# Colliding with a spinning ball: linear momentum question

Hi all, I have a question. First, a moving ball, ball A, slides straight into a ball which is standing still, ball B. The balls then collide, and ball B rebounds along, say, the Y-axis at a certain speed.

Now suppose we recreate the collision, except this time, ball B is spinning on its Z-axis. Afterwards, the Y-component of ball B's rebound velocity will be the same as the previous time, correct? This is an example of linear momentum always being conserved in a collision?

To make things simpler, the ground is frictionless, and the collision is non-elastic with no deformation when the balls collide.

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Yes, for an elastic collision (no energy lost to deformations), ball B's final linear motion is the same whether or not it is spinning. Like you said, linear momentum is conserved. So is the angular momentum of ball B.

Billiard balls provide the classic example of (nearly) elastic collisions. But if you play pool much, you know that accidently putting side spin on the cue (white) ball can cause you to miss even if you had perfect aim. This is caused by, among other things, the inelasticity of the collision. There is some friction between the balls, and they are in contact for a finite time. I've heard that when you hit the ball extremely hard, as on a break shot in 8-ball, the area of contact between the balls can be the size of a quarter! Don't know if that's true though.

Andrew Mason
Homework Helper
Nickel said:
Now suppose we recreate the collision, except this time, ball B is spinning on its Z-axis. Afterwards, the Y-component of ball B's rebound velocity will be the same as the previous time, correct?
Not necessarily. But the total momentum of A and B after the collision will always add to the initial momentum of A.

If some of the energy of A is converted into rotational energy, the translational kinetic energy of the balls after the collision cannot add up to A's initial translational kinetic energy. If this occurs, B's velocity will be less than in the first case and A's will be more.

AM

Andrew Mason said:
Not necessarily. But the total momentum of A and B after the collision will always add to the initial momentum of A.

If some of the energy of A is converted into rotational energy, the translational kinetic energy of the balls after the collision cannot add up to A's initial translational kinetic energy. If this occurs, B's velocity will be less than in the first case and A's will be more.
I see where you're coming from, but are you sure that when ball A is rotating afterwards, that rotational energy will have come out of A's translational energy? I had been thinking that A would be getting its rotational energy from B's rotational energy during their moment of contact due to the friction, and its translation energy would be left intact and still be transferred fully to B.

Yes, for an elastic collision (no energy lost to deformations), ball B's final linear motion is the same whether or not it is spinning. Like you said, linear momentum is conserved. So is the angular momentum of ball B.
Yes, I meant elastic collision, a total brainfart on my part. I'm not a physics major, so bear with me, guys. :)

Andrew Mason said:
Not necessarily. But the total momentum of A and B after the collision will always add to the initial momentum of A.

If some of the energy of A is converted into rotational energy, the translational kinetic energy of the balls after the collision cannot add up to A's initial translational kinetic energy. If this occurs, B's velocity will be less than in the first case and A's will be more.

AM

How does translational energy get converted into rotational energy in an elastic collision between balls?

By the way, I had assumed Nickel meant B was spinning before A hit it.

Andrew Mason
Homework Helper
Nickel said:
I see where you're coming from, but are you sure that when ball A is rotating afterwards, that rotational energy will have come out of A's translational energy? I had been thinking that A would be getting its rotational energy from B's rotational energy during their moment of contact due to the friction, and its translation energy would be left intact and still be transferred fully to B.
If you are talking about elastic head-on collisions in both cases and the masses are equal, then the incident ball, A, stops and B moves with A's incident velocity. The energy and momentum of A are transferred to B. Since the centres of mass are aligned, there is no torque added by the collision, so total angular momentum is unchanged. The spin of B has no effect on translational motion.

If the centres of mass of the two balls are not aligned (ie. with the incident momentum vector of A) both balls experience a torque during the time of the collision, so the angular momentum of each changes. Some of A's translational energy (but not momentum) would be converted to rotational energy. This would be, in effect, an inelastic collision for purposes of determining resultant momenta of A and B.

Since you refer to B moving in the Y direction in the first case and you refer to the Y component of B's motion in the second I am not clear on what it is you are describing in either case.

AM

Andrew Mason said:
If you are talking about elastic head-on collisions in both cases and the masses are equal, then the incident ball, A, stops and B moves with A's incident velocity. The energy and momentum of A are transferred to B. Since the centres of mass are aligned, there is no torque added by the collision, so total angular momentum is unchanged. The spin of B has no effect on translational motion.

If the centres of mass of the two balls are not aligned (ie. with the incident momentum vector of A) both balls experience a torque during the time of the collision, so the angular momentum of each changes. Some of A's translational energy (but not momentum) would be converted to rotational energy. This would be, in effect, an inelastic collision for purposes of determining resultant momenta of A and B.

Since you refer to B moving in the Y direction in the first case and you refer to the Y component of B's motion in the second I am not clear on what it is you are describing in either case.

AM
Ah I see. Sorry guys, I guess I should have put the question in the context of a non elastic collision then. Then this would mean there is friction between the balls, and that some of B's rotational energy would be transferred over to A and cause A to start rotating, correct? Then A and B would move away at different angles from A's initial line of motion, I think. While B's Y-component would still be the same like I hypothesized in my initial question?

Also yes, I'm assuming that the two balls' centers of mass are aligned during the collision. I mentioned the Y component in my initial question because I had already gone and assumed that A and B would be moving away at an angle from A's initial line of motion. My bad. 