Show that kinetic energy of the earth after a collision can be ignored

I can't quite follow what you're doing there, and it's really late where I am, so I'll just start you off the right way.

This problem basically hinges on the Earth being much more massive than the ball, so the ratio of the mass of the ball to that of the earth is very close to zero.

Let's use some simpler symbols. Up is positive, down is negative. m is the mass of the ball, M is the mass of the earth. u is the initial speed of the ball, v is the final speed of the ball. The initial (vertical) speed of the earth is zero, the final speed is w.

So ##-mu = mv - Mw## ---eqn 1

##\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + \frac{1}{2}Mw^2## --eqn 2

The first eqn is conservation of linear momentum, second is conservation of K.E.

Rearrange eqn 1 to isolate w in terms of the other two variables, then put that in eqn 2. You should obtain an equation relating the ratio ##\displaystyle \frac{u}{v}## to an expression in terms of the ratio ##\displaystyle \frac{m}{M}##. When you let ##\displaystyle \frac{m}{M} \rightarrow 0##, what can you say about u and v?

Now consider what that means in terms of equation 2. Observe that the second term on the RHS represents the K.E. of the earth. Remember that the only kinetic energy here comes from the ball. If it bounces up with (almost) the same kinetic energy as it came down with, what can you say about the kinetic energy imparted to the earth in the collision?

Source: http://www.physicsforums.com/showthread.php?t=704192&goto=newpost

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