- #1

merzperson

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__SOLVED__## 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

U

_{g}=mgh

K=0.5mv

^{2}

P=mv

## The Attempt at a Solution

I first calculated the momentum of the steel block (block 2):

P=mv

P

_{2}=(.225)(1.85)

P

_{2}=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 (P

_{1}=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=0.5mv

^{2}

K=0.5(0.120)(3.46875)

^{2}

K=0.722J

Since energy is conserved, U

_{g}at the top of the ramp is equal to K at the bottom of the ramp (U

_{g}=0.722J). We can now find the height of the block:

U

_{g}=mgh

0.722=(0.120)(9.8)h

0.722=1.176h

h=0.6139m=61.39cm

**Where did I go wrong?**

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