Question

A 41 g marble moving at 2.0 m/s strikes a 21 g marble at rest. Assume the collision is perfectly elastic and the marbles collide head-on. a:What is the speed of the first marble immediately after the collision? b:What is the speed of the second marble immediately after the collision?

Answer 1

masses m = 41 g

M = 21 g

Initial velocities u = 2 m/s

U = 0 m/s

Coefficient of restitution e = 1

(V-v) /(u -U ) = 1

V - v = u - U

= 2-0

V = v+2 ------------( 1)

From law of conservation of momentum ,

mu + MU = mv + MV

(41x2) + 0 = 41 v + 21(v+2)

82 = 41v +21v+42

= 62 v + 42

62 v = 40

v = 0.645 m/s

V = v + 2

= 2.645 m/s

a):the speed of the first marble immediately after the collision = 0.645 m/s

b):the speed of the second marble immediately after the collision = 2.645 m/s