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 )

Okay, fine.

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 )