TS EAMCET 18-July-2022 Shift 2 Solved Paper

Let the mass of the moving particle =m
∴ The mass of the stationary particle =‌

1

n

×m=‌

m

n

Also, let the speed of moving particle =v Now, by the law of conservation of linear momentum

Situation after collision
∴‌‌mv+(‌

m

n

×0)=mv′+‌

m

n

v′ .
⇒‌‌mv=m(v′+‌

v′

n

) ⇒
v=v′+‌

v′

n

‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
Also, coefficient of restitution,
e=‌

−(v′′−v′)

(0−v′)

Since, it is elastic collision,
∴‌‌e=1
⇒‌‌1=‌

−(v′′−v′)

−v

⇒v′′=v+v′
⇒‌‌v′=v′′−v‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
From Eqs. (i) and (ii), we get
v=v′′+‌

v′′

n

v′=[‌

2v

1+‌ 1 n

]‌‌‌⋅⋅⋅⋅⋅⋅⋅(iii)
Now, kinetic energy transferred to the stationary particle
=(‌ kinetic energy ‌)‌final ‌−(‌ kinetic energy ‌)‌initial ‌
⇒‌‌Kf′−Ki′=‌

1

2

×‌

m

n

×v′′2‌‌‌⋅⋅⋅⋅⋅⋅⋅(iv)
Also, initial kinetic energy of mass m
=‌

1

2

mv2‌‌‌⋅⋅⋅⋅⋅⋅⋅(v)
∴ The fraction of KE that gets transferred to the stationary particle from the moving particle
‌=‌

‌ 1 2 × m n ×v′2

1 2 mv2

=‌

v′2

nv2

=‌

1

n

(‌

v′

v

)2
‌=‌

1

n

(‌

2v

(1+‌ 1 n )v

)2
‌=‌

1

n

‌

4v2

(1+‌ 1 n )2v2

=‌

4

n(‌ n+1 n )2

=‌

4n

(1+n)2