Why does a ball bounce back when thrown at a wall?

• Eolill
In summary, the confusion regarding why an object bounces back when thrown at a wall can be explained by the concept of conservation of energy and the coefficient of restitution. The wall exerts a normal force on the object, causing it to compress and then decompress, resulting in a bounce. This effect is a property of the materials involved and can vary depending on their coefficients of restitution. However, if both the object and the wall were infinitely hard and did not deform, there would be no bounce.
Eolill
All right, so, we just went over linear momentum in school, and one thing is confusing me:

If I throw something at a wall, why does it bounce back? (I'm assuming both the object and the wall are infinitely hard and don't get deformed, and the wall won't move)

It seems pretty straight-forward, but I can't get my mind around it, which makes me feel dumb. My dad tries to explain it with conservation of energy -- it has kinetic energy, and it can't just lose all that energy, so it has to keep some being kinetic energy, and that makes it go back. All right, so it has to. But I still don't understand how. I tried to ask him to explain what was happening in terms of forces instead, but he couldn't.

The way I see it, is that the wall exerts normal force on the object thrown (ball). The normal force is a reaction force, right, so it ought only exert so much force on the ball that the ball stops and doesn't go through it. Why would it go ahead and exert extra force on the ball to make it bounce back? Where would it come from?

I realize I'm probably just missing something simple, but could someone try to explain to me? It's really frustrating to not understand.

Nothing forces it to come back --- it's really a property of the material. Imagine throwing a sponge --- it wouldn't bounce much...

To elaborate ganneth's point:
An object bounces back because the initial normal force compresses it a bit. Then the object will try to decompress after it stopped and thus extern another normal force on the wall. Via Newton's third law, the wall pushes back and makes it bounce. That is why a basketball will bounce back, while your "infinitely hard" object will not.

Eolill said:
All right, so, we just went over linear momentum in school, and one thing is confusing me:

If I throw something at a wall, why does it bounce back? (I'm assuming both the object and the wall are infinitely hard and don't get deformed, and the wall won't move)

It seems pretty straight-forward, but I can't get my mind around it, which makes me feel dumb. My dad tries to explain it with conservation of energy -- it has kinetic energy, and it can't just lose all that energy, so it has to keep some being kinetic energy, and that makes it go back. All right, so it has to. But I still don't understand how. I tried to ask him to explain what was happening in terms of forces instead, but he couldn't.

The way I see it, is that the wall exerts normal force on the object thrown (ball). The normal force is a reaction force, right, so it ought only exert so much force on the ball that the ball stops and doesn't go through it. Why would it go ahead and exert extra force on the ball to make it bounce back? Where would it come from?

I realize I'm probably just missing something simple, but could someone try to explain to me? It's really frustrating to not understand.
The infinitely hard object would just fall to the ground. This is to say the object would lose all it's kinetic energy but then gravity will accelerate it towards the ground. However energy is conserved because that wall is planted in the ground. Momentum transfers to the wall which will transfer it to the ground. Some of the objects kinetic energy will also be dissipated in the form of heat energy.

The bounce effect you talk about is an extrinsic property of an object. It is called the coefficient of restitution: http://en.wikipedia.org/wiki/Coefficient_of_restitution. It really a measure of how much energy is lost in a collision of a particular object.

Feldoh said:
However energy is conserved because that wall is planted in the ground. Momentum transfers to the wall which will transfer it to the ground. Some of the objects kinetic energy will also be dissipated in the form of heat energy.

But if the wall won't move, it can't have any momentum, can it? And if both ball and wall are infinitely hard, there will be no heat or sound upon impact, will there?

What was said about the ball not bouncing back if it were infinitely hard, was very helpful, though. Thanks! <3

no, the wall HAS momentum

A perfectly elastic body is that which regains its shape completely after the deforming force is removed, which implies that perfectly rigid bodies are perfectly elastic bodies in the limit of deformation tending to zero. Collision between two perfectly rigid bodies would be a perfectly elastic collision. The KE and momentum would both be conserved. The molecules in an ideal gas are indeed modeled after such a concept.

An “infinitely hard” sphere, meaning a perfectly rigid body, after hitting the wall, would not fall to the ground. How much it rebounds will of course depend on the properties of both the wall and the ball, i.e., on the co-efficient of restitution of these two materials, but it will rebound considerably if the wall is not very plastic. On the other extreme, a perfectly plastic body would indeed splat on the wall and fall to the ground.

To answer the original question, when a ball hits a wall, both the ball and the wall are deformed like “springs”. After the relative motion becomes zero, both of them tries to come back to their original shapes, and this makes the ball rebound. A part of the original energy is permanently absorbed by both the bodies as heat and sound or used for permanent deformation.

Shooting Star said:
A perfectly elastic body is that which regains its shape completely after the deforming force is removed, which implies that perfectly rigid bodies are perfectly elastic bodies in the limit of deformation tending to zero. Collision between two perfectly rigid bodies would be a perfectly elastic collision. The KE and momentum would both be conserved. The molecules in an ideal gas are indeed modeled after such a concept.

An “infinitely hard” sphere, meaning a perfectly rigid body, after hitting the wall, would not fall to the ground. How much it rebounds will of course depend on the properties of both the wall and the ball, i.e., on the co-efficient of restitution of these two materials, but it will rebound considerably if the wall is not very plastic. On the other extreme, a perfectly plastic body would indeed splat on the wall and fall to the ground.

To answer the original question, when a ball hits a wall, both the ball and the wall are deformed like “springs”. After the relative motion becomes zero, both of them tries to come back to their original shapes, and this makes the ball rebound. A part of the original energy is permanently absorbed by both the bodies as heat and sound or used for permanent deformation.

There wouldn't be a deforming force to begin with if it is "completely hard"

Feldoh said:
There wouldn't be a deforming force to begin with if it is "completely hard"

"Completely hard" things don't exist in nature. When the deformation is very small, the time of impact is very small and the co-eff of restitution is almost equal to 1, we idealize something as perfectly rigid. Very little energy is lost in such macroscopic collisions. They are like almost perfectly elastic collisions.

EDIT: Saw the above link after I had submitted my post. Overall, I feel, that thread is saying the same thing as me. But this sentence, form post #5 in the above thread, is erroneous:
"Ah, ok, so the force that causes the ball to bounce is contained in the ball, and is not related to the surface it is bouncing against."

The amount of bounce depends on both the objects.

Last edited:
One interesting point, marginally related with the issue, is that infinitely rigid objects are incompatible with special relativity. If you had a 1-km infinitely rigid rod, you could push one end to send a signal instantaneously to the other end.

ahrkron said:
One interesting point, marginally related with the issue, is that infinitely rigid objects are incompatible with special relativity. If you had a 1-km infinitely rigid rod, you could push one end to send a signal instantaneously to the other end.

Hi ahrkron,

I feel that a thread should be kept focused on the original discussion, at least until it is resolved to a satisfactory degree, instead of digressing. This has become a problem in PF, where a thread may gradually veer off to an absolutely unrelated subject, or is sometimes hijacked by people just because it contains a particular term which they wish to discuss. I have voiced my concerns over this to certain mentors.

Please understand that I am NOT saying that you are one of these people, and the point you raised is extremely interesting, but I'm afraid that the original questions will never be cleared if new and more sophisticated points are inserted continually in a discussion. A new thread should be started for that.

Thank you.

What is linear momentum ball bounce?

Linear momentum ball bounce refers to the conservation of momentum that occurs when a ball bounces off a surface. This means that the total momentum of the ball before and after the bounce remains the same.

How does the mass of the ball affect its bounce?

The mass of the ball does not affect its bounce, as long as the surface it is bouncing on is the same. This is because the mass of the ball cancels out in the momentum equation, leaving only the velocity and the angle of the bounce to determine the height and distance of the bounce.

What factors affect the height and distance of a ball bounce?

The height and distance of a ball bounce are affected by the velocity and angle at which the ball hits the surface, as well as the elasticity of the surface. A more elastic surface will result in a higher bounce, while a less elastic surface will result in a lower bounce.

Can the initial velocity of the ball be calculated using the height and distance of the bounce?

Yes, the initial velocity of the ball can be calculated using the height and distance of the bounce, as well as the angle at which the ball was thrown. This can be done using the equation v = sqrt(gh/2sin^2θ), where v is the initial velocity, g is the acceleration due to gravity, h is the height of the bounce, and θ is the angle at which the ball was thrown.

How does air resistance affect the bounce of a ball?

Air resistance can affect the bounce of a ball by reducing its velocity and therefore the height and distance of the bounce. This is because air resistance acts as a force opposing the motion of the ball, and can absorb some of its kinetic energy. However, for most cases, the effect of air resistance on the bounce of a ball is minimal.

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