Can anyone answer a simple question on the laws of motion?

Question

When a pool/snooker ball strikes another ball full on what determines how much of the momentum of the cue ball is passed to the object ball as acceleration and how much is retained as momentum in the cue ball following through?

My question is more to do with contact and propulsion than the actual spin on the balls.

I assume that if the object ball isn't hit full on - flicked at an angle - the comparatively lesser acceleration passed on will be proportional to the angle from the direction of the cue ball.

Answer 1

Pool/snooker is probably the finest demonstration there is of 2-dimensional Newtonian kinematics ☺

In the case of a straight-on collision, if you assume that the collision is perfectly elastic (it isn't, but it's --real-- close) and the balls of identical mass (again, not perfect but very close) then the amount of linear momentum coupled to the object ball is exactly equal to the amount of linear momentum contained in the cue ball, and anything that the cue does after the collision is due to conversion of angular momentum of the cue ball. And, of course, once the object ball begins to move, some of its linear mamentum is converted to angular momentum as it begins to 'roll' rather than 'slide' over the felt.

Doug

Answer 2

It's all to do with how much energy is dissipated in the collision. Let's assume to start with that the two balls are the same mass and that there is no spin involved. If no energy is dissipated (ie all the kinetic energy of the cue ball ends up in one or other of the balls) then the cue ball will stop dead and the struck ball will move off, having acquired all the energy of the cue ball. So its energy amd momentum will both be what the cue ball had before the collision. If some energy is lost (as sound or heat or whatever), then the cue ball will still move slowly forward, and the struck ball will acquire only a proportion of the momentum of the cue ball. If you know what proportion of the energy is dissipated as noise etc. you can work out the velocities of the two balls after the collision.

If the two balls are different masses then, in the ideal case of no dissipation, you could solve for conservation of energy and conservation of momentum simultaneously to find the two velocities after the collision.

Answer 3

If hit head on with no spin, the cue ball stops and the velocity of the object ball after collision equals the initial velocity of the cue ball.

You need to use equations for both conservation of energy
and conservation of momentum.

Momentum: m.u1 = m(u2 + v2)
Energy: m.u1^2 = m(u2^2 + v2^2)
Substitute value of u1 from 1st, into 2nd equation:
u2^2 + 2.u2.v2 + v2^2 = u2^2 + v2^2
So: u2.v2 = 0
i.e. either u2 = 0 or v2 = 0

Clearly it must be u2 = 0 (velocity of cue ball after impact)
and velocity of object ball after impact, v2 = u1 [QED]

P.S. The object ball will acquire some spin in practice, since hitting it mid-height pushes the ball bodily and it only starts rolling because the cloth slows down the bottom of the ball.

Similarly with a cue - you need to strike the ball above its mid-point (at 0.6 x radius, I think it is - check out height of cushion) for it to have no spin - i.e. the roll exactly matches forward motion - no slippage on the cloth.


[If you include friction, it gets complicated with a need to consider angular momentum of balls before and after contact. So the above is theoretical, but pretty close in reality]

Answer 4

if the white ball and target ball are the same mass and size and they strike each other dead on I think the white ball is supposed to stop and the target ball is supposed to continue with the white balls speed - the energy lost in the collision.

you need to look up applied maths ( maths physics) and study collisions

Answer 5

if the ball hits it strait and and the white one stops the new moving ball will not get 100% of the energy. so it will not go as fast as the white ball hit it. some energy will have been tacken of as sound

Answer 6

The answer is the striking force. What the hell do you think. I play pool and I am brainy so I take it you want the answer for an exam or for your own furtherance of ecucation. but it is self explanitory. My MAIN point is that yo0u are talking about hitting the ball dead centre. Nobody can know the strenght of a mans hand through maths but you can calcute String force.................... WHO is hitting the ball. THATS YOUR ANSWER.