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## Homework Statement

A block of mass m=2.2 kg slides down a 30 degree incline which is 3.6 m high. At the bottom, it strikes a block of mass M=7 kg which is at rest on a horizontal surface. If the collision is elastic, and friction can be ignored, determine (a) the speeds of the two blocks after the collision, (b) how far back up the incline the smaller mass will go, and (c) what is the upper limit on mass m if it is to rebound from M, slide up the incline, stop, slide down the incline, and collide with M again?

## Homework Equations

mgh=1/2mv^2

mghsintheta=1/2mv^2

mv1+mv2=mv1'+mv2'

## The Attempt at a Solution

a) mgh=1/2mv^2

v=8.4m/s

v'a=(2.2-7/2.2+7)(8.4)=-4.38 m/s

v'b=(2*2.2/7+2.2)(8.4)= 4.02 m/s

b) mghsintheta=1/2mv^2

h=1.95 m

I am not sure if I did a and b right and I have no idea how to set c up. Any help would be appreciated!