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## Homework Statement

A block of

*m*mass slides along a frictionless horizontal surface with a speed of

*v*m/s. After sliding a distance of

*d*, the block makes a smooth transition to a frictionless ramp inclined at an angle of theta ° to the horizontal. The ramp is not fixed to the frictionless surface and is able to move. How far up the ramp does the block slide before sliding back down? What is the velocity of the ramp and block after the block slides back down?

## Homework Equations

momentum

_{initial}= momentum

_{final}

E

_{potential}= mgh

E

_{kineticl}= 0.5mv

^{2}

E

_{initial}=E

_{final}

## The Attempt at a Solution

When the block hits the ramp, momentum is conserved and the block-ramp system initially moves at:

m-block*v-block = (m-block+v-block)*v-final

v-final = m-block*v-block/(m-block+v-block)

With conservation of energy, we see that:

0.5m-block*v-block

^{2}=m-block*g*h+0.5m-ramp*v-ramp

^{2}

Since the ramp's velocity is the same as v-final,

0.5m-block*v-block

^{2}=m-block*g*h+0.5m-ramp*v-final

^{2}

0.5m-block*v-block

^{2}=m-block*g*h+0.5m-ramp*[m-block*v-block/(m-block+v-block)]

^{2}

Following this, the isolation for h becomes trivial.

What I am having trouble with is the second part of the question that asks for the velocity of the block and ramp after the block slides down. When the block moves up the ramp, the system's centre of mass changes (does this change the velocity?) Is there any easier way to do this problem? Is my solution even correct?

-PL