block 1 of mass 1 slides from rest along a frictionless ramp from height h=2.50m and then collides with stationary block 2, which has mass2=2.00 m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction is 0.500 and comes to a stop in distance d within the region. What is the value od distance d if the collision is a)elastic and b)completely elastic.

How would you find the collision if It is elastic and inelastic? My professor really didn't explain how to do it very well.

It has to do with whether or not energy is conserved. When the block has moved down a height of h=2.50m, it will have a velocity of

Squareroot[2.50m*2*g] = 7 m/s

It is convenient in these problems to view the collision in the center of mass frame where the total momentum is zero.

In the "lab" frame, just before the collision, one block is moving at a velocity of 7 m/s while the other block of twice the mass is at rest.

Suppose you move in the same direction as the block with velocity v. Then, in your frame the block is moving with velocity

v1 = 7 m/s - v

and the other block that is at rest in the lab frame is moving with velocity

v2 = -v

P = (7 m/s - v)M1 - 2M1 v

This is zero if

v = 7/3 m/s.

The momentum of block1 in this frame is

P1 = (7 m/s - 7/3 m/s) M1 =
4.666 M1 m/s

and block 2 has a momentum of:

P2 = - 7/3*2M m/s

which is indeed -P1.

The total energy before the collision in the center of mass frame is:

E = P1^2/(2M1) + P2^2/(2M2) =

P1^2/(2M1) + P1^2/(4M1) =

P1^2/(2M1)*3/2

Note that you can write 1/2 mV^2 as momentum squared divided by (2m).

In the collision the total momentum must be conserved. This is zero in the center of mass frame. Therefore, if one block has a momentum of P after the collision, then the other block must
have a momentum of -P after the collision. So, after the collision the total energy will then be:

E_{after} = P^2/(2M1) + (-P)^2/(2M2) =
P^2/(2M1)*3/2

So, if the total energy is to be conserved, then

P^2/(2M1)*3/2 =P1^2/(2M1)*3/2 -->

P = +/- P1

The collision will have the effect of reversing velocities, so you must choose the minus sign here.

In inelastic collisons energy is lost. The maximum amount of energy loss happens when P = 0. Then
E_{after} = 0.

In the elastic case the momentum of the second block after the collision is thus:

4.666 M1 m/s

The velocity in the center of mass frame is:

4.666 M1 m/s/(M2) = 2.333 m/s

In the lab frame the velocity is:

2.333 m/s + 7 m/s = 9.333 m/s

In the completely inelastic case you lose the maximum posible amount of energy (which is the entire center of mass energy). This means that in the center of mass frame both momenta are zero. Therefore the velocity of the second block is zero in this frame. This means that in the lab frame the block moves at a velocity of 7 m/s after the collision.

Sorry, in the last part you must add 2.33 m/s to go back to the lab frame not 7 m/s.

so, in the lab frame the velocity of the second blck after the collision is

In the elastic case the momentum of the second block after the collision is thus:

4.666 M1 m/s

The velocity in the center of mass frame is:

4.666 M1 m/s/(M2) = 2.333 m/s

In the lab frame the velocity of the second block in the elestic case is:

4.666 m/s

and 2.333 m/s in the inelastic case.

in

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