When a moving body collides with a stationary body, of n times its mass, then the amount of kinetic energy transferred to the stationary body is __________
(a) Suppose, mass of moving body is m1. Mass of stationary body, M2=nm1 For elastic collision, velocity of separation = velocity of approach ‌v2−v1=u−0 ‌v2−u=v1‌‌‌.........(i) By the law of conservation of momentum, m1u‌‌=m1v1+M2v2 m1u‌‌=m1v1+nm1v2 u‌‌=v1+nv2‌‌‌......(ii) u‌‌=v2−u+nv2‌‌‌ [From Eq. (i) ] ‌ 2u‌‌=(n+1)v2 v2‌‌=‌
2u
n+1
Putting this value in Eq. (i) v1=v2−u=‌
2u
n+1
−u=‌
2u−nu−u
n+1
=‌
(1−n)u
n+1
∴ Kinetic energy of moving body of mass m1, before collision, K1=‌
1
2
m1u2 Kinetic energy of stationary body after collision, K2‌‌=‌
1
2
M2v22=‌
1
2
nm1(‌
2u
n+1
)2 ‌‌=‌
1
2
m1u2⋅‌
4n
(n+1)2
∴ Amount of KE transferred to stationary body =‌