# Momentum and Energy- Major help needed

1. Oct 25, 2006

### parwana

A small block of mass m1 = 0.300 kg is released from rest at the top of a curved wedge of mass m2 = 2.00 kg, which sits on a frictionless horizontal surface as in Figure P6.59a. When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right, as in Figure P6.59b.

(a) What is the velocity of the wedge after the block reaches the horizontal surface?
I found the velocity to be -.6 m/s. I did 0=m2v2 + M(block)V(block)

(b) What is the height h of the wedge?

I know the energy equation will be

mgh= .5mv^2

2. Oct 25, 2006

### parwana

3. Oct 25, 2006

### AVD

If you find the velocity of the block and the wedge, you should be able to apply concervation of energy to determine the height of the wedge. This means one point you have to apply the formula to is when M1 is at rest at the top of the wedge. use mgh = (whatever you decide) to determine the height of M1 relative to the ground. This is the same as the height of the wedge.

4. Oct 26, 2006

### parwana

I still dont understand. Can you show me how to set up the equation

Would it be

mgh= .5mv^2 Now which mass and which velocity should I use to find h here

5. Oct 26, 2006

As AVD said, you have to use conservation of energy on the system consisting of the block and the wedge. So, write down the sum of kinetic and potential energy for these two momens and set them to equal one another; a) the block rests on the wedge, b) the block is on the horizontal surface.

6. Oct 26, 2006

### parwana

^ I am really sorry, I tried setting it up but I am getting it wrong.

7. Oct 26, 2006

### rkslperez04

would it be?

KE + PE = KE + PE

and go from there?

8. Oct 26, 2006

### parwana

I tried that

.5(.3)(4^2) + .3(9.8)h= .5(32)(.6^2) + 2(9.8)h

is that right

It didnt work for some reason, I got .112 is height, or am I doing something wrong

9. Oct 26, 2006

### rkslperez04

am I showing in notes from that this may be an inelastic collision (m1v1i + m2v2i = (m1 + m2)Vf)... sorry Im not help to you :(

10. Oct 26, 2006

PE(block) = KE(block) + KE(wedge), where PE(block) is for the moment the block is resting on the wedge, and KE(block) and KE(wedge) are for the moment when the block is on the horizontal surface, and the wedge has the velocity you calculated in part 1.

11. Oct 26, 2006

### parwana

^ Its ok, I hope someone else helps

12. Oct 26, 2006

### parwana

Thanks so much radou, but why dont we use the PE of the wedge?

13. Oct 26, 2006