Question

Two billiard balls, a cue ball and an eight ball, of equal mass undergo a perfectly elastic head-on collision. If the cue ball’s initial speed was 8.8 m/s, and the eight ball's was 8.3 m/s in the opposite direction, what will be the velocity of the eight ball after the collision? Consider the initial direction of the cue ball to be the positive direction.

Answer 1

Now it is elastic collision, so
using momentum conservation
m1u1 + m2u2 = m1v1 + m2v2
u2 = -8.3

u1 = 8.8
m*8.8 - m*8.3 = m*v1 + m*v2

0.5 = v1 + v2

another condition for elastic collision
u1 - u2 = v2 - v1
v2 - v1 = 8.8 - (-8.3)

v2 - v1 = 17.1

v1 = speed of cue ball after collision

v2 = speed of eight ball after collision

Add both equation

2*v2 = 17.1 + 0.5

v2 = 17.6/2 = +8.8 m/sec