If a ball collides with a ball of same mass the first ball stops and the second ball gets the velocity of first ball.The first ball stops due to the reaction force acting on it. But when a ball collides with a wall why doesn't it stop due to the reaction force?

  • $\begingroup$ The mass of the ball and the wall aren't the same. Also in real-life situations, the collision isn't elastic. That explains why the ball doesn't stop. The one you've described where the velocities get swapped is only true if the collision is between two equal masses and is elastic. $\endgroup$ Apr 3, 2020 at 17:47
  • $\begingroup$ Can you explain why you think it should stop? $\endgroup$ Apr 3, 2020 at 17:52
  • $\begingroup$ When a ball collides with another ball of same mass it stops due to reaction force.When it collides with a wall there is also a reaction force.Then why doesnt it stop? $\endgroup$ Apr 3, 2020 at 18:08
  • $\begingroup$ oh ok, so your thought process is just "Reaction forces should stop things"? $\endgroup$ Apr 3, 2020 at 18:21
  • 1
    $\begingroup$ But the action and the reaction forces aren't on the same object, therefore, don't expect them to always cancel out. You've misinterpreted Newton's 3rd Law. $\endgroup$ Apr 3, 2020 at 18:26

3 Answers 3


GIF showing the two different scenarios I'll let the animation speak for itself.

The blue arrow shows the force on the object.

For this scenario to happen it is important that the collision is elastic (all energy is conserved). I used a force that is proportional to penetration depth. This way the balls feel a force that is the same during the deceleration and recoil. In inelastic collisions it is possible for two balls to stick together after the collision. Imagine two pieces of clay colliding in mid air. In inelastic collisions the force is smaller (or zero) during the recoil than during the deceleration.

  • $\begingroup$ Is the force larger at second scenario? $\endgroup$ Apr 4, 2020 at 12:56
  • 1
    $\begingroup$ @AbdullahAlZami Both cases have the same spring constant so for a given penetration depth they experience the same force. The maximal force is larger though for the second scenario. This is because in the first case the second ball moves starts moving during the collision, decreasing the duration of the deceleration. $\endgroup$ Apr 4, 2020 at 13:09
  • $\begingroup$ In the second case if the ball hits the wall with a velocity of 1 ms^-1(assume the mass is 1 kg).Then what will be amount of the force for reflecting the ball?Will it be (2xForce at first case)? $\endgroup$ Apr 4, 2020 at 13:12
  • $\begingroup$ @AbdullahAlZami For a general collision you can't know this. The change in momentum is $\Delta p=2 $kgm/s though. You can calculate the force for the case where the force is like a spring (like I did) but this would be too long for a comment. Do you really want to know this? $\endgroup$ Apr 4, 2020 at 13:19
  • $\begingroup$ Is the force at second case = 2x(force at first case)? $\endgroup$ Apr 4, 2020 at 13:23

Because the reaction force is larger.

The force needed to make an object reach a certain speed is the same as the force needed to slow it down from that speed to zero over the same time. This is Newton's 2nd law.

So, when two equal and movable objects collide, the (action) force that is required to make one object speed up to a certain speed is exactly the same as the (reaction) force required to slow the impacting object down to precisely zero.

When two non-equal objects collide, this is not necessarily so, and thus you will not see any of the objects stopping. Instead you might see that when

  • a lighter object hits a heavier object, the action/reaction forces between them are larger than they would be between equal objects. And thus, the reaction force that stops the lighter object is larger than what is needed to precisely slow it down to zero. Thus, the lighter object slows down to zero and then start moving backwards, as if it bounces off.
  • When an object hits a wall, you can consider this a collision between a light object and a very heavy object, and so, this bouncing-back effect is much stronger (the reaction force is much bigger). It might actually be so strong that the impacting light object bounces backwards with exactly the speed it came with, just directed oppositely.
  • 2
    $\begingroup$ Doesn't this beg the question as to why the reaction force is different in different scenarios? $\endgroup$ Apr 3, 2020 at 18:06
  • $\begingroup$ Yeah!Are they different or not? $\endgroup$ Apr 3, 2020 at 18:12
  • 1
    $\begingroup$ The action-reaction terminology here is very misleading. I strongly recommend using different words. Action-reaction is almost universally associated with Newton’s 3rd law, and there the action-reaction forces are always equal and opposite $\endgroup$
    – Dale
    Apr 3, 2020 at 19:07
  • $\begingroup$ Action-reaction forces are always equal and opposite. But Newton's 2nd law says that a body's acceleration is inversely proportional to its mass, so during the time while the forces are applied, the lighter mass is accelerated to a higher speed (or change of speed to be precise) than the heavier one. $\endgroup$
    – Cuspy Code
    Apr 3, 2020 at 19:58

Try considering momentum conservation. The wall won’t be moving before or after the collision with the ball, and presuming the collision is elastic (no energy lost to heat, sound etc.) then the sum of the momenta will be conserved before and after the collision and you should find that the ball must have the same momentum before and after the collision. So it must be in moving too.


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.