Momentum and energy of rebounding balls

In summary, the ball gains more momentum than it loses when it rebounds off the wall. The momentum is reversed because the ball is elastic.
  • #1
d-richard
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Greetings,
I was thinking about bouncing a tennis ball against a wall and how its momentum and kinetic energy would change. I asked a friend of mine and he answered that the ball would transfer more forward momentum than it had to the wall but its kinetic energy would remain constant. How is that possible? I know that the ball will strike the wall with momentum p and and bounce back with a momentum -p ignoring any forces (including gravity) that affect my ball-wall system. The energy should be conserved as the wall is assumed to be frictionless. How is it then, that the ball rebounds off the wall by transferring more momentum than it had? I assumed that this is a closed system, so therefore this should be an elastic collision. Am I wrong in doing so? And if so, how does the momentum actually reverse direction? Thanks for any answers
 
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  • #2
The ball strikes the wall with forward momentum p_forward = mv.

Assuming a perfect ball, it bounces backward with momentum p_backward = -mv.

The change in momentum is (p_forward - p_backward) = (mv) - (-mv) = 2*mv.

Hence the impulse felt by the wall is double the momentum of the ball. But if the bounce is perfect, the speed is unchanged - thus the kinetic energy of the ball is unchanged: only the direction changes.

For imperfect balls, walls, and bounces you get something less for both.

PS: the ball reverses direction because it is _elastic_.
 
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  • #3
Thanks a lot. Makes a lot more sense
 
  • #4
Also, total momentum of the Earth and ball and wall is conserved. The ball has gained -2P momentum, and the Earth and wall will gain 2P momentum (since the wall is firmly stuck in the earth). say the mass of Earth and wall is ##M_E + M_W## this will be of the order of 10^24 kg, and the momentum 2P will be of the order of 10 kg m/s So the velocity of the Earth and wall, which results from the transfer of momentum will be of the order of 10^(-24) m/s In other words, you just wouldn't notice it.
 
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  • #5
I see. Very useful. Thanks
 

1. What is the relationship between momentum and energy in rebounding balls?

Momentum and energy are closely related in rebounding balls. When a ball collides with a surface, it experiences a change in momentum due to the force applied by the surface. This change in momentum also results in a transfer of energy between the ball and the surface. The total amount of energy in the system (ball and surface) remains constant, but the distribution of energy between kinetic and potential energy may change.

2. How does the mass of a ball affect its momentum and energy during a rebound?

The mass of a ball affects both its momentum and energy during a rebound. A heavier ball will have a greater momentum and will transfer more energy to the surface upon impact. However, the amount of energy transferred also depends on the elasticity of the ball and surface.

3. What is the difference between elastic and inelastic collisions in rebounding balls?

Elastic collisions occur when the kinetic energy of the ball is conserved during a rebound, meaning that the ball bounces back with the same speed and energy as before the collision. Inelastic collisions, on the other hand, involve a loss of kinetic energy as the ball deforms or sticks to the surface upon impact. This results in a decrease in both the momentum and energy of the ball after the collision.

4. Can the momentum and energy of a rebounding ball be calculated?

Yes, the momentum and energy of a rebounding ball can be calculated using the equations p=mv (momentum) and E=mv^2/2 (kinetic energy), where p is momentum, m is mass, v is velocity, and E is energy. However, the calculations may be more complex for inelastic collisions due to the loss of energy.

5. How does the surface on which the ball rebounds affect its momentum and energy?

The surface on which the ball rebounds can greatly affect its momentum and energy. A harder and more rigid surface will result in a greater transfer of momentum and energy, while a softer and more elastic surface may cause less energy to be transferred. This is why a rubber ball bounces higher on a hard surface compared to a soft surface.

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