

mv=mv?
 

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  • #2
Doc Al
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What exactly is the problem?
 
  • #3
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oops! =]

**Two people are standing on a very light board that is balanced on a fulcrum. The lighter person suddenly jumps straight up at 1.5m/s
Just after he jumps, how fast will the heavier person be moving?**
 
  • #4
Doc Al
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Consider conservation of angular momentum.
 
  • #5
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mrv=mrv but it doesnt rotate?
 
  • #6
Doc Al
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Assuming you mean that to be [itex]m_1 r_1 v_1 = m_2 r_2 v_2[/itex], then that is correct. What makes you think it doesn't rotate? (If the board doesn't rotate, the second man would have speed = 0.)

A bit more explanation may help.

The total angular momentum of the system about the fulcrum is conserved. It starts out at zero before the man jumps and remains at zero after the man jumps. The system consists of both men and the board. (Since the board is light, we can ignore its mass and angular momentum.)

What is the angular momentum of the man after he jumps? Then what must be the angular momentum of the other man? (Don't forget that angular momentum has an orientation; think clockwise versus counterclockwise.) Use that to figure out the second man's speed.
 

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