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Why does a billiards ball stop when it hits another billiards ball head on?
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(I'm repeating myself alot here but its 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 than 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 looses.
$$F_1->2 = F_2->1$$
Thats 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?
newtonian-mechanics forces momentum conservation-laws collision
edited 14 mins ago
Qmechanic♦
97.1k121631034
asked 42 mins ago
Joshua Ronis
181110
add a comment |Â
up vote
1
down vote
favorite
(I'm repeating myself alot here but its 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 than 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 looses.
$$F_1->2 = F_2->1$$
Thats 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?
newtonian-mechanics forces momentum conservation-laws collision
edited 14 mins ago
Qmechanic♦
97.1k121631034
asked 42 mins ago
Joshua Ronis
181110
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
(I'm repeating myself alot here but its 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 than 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 looses.
$$F_1->2 = F_2->1$$
Thats 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?
newtonian-mechanics forces momentum conservation-laws collision
edited 14 mins ago
Qmechanic♦
97.1k121631034
asked 42 mins ago
Joshua Ronis
181110
(I'm repeating myself alot here but its 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 than 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 looses.
$$F_1->2 = F_2->1$$
Thats 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?
newtonian-mechanics forces momentum conservation-laws collision
newtonian-mechanics forces momentum conservation-laws collision
edited 14 mins ago
Qmechanic♦
97.1k121631034
asked 42 mins ago
Joshua Ronis
181110
edited 14 mins ago
Qmechanic♦
97.1k121631034
asked 42 mins ago
Joshua Ronis
181110
edited 14 mins ago
Qmechanic♦
97.1k121631034
edited 14 mins ago
Qmechanic♦
97.1k121631034
edited 14 mins ago
Qmechanic♦
97.1k121631034
97.1k121631034
asked 42 mins ago
Joshua Ronis
181110
asked 42 mins ago
Joshua Ronis
181110
asked 42 mins ago
Joshua Ronis
181110
181110
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
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 19 mins ago
physicsguy19
202111
add a comment |Â
up vote
3
down vote
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 19 mins ago
Tausif Hossain
2,3081618
answered 26 mins ago
nasu
1,06259
1
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
20 mins ago
add a comment |Â
up vote
1
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 33 mins ago
FrodCube
1,0231514
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
29 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
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 19 mins ago
physicsguy19
202111
add a comment |Â
up vote
3
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 19 mins ago
physicsguy19
202111
add a comment |Â
up vote
3
down vote
up vote
3
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 19 mins ago
physicsguy19
202111
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 19 mins ago
physicsguy19
202111
answered 19 mins ago
physicsguy19
202111
answered 19 mins ago
physicsguy19
202111
answered 19 mins ago
physicsguy19
202111
202111
add a comment |Â
add a comment |Â
up vote
3
down vote
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 19 mins ago
Tausif Hossain
2,3081618
answered 26 mins ago
nasu
1,06259
1
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
20 mins ago
add a comment |Â
up vote
3
down vote
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 19 mins ago
Tausif Hossain
2,3081618
answered 26 mins ago
nasu
1,06259
1
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
20 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
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 19 mins ago
Tausif Hossain
2,3081618
answered 26 mins ago
nasu
1,06259
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 19 mins ago
Tausif Hossain
2,3081618
answered 26 mins ago
nasu
1,06259
edited 19 mins ago
Tausif Hossain
2,3081618
edited 19 mins ago
Tausif Hossain
2,3081618
edited 19 mins ago
Tausif Hossain
2,3081618
2,3081618
answered 26 mins ago
nasu
1,06259
answered 26 mins ago
nasu
1,06259
answered 26 mins ago
nasu
1,06259
1,06259
1
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
20 mins ago
add a comment |Â
1
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
20 mins ago
1
1
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
20 mins ago
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
20 mins ago
add a comment |Â
up vote
1
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 33 mins ago
FrodCube
1,0231514
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
29 mins ago
add a comment |Â
up vote
1
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 33 mins ago
FrodCube
1,0231514
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
29 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
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 33 mins ago
FrodCube
1,0231514
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 33 mins ago
FrodCube
1,0231514
answered 33 mins ago
FrodCube
1,0231514
answered 33 mins ago
FrodCube
1,0231514
answered 33 mins ago
FrodCube
1,0231514
1,0231514
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
29 mins ago
add a comment |Â
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
29 mins ago
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
29 mins ago
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
29 mins ago
add a comment |Â
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