Calculating Maximum Elevation of Top Ball After Collision

[Crebble32]

Homework Statement

Two balls fall one above the other from height h= 5m. The bottom one has m= .1kg and the top one has m= .01kg. What will be the maximum elevation of the TOP ball after the balls hit the floor? Assume that collisions with the floor and between the balls are elastic.

Homework Equations

I tried this with: V1f= V1i* ((m1-m2)/(m1+m2)) + ((2m2)/(m1+m2))*V2i
V2f= V1i* ((2m1)/(m1+m2)) + ((m2-m1)/(m1+m2))*V2i

Then use: h= (1/2 m(V2)^2)/(mg)

The Attempt at a Solution

The issue is, the Vf is when the bottom ball has hit the floor and has changed directions and the m2 ball is still falling down... They collide and then the m2 ball changes direction.

I am so confused on what to do next. I appreciate any help possible! Thanks

 

[Filip Larsen]

I assume that you are not given the sizes of the balls, in which case you can assume that the balls are small compared to the height of 5 meter. You may also assume that the two balls are placed and released with one just above the other and ignoring air resistance they will fall with same acceleration and therefore maintain their relative position until just before they reach the floor. During the collisions you can assume the bottom ball collides with the floor and reverses direction (with same speed as before since the floor does not move) and thereafter collides with the top ball, but that both collisions happens so fast after each other that the bottom ball hits the top ball with same speed as it hit the floor.

With all these assumptions and the equations you have correctly specified, you should be able to determine the two initial speeds in the 2nd collision from the height, from this you can get the speed of the top ball and from this again you finally get the maximum height the top ball can obtain.

 

[Crebble32]

Thank you for your help!