Question

A ball of mass 0.198 kg with a velocity of 1.60 î m/s meets a ball of mass 0.306 kg with a velocity of -0.408 î m/s in a head-on, elastic collision.

(a) Find their velocities after the collision.
V1?

V2?

(b) Find the velocity of their center of mass before and after the collision.
Before?

After?

Answer 1

Let
m1 = 0.198 kg
u1 = 1.60 m/s
m2 = 0.306 kg
u2 = -0.408 m/s

a) After the collision,
v1 = ( (m1 - m2)*u1 + 2*m2*u2 )/(m1 + m2)

= ( (0.198 - 0.306)*1.60 + 2*0.306*(-0.408) )/(0.198 + 0.306)

= -0.838 m/s <<<<<<<<<<---------------Answer

v2 = ( (m2 - m1)*u2 + 2*m1*u1 )/(m1 + m2)

= ( (0.306 - 0.198)*(-0.408) + 2*0.198*1.60 )/(0.198 + 0.306)

= 1.17 m/s <<<<<<<<<<---------------Answer

b) velocity of center of mass before the collision,

v_cm(before) = (m1*u1 + m2*u2)/(m1 + m2)

= (0.198*1.6 + 0.306*(-0.408))/(0.198 + 0.306)

= 0.381 m/s <<<<<<<<<<---------------Answer

velocity of center of mass after the collision,

v_cm(after) = (m1*v1 + m2*v2)/(m1 + m2)

= (0.198*(-0.838) + 0.306*1.17)/(0.198 + 0.306)

= 0.381 m/s <<<<<<<<<<---------------Answer