Question

A 48 g marble moving at 1.9 m/s strikes a 24 g marble at rest. Assume the collision is perfectly elastic and the marbles collide head-on.

What is the speed of the first marble immediately after the collision?

What is the speed of the second marble immediately after the collision?

Answer 1

let the speed of first marble after collision be v1 and second marble after collision be v2

By the conservation of momentum we have

m1u1 + m2u2 = m1v1 + m2v2

48*1.9 + 0 = 48v1 + 24v2

2v1 + v2 = 3.8 ------------------ (1)

velocity of approach = velocity of recess

v2 - v1 = u1 - u2

v2 - v1 = 1.9 - 0

v2 - v1 = 1.9 ----------(2)

subtracting (2) from (1) we have

3v1 = 1.9

v1 = 0.63 m/s

v2 = 1.9 + v1 = 1.9+0.63 = 2.53 m/s

so the speed of first marble after collision = 0.63 m/s

sp speed of second marble after collision = 2.53 m/s