Question

A 49.0 g marble moving at 2.10 m/s strikes a 21.0 g marble at rest. Note that the collision is elastic and that it is a "head-on" collision so all motion is along a line. What is the speed of 49.0 g marble immediately after the collision? Express your answer with the appropriate units. What is the speed of 21.0 g marble immediately after the collision? Express your answer with the appropriate units.

Answer 1

let the velocities after collision be v1 and v2

m1u1 + m2u2 = m1v1 + m2v2

0.049*2.1 + 0 = 0.049v1 + 0.029v2

49v1 + 29v2 = 102.9 ---------(1)

velocity of approach = velocity of recess

v2 - v1 = u1-u2

v2 - v1 = 2.1 - 0

v2 - v1 = 2.1 -------(2)

mutilpy (2) by 49 and adding we have

78v2 = 205.8

v2 = 2.63 m/s

v1 = v2 - 2.1 = 2.63 - 2.1 = 0.538 m/s

speed of 49 g marble = 2.63 m/s

speed of 21 g marble = 0.538 m/s