Motion
The motion of the balls can be catagorized into two general catagories which are collisions and spin or
rotation. Here the focus will be on collisions between the balls. The spin can have a
significant effect on the motion of the ball, but due to time constraints and complexity of the science of
it it will not be addressed to much detail.
"An elastic collision between two objects is one in which the total kinetic energy (as well as total momentum) of the system is the same before and after the collision." (Physics for Scientists and Engineers)
In a collision between the cue ball and another ball, since the masses are so similar the resulting angle
between the vectors of the balls is approximately ninety degrees. This is due to the Conservation of
Energy:
1/2 m(cue) * v(cue(i))^2 = 1/2 m(cue) * v(cue(f))^2 + 1/2 m(ball) * v(ball(f))^2
or by dividing out the 1/2 m => v(cue(i))^2 = v(cue(f))^2 + v(ball(f))^2;
where m(cue) is the mass of the cue ball, v(cue(i)) is the initial velocity of the cue ball, m(ball) is the
mass of the other ball, and v(ball(f)) is the velocity of the ball.
Since the velocities we are dealing with occure in a two dimentional plane they can be represented in
vector form which allows us to determine the angle between the two balls using the dot product rule.
Dot Product: V(cue(f))
• V(ball(f)) = v(cue(f)) * v(ball(f)) cos (Ø
+ O)
v(cue(i))^2 = v(cue(f))^2 + v(ball(f))^2
+ 2 * v(cue(f)) * v(ball(f)) cos (Ø + O)
Subtracting the first equation from the above yields the following:
0 = 2 * v(cue(f)) * v(ball(f)) cos
(Ø + O)
dividing by 2 * v(cue(f)) * v(ball(f))
=> 0 = cos (Ø + O)
Therefore Ø + O = 90
degrees
This site was created by:
Erik Crawford
Physics 212 - Spring 2005
University of Alaska Fairbanks