A steel ball of mass 1 kg travelling along a frictionless track at 4 m.s-1 collides elastically with another stationary steel ball of mass 2 kg and then rebounds in the opposite direction with a speed of 1.33 m.s-1. What is the velocity of the 2 kg ball after the collision?

Solution: Let the initial direction of travel of the 1 kg ball represent positive velocity. Then we have

u1 = 4.0 m.s-1

v1 = -1.33 m.s-1

u2 = 0 m.s-1

v2 = ? m.s-1

Since momentum is conserved,

Elastic collisions equations

(1 kg x 4.0 m.s-1) + (2 kg x 0 m.s-1) = (1 kg x -1.33 m.s-1) + (2 kg x v2 m.s-1)

from which v2 = 5.33 kg.m.s-1 / 2 kg = 2.67 m.s-1.