AP EAMCET Engineering 21 Sep 2020 Shift 2 Paper

(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
=‌

K2

K1

=‌

‌ 1 2 m1u2× 4n (n+1)2

1 2 m1u2

=‌

4n

(n+1)2