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.(adsbygoogle = window.adsbygoogle || []).push({});

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

Plug in your values.

(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

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# Homework Help: Block Moving Down Slide

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