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First, I don't have a strong knowledge of physics, so please forgive my lack of precision in defining my question.

Consider an airless free-fall situation where a steel ball and a balsa ball (with the expected great difference in mass) were dropped simultaneously. They fall, say, 1 meter and hit the ground, which is (for example) a 10 cm thick dry hard concrete slab.

Now consider the above experiment, performed on the earth, and performed again on the moon. What are the differences that would be observed, based on the differing gravities?

At first I was thinking that on the moon, the recoil or bounce from the object striking the ground would lift the objects higher up. But now I think that the reduced force of gravity on the moon would result in less impact force, and therefore an identical rebound, in height.

But the time spent falling should be different, correct? And would the "hang time" of the object after the bounce be longer on the moon?

Thanks for your help with this.

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You could start by setting up (1) the forces acting on a ball that is at rest but released at some height, (2) the forces acting on a ball that has gained some velocity while falling (3) the forces acting when the ball, at some velocity, hits the surface –  Michiel Feb 26 '13 at 19:35
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2 Answers

up vote 1 down vote accepted

Assuming the wooden ball and the steel ball were both 100% elastic, the only difference between the earth and the moon would be the time it takes.

If you used a slow-motion camera on the earth, it could look just like it does on the moon.

(BTW, "hang time" is not a physical quantity - it's just something you perceive :)

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I always took 'hang time' to mean "time between bounces" or "time in air" –  Jerry Schirmer Feb 26 '13 at 20:47
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Let's not assume that the collisions between the balls and the surface are elastic. In other words, let's assume that there is some energy lost during the collision which causes the balls to bounce to a smaller height than that from which they were first released. If we call $v_1$ the speed of a ball right before it hits the surface and $v_2$ the speed of a ball right after it hits the surface, then this inelasticity corresponds to $v_2$ being less than $v_1$ by some amount. Let's assume further that the ratio $v_2/v_1$ is some constant for a ball of given material that is independent on the speed with which the ball hits the surface. I'm not sure to what extent this last assumption is generally true, but I think it's true to good approximation for generic, reasonably elastic materials since it has a special name: the coefficient of restitution. $$ C_R = \frac{v_2}{v_1} $$ Notice that since $v_2<v_1$, $C_R <1$. Now let's notice something interesting. Near the surface of a given planet with acceleration due to gravity $g$, the original height $h_1$ from which the ball is dropped can be related to $v_1$ through conservation of energy, and likewise for $v_2$; $$ mgh_1 = \frac{1}{2} mv_1^2, \qquad mgh_2 = \frac{1}{2}mv_2^2 $$ where $m$ is the mass of the ball being dropped. Solving for $v_1$ and $v_2$ in terms of $h_1$ and $h_2$, and then plugging these expression back into the expression for the coefficient of restitution, we find that $$ C_R = \sqrt{\frac{h_2}{h_1}} $$ What's cool is that this result tells us that (under the stated assumptions)

If we bounce a ball from a certain height on a planet, then the fraction of its height to which it returns doesn't depend on the value of the acceleration due to gravity near the surface

In particular, even when you include a basic model of inelasticity for the collision, the bouncing process will look exactly the same on any two given process except that one will occur more slowly than the other because the acceleration due to gravity is weaker in one case. The conclusion of Mike Dunlavey in his response is therefore seen to hold even when the collision isn't 100% elastic.

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