How High Does the Lighter Ball Rise After Collision?

[JennV]

Homework Statement

Amplified Rebound Height Two small rubber balls are dropped from rest at a height h above a hard floor. When the balls are released, the lighter ball (with mass m) is directly above the heavier ball (with mass M). Assume the heavier ball reaches the floor first and bounces elastically; thus, when the balls collide, the ball of mass M is moving upward with a speed v and the ball of mass m is moving downward with essentially the same speed.

In terms of h, find the height to which the ball of mass m rises after the collision. (Use the equations for final velocities after elastic collision of two objects with masses m_1 and m_2:
v1,f = (m1-m2 / m1+m2)v1 + (2m2 / m1+m2)v2
v2,f = (2m1 / m1+m2)v1 + (m2-m1 / m1+m2)v2
and assume the balls collide at ground level.)

Express your answer in terms of h, m and M.

Homework Equations

The Attempt at a Solution

So something like this?

|
| m (the lighter ball)
|
V
^
|
| M (the heavier ball)
|

AFTER:
The lighter ball bounces back up (elastic) after the collision, so I'm trying to find the height that it bounces back up before it hits the ground?

But how would I find the expression of h? =S
Thanks in advance.

 

[MasterX]

First of all, v=sqrt(2*g*h)

Since they collide at ground level, we could assume that the velocity of ball M is zero after the collision. Thus, the velocity,v1, of ball m after the collision is

v1^2=(m+M)V^2/m

Then use Newtons' 3rd law to find the position of the ball. The only force that is exerted on that ball is gravity.