# Cart Momentum Question

1. Oct 12, 2008

### MightyMan11

1. The problem statement, all variables and given/known data
A 478.0 g cart is released from rest 1.16 m from the bottom of a frictionless, 31.0° ramp. The cart rolls down the ramp and undergoes a collision with a rubber block at the bottom. After the cart bounces, how far does it roll back up the ramp if the maximum force applied is 302.0 N for 22.9 ms?

2. Relevant equations

pf=pi+impulse
p=m*v

3. The attempt at a solution
First I found my initial impulse:
Vf2=0+2(9.8cos31)(1.16)
Vf=4.41m/s
but since its moving in the negative direction, it is -4.41m/s

p = m*v
p = (0.478)(-4.41)
=-2.11

The I calculate my impulse (the area under the graph):
0.5(302)(0.0229)
=3.46 kg m/s

Then I calculated my final impulse:
pf=pi+impulse
pf=-2.11+3.46
=1.352

I then found the distance it travels up after it gains that momentum from hitting the rubber block:
p=m*Vi
1.352=0.478*Vi
Vi=2.83

0=2.832+2(-9.81cos31)d
d=0.477m

Last edited: Oct 12, 2008
2. Oct 12, 2008

### tiny-tim

Welcome to PF!

Hi MightyMan11! Welcome to PF!

Isn't it sin31º?

3. Oct 13, 2008

### MightyMan11

Re: Welcome to PF!

AHHH...
Looks like it.

Thanks for the help and the welcome