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Why does a billiard ball stop when it hits another billiard ball head on?
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(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.
Consider:
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?
EDIT:
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
newtonian-mechanics forces momentum conservation-laws collision
edited 28 mins ago
CJ Dennis
430413
asked Sep 20 at 17:08
Joshua Ronis
6731414
 |Â
show 6 more comments
up vote
60
down vote
favorite
(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.
Consider:
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?
EDIT:
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
newtonian-mechanics forces momentum conservation-laws collision
edited 28 mins ago
CJ Dennis
430413
asked Sep 20 at 17:08
Joshua Ronis
6731414
3
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.
– David White
Sep 20 at 19:33
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.
– robert bristow-johnson
Sep 21 at 0:07
1
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.
– Keith
Sep 21 at 3:47
1
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.
– Stian Yttervik
Sep 21 at 7:08
1
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...
– Stian Yttervik
2 days ago
 |Â
show 6 more comments
up vote
60
down vote
favorite
up vote
60
down vote
favorite
(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.
Consider:
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?
EDIT:
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
newtonian-mechanics forces momentum conservation-laws collision
edited 28 mins ago
CJ Dennis
430413
asked Sep 20 at 17:08
Joshua Ronis
6731414
(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.
Consider:
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?
EDIT:
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
newtonian-mechanics forces momentum conservation-laws collision
newtonian-mechanics forces momentum conservation-laws collision
edited 28 mins ago
CJ Dennis
430413
asked Sep 20 at 17:08
Joshua Ronis
6731414
edited 28 mins ago
CJ Dennis
430413
asked Sep 20 at 17:08
Joshua Ronis
6731414
edited 28 mins ago
CJ Dennis
430413
edited 28 mins ago
CJ Dennis
430413
edited 28 mins ago
CJ Dennis
430413
430413
asked Sep 20 at 17:08
Joshua Ronis
6731414
asked Sep 20 at 17:08
Joshua Ronis
6731414
asked Sep 20 at 17:08
Joshua Ronis
6731414
6731414
3
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.
– David White
Sep 20 at 19:33
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.
– robert bristow-johnson
Sep 21 at 0:07
1
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.
– Keith
Sep 21 at 3:47
1
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.
– Stian Yttervik
Sep 21 at 7:08
1
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...
– Stian Yttervik
2 days ago
 |Â
show 6 more comments
3
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.
– David White
Sep 20 at 19:33
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.
– robert bristow-johnson
Sep 21 at 0:07
1
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.
– Keith
Sep 21 at 3:47
1
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.
– Stian Yttervik
Sep 21 at 7:08
1
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...
– Stian Yttervik
2 days ago
3
3
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.
– David White
Sep 20 at 19:33
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.
– David White
Sep 20 at 19:33
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.
– robert bristow-johnson
Sep 21 at 0:07
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.
– robert bristow-johnson
Sep 21 at 0:07
1
1
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.
– Keith
Sep 21 at 3:47
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.
– Keith
Sep 21 at 3:47
1
1
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.
– Stian Yttervik
Sep 21 at 7:08
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.
– Stian Yttervik
Sep 21 at 7:08
1
1
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...
– Stian Yttervik
2 days ago
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...
– Stian Yttervik
2 days ago
 |Â
show 6 more comments
5 Answers
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up vote
75
down vote
accepted
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.
edited Sep 20 at 17:32
Tausif Hossain
2,3381618
answered Sep 20 at 17:25
nasu
1,597711
3
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.
– Tausif Hossain
Sep 20 at 17:30
4
This is why cars are not elastic. Otherwise, a crash would cause much stronger forces.
– rexkogitans
Sep 20 at 19:32
3
@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.
– user128216
Sep 21 at 2:21
2
+1 This great answer would be tremendous if it linked to a hi-speed-camera video of a "rigid" body deforming in a collision. Does anyone know a good one?
– JiK
2 days ago
5
I think en.wikipedia.org/wiki/Newton%27s_cradle needs to be mentioned
– UKMonkey
2 days ago
 |Â
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20
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You are forgetting about conservation of energy. You need to impose that
$$mv = mv_1 + mv_2$$
and
$$frac12mv^2 = frac12mv_1^2 + frac12mv_2^2$$
And this is only solved by $v_1=0$ and $v_2=v$.
answered Sep 20 at 17:18
FrodCube
1,2921515
22
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.
– Tausif Hossain
Sep 20 at 17:22
1
i think this is the best answer.
– robert bristow-johnson
Sep 21 at 0:05
4
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 $c=(1+sqrt5)/2$ that conserves both energy and momentum.
– user21820
Sep 21 at 3:18
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!
– Joshua Ronis
Sep 21 at 9:24
@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 bassâ€Â. 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.
– leftaroundabout
2 days ago
 |Â
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15
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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.
answered Sep 20 at 17:32
physicsguy19
331112
add a comment |Â
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6
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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
$$m_amathbfv_a1 + m_bmathbfv_b1=m_amathbfv_a2+m_bmathbfv_b2$$
In this case, $mathbfv_b1=mathbf0$ and $m_a=m_b$, so
$$mathbfv_a1=mathbfv_a2+mathbfv_b2$$
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.)
$$fracm_amathbfv_a1^22+fracm_bmathbfv_a1^22=fracm_amathbfv_a2^22+fracm_bmathbfv_b2^22 + Delta H$$
$$mathbfv_a1^2=mathbfv_a2^2+mathbfv_b2^2 + frac2Delta Hm$$
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
$$mathbfv_a1^2=mathbfv_a2^2+mathbfv_b2^2$$
$$mathbfv_a1^2=(mathbfv_a1 - mathbfv_b2)^2+mathbfv_b2^2$$
$$mathbfv_a1^2=mathbfv_a1^2 - 2mathbfv_a1mathbfv_b2 + mathbfv_b2^2 + mathbfv_b2^2$$
$$mathbfv_a1mathbfv_b2 = mathbfv_b2^2$$
For a straight-on collision, $mathbfv_a1$ and $mathbfv_b2$ are parallel, so
$$|mathbfv_a1||mathbfv_b2| = |mathbfv_b2|^2$$
Since billards cannot pass through one another $mathbfv_b2 neq 0$, and thus $mathbfv_b2 = mathbfv_a1$.
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.,
answered 2 days ago
Paul Draper
26115
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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.
answered 14 hours ago
Anthony X
2,44311218
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5 Answers
5
active
oldest
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5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
75
down vote
accepted
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.
edited Sep 20 at 17:32
Tausif Hossain
2,3381618
answered Sep 20 at 17:25
nasu
1,597711
3
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.
– Tausif Hossain
Sep 20 at 17:30
4
This is why cars are not elastic. Otherwise, a crash would cause much stronger forces.
– rexkogitans
Sep 20 at 19:32
3
@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.
– user128216
Sep 21 at 2:21
2
+1 This great answer would be tremendous if it linked to a hi-speed-camera video of a "rigid" body deforming in a collision. Does anyone know a good one?
– JiK
2 days ago
5
I think en.wikipedia.org/wiki/Newton%27s_cradle needs to be mentioned
– UKMonkey
2 days ago
 |Â
show 4 more comments
up vote
75
down vote
accepted
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.
edited Sep 20 at 17:32
Tausif Hossain
2,3381618
answered Sep 20 at 17:25
nasu
1,597711
3
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.
– Tausif Hossain
Sep 20 at 17:30
4
This is why cars are not elastic. Otherwise, a crash would cause much stronger forces.
– rexkogitans
Sep 20 at 19:32
3
@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.
– user128216
Sep 21 at 2:21
2
+1 This great answer would be tremendous if it linked to a hi-speed-camera video of a "rigid" body deforming in a collision. Does anyone know a good one?
– JiK
2 days ago
5
I think en.wikipedia.org/wiki/Newton%27s_cradle needs to be mentioned
– UKMonkey
2 days ago
 |Â
show 4 more comments
up vote
75
down vote
accepted
up vote
75
down vote
accepted
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.
edited Sep 20 at 17:32
Tausif Hossain
2,3381618
answered Sep 20 at 17:25
nasu
1,597711
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.
edited Sep 20 at 17:32
Tausif Hossain
2,3381618
answered Sep 20 at 17:25
nasu
1,597711
edited Sep 20 at 17:32
Tausif Hossain
2,3381618
edited Sep 20 at 17:32
Tausif Hossain
2,3381618
edited Sep 20 at 17:32
Tausif Hossain
2,3381618
2,3381618
answered Sep 20 at 17:25
nasu
1,597711
answered Sep 20 at 17:25
nasu
1,597711
answered Sep 20 at 17:25
nasu
1,597711
1,597711
3
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.
– Tausif Hossain
Sep 20 at 17:30
4
This is why cars are not elastic. Otherwise, a crash would cause much stronger forces.
– rexkogitans
Sep 20 at 19:32
3
@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.
– user128216
Sep 21 at 2:21
2
+1 This great answer would be tremendous if it linked to a hi-speed-camera video of a "rigid" body deforming in a collision. Does anyone know a good one?
– JiK
2 days ago
5
I think en.wikipedia.org/wiki/Newton%27s_cradle needs to be mentioned
– UKMonkey
2 days ago
 |Â
show 4 more comments
3
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.
– Tausif Hossain
Sep 20 at 17:30
4
This is why cars are not elastic. Otherwise, a crash would cause much stronger forces.
– rexkogitans
Sep 20 at 19:32
3
@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.
– user128216
Sep 21 at 2:21
2
+1 This great answer would be tremendous if it linked to a hi-speed-camera video of a "rigid" body deforming in a collision. Does anyone know a good one?
– JiK
2 days ago
5
I think en.wikipedia.org/wiki/Newton%27s_cradle needs to be mentioned
– UKMonkey
2 days ago
3
3
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.
– Tausif Hossain
Sep 20 at 17:30
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.
– Tausif Hossain
Sep 20 at 17:30
4
4
This is why cars are not elastic. Otherwise, a crash would cause much stronger forces.
– rexkogitans
Sep 20 at 19:32
This is why cars are not elastic. Otherwise, a crash would cause much stronger forces.
– rexkogitans
Sep 20 at 19:32
3
3
@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.
– user128216
Sep 21 at 2:21
@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.
– user128216
Sep 21 at 2:21
2
2
+1 This great answer would be tremendous if it linked to a hi-speed-camera video of a "rigid" body deforming in a collision. Does anyone know a good one?
– JiK
2 days ago
+1 This great answer would be tremendous if it linked to a hi-speed-camera video of a "rigid" body deforming in a collision. Does anyone know a good one?
– JiK
2 days ago
5
5
I think en.wikipedia.org/wiki/Newton%27s_cradle needs to be mentioned
– UKMonkey
2 days ago
I think en.wikipedia.org/wiki/Newton%27s_cradle needs to be mentioned
– UKMonkey
2 days ago
 |Â
show 4 more comments
up vote
20
down vote
You are forgetting about conservation of energy. You need to impose that
$$mv = mv_1 + mv_2$$
and
$$frac12mv^2 = frac12mv_1^2 + frac12mv_2^2$$
And this is only solved by $v_1=0$ and $v_2=v$.
answered Sep 20 at 17:18
FrodCube
1,2921515
22
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.
– Tausif Hossain
Sep 20 at 17:22
1
i think this is the best answer.
– robert bristow-johnson
Sep 21 at 0:05
4
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 $c=(1+sqrt5)/2$ that conserves both energy and momentum.
– user21820
Sep 21 at 3:18
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!
– Joshua Ronis
Sep 21 at 9:24
@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 bassâ€Â. 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.
– leftaroundabout
2 days ago
 |Â
show 5 more comments
up vote
20
down vote
You are forgetting about conservation of energy. You need to impose that
$$mv = mv_1 + mv_2$$
and
$$frac12mv^2 = frac12mv_1^2 + frac12mv_2^2$$
And this is only solved by $v_1=0$ and $v_2=v$.
answered Sep 20 at 17:18
FrodCube
1,2921515
22
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.
– Tausif Hossain
Sep 20 at 17:22
1
i think this is the best answer.
– robert bristow-johnson
Sep 21 at 0:05
4
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 $c=(1+sqrt5)/2$ that conserves both energy and momentum.
– user21820
Sep 21 at 3:18
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!
– Joshua Ronis
Sep 21 at 9:24
@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 bassâ€Â. 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.
– leftaroundabout
2 days ago
 |Â
show 5 more comments
up vote
20
down vote
up vote
20
down vote
You are forgetting about conservation of energy. You need to impose that
$$mv = mv_1 + mv_2$$
and
$$frac12mv^2 = frac12mv_1^2 + frac12mv_2^2$$
And this is only solved by $v_1=0$ and $v_2=v$.
answered Sep 20 at 17:18
FrodCube
1,2921515
You are forgetting about conservation of energy. You need to impose that
$$mv = mv_1 + mv_2$$
and
$$frac12mv^2 = frac12mv_1^2 + frac12mv_2^2$$
And this is only solved by $v_1=0$ and $v_2=v$.
answered Sep 20 at 17:18
FrodCube
1,2921515
answered Sep 20 at 17:18
FrodCube
1,2921515
answered Sep 20 at 17:18
FrodCube
1,2921515
answered Sep 20 at 17:18
FrodCube
1,2921515
1,2921515
22
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.
– Tausif Hossain
Sep 20 at 17:22
1
i think this is the best answer.
– robert bristow-johnson
Sep 21 at 0:05
4
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 $c=(1+sqrt5)/2$ that conserves both energy and momentum.
– user21820
Sep 21 at 3:18
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!
– Joshua Ronis
Sep 21 at 9:24
@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 bassâ€Â. 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.
– leftaroundabout
2 days ago
 |Â
show 5 more comments
22
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.
– Tausif Hossain
Sep 20 at 17:22
1
i think this is the best answer.
– robert bristow-johnson
Sep 21 at 0:05
4
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 $c=(1+sqrt5)/2$ that conserves both energy and momentum.
– user21820
Sep 21 at 3:18
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!
– Joshua Ronis
Sep 21 at 9:24
@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 bassâ€Â. 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.
– leftaroundabout
2 days ago
22
22
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.
– Tausif Hossain
Sep 20 at 17:22
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.
– Tausif Hossain
Sep 20 at 17:22
1
1
i think this is the best answer.
– robert bristow-johnson
Sep 21 at 0:05
i think this is the best answer.
– robert bristow-johnson
Sep 21 at 0:05
4
4
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 $c=(1+sqrt5)/2$ that conserves both energy and momentum.
– user21820
Sep 21 at 3:18
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 $c=(1+sqrt5)/2$ that conserves both energy and momentum.
– user21820
Sep 21 at 3:18
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!
– Joshua Ronis
Sep 21 at 9:24
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!
– Joshua Ronis
Sep 21 at 9:24
@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 bassâ€Â. 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.
– leftaroundabout
2 days ago
@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 bassâ€Â. 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.
– leftaroundabout
2 days ago
 |Â
show 5 more comments
up vote
15
down vote
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.
answered Sep 20 at 17:32
physicsguy19
331112
add a comment |Â
up vote
15
down vote
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.
answered Sep 20 at 17:32
physicsguy19
331112
add a comment |Â
up vote
15
down vote
up vote
15
down vote
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.
answered Sep 20 at 17:32
physicsguy19
331112
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.
answered Sep 20 at 17:32
physicsguy19
331112
answered Sep 20 at 17:32
physicsguy19
331112
answered Sep 20 at 17:32
physicsguy19
331112
answered Sep 20 at 17:32
physicsguy19
331112
331112
add a comment |Â
add a comment |Â
up vote
6
down vote
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
$$m_amathbfv_a1 + m_bmathbfv_b1=m_amathbfv_a2+m_bmathbfv_b2$$
In this case, $mathbfv_b1=mathbf0$ and $m_a=m_b$, so
$$mathbfv_a1=mathbfv_a2+mathbfv_b2$$
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.)
$$fracm_amathbfv_a1^22+fracm_bmathbfv_a1^22=fracm_amathbfv_a2^22+fracm_bmathbfv_b2^22 + Delta H$$
$$mathbfv_a1^2=mathbfv_a2^2+mathbfv_b2^2 + frac2Delta Hm$$
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
$$mathbfv_a1^2=mathbfv_a2^2+mathbfv_b2^2$$
$$mathbfv_a1^2=(mathbfv_a1 - mathbfv_b2)^2+mathbfv_b2^2$$
$$mathbfv_a1^2=mathbfv_a1^2 - 2mathbfv_a1mathbfv_b2 + mathbfv_b2^2 + mathbfv_b2^2$$
$$mathbfv_a1mathbfv_b2 = mathbfv_b2^2$$
For a straight-on collision, $mathbfv_a1$ and $mathbfv_b2$ are parallel, so
$$|mathbfv_a1||mathbfv_b2| = |mathbfv_b2|^2$$
Since billards cannot pass through one another $mathbfv_b2 neq 0$, and thus $mathbfv_b2 = mathbfv_a1$.
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.,
answered 2 days ago
Paul Draper
26115
add a comment |Â
up vote
6
down vote
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
$$m_amathbfv_a1 + m_bmathbfv_b1=m_amathbfv_a2+m_bmathbfv_b2$$
In this case, $mathbfv_b1=mathbf0$ and $m_a=m_b$, so
$$mathbfv_a1=mathbfv_a2+mathbfv_b2$$
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.)
$$fracm_amathbfv_a1^22+fracm_bmathbfv_a1^22=fracm_amathbfv_a2^22+fracm_bmathbfv_b2^22 + Delta H$$
$$mathbfv_a1^2=mathbfv_a2^2+mathbfv_b2^2 + frac2Delta Hm$$
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
$$mathbfv_a1^2=mathbfv_a2^2+mathbfv_b2^2$$
$$mathbfv_a1^2=(mathbfv_a1 - mathbfv_b2)^2+mathbfv_b2^2$$
$$mathbfv_a1^2=mathbfv_a1^2 - 2mathbfv_a1mathbfv_b2 + mathbfv_b2^2 + mathbfv_b2^2$$
$$mathbfv_a1mathbfv_b2 = mathbfv_b2^2$$
For a straight-on collision, $mathbfv_a1$ and $mathbfv_b2$ are parallel, so
$$|mathbfv_a1||mathbfv_b2| = |mathbfv_b2|^2$$
Since billards cannot pass through one another $mathbfv_b2 neq 0$, and thus $mathbfv_b2 = mathbfv_a1$.
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.,
answered 2 days ago
Paul Draper
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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
$$m_amathbfv_a1 + m_bmathbfv_b1=m_amathbfv_a2+m_bmathbfv_b2$$
In this case, $mathbfv_b1=mathbf0$ and $m_a=m_b$, so
$$mathbfv_a1=mathbfv_a2+mathbfv_b2$$
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.)
$$fracm_amathbfv_a1^22+fracm_bmathbfv_a1^22=fracm_amathbfv_a2^22+fracm_bmathbfv_b2^22 + Delta H$$
$$mathbfv_a1^2=mathbfv_a2^2+mathbfv_b2^2 + frac2Delta Hm$$
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
$$mathbfv_a1^2=mathbfv_a2^2+mathbfv_b2^2$$
$$mathbfv_a1^2=(mathbfv_a1 - mathbfv_b2)^2+mathbfv_b2^2$$
$$mathbfv_a1^2=mathbfv_a1^2 - 2mathbfv_a1mathbfv_b2 + mathbfv_b2^2 + mathbfv_b2^2$$
$$mathbfv_a1mathbfv_b2 = mathbfv_b2^2$$
For a straight-on collision, $mathbfv_a1$ and $mathbfv_b2$ are parallel, so
$$|mathbfv_a1||mathbfv_b2| = |mathbfv_b2|^2$$
Since billards cannot pass through one another $mathbfv_b2 neq 0$, and thus $mathbfv_b2 = mathbfv_a1$.
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.,
answered 2 days ago
Paul Draper
26115
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
$$m_amathbfv_a1 + m_bmathbfv_b1=m_amathbfv_a2+m_bmathbfv_b2$$
In this case, $mathbfv_b1=mathbf0$ and $m_a=m_b$, so
$$mathbfv_a1=mathbfv_a2+mathbfv_b2$$
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.)
$$fracm_amathbfv_a1^22+fracm_bmathbfv_a1^22=fracm_amathbfv_a2^22+fracm_bmathbfv_b2^22 + Delta H$$
$$mathbfv_a1^2=mathbfv_a2^2+mathbfv_b2^2 + frac2Delta Hm$$
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
$$mathbfv_a1^2=mathbfv_a2^2+mathbfv_b2^2$$
$$mathbfv_a1^2=(mathbfv_a1 - mathbfv_b2)^2+mathbfv_b2^2$$
$$mathbfv_a1^2=mathbfv_a1^2 - 2mathbfv_a1mathbfv_b2 + mathbfv_b2^2 + mathbfv_b2^2$$
$$mathbfv_a1mathbfv_b2 = mathbfv_b2^2$$
For a straight-on collision, $mathbfv_a1$ and $mathbfv_b2$ are parallel, so
$$|mathbfv_a1||mathbfv_b2| = |mathbfv_b2|^2$$
Since billards cannot pass through one another $mathbfv_b2 neq 0$, and thus $mathbfv_b2 = mathbfv_a1$.
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.,
answered 2 days ago
Paul Draper
26115
edited 2 days ago
answered 2 days ago
Paul Draper
26115
answered 2 days ago
Paul Draper
26115
answered 2 days ago
Paul Draper
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26115
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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.
answered 14 hours ago
Anthony X
2,44311218
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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.
answered 14 hours ago
Anthony X
2,44311218
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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.
answered 14 hours ago
Anthony X
2,44311218
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.
answered 14 hours ago
Anthony X
2,44311218
edited 12 hours ago
answered 14 hours ago
Anthony X
2,44311218
answered 14 hours ago
Anthony X
2,44311218
answered 14 hours ago
Anthony X
2,44311218
2,44311218
add a comment |Â
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3
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.
– David White
Sep 20 at 19:33
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.
– robert bristow-johnson
Sep 21 at 0:07
1
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.
– Keith
Sep 21 at 3:47
1
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.
– Stian Yttervik
Sep 21 at 7:08
1
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...
– Stian Yttervik
2 days ago