We attach a ball of mass m to a rope of length L. We hold the rope horizontal, above a frictionless floor. On the floor is a box of mass M.
The total energy of the system is
total E = KE + GPE = 0 + mgL
We then release the ball, so that it swings down towards the box. Just before it reaches the box, at the bottom of its swing, we can figure out the velocity v0 of the ball. We use the conservation of energy.
total E = KE + GPE = mgL = 1/2 m v0^2 + 0 = mgL so v0 = sqrt( 2gL )
Now, the ball strikes the box and bounces back. We are told that this is an ELASTIC collision, which means that kinetic energy is conserved; and momemtum is also conserved, as it always is in the absence of external net forces.
Let's call the final speed of the ball, moving backwards, v1, and the final speed of the box, moving forwards, v2.
From the conservation of momentum, we have
m v0 = - m v1 + M v2
From the conservation of energy, we have
1/2 m v0^2 = 1/2 m v1^2 + 1/2 M v2^2
Can we find the values of the final speeds v1 and v2? Yes, we can: they are two unknown quantities, but we have two equations connecting them.
Use the momentum equation to express v1 in terms of v2, so we can get rid of v1.
M v2 - m v0 M v1 = ----------------- = --- v2 - v0 m m
Now, substitute this expression for v1 into the kinetic energy equation. We should end up with a single big equation with just a single unknown quantity, v2.
2 2 ( M ) 2 1/2 m v0 = 1/2 m ( --- v2 - v0 ) + 1/2 M v2 ( m )
We can work out the square and do some algebra to simplify things.
2 2 M 2 2 2 1/2 m v0 = 1/2 --- v2 - M v0 v2 + 1/2 m v0 + 1/2 M v2 m 2 M 2 2 0 = 1/2 --- v2 - M v0 v2 + 1/2 M v2 m 2 M = 1/2 --- v2 - M v0 + 1/2 M v2 m
Now just solve for v2 in terms of the masses and the initial speed of the ball.
( 2 ) v2 = ( -------- ) v0 ( M/m + 1 )
You can then place this result into the equation we derived early on for v1 to find
( 1 - m/M ) v1 = ( -------- ) v0 ( 1 + m/M )