Question

A ball of mass 0.195 kg with a velocity of 1.5 m/s meets a ball of mass 0.299 kg with a velocity of-039i m/s in a head-on, elastic collision (a) Find their velocities after the collision. m/s m/s (b) Find the velocity of their center of mass before and after the collision. Vom, bfore 0.383 aat0.383 m/s m/s

Answer 1

(a) Use the Principle of Conservation of Momentum :

m1u1 + m2u2 = m1v1 + m2v2

(0.195)(1.52) + (0.299)(-0.393) = (0.195)v1 + (0.299)v2 .........(1)

Also

since the collision is elastic, e = 1

So

(u1 - u2) = (v2 - v1)

v2 - v1 = 1.913 ...............(2)

Solve (1) and (2) to get v1 and v2

Continuing with (1):

0.178893 = 0.195v1 + 0.299v2 = 0.195v1 + 0.299(v1+1.913) (from 2)

=> v1 = -0.796i m/s

So, from (2): v2 = 1.117i m/s

(b) Velocity of Center of Mass before collision

V(before) = (m1u1 + m2u2) / (m1 + m2)

= 0.362 m/s

Since there is no external force on the system, the velocity of the center of mass remains unchanged after the collision.

So:

V(after) = 0.362 m/s