Physics Puzzle: Stuns Shots and the 90-Degrees Rule in Billiard
June 25th, 2009
| Categories: Mechanics, Physics Puzzles
Level of Difficulty: Undergraduate
In billiards a stun shot is one in which the ball doesn’t roll, or, more correctly, doesn’t get a chance to start rolling before it hits its target ball. In such cases, and assuming the ball isn’t spinning at all (no English, Top Spin, Bottom Spin and all that jazz), a well known rule of thumb says that after impact, the angle between the cue ball (the one you hit with the stick) and target ball (the one that gets hit by the cue ball) is 90 degrees, as shown below. Can you prove this?
EDITED, 26/June/2009: hats off to our reader peter for his solution. A slightly more detailed one can be found here.
i’m a little confused, this obviously doesn’t hold for when the ball strikes the other ball head on?
Excellent observation, but I’m going to throw it right back at you - can you reconcile reality (the 90 degrees rule, which - as any pool player will tell you - holds quite nicely for stun shots) with this seemingly paradoxical state of affairs?
Hint: what makes a head-on collision special?
ok i got it, using vectors, not sure how rigourous this is:
first ball has initial velocity = v*i
after collision:
2nd ball will have velocity = ai+bj
1st ball will have velocity = (v-a)i -bj
dot them, and you get va-a^2-b^2
but, by conservation of energy, v^2 = (a^2+b^2) + (v-a)^2 + b^2
and you find out that the dot product is equal to 0
hm, well in a head on collision the first ball remains still, so i guess you could say the velocity of 2nd ball is perpendicular to the velocity of the 1st ball (in fact any vector will be perpendicular to it). Is this right?
Your observation is once again correct, although I’d change the wording a bit. As a collision becomes more and more head-on the cue ball emerges with smaller and smaller velocities. In a head-on collision the cue ball simply comes to a rest, so it has no direction, making the statement “X degrees between the first and second ball” meaningless.
The exact offset of the 2 balls and, in turn, the angle between them at the point of impact completely determine the direction of the second ball’s velocity. We should have our y-axis line up with the second ball’s velocity. So in the y-axis direction, there’s essentially a 1-dimensional collision. And since collisions with pool balls are essentially perfectly elastic, all of the momentum in that direction is transferred from the first ball to the second ball.
This leaves only the momentum in the x direction for the first ball, meaning the velocities are perpendicular. I’m not exactly sure how the spin of the ball would affect the collision in other situations. Perhaps it makes the collision much less elastic so the first ball ends up with some momentum in the y direction.
I’m not sure if I explained it very well, but I’m pretty sure that’s what’s going on.
assuming that every billiard ball has the same mass,and the collisions are perfectly elastic,
we could resolve the components of the velocity along two mutually perpendicular directions.one along the common normal to both the balls and the other perpendicular to it.usin momentum conservation one can say that the component of velocity of the ball initially in motion,along the common normal would be 0 after the collision while the ball initially at rest,would move with a velocity equal to the velocity component of the previous ball .
mathematically,
let us denote the 2 perpendicular directions using r^ and n^.
initially,
ball one-v(r)r^+v(n)n^
ball 2-0r^+0N^
using conservation of momentum and the coeff. of restitution,
ball2-0r^+v(n)n^
ball1-v(r)r^+0n^
these are two mutually perpendicular directions.
hence,the proof.