Rolls down a ramp and bounces back up

[Captain Zappo]

Hi guys (and girls), I'm kinda new to these forums. What I mean by that is I haven't yet posted anything, but I have been lurking around for a few weeks. Anyways, I'm stuck on this homework problem.


A 500 g cart is released from rest 1.0 m from the bottom of a frictionless, 30 degree ramp. The cart rolls down the ramp and bounces off a rubber block at the bottom. The figure shows the force during the collision.

I have setup my force diagram and found the following information:

Fnet(x): 2.45N
Velocity at bottom of ramp: 3.13m/s
Time to reach bottom of ramp: 0.64s



Unfortunately, I don't know what to do next.

This problem is very elementary I know, but all help will be appreciated.

 

[SpaceTiger]

So...what is it asking you to do?

 

[Captain Zappo]

Sorry guys. I forgot to ask the question...haha. But anyways, I figured it out. I forgot an important part of the conservation of momentum theorem.

Just for the record, the question was "How far up the ramp does the cart rebound?"

 

[kite7]

Hey guys, I have the same problem and I just happened to find this thread

I took J which is the area under the force curve in the diagram so I have 26.7N

Then P2x = p1x + Jx = (3.13m/s * 0.500kg) + 26.7N = 28.265 kgm/s

Then V2x = 28.265kgm/s / 0.500kg = 56.53m/s? I think I might have screwed up because I don't think it can have that velocity.

Now I don't know where to go once I have the velocity

nevermind I didnt convert ms to s correctly, now I have a post collision velocity of 8.47m/s but I don't know what to do after

 

[kirbykirbykirby]

Hi,

I have the same question but where's the 3.13m/s coming from?
Here I try to do it the fancy way.
I did
$$\[ v^2_f = v^2_i + 2ad \] \[ v^2_f = 0+ 2(mgsinx)(1)\] so /[v_f = 2.21 m/s \]$$

I took area like the last guy expcept I got 2.67Ns

Then I did whole bunch of other stuff to get wrong answer so can somebody help me with the first part?