i need a site that can explain the concept of the bouncing ball for my AS physics?

Question

please help me find a physics site that has all the explanation needed for my AS physics exam

Answer 1

SUVAT: distance, initial velocity, final velocity, acceleration, and time equations. These equations, plus the sum of forces acting on the ball, can be used to "explain the concept" of a BB. [See source.]

Assume the ball is in your hand at some altitude s. What are the forces acting on it? Weight (W) is an obvious one, but if that were the only force, the ball would be accelerating downward at a rate of g = 9.81 m/sec^2. The ball is static; the only way that can happen is for a counterforce to W to exist and that's the force of your hand pushing upward.

So the net force (f) on the ball is f = W - F; where F is the upward force of your hand. As long as you hold that ball, f = W - F = 0; so that W = F and the force of your hand pushing up is exactly equal to the weight of the ball.

Now drop (don't throw) the ball. What are the forces now (discounting air drag)? Weight, that's all; so f = W - F = W = mg; where m is the mass of the ball. So the ball has to drop at an acceleration of g = 9.81 m/sec^2.

The potential energy of the ball at s meters is PE = mgs = Ws. As the ball drops, it acclerates so its initial velocity (U = 0) starts to speed up. When the ball hits the cement, re one of the SUVAT equations, it has terminal velocity (V^2 = U^2 + 2gs); so that V = sqrt(U^2 + 2gs) = sqrt(2gs).

Which means the ball will have kinetic energy KE = 1/2 m V^2 = 1/2 m 2gs = mgs = PE, which means the kinetic energy at the bottom of the fall is exactly equal to the potential energy the ball had at the top of the fall from a height of s.

This is a great example of the conversion of energy and the conservation of energy. The conversion is from PE to KE, the conservation is that PE = KE, no energy was lost, just converted.

Now the ball bounces. First, it smooshes down because of its sudden deceleration upon impact; so there is a smoosh force acting down on it. (Smoosh is a technical term I just made up.)

When a ball smooshes, its like compressing a spring...energy is stored, which is another form of potential energy. In this case PE = 1/2 k (x^2); where k is the coefficient of restitution or spring constant and x is the difference between the unsmooshed diameter and the smooshed diameter of the ball. [See source.]

Now, like the previous PE, this smoosh PE is converted to kinetic energy as the ball's shape restores. So as the ball's shape returns to spherical the conversion of PE into KE causes an upward force (F) so that f = F - W and the net force on the ball is upward while the ball is restoring its shape off the cement.

But once the ball is in the air and its shape is restored, we are back to f = W since there is no more upward push off the cement. Now, if there were no losses in energy due to inelasticity or heat, the ball would have the same KE on take off that it had on impact. This results from the conservation of energy.

In the real world, there will be less KE because of those losses. so, the ball will lose KE as its inital velocity (U') slows down according to U' = sqrt(2gs') and s' = U^2/2g; where s' < s and s' is the new, but lower height where V' = 0, the final velocity upward. s' is the height to which the ball bounces. At this point PE' = mgs' < PE = mgs.

Unless you catch the ball at s', it will turn around to fall once again. It will continue to do this until such time the PE in the ball can no longer be used to create an upward force greater than the weight of the ball.