# Homework Help: Block Moving Down Slide

1. May 30, 2007

### MJC8719

A small block of mass m1 = 0.480 kg is released from rest at the top of a curved-shaped frictionless wedge of mass m2 = 3.00 kg, which sits on a frictionless horizontal surface as in Figure P9.60a. When the block leaves the wedge, its velocity is measured to be v1 = 3.00 m/s to the right, as in Figure P9.60b.

Figure P9.60

(a) What is the velocity of the wedge after the block reaches the horizontal surface?
m/s to the left

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

You must use conservation of momentum:

m1v1 = m2v2

m1 - mass of block
m2 - mass of wedge
v1 - velocity of block
v2 - velocity of wedge

(0.480 kg)(3.00 m/s) = (3.00 kg)v2

Multiply.

1.44 kg m/s = (3.00 kg)v2

Divide both sides of the equation by 3.00 kg.

v2 = 0.48 m/s to the left

Part B

PEi + KEi = PEf + KEf

mgh + 0.5mvi2 = mgh + 0.5mvf2

mgh + .5mvi^2 = mgh + .5 mvf^2

(0.48)(9.8)(h) + .5(.48)(3^2) = 0 (as the height is now zero) + .5(.48)(.48^2)

When i solve for this I get -.4474 which is incorrect.

What am I doing wrong

2. May 30, 2007

### MJC8719

Wow, Am I an idiot at times.....

I figured it out...Here is the correct way to solve the second half:

(0.480)(9.8)h = (.5)(.48)(3^2) + (.5)(3)(.48^2)

h = 0.53265

So Simple lol