(I'm repeating myself a lot here, but it's because I want to make my confusion clear.)

If 2 billiard balls are the same exact mass, and one hits another stationary one head on, I have heard that the hitting ball will often stop entirely while the one which got hit will go off at the original velocity of the first one (ignoring friction and heat and other potential loss of energy).

I understand that this agrees with conservation of momentum. However, now that I am thinking about it, I am a little confused as to how it is possible.


Let's say we have 2 balls, Ball 1 and Ball 2. The balls have equal mass. $$M_{1} = M_2$$

Ball 1 is the hitting ball, with an original velocity of $V_1$, and Ball 2 is getting hit, originally stationary.

Once Ball 1 hits Ball 2, it immediately starts accelerating it at the same rate that it decelerates. In other words, Ball 1 exerts the same force on Ball 2 that Ball 2 exerts on Ball 1. In this way, momentum is conserved, so that for any amount of momentum that Ball 2 gains, Ball 1 loses.

$$F_{1 \to 2} = F_{2 \to 1}$$

That's just Newton's third law.

Since they have the same mass, Ball 1 will decelerate at the same rate Ball 2 accelerates.

However, after a certain amount of time, both balls will have the same amount of momentum in the same direction. That is, Ball 1 will have been decelerated to ${V_{i}/2}$ and ball 2 will have been accelerated to ${V_{i}/2}$.

Ball 1 and Ball 2 at this point of equal momentum must have the same velocity, since they have the same mass.

Now, the only way for Ball 1 to exert a force on Ball 2 is for them to be in contact. However, the instant after they gain the same velocity, the 2 balls would no longer be in contact, as Ball 2 would now be moving away from Ball 1, or at least at the same rate as Ball 1.

That being said, it seems impossible that Ball 2 would ever become faster than Ball 1, since Ball 2 would only be able to be accelerated up to the point where it is going at the same velocity as Ball 1.

And it seems even more impossible for Ball 1 to stop completely and Ball 2 to go off at the original velocity of Ball 1.

What am I missing?


After reading some scrupulous comments, (not in a bad way though, I love it when there are new variables introduced and things pointed out, and all the "what if" questions allways make for awesome discussion, thanks for the comments!) I've realized its better to imagine the billiard balls as floating through space than on a pool table, to get rid of the possibility that spin will affect the collision.

Or we could just imagine a frictionless pool table.

But better to imagine them floating through space, because everyone loves space

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    $\begingroup$ I think the original OP statement needs some qualifications. For the case of a cue ball striking an object ball, the cue ball must not be spinning when it hits the object ball, if it is to stop "dead". If the cue ball has top spin on it, it will continue moving forward after contact, albeit slower, and if the cue ball has back spin on it, it will "draw back" after the collision. $\endgroup$ Commented Sep 20, 2018 at 19:33
  • $\begingroup$ it's because it's a perfectly elastic collision. so momentum is conserved as in all collisions. but in a perfectly elastic collision, so also is kinetic energy conserved. @FrodCube 's answer spells out the solution. $\endgroup$ Commented Sep 21, 2018 at 0:07
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    $\begingroup$ Can we also note that 'billiard ball' is misleading here; in the real world they have angular momentum. This scenario applies better to non-rotating block collisions. $\endgroup$
    – Keith
    Commented Sep 21, 2018 at 3:47
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    $\begingroup$ It does not stop dead on, it will roll a little. In order to get the effect of dead stop, the player must apply spin to the cueball. $\endgroup$
    – Stian
    Commented Sep 21, 2018 at 7:08
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    $\begingroup$ It would still do so. There is a difference between reality and theory, in that reality ignores assumptions of ideality while theory quite often relies on them... $\endgroup$
    – Stian
    Commented Sep 21, 2018 at 13:59

7 Answers 7


Your analysis is correct up to the point when the two speeds are equal. This is also the point when the relative velocity is zero. But at this point the two balls are deformed (elastically). The forces acting on the two balls during approach is an elastic force. However small the deformation is, it is there. So what happens now is that indeed the two bodies start to fall apart (the distance between them increases) and they start to expand back to the original shape. The interaction force starts to decrease (does not drop to zero instantly). The work done during this second part of the process is the same as during the first part so the kinetic energies and velocities change by another on half. This is the difference between perfect elastic and non-elastic collisions. In the later case the two bodies do not expand back to the same shape (or not at all) so the exchange of velocities is not complete.

It may help if imagine a spring between the two bodies. When the velocities are equal, the spring is still compressed. Actually it has maximum compression. So obviously the force exerted by the spring is not zero as it starts to expand.

Note:A nice example is considering a rubber ball bouncing off a ground(suppose perfectly elastically), in that situation there is also a point where the relative velocity with the ground is zero(when the ball is momentarily at rest with the ground) but still the ball rises up due to elastic forces as it gains kinetic energy due to the energy stored through the deformation it had initially endured.

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    $\begingroup$ Brilliant answer, you beat me to it. A nice example to add would be a rubber ball bouncing off a ground(suppose perfectly elastically), in that situation there is also a point where the relative velocity with the ground is zero but still the ball rises up due to the energy stored through the deformation it had initially endured. $\endgroup$ Commented Sep 20, 2018 at 17:30
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    $\begingroup$ This is why cars are not elastic. Otherwise, a crash would cause much stronger forces. $\endgroup$ Commented Sep 20, 2018 at 19:32
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    $\begingroup$ @stannius, I think crumple zones are more a matter of spreading the collision out over a longer period of time, which lessens the rate of deceleration. $\endgroup$
    – user205805
    Commented Sep 21, 2018 at 2:21
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    $\begingroup$ I think en.wikipedia.org/wiki/Newton%27s_cradle needs to be mentioned $\endgroup$
    – UKMonkey
    Commented Sep 21, 2018 at 10:36
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    $\begingroup$ The problem with observing the deformation directly is not just the speed. The change in size may be too small to be easily visible in a video, even if it is slowed down. Here are some calculations and videos. dbkcues.ru/articles-2/collision-of-billiard-balls/?lang=en The deformation is just a fraction of a mm, less than 1% of the diameter of the ball. This will be difficult to see in a video. You'll have to examine individual frames and do careful measurements. $\endgroup$
    – nasu
    Commented Sep 24, 2018 at 15:39

You are forgetting about conservation of energy. You need to impose that

$$mv = mv_1 + mv_2$$ and $$\frac{1}{2}mv^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2$$

And this is only solved by $v_1=0$ and $v_2=v$.

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    $\begingroup$ This is of-course true mathematically but it does not answer OP's question on how a collision is maintained even when there seems to be no relative velocity between the objects. $\endgroup$ Commented Sep 20, 2018 at 17:22
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    $\begingroup$ i think this is the best answer. $\endgroup$ Commented Sep 21, 2018 at 0:05
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    $\begingroup$ Note that conservation of energy and momentum is insufficient to solve some 3-body collisions, and so the exact kinematic details (as per nasu's answer) would have to be examined. For example Newton's cradle with 3 balls already has more than one solution that conserves both momentum and energy in most cases, such as where 2 balls are in contact and move together at unit speed to the right to hit 1 ball moving at unit speed to the left, in which the actual resulting speeds are $(-1,1,1)$ but there is another solution $(1-c,0,c)$ with $c=(1+\sqrt{5})/2$ that conserves both energy and momentum. $\endgroup$
    – user21820
    Commented Sep 21, 2018 at 3:18
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    $\begingroup$ Thanks so much for the answer, the problem with the energy equations is that although they are true, they allways seem to me to be derived from intuitive analysis but not intuitive in themselves, if that makes any sense. However, that may just be me; appreciate the energy answer regardless! Thanks! $\endgroup$ Commented Sep 21, 2018 at 9:24
  • $\begingroup$ @user21820 right. And perhaps more relevantly here, all billiard games crucially depend on the fact that the collision can come out differently from just “cue ball stops, giving all its energy&momentum to the object bass”. Such straight-on shots are a quite unusual situation, usually you'll want to hit the object ball at an angle and then only part of the energy is transferred, and the momentum split up in two diverging directions. Fine-tuning this momentum- and thus direction-splitting is a large part of a billiard player's skill. $\endgroup$ Commented Sep 21, 2018 at 10:50

I think when you visualize the impact, it's a bad idea to think of the balls as completely rigid objects. We can think of the impact as if there's a massless spring between them. Even when they get to the same velocity, the spring will continue pushing the balls until the incoming ball loses all its speed and the speed of the other ball gets to maximum. That will be the moment when they lose contact.


Short answer: During the collision, while the first ball is going faster than the second, the balls are compressing. Once the second ball begins moving as fast as first, the balls start uncompressing, pushing the balls apart. Once they fully separate, they are moving different speeds.

Medium answer:

The first half of the collision pushes the billards together and deforms them, and in the second half of the collision, the energy stored in the deformation is released back into mechanical energy, and the billards undeform and push apart, resulting in different velocities. Billards are "elastic", or like springs: almost all of the energy when deforming them is stored and mechanically released.

This elastic deformation is the same reason when you throw a billard against a wall, it doesn't stick to the wall; it bounces back. Inelastic materials, like a ball of Play-Doh, will stick to the wall. (Or other balls of Play-Doh.)

Long answer: In addition, the earlier answers, you can reach the result mathematically with physical laws and one assumption.

A result of Newton's Laws is the conservation of momentum

$$m_a\mathbf{v_{a1}} + m_b\mathbf{v_{b1}}=m_a\mathbf{v_{a2}}+m_b\mathbf{v_{b2}}$$

In this case, $\mathbf{v_{b1}}=\mathbf{0}$ and $m_a=m_b$, so


Obviously, there are many values that satisfy this equation. So we need add another physical law: the Law of Conservation of Energy. (Assumption: there is negligible rotation of the balls.)

$$\frac{m_a\mathbf{v_{a1}}^2}{2}+\frac{m_b\mathbf{v_{a1}}^2}{2}=\frac{m_a\mathbf{v_{a2}}^2}{2}+\frac{m_b\mathbf{v_{b2}}^2}{2} + \Delta H$$ $$\mathbf{v_{a1}}^2=\mathbf{v_{a2}}^2+\mathbf{v_{b2}}^2 + \frac{2\Delta H}m$$

Because billards are elastic, none of the energy is converted into heat (this in fact, is what is meant by elastic...you could go further into materials science and find out exactly what about the chemistry of the billards causes this property, but at the end of the day, billards are elastic, like bouncy balls or springs), and

$$\mathbf{v_{a1}}^2=\mathbf{v_{a2}}^2+\mathbf{v_{b2}}^2$$ $$\mathbf{v_{a1}}^2=(\mathbf{v_{a1}} - \mathbf{v_{b2}})^2+\mathbf{v_{b2}}^2$$ $$\mathbf{v_{a1}}^2=\mathbf{v_{a1}}^2 - 2\mathbf{v_{a1}}\mathbf{v_{b2}} + \mathbf{v_{b2}}^2 + \mathbf{v_{b2}}^2$$ $$\mathbf{v_{a1}}\mathbf{v_{b2}} = \mathbf{v_{b2}}^2$$

For a straight-on collision, $\mathbf{v_{a1}}$ and $\mathbf{v_{b2}}$ are parallel, so

$$|\mathbf{v_{a1}}||\mathbf{v_{b2}}| = |\mathbf{v_{b2}}|^2$$

Since billards cannot pass through one another $\mathbf{v_{b2}} \neq 0$, and thus $\mathbf{v_{b2}} = \mathbf{v_{a1}}$.

Note that this result depends on the assumption $\Delta H = 0$; i.e. the balls un-deformation converted 100% of the energy from the deformation back into mechanical energy. I.e. an elastic collision.,


Let's look at it from a different frame of reference and use a simple relativistic way of thinking:

Consider a frame of reference that has the origin $(0,0)$ at the midpoint between the 2 balls, and moves with velocity $\frac{v}{2}$ in the same direction as the first ball.

Then in this frame of reference, both balls move with velocity $\frac{v}{2}$. In this frame of reference the situation is symmetric, so the result must also be symmetric: both balls will bounce and recede with the same speed (or pass through each other, which is not possible) to conserve the momentum.

When we go back to the original frame of reference this gives us the ball that was moving initially as now remaining in rest, and the ball that was previously resting as moving with velocity $v$.

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    $\begingroup$ This is by far the best way to accurately deal with the difficult problem especially if there Is energy loss and inelastic collision..it will still be perfectly symmetrical in the centre of mass frame and we can still say a lot and analyse best therein. In CM frame we have left right symmetry obviously. Also time symmetry is highly likely during interaction whereupon we see the “release” paradox raised in the question is also an “absorption” paradox. $\endgroup$
    – blanci
    Commented Mar 29, 2021 at 22:17

Collisions between billiard balls are essentially elastic collisions. Although billiard balls seem rigid, they are more-or-less elastic - the material responds to compressive stress (pressure) with some amount of elastic strain (deformation). That strain stores energy; work is done to deform the material, which is released back as the stress is removed and the material restores to its original shape. The strain is tiny, but it is non-zero and, except for perhaps exotic or theoretical materials, no material is perfectly elastic, so some energy will be lost from the collision. Nevertheless, billiard balls are a pretty good approximation of elastic collision.

Both momentum and energy must be conserved in the collision. The energy of the collision comes from the relative velocity between the two balls; conservation of this energy means that the magnitude of this relative velocity must be the same after the collision as before. The only way to satisfy this condition as well as conservation of momentum is if all of the striking ball's velocity is transferred to the struck ball - the striking ball comes to a stop.

Collisions between real billiard balls on a real playing surface are more complicated for all sorts of reasons. Here are just a few:

  • A rolling ball has both the linear momentum of its motion along the table and the angular momentum of its roll, so the striking ball may "follow" the struck ball. The collision primarily transfers linear momentum, so while the striking ball should initially stop after the collision, its angular momentum can cause it to start rolling forward again as it "drives" against the playing surface.
  • The "click" of the collision is acoustic energy radiated away from the balls in collision; that energy came from the kinetic energy of the moving ball and is gone from the "system" (the two colliding balls). In the real world, it's tiny and pretty much inconsequential, but it's just one example of the many ways in which the elastic collision is imperfect.
  • The playing surface imposes drag on any moving ball. If the ball is sliding, there is friction causing the ball to both slow down and start rolling. If the ball is rolling, it resists the rolling. If a ball is spinning in one spot, it will start moving in some direction (depending on the orientation of its spin axis). If this were not so, moving balls would never come to a stop.

The stopping of the ball which hits another ball depends upon the elasticity of the balls or even one of the two balls. If the balls are very hard like if made of wood or metal, when one ball hits another there is no "waiting time" for the transfer of kinetic energy from one to another. It is near instantaneous and hence one ball transfers nearly all of it's kinetic energy to the ball it hits. Hence the ball "nearly" stops.

Eg. Newtons Cradle

enter image description here

This is not the case in case of elastic balls.

enter image description here

Consider two equal balls of rubber,

When the ball A hits B, the shape of A changes due the inertia of the ball B and also due to it's own elasticity. When the ball A relaxes it transfers it's kinetic energy to the ball B minus the internal friction in the ball A due to shape change and other external frictional losses. Now the ball B deforms a little due to ball A pushing on it(further waste of energy) and moves forward by pushing upon ball A when it relaxes + the energy imparted by the collision itself.

Now the fate of ball A is either it can move forward using the little energy it has left after the collision or roll backwards depending upon the force with which it has hit the other ball.

And it seems even more impossible for Ball 1 to stop completely and Ball 2 to go off at the original velocity of Ball 1.

What am I missing?

If the balls A and B in the above picture were rigid,

The ball B doesn't go off in the "original velocity", it can be near the original velocity but never the velocity of the ball A itself.

And the ball A stops due to the nearly total transfer of kinetic energy.

Hope this helps.


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