# S Help Solve Block Slides Down Ramp Problem

• inutard
I got it now. So to solve for the velocity of both the block and ramp after the block slides back down, I use the conservation of momentum and energy equations just like what you did? Yes. The total kinetic energy of the system is conserved, and the total momentum is conserved too. Since the block and the ramp are together, the velocity of the block with respect to the ramp is the same as the velocity of the block with respect to the ground. You can use the same equations, just substitute vfinal instead of v. In summary, the block of mass m slides along a frictionless surface with a velocity of v m/s before transitioning to a frictionless ramp inclined at an angle of theta ° to the horizontal

## 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

momentuminitial = momentumfinal
Epotential = mgh
Ekineticl = 0.5mv2
Einitial=Efinal

## 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-block2=m-block*g*h+0.5m-ramp*v-ramp2
Since the ramp's velocity is the same as v-final,

0.5m-block*v-block2=m-block*g*h+0.5m-ramp*v-final2
0.5m-block*v-block2=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

Edit: The mass of the ramp is known.

When the block moves on the ramp, at that instant the velocity of the combined system is
v' = mv/(m+M) Now the system is moving with uniform velocity v'. Since it an inertial frame of reference, the the velocity of the block on the ramp does not affect the velocity of the system. the retardation of the block is gsinθ. You can use the conservation of energy to determine how far up the ramp does the block slide before it starts sliding down.

What is the velocity of the ramp and block after the block slides back down?
Just before the block and ramp separate, the velocity of the system is v'. So apply the conservation of momentum,
(M+m)v' = mv1 - Mv2.
Apply the conservation of energy to find v1 and v2.

rl.bhat said:
When the block moves on the ramp, at that instant the velocity of the combined system is
v' = mv/(m+M) Now the system is moving with uniform velocity v'. Since it an inertial frame of reference, the the velocity of the block on the ramp does not affect the velocity of the system. the retardation of the block is gsinθ. You can use the conservation of energy to determine how far up the ramp does the block slide before it starts sliding down.

So what do I use for the initial velocity of the block on the ramp? It can't be v', can it? Also, the v' refers to the velocity of the system, but how do you determine the velocity of the ramp itself to the ground?

inutard said:
So what do I use for the initial velocity of the block on the ramp? It can't be v', can it? Also, the v' refers to the velocity of the system, but how do you determine the velocity of the ramp itself to the ground?
At the instant the block is on the ramp, the velocity of the block and the ramp is v'.Later on the ramp continues its motion with velocity v'. But the block's velocity decreases as it moves up until it stops momentarily.

inutard said:
1.

## 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-block2=m-block*g*h+0.5m-ramp*v-ramp2
Since the ramp's velocity is the same as v-final,

0.5m-block*v-block2=m-block*g*h+0.5m-ramp*v-final2
0.5m-block*v-block2=m-block*g*h+0.5m-ramp*[m-block*v-block/(m-block+v-block)]2

At its highest position, the block is in rest with respect to the ramp and they move together with velocity vfinal. It is the same situation as in case of a totally inelastic collision. So the total kinetic energy of the system is
KE=0.5(mblock+mramp))vfinal2.

inutard said:
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

The block is back, the energy is conserved, both the ramp and the block can move only horizontally, and there is no horizontal external force that could have changed the overall momentum. The only change is that the block moves with respect to the ramp with a relative velocity opposite that it was at the beginning. It is the situation for an elastic collision.

ehild

Hm.. Thanks guys