Ball bounces back to same height after Angular Impulse?

AI Thread Summary
A ball moving off a frictionless cliff experiences a reduction in its x-component velocity upon collision with a frictional surface, while its y-velocity is fully reflected upwards. The discussion explores whether the ball can rebound higher than its original drop height, particularly if it is not perfectly rigid and can compress during the collision. The idea of converting rotational energy into vertical energy through a pulley and hammer system is proposed as a potential mechanism for achieving a higher rebound. This scenario hinges on the timing of the hammer's impact with the ball immediately after the initial collision. Overall, the conversation delves into the complexities of energy transfer during the collision and rebound process.
FallenApple
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So Imagine a ball moving off a frictionless cliff with a velocity v to the right, at the bottom of the cliff the surface has friction.

So after the collision, the x component of the velocity would be reduced, and the angular momentum would be fixed clockwise because the frictional impulse acted to the left.

Now clearly the bounce would not be symmetric. But I'm thinking only in the x. The maximum height of the rebound should still be the same, its only the x velocity that was reduced and convered to rotational motion. The y velcocity gets completely reflected back upwards.

This is assuming that the ball is rigid enough that it doesn't deform at all during the collision. It just get torqued up.

Is this correct?
 
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FallenApple said:
Is this correct?
yup
 
It gets interesting when you ask, if the ball can rebound higher than the drop, given the right initial spin.
 
A.T. said:
It gets interesting when you ask, if the ball can rebound higher than the drop, given the right initial spin.

Interesting. I suppose this is possible. I don't know if it can happen if the ball is perfectly rigid since friction horizontal only, the ball's y velocity would still be perfectly reflected regardless of original spin.

Unless the ball wasn't perfectly rigid and is spring like in constitution. So during the collision the ball would be compressed and somehow the loss of spin would further contribute to the vertical compression, of which finally, the compression is released and the ball flies up higher than it started.

I can think of one way to convert the original spin to a higher height though. If the ball has a original spin clockwise and during the collision, the ball torques down on a rope connected to pulley that has the other side of the rope in a tunnel under the floor connected to a vertically hanging hammer( also under the floor. The ball's spin would slow down, the rope would be pulled by Newton's third law, and that rotational energy lost by the ball would be transmitted to rotating the hammer like pendulum, and finally hitting the ball upwards.

If the hammer hits the ball the instant after the floor itself has completed it's impulse( assume that either the floor opens right after or that the hammer is strong enough to break through the floor), the ball would receive another impulse(without any pause following the first one) and therefore reach a final higher height.

The horizontal energy and rotational energy lost by the ball was converted to vertical energy via the pulley hammer contraption.
 
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