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Regarding problem 6.5: can someone please explain why tennis ball velocity is 3v after colliding with basketball.

MIT 8.01 Physics I Classical Mechanics, Fall 1999
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The basketball clearly bounce back (up) with a speed of v and then collides with the tennis ball. From the frame moving with the basketball the tennis ball appears to collide with the basketball at a speed of 2v. Lewin then says that in the original frame the tennis ball has speed 2v+v=3v! Why are we adding? All help appreciated.
All collisions are considered perfectly elastic. In the approximation, the tennis ball and the basketball collide at about the same height so by conservation of energy P.E.=K.E. and they have about the same speed v, but in opposite directions. so the difference in their relative velocities is 2v, i.e. If you were on the basket ball, you would see the tennis ball approaching and rebounding with a speed of 2v. But the basket ball is moving upward at a speed of v, so by gallilean transformaoin of frames, the tennis ball is moving upward with a speed of 3v in the frame of reference of the ground. A key assumption here is that the mass of the basketball is much larger than the mass of the tennis ball, so that the speed of the basketball effectively remains unchanged after the collision. Then by using mgh=1/2v^2, you can see that the tennis ball bounces off the basketball to a height approximately 9 times than if it were to just bounce off the ground

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