Answer to the Question 11/98
BOUNCING BRICKThe question was:
A heavy brick is falling from a height of 1 meter, hits a tennis
ball and jumps back to the height of (almost) 1 meter. (We assume
that the mass of the tennis ball is negligible compared to the mass
of the brick. The collision with the ball is assumed to be elastic,
i.e. there is no energy loss.) To what height will the tennis ball
jump?(2/1999) The problem has been solved correctly by William Chess (e-mail chess@kennett.net), by Eitan Federovsky (e-mail eitan@peach-networks.com), by Bill Bruml (e-mail bruml@alum.mit.edu), by Age Sjastad (e-mail asj@hials.no), and by Javier Groshaus (e-mail jgros@techunix.technion.ac.il) M.Sc. student from Technion (Haifa, Israel).
The answer: The tennis ball will jump to the height of 1/4 meter.
The solution:
The brick will fall and strike the tennis ball, which will compress symmetrically and then return to its original shape. During this compression, the center of mass of the tennis ball moves exactly half the distance that the topmost point of tennis ball moves (the point first contacted by the brick). Since the distances moved by these two points on the ball occur during the same time period, the velocity of the center of mass of the ball is exactly one-half the velocity of the top of the ball. During the rebound, at the moment the tennis ball returns to its original shape, the brick (and the top of the tennis ball) will have reached its maximum upward velocity, which, according to the problem statement, is sufficient to propel the brick to a maximum height of almost 1 m. Since the maximum height of a projectile is proportional to the square of the vertical component of its initial velocity (H=v2/(2g)), and since the center of mass of the tennis ball was moving at one-half the velocity of the brick at the moment of restoration, the tennis ball will reach a maximum height equal to one-quarter of the maximum height of the brick.
(11/2000) A very interesting and important remark has been made by Bill Unruh (e-mail unruh@physics.ubc.ca). He noticed that, although it is claimed that everything in the problem is elastic, the solution does not seem to be reversible in time: Consider the brick at the lower-most point with the ball maximally compressed. We claimed that if the time "runs forward" then both the brick and the ball will jump. However, if we "run the time backward" then the brick returns its it original position, while the ball is still on the ground. So the solution is not "time reversible". (Actually the time reversibility should be slightly more complicated, since there is a third body ("the ground") involved. However, the conclusion regarding ireversibility is not changed.) Obviously, both the formulation of the problem and its solution neglected many things; some of them are more important, while the others are negligible. E.g., the fact that the "sound" or compression wave in the ball propagates at a finite speed seems to be not particularly relevant, provided this speed is much larger than the velocity of the brick. However, certain "inelastic" aspects may be important. We welcome a more detailed analysis of the interaction between the ball and the brick.
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