Maximum Height of Colliding Balls

[iamalexalright]

Homework Statement

A superball of mass M and a marble of mass m are dropped from a height h with the marble just on top of the superball. The superball collides with the floor, rebounds, and hits the marble. How high does the marble go? How high does the superball go? Ignore sizes of the superball and marble and assume perfectly elastic collisions.


2. The attempt at a solution

Well, it's been awhile since I've done anything like this so I have a feeling what I did is wrong...

At a height h we have U = .5(M+m)g
When they hit the ground T= .5(M+m)v^2

I figure after they hit the ground, the force on the marble is equal and opposite to its weight.

F = -mg = ma
solving for v(t) = -gt + (2gh)^(1/2)
solving for time t when v(t) = 0
t = (2h/g)^(1/2)

solving for x(t) = -.5gt^2 + t(2gh)^(1/2)
x(t*) = -.5g*2h/g + (2h2gh/g)^(1/2) =
2h - h = h

This doesn't seem correct...

 

[tiny-tim]

Hi iamalexalright!

(try using the X² tag just above the Reply box )

F = -mg = ma
solving for v(t) = -gt + (2gh)^(1/2)
solving for time t when v(t) = 0
t = (2h/g)^(1/2)

solving for x(t) = -.5gt^2 + t(2gh)^(1/2)
x(t*) = -.5g*2h/g + (2h2gh/g)^(1/2) =
2h - h = h

sorry, i don't understand any of this

to find v, just use KE + PE = constant, then treat it like an ordinary collision, then use KE + PE = constant again