Question

A 4.80-kg ball, moving to the right at a velocity of +1.26 m/s on a frictionless table, collides head-on with a stationary 8.20-kg ball. Find the final velocities of (a) the 4.80-kg ball and of (b) the 8.20-kg ball if the collision is elastic. (c)Find the magnitude and direction of the final velocity of the two balls if the collision is completely inelastic.

Answer 1

(a) For the elastic collision -  
The expression for the final velocity of 4.80 kg ball is

v1' = 2(m1v1 + m2v2)/(m1+m2) - v1
v1' = 2(4.8*1.26 + 8.20*0)/(13.0) - 1.26 = 0.93 - 1.26 = - 0.33 m/s

The negative sign shows that the ball (4.80 kg) bounces back with 0.33 m/s.

(b) The expression for the final velocity of the second ball is -

v2' = 2(m1v1 + m2v2)/(m1+m2) - v2
v2' = 2(4.8*1.26 + 8.2*0)/(13.0) - 0
v2' = 0.93 m/s

The positive sign shows that the second ball will move in the initial direction of first ball after the before collision.

(c) For inelastic collision:

v' = (v1 m1 + v2m2) / (m1+m2)
v' = (1.26*4.80)/(13.0) = 0.47 m/s

The two ball will stick togather and move with a velocity of 0.47 m/s after the collision.