A 1kg ball moving at 12m/s collides head on with 2kg ball moving in opposite direction at 24m/s. If the coefficient of restitution is 2/3,then how much energy is lost in collision

we are given

for ball 1

m1 = 1 kg

u1 = 12 m/s

and

for ball 2

m2 = 2 kg

u2 = 24 m/s

now,

the coefficient of restitution is given as

e = (v2 - v1) / (u1 - u2) = 2/3 = 0.66

so,

(v2 - v1) = 0.66 x (u1 - u2) = 0.66 x (12 - 24)

or

(v2 - v1) = -7.92

or

v2 = v1 - 7.92  ..............................(1)

now, as e

m1u1 + m2u2  = m1v1 + m2v2

or

1x12 + 2x24 = 1xv1 + 2xv2

or

60 = v1 + 2v2

so,

v2 =  (60 - v1) / 2  ...........................(2)

by equating (1) and (2), we get

v1 - 7.92 = (60 - v1) / 2

or by rearranging 

2v1 + v1 = 60 + 2x7.92

3 v1 = 75.48

thus,

v1 = 25.16 m/s

and

v2 = 25.16 - 7.92 = 17.24 m/s

so,

loss in kinetic energy will be

dE = Ei - Ef = [(1/2)m1u12  +(1/2)m2u22] - [(1/2)m1v12  +(1/2)m1v22]

which can be calculated...