Is there mathematical proof that when a cue ball hits an object ball at an angle the paths they follow are 90*

Question

basically i know that in snooker or billiards or pool when you hit the cue ball at a stationary object ball the paths they follow after collision are at right angles to each other. i was wondering if anybody knew any mathematical proof that shows that this is true.

im just saying if there is no spin and both balls have perfect elasticity and there is little to no friction

im saying that the natural angle providing the cue ball hits it at angle and not straight on is 90*. IT never happens because of various factors such as spin and friction but discounting these things and the cue balls paths will go off perpendicular to each other

Answer 1

Yes, there is such a proof, provided you make the following assumptions:
1. a perfectly elastic collision with no friction;
2. No "english";
3. the two balls have the same mass;
4. Both balls actually move after the collision (an alternate solution is that the cue ball stops dead)

Let's use capital leters for velocity vectors, and small letters for speed:
V = initial velocity vector of cue ball; v = |V|
V1 = final velocity vector of cue ball; v1 = |V1|
V2 = final velocity vector of other ball; v2 = |V2|

By conservation of momentum:
Total momentum before = total momentum after:
mV = mV1 + mV2
or (dividing by m):
V = V1 + V2

Now, a general theorem about adding vectors says that:
|A+B|² = |A|² + |B|² + 2|A||B|cosθ,
where θ is the angle between the vectors. Applying that to the above, we have:

|V|² = |V1|² + |V2|² + 2|V1||V2|cosθ
or (using lower case letters):
v² = (v1)² + (v2)² + 2(v1)(v2)cosθ

where θ is the angle at which the two balls separate.

Now, if the collision is elastic, energy is also conserved:
mv²/2 = m(v1)²/2 + m(v2)²/2

or (multiplying by 2/m):
v² = (v1)² + (v2)²

Combining this with the previous equation for v² gives:
2(v1)(v2)cosθ = 0

This implies that either v1 is zero, or v2 is zero, or cosθ is zero. If cosθ is zero then θ = 90 degrees.

Answer 2

The paths are not necessarily at right angles to each other. In such a collision, there are two properties involved, energy and momentum. Energy is a scalar property, i.e it has magnitude, but no direction. Momentum is a vector property, i.e. it has both magnitude and direction. In the collision of the two balls, energy is not a conserved quantity, as some of the kinetice energy of the moving ball will be converted to sound and some to heat in the very small distortion of the material of the balls. The combined energy of the two balls after the collision is less than that of the moving ball before the collision. Momentum is however a conserved property. It has the value mv, where m is mass and v is the velocity. The vector sum of the momentum of the two balls after the collision will be equal to the momentum of the moving ball before the collision.
The angle between the lines of travel of the two balls will be determined by the lateral offset between the centres of the two balls,measured perpendicular to the direction of travel of the moving ball. If the two centres are both on the line of travel, then in theory, both balls will continue travel in the same direction at the same speed, i.e. with the same momentum. This is unlikely to happen in reality, due to the small differences in mass, diameter, elasticity, and effects of the nap on the table covering

Answer 3

When the two balls hit each other, they apply a force to each other, that acts along the line that passes through the centre of both balls. The target ball will move off in the direction of this line. Where the cue ball goes will depend on its spin at the moment of impact.

Way back when I did A level maths, exercises of two balls hitting each other were used to understand the conservation of momentum and vector algebra. These were approximations, as the balls were considered to be moving without any rotation, on a frictionless surface.

This may be why I'm pants at snooker.

Answer 4

Well I think if balls hit each other in space and there was no gravity and the cue ball didn't rotate you might be right . The object ball will move along the angle of contact and absorb all the momentum resolved in the direction of that line so the other ball will move at right angles to the angle of contact. I am thinking this is so from the toy Newton's cradle.

I am not sure why and if it is true that the object ball absorbs all the momentum along the line of contact though.

Answer 5

You obviously haven't tried to pot a straight blue into the middle pocket. The white ball will follow the blue (more if you put topspin or even pull back into the opposite pocket if you put backspin on it). It's rarely that both balls leave at exactly 90 degrees.

Answer 6

Conservation of Ek> m1*v1^2=m1*v2^2+m2*v3v^2
where m1= mass of cue ball, m2=mass of target ball and vi=speeds b4 and after collision.
If m1=m2 then
v1^2=v2^2+v3^2
conservation of linear momentum (2-d vectors)
V1=V2+V3 where Vi= vectors
So by Pi theorem angle between vectors V2 and V3= 90deg!
If collision is inelastic (real world) then angle <90deg

Answer 7

what are you on about!! it's almost impossible to get a 90 degree angle coz the object ball would have to be hit so thinly to achieve that angle. all angles are over 90 degrees so there would not be any mathematical proof.

Answer 8

Yeah.

Sum of the forces in x = ma
Sum of the forces in y = ma = 0

The que ball will never travel in exactly a 90* direction becuase of these equations. Assuming y is the 90* direction.

Answer 9

This is not the case. Consider if you hit the que ball
directly at the object ball.

Also if you put spin on the que ball you can vary
the angle.