4
$\begingroup$

I understand that when a ball collides with a wall it exerts a force. By Newton's 3rd law, the wall exerts an equal and opposite force back on the ball. As a result in an ideal world, the ball will bounce back at the same speed as before, conserving momentum. Is there a mistake in this intuition?

In addition, I don't understand how the ball can travel back at the same speed as before when hitting the wall. Wouldn't the wall have to exert twice the force ( 2 times the force the ball exerts on the wall ) to do this? The way I understand this, is when the wall exerts a force equal in magnitude to the force exerted by the ball on the wall, it should bring the ball to halt. It should then exert this same force again to send the ball back in the opposite direction. So shouldn't the wall exert twice the force?

Could someone please explain the flaw in my intuition?

$\endgroup$
1
  • $\begingroup$ Your $2 \times$ force proposal breaks every physics law,- starting from momentum conservation, then energy conservation law, i.e. first law of thermodynamics. If that would be the case as you say, then constantly bumping on walls and doubling previous momentum, body could raise kinetic energy to infinity for no reason. $\endgroup$ Commented Apr 29, 2022 at 10:25

4 Answers 4

2
$\begingroup$

I understand that when a ball collides with a wall it exerts a force. By Newton's 3rd law, the wall exerts an equal and opposite force back on the ball. As a result in an ideal world, the ball will bounce back at the same speed as before, conserving momentum. Is there a mistake in this intuition?

Unfortunately yes. You are not thinking clearly about system, internal forces and external forces.

Momentum is conserved when external force on system is zero.

What is the system here. There are two options.

If you consider ball alone, then momentum is not conserved. The ball experiences an external force during collision. Momentum conservation is not valid. In any case, we can see the momentum has changed. The direction of momentum has changed.

The other option is to consider ball + wall as system. Here wall is connected to rest of building. So again momentum conservation is not valid. Even if the wall was lying on a friction-less surface, huge mass of wall would make applying momentum conservation impractical.

In addition, I don't understand how the ball can travel back at the same speed as before when hitting the wall. Wouldn't the wall have to exert twice the force ( 2 times the force the ball exerts on the wall ) to do this?...

Here you are mixing up force, impulse, momentum and speed. You are thinking about mechanism of ball stopping and returning correctly.

Impulse is change in momentum. Force is impulse / time.

...The way I understand this, is when the wall exerts a force equal in magnitude to the force exerted by the ball on the wall, it should bring the ball to halt. ...

Correct idea with wrong terms. When ball stops, wall applies impulse equal to momentum of ball.

...It should then exert this same force again to send the ball back in the opposite direction. So shouldn't the wall exert twice the force?

Not same force, same impulse! So net impulse by wall would be twice the momentum of ball.

Of course this double impulse will act on wall also (as per Newton's third Law). Hopefully our wall is strong enough to handle that.

$\endgroup$
0
$\begingroup$

For the ball to bounce back, all that it needs to do is to accelerate in the opposite direction to its initial motion for some period of time (i.e., reduce its velocity to zero and then accelerate back the other way).

Acceleration is given by $a = F/m$ and velocity $v = at$ or $v = Ft/m$.

If the initial velocity is $u$ then $Ft/m$ needs to be $-2u$ to stop and reverse direction with the same speed.

Since we have time $t$ in the equation, then $F$ can be any negative value and as long as the bounce takes long enough, it will reverse (at least mathematically - we have to stop the ball before it goes through the wall). $F$ can also vary over time; it doesn't have to be the same throughout the whole contact with the wall.

In practice, the force will vary. The force is usually mostly caused by the ball deforming (assuming it is elastic) and partially by the wall deforming also.

As the ball is compressed, it acts like a spring. The more it is compressed the greater the force. So on initial contact there will be zero force and when the ball comes to a stop the force will be at a maximum, but we can't really say exactly what it is.

A hard ball will generate a greater force as it stops and reverses in a shorter amount of time whereas a soft ball will generate a smaller force as it compresses more and takes longer to reverse.

Whatever force is generated against the wall, it is equal and opposite. The ball pushes against the wall and the wall pushes against the ball.

$\endgroup$
0
$\begingroup$

Of some reason you are equating the force that the ball exerts with its speed. So that it takes that amount of force to exactly reduce the speed of the ball to zero, and thus it should require double as much force to caues it to revert backwards at that same speed again.

But there is no reason to tie a force to the speed. No law of physics ties force to speed. Rather, we have a law that ties force to the change in speed (to acceleration). We call it Newton's 2nd law of motion:

$$\sum F=ma\quad\Leftrightarrow \quad \sum F=m \frac{\mathrm dv}{\mathrm dt}.$$

With the same impact speed, you might see different forces depending on how much time the speed reducing takes place over. Slowing down the ball fast implies a much higher speed.

So, clearly, you impact force by ball-on-wall is not tied to the value of the speed that the ball happens to impact with. Instead, the force causes what is necessary for other laws of physics to be upheld. A law such as momentum conservation and energy conservation. The force that turns out to be exerted, turns out to exactly match what is needed for the ball to perfectly slow down to zero and then speed up again to the same speed in the opposite direction (assuming perpendicular impact).


The reason for this is the mechanics of the ball and wall. The ball might ideally be modelled as a fully elastic system and the wall as a perfectly rigid system.

  • The ball will at impact apply a force on the wall which the wall counters with an eqal force backwards. This will then compress the ball due to the ball's inertia that keeps it moving forwards. This compression will store the kinetic energy as elastic energy within it as was it a spring. At its maximum compression, the stored elastic energy exactly equals the original kinetic energy.

  • Now, since we model this as an ideal elastic (spring) system, all this energy will expectedly be delivered back without loss by the ball decompressing/elongating again. During that decompression, the ball now pushes on the wall once more and the wall correspondingly pushes on the ball. This will cause the ball to move away from the wall. Since the wall doesn't move, all the stored elastic energy can only be converted back into kinetic energy in the ball again. The exerted forces exactly take care of that.

As you can see we are not dealing with just one force but rather with a bundle of forces over the duration of the impact. The force at each moment takes care of causing the speed change at that moment. And this force is not tied to the speed at all.

$\endgroup$
0
$\begingroup$

Your question about Newton's third law is so basic. Various other laws depends on it. You have to use your common senses here. The ball exerts all it's force on the wall and it's velocity becomes 0. The wall doesn't move due to friction but applies the same and opposite force on the ball.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.