Momentum Ques.

  • Thread starter robbondo
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  • #1
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Homework Statement


A ball with mass m with an initial velocity of 5 m/s strikes a ball with a mass of 3m hanging at rest from a string 50 cm long. Find the maximum angle(x) with which the block swings after its hit.


Homework Equations


k=1/2mv^2
p=mv


The Attempt at a Solution


Well first I solved the momentum

mAvA1+mBvB1=mAvA2+mBvB2

to get vB2=5-vA2

Then I used conservation of energy to get

[tex]sqrt {25 - v_{A2}/3} [\tex]

I solved the two equations for vB2 and got 3.53 m/s

So, since the kinetic energy from the start of the mass on the pendelum moving to its peak is h=(.5-.5cosx)

1/2m(vB2^2)=mg(.5-.5cosx)

I solved for the angle and got 50.2 degrees.

Anyone see anything wrong with my math or my logic? I can only attempt the problem one more time before the program gives me no credit.
 

Answers and Replies

  • #2
radou
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Couldn't you just use energy conservation, since the block's kinetic energy at the maximal angle equals zero? It posesses only potential energy at that point, and at the impact point, there is only kinetic energy from the ball.
 
  • #3
Dick
Science Advisor
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Couldn't you just use energy conservation, since the block's kinetic energy at the maximal angle equals zero? It posesses only potential energy at that point, and at the impact point, there is only kinetic energy from the ball.
The other ball carries away some of the kinetic energy. I think he's doing it right. I haven't checked the answer though.
 
  • #4
radou
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The other ball carries away some of the kinetic energy. I think he's doing it right. I haven't checked the answer though.
Good point, so after reconsidering...

Well first I solved the momentum

mAvA1+mBvB1=mAvA2+mBvB2

to get vB2=5-vA2
...shouldn't this be VB2 = (5 - VA2)/3 ?
 

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