Cart Collision Homework: Calculate Distance Traveled After Bounce

[MightyMan11]

Homework Statement

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?

Homework Equations

p_f=p_i+impulse
V_f²=V_i²+2ad
p=m*v

The Attempt at a Solution

First I found my initial impulse:
V_f²=V_i²+2ad
V_f²=0+2(9.8cos31)(1.16)
V_f=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:
p_f=p_i+impulse
p_f=-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*V_i
1.352=0.478*V_i
V_i=2.83

V_f²=V_i²+2ad
0=2.83²+2(-9.81cos31)d
d=0.477m

However, this answer does not match my answer key...

Thanks in advanced

 

[tiny-tim]

Welcome to PF!

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?
…
First I found my initial impulse:
V_f²=V_i²+2ad
V_f²=0+2(9.8cos31)(1.16)

Hi MightyMan11! Welcome to PF!

Isn't it sin31º?

 

[MightyMan11]

Hi MightyMan11! Welcome to PF!

Isn't it sin31º?

AHHH...
Looks like it.

Thanks for the help and the welcome