# Bouncing ball and Newton's third law

If I throw a ball to a wall, what makes the bouncing ball to compress? Is it the equal and opposite force from the wall or is it some molecular level interaction the ball has within that makes it squish?

If it is the equal and opposite force, what happens if a bouncing ball gets to a perfectly elastic collision with a wall by which I mean the ball collides with the wall at x velocity and bounces back with the same velocity.

From what I understand the ball has to transfer energy to the wall for the wall to provide an equal and opposite force, but in this case the ball retains its motion without a bit of energy loss. Doesn't this imply that there weren't any energy transfer from the ball to the wall? If it had wouldn't the ball slow down and makes the wall move atleast a bit?

I understand from reading other answers that a perfect elastic collision might be impossible if a ball hits a wall the whole system (wall + earth moves at least an atomic level distance). But hypothetically, what allows the ball to flip its direction of the velocity after the collision without any energy transfer between the ball and the wall (by not having the wall providing the equal and opposite force to compress the ball)?

Am I missing on some key insight here about how Newton's third law works?

• This is an interesting statement: "the ball has to transfer energy to the wall for the wall to provide an equal and opposite force". As it stands it is not true. However, you may have some context in mind that is unsaid. Can you explain what you mean? Commented May 15, 2021 at 14:04
• I was thinking the ball slowing down when collided is because the kinetic energy is getting transferred to the wall making it move ever so slightly. Was I wrong in that assumption? Commented May 15, 2021 at 14:10

When the ball hits a perfectly rigid (immovable) wall, it's the ball's inertia that causes the ball to compress to a minor or greater extent.

As a simple model consider the ball to be a Hookean spring. Inertia causes compression of the ball and development of a restorative force $$F_{rest}$$ and with Newton's 2nd:

$$ma=-F_{rest}$$

For a Hookean spring:

$$m\ddot{x}=-kx$$

where $$x$$ is the elastic deformation (displacement).

$$\Rightarrow m\ddot{x}+kx=0$$

which is the Newtonian equation of motion of the ball, during the collision.

The compression of the ball continues until all its kinetic energy $$K$$ is converted to spring elastic energy $$U$$, at maximum deformation $$x_{max}$$:

$$K=U$$

or:

$$\frac12mv_0^2=\frac12 kx_{max}^2\Rightarrow x_{max}=\sqrt{\frac{mv_0^2}{k}}$$

where $$v_0$$ is the ball's velocity just prior to the collision.

The restoring force then returns the stored elastic energy $$U$$ to kinetic energy $$K$$.

or is it some molecular level interaction

Ultimately, the ball's behaviour as a spring in compression is indeed due to atomic/molecular interactions.

Of course the above model deals only with perfectly elastic collisions. For Real World collisions some loss factor has to be built in.

• Thanks for the answer! How does a perfectly elastic ball detect the collision without transferring some energy to the wall (causing a reduction in the energy it holds and making it slow down a little bit), Isn't perfect elasticity an impossibility? Shouldn't there be some force > 0 (however low it is) from the wall that initiate this compressing? Commented May 15, 2021 at 14:03
• Perfect elasticity is indeed impossible. But imperfect elasticity is not necessarily caused by energy absorbed by the wall (although it can be). Imagine a ball made from Blutack! Its deformation is partially permanent, so lots of energy permanently lost there.
– Gert
Commented May 15, 2021 at 14:06
• But isn't the initial deformation (although permanent) of the Blutack from some equal and opposite force exerted by some other object/wall? Commented May 15, 2021 at 14:13
• Yes but that doesn't change anything. Assume the wall is perfectly rigid then it can not do any work because its deformation is $0$: $dW=Fds=F\times 0=0$
– Gert
Commented May 15, 2021 at 14:15
• Can a perfectly rigid wall make an oncoming elastic ball to convert its kinetic energy to internal (compressed) energy without exerting a force at all? Isn't this in some sense like the ball is kind of stopping and compressing itself from the inside? Commented May 15, 2021 at 14:20

The ball gets "squashed" because of distortions in its molecular structure due to the sudden impulsive force (forces of very large magnitude acting for a small duration) that the wall exerts on the ball. In a perfectly elastic collision, these deformations get perfectly recovered after the duration of collision is over, and all of the potential energy that the body had gained due to the squashing gets converted to kinetic energy. So, the kinetic energy of the ball remains conserved. Of course, here I am ignoring any dissipative forces, such as friction, which could have been present. So the flow of energy in elastic collision is:

$$KE$$ of ball $$\to$$ $$PE$$ of ball due to distortion $$\to$$ $$KE$$ of ball returning after collision

You are correct that ideally no energy transfer is done from ball to wall. Rather, the kinetic energy of the ball is converted into elastic potential energy in the ball and then again back to kinetic energy of the ball.

When the ball hits the wall, the wall exerts a force (due to Newton's third law). This force is initially applied to the first few particles that touch. It is a force strong enough to slow down those particles immediately.

The particles behind the first ones are still moving forwards. Due to the elastic material the force on the first particles is not propagated to the next particles in full. Rather, these next particles continue forwards and get closer to the first particles. While getting closer, the elastic forces increase.

Soon the elastic forces are large enough to fully stop those next particles. And this continues for the next layer of particles and the next layer and the next etc. Soon they have all been fully stopped.

In this stopped moment, the particles are squeezed together more than usual. They are not at their relaxed separation. The elastic forces between them are pushing backwards, and this is what we define as elastic energy being stored.

These elastic forces now again push the particles apart again, and since the first particles can't move forwards, then the wall will again push to hold them back. The elastic force is thus causing motion purely backwards.

In the ideal scenario, with no energy transferred to the wall, all this elastic energy is again converted to kinetic energy so the speed is the same as before the impact - just the opposite way.

• Thanks a lot, this really helped me visualise what's happening physically. But wouldn't this collision with the wall make this not perfectly elastic because some kind of energy transfer occured atleast in the initial contact, however little it was. "When the ball hits the wall, the wall exerts a force (due to Newton's third law)." So isn't a perfectly elastic bouncing ball collision impossible, cause basically the ball without hitting the wall is kind of stopping and compressing itself from the inside in some sense? Commented May 15, 2021 at 13:34
• @Tangent Sure, no collusion is perfectly ideal and fully elastic in real life. There will always be some kind of energy transfer of loss. Not least within the flexing material itself. But theoretically there doesn't have to be. Remember that in general it doesn't require energy to exert a force. Commented May 15, 2021 at 14:04
• Can you expand on not requiring energy to exert a force in the context of this bouncing ball example? I am a beginner trying to get my head around this. Commented May 15, 2021 at 14:08
• @Tangent It's just a general thing. Force being applied does not necessarily mean energy spent. For instance, the table can hold up the plate with a constant upwards normal force. No energy is exchanged in this act. It can hold it up forever. So, in general there is no reason to expect energy to be transferred just because a force is applied from the wall. What actually happens is, as I tried to outline in the answer, that energy is being converted between various forms within the ball itself. The wall force facilitates the energy conversions, sure, but no energy actually exits the ball. Commented May 15, 2021 at 14:57
• Ok but isn't the instance of the table holding a plate just a case when the objects are at rest and in equilibrium? Instead in this case wouldn't the wall exert a force to slow down the ball to rest? Commented May 15, 2021 at 19:36