Ratio of masses for maximum energy transfer in elastic collision?

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The discussion revolves around determining the mass ratio m1/m2 for two elastic balls to maximize energy transfer during an elastic collision. The lower ball (m2) hits the ground first, and the upper ball (m1) is affected by the collision dynamics. Momentum conservation is emphasized, but participants note the need for a second equation to relate the final velocities of both balls. The key focus is on deriving an expression for the kinetic energy gained by the upper ball after the collision and finding the mass ratio that maximizes this energy. Ultimately, the conclusion suggests that the ratio m1/m2 should be 1/3 for optimal energy transfer.
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Homework Statement


Two elastic balls of mass m1 and m2 are placed on top of each other with a small gap between them and then dropped onto the ground. What is ratio m1/m2, for which the upper ball m1 ultimately receives the largest fraction of the total energy? (m1<m2)

Homework Equations


The Attempt at a Solution


At the instant when lower ball(m2) hit the ground, both balls will have the same velocity u1. And when the lower ball bounces up, it will have a velocity of -u1. However, the problem is that there is a gap between the two balls, and if I introduce another variable h, I don't think this variable can be eliminated at the end to give the answer m1/m2 = 1/3.
 
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Don't worry about the gap. The approximation here is that the collision is over very fast, i.e. fast enough so that gravity does not affect the motion during the collision. Then we can say that momentum is conserved because no (appreciable) external forces act on the system. Treat the problem so that the lower ball hits the ground first, reverses direction and collides with the upper ball while it is still moving down.
 
Then I think it is correct to say that downward velocity of the upper ball is u1 and upward velocity of lower ball is -u1. Then,

m1u1-m2u1=m1v1 + m2v2

u1-u1=v2-v1
Since we are interested only on the velocity of the upper ball,so
v2=2u1+v1
Substitute this equation into previous equation
u1 (m1-m2) = m1v1 + m2(2u1+v1)
u1 (m1-3m2)=v1(m1+m2)
v1=u1(m1-3m2)/(m1+m2)
But I don't know how to continue from here to find maximum kinetic energy gained?
 
gaobo9109 said:
Then I think it is correct to say that downward velocity of the upper ball is u1 and upward velocity of lower ball is -u1.
I like to think of "down" as negative and "up" as positive. Please humor me.
m1u1-m2u1=m1v1 + m2v2

u1-u1=v2-v1
The first equation is OK. The second equation does not follow because the masses are not equal so you cannot cancel them. Besides, isn't u1-u1 always zero?

Look, you have two unknowns, the final velocities of each ball. Momentum conservation is only one equation relating these unknowns. You need a second equation. What is that equation?
 
Sorry that is a typing error. The actual equation is supposed to be like this
u1-(-u1)=v2-v1

Initial relative velocity = final relative velocity
 
gaobo9109 said:
v1=u1(m1-3m2)/(m1+m2)
But I don't know how to continue from here to find maximum kinetic energy gained?
Correct so far for the velocity of the top ball. Now you need to find the kinetic energy gained by it after the collision, then worry about its maximum. Can you find an expression for the KE gained? If "yes", write it in terms of the ratio of the masses μ = m1/m2, then see for what value of μ your expression is maximum.
 

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