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Elastic Collision and Conservation of Energy

  • Thread starter merzperson
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Homework Statement



A 120g granite cube slides down a 40 degree frictionless ramp. At the bottom, just as it exits onto a horizontal table, it collides with a 225g steel cube at rest. How high above the table should the granite cube be released to give the steel cube a speed of 185cm/s?

Homework Equations



Ug=mgh
K=0.5mv2
P=mv

The Attempt at a Solution



I first calculated the momentum of the steel block (block 2):
P=mv
P2=(.225)(1.85)
P2=0.41625kg*m/s

Then, using conservation of momentum I deduced that the momentum of the granite block (block 1) before the collision must be equal to the momentum of the second block after the collision (P1=0.41625kg*m/s). Using this I can find the velocity of block 1 just before the collision:
P=mv
0.41625=(0.120)v
v=3.46875m/s

Now I can find the kinetic energy of block 1 right before it hit block 2:
K=0.5mv2
K=0.5(0.120)(3.46875)2
K=0.722J

Since energy is conserved, Ug at the top of the ramp is equal to K at the bottom of the ramp (Ug=0.722J). We can now find the height of the block:
Ug=mgh
0.722=(0.120)(9.8)h
0.722=1.176h
h=0.6139m=61.39cm

Where did I go wrong?
 
Last edited:

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