Originally Posted by **PhysicsKidd** A superball of mass M and a much lighter superball of mass m are dropped together vertically, with the light ball directly above the heavy ball, but not in contact. Show that the lighter ball will fly just over 8 times the height from which it was dropped.
I know that I need to use momentum and different frames of reference, but I'm very confused as to how... Thank you |

I'm getting closer to 9x but that will all depend on how the masses are chosen.

Here are some highlights to get you started. You will need to verify the steps on your own.

I am going to assume all collisions are elastic. That means when the larger ball hits the ground it is perfectly reflected. Let v1 = sqrt(2gh) downward be the velocity of the large ball when it hits the floor. (Prove this.) So immediately after it hits the floor it has a velocity v1 upward. Additionally the small ball with also have a velocity of v1 downward just before the collision with the large ball. (Why?)

Calling +y upward, we have from momentum conservation:

Mv1 - mv1 = MV + mv

where V and v are unknowns.

Likewise energy is conserved in elastic collisions, so

(1/2)Mv1^2 + (1/2)mv1^2 = (1/2)MV^2 + (1/2)mv^2

So we have two equations in the unknowns v and V. We wish to find v.

First solve the momentum equation for V:

V = (1/M)(Mv1 - mv1 - mv)

Now put that value of V into the energy equation. Skipping along a bit I get the following result. (See the attachment below.)

As ugly as this looks, it's a quadratic. Now, I'm not going to do it, but you can get some numbers in to do an approximation. That'll make it easier to work with. I don't know if your instructor would like it that way. You'll have to make a judgement call.

Okay, solving the quadratic equations is a bear, but possible. Just do it one step at a time. I get two possible results:

1) v = -v1. This implies the smaller ball as passed right through the larger one. Not possible.

2) v = (3M/(M + m))v1 - m/(M + m)v1 As this gives an upward velocity for the small ball we are good with this.

Now, to see how high (H) the smaller ball is after the collision we get H = v^2/(2g). I'm going to use M = 100 kg and m = 1 kg. A little extreme I suppose. You can use any mass ratio you like. I get v = 3*v1 (approximately). This gives H = 9h.

-Dan