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- Homework Statement
- Two elastic balls of masses m1 and m2 are placed one above the other (leaving a very small gap between them) and

are released to fall freely on a height from a height h. Assuming that the impact on the floor is elastic, calculate:

a. what must be the ratio m1/m2 of the balls so that the top ball gets the biggest part of total balls energy after impact?

b. What is the maximum height a ball of mass m1 can rise after impact?

h is much bigger than dimension the of balls.

- Relevant Equations
- Conservation of energy and momentum.

With given information in the problem I can't decide which direction balls will go after collision. I assumed they will go in opposite directions. I know that is not a full solution, but I can't come up with anything else.

From conservation of energy we have:

$$m_{2}{v}_{0}^{2}+m_{1}{v}_{0}^{2}=m_{2}{v}_{2}^{2}+m_{1}{v}_{1}^{2} (1)$$

$${v}_{0}=\sqrt{2gh}$$

Conservation of momentum:

$$m_{2}{v}_{0}-m_{1}{v}_{0}=-m_{2}{v}_{2}+m_{1}{v}_{1} (2)$$

from these equations I get:

$$2{v}_{0}={v}_{1}+{v}_{2} (3)$$

$$m_{1}gh+m_{2}gh=\frac{m_{1}{v}_{1}^{2}+m_{2}{v}_{2}^{2}}{2} (4)$$

dividing by m2:

$$2\frac{m_{1}}{m_{2}}gh+2gh={\frac{m_{1}}{m_{2}}{v}_{1}^{2}+{v}_{2}^{2}} (5)$$

Then I replaced v2 from (3), solved quadratic and took derivative with respect to ##m1/m2## , but that gives me absolutely nothing.

I suppose I am making mistake in (2).

I also do not understand derivation of equation of balls speeds in perfectly elastic collision. There is equation:

$$m_{1}{v}_{1}+m_{2}{v}_{2}=m_{1}{v}_{1}^{'}+m_{2}{v}_{2}^{'}$$

And I do not understand how can we take them all positive if we don't know which direction after collision will they go.

If I take all momentum terms positive in equation (2), I get ##v_1=v_2##, which can't be true.

Please explain it to me.