Two billiard balls of equal mass (m) collide elastically. Initially the 13 ball has speed (v) and the 9 ball is at rest. After the collision the 13 ball has speed (1/2v) and the 9 ball has speed Vf. What is Vf?
The professor says that Vf is (sqrt(3))v/2. Can someone help me with the steps it takes to reach this answer?
- Old Science GuyLv 74 years agoFavorite Answer
if they collided head on then the 13 would be at 0 and 9 would be at v
owing to their equal masses
but that is not the given case so they must have glanced off of each other at a non-zero angle
with elastic collisions both momentum and kinetic energy are conserved
1/2 m V1^2 + 1/2 m V2^2 = 1/2 m V3^2 + 1/2 m V4^2
1/2's cancel as do all m's giving
V1^2 + V2^2 = V3^2 + V4^2
substituting given information we get
v^2 + 0 = (1/2v)^2 + vf^2 = 1/4 v^2 + vf^2
Vf^2 = 3/4 v^2
vf = sqrt(3/4 v^2) = sqrt(3)/2 v
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