conservation of kinetic energy, and conservation of momentum?
crazecaf asked:
Ana-Lucia needs to sink the 8-ball to win a pool match. She can sink it by hitting the 8-ball so it travels 35° away from the original path of the cue-ball. At what angle does the cue-ball deflect after hitting the 8-ball? Show the angle by using the conservation of kinetic energy, conservation of momentum and assume the masses are equal.
Powered by Yahoo Answers
Ana-Lucia needs to sink the 8-ball to win a pool match. She can sink it by hitting the 8-ball so it travels 35° away from the original path of the cue-ball. At what angle does the cue-ball deflect after hitting the 8-ball? Show the angle by using the conservation of kinetic energy, conservation of momentum and assume the masses are equal.
Powered by Yahoo Answers
Kind of hard to describe, I could draw this out better.
As the cue ball comes into contact with the 8 ball, the 8 ball goes away from the CB (cue ball) at a 35 degree angle derivative of the CBs original path. The 8B (8 ball) now goes on its own path towards the pocket, but the CBs path has now changed. If the 8B has gone 35 degrees off the CBs original path of motion, and assuming both are equal in mass, the CB will go on a new path that is 55 degrees off the path it was originally taking.
When drawn out, it’s easier to explain, but basically, imagine the CB on its original path, then coming into contact with the 8B. The 8B goes DOWN and to the RIGHT 35 degrees, while the CB goes UP and to the RIGHT 55 degrees. The CB will continue on its path, but its thrown 55 degrees off course.
Comment by zsamd425 — December 20, 2008 @ 7:16 am
Assume that the cue ball has mass m and initial velocity v, and that after the collision, the cue ball has velocity v1 at an angle of x to its original path, and the 8 ball has velocity v2.
x-direction: p1 = p2: mv = mv2 cos 35 + mv1 cos x
y-direction: p1 = p2: 0 = v2 sin 35 - v1 sin x
v = v2 cos 35 + v1 cos x
v1 = v2 sin 35 / sin x
v = v2 cos 35 + v2 sin 35 cot x
Kinetic energy: 1/2 mv^2 = 1/2 m (v1^2 + v2^2)
v^2 = v1^2 + v2^2
(v2 cos 35 + v2 sin 35 cot x)^2 = (v2 sin 35 / sin x)^2 + v2^2
(cos 35 + sin 35 cot x)^2 = (sin 35 csc x)^2 + 1
cos^2 35 + (2 cos 35 sin 35) cot x + (sin ^2 35) cot^2 x = (sin^2 35) csc^2 x + 1
cos^2 35 + (2 cos 35 sin 35) cot x = (sin^2 35)( csc^2 x - cot^2 x) + 1
[Remember that 1 + cot^2 x = csc^2 x]
cos^2 35 + (2 cos 35 sin 35) cot x = (sin^2 35)( 1) + 1
(2 cos 35 sin 35) cot x = sin^2 35 + (1 - cos^2 35)
(2 cos 35 sin 35) cot x = 2 sin^2 35
cot x = 2 sin^2 35 / (2 cos 35 sin 35) = tan 35 = cot 55
x = 55 degrees
Thanks for giving me something to do.
Comment by Clueless — December 21, 2008 @ 11:48 pm