Question

A body is hanging from rigid support by an inextensible string of length ‘$l$’. It is struck inelastically by an identical body of mass $m$ with horizontal velocity $v=2gl$, the tension in the string increases just after the striking by:
(A) $mg$
(B) $3 mg$
(C) $2 mg$
(D) none of these

Answer 1

Hint
The law of conservation momentum will provide the velocity of the body after a collision. The resultant force on the string will provide the change in tension in the string.

Complete step by step solution
Given, The length of an inextensible string is ${\rm{'l'}}$.
The mass of the body is ‘m’.
The horizontal velocity of the body is ${\rm{v}} = \sqrt {2{\rm{gl}}} $
According to the law of conservation of momentum, the momentum of two or more bodies will be equal to the momentum of those two or more bodies after a collision. Thus, the momentum of the bodies remains constant during an elastic collision.
Since the mass of the bodies after the collision is given by, ${\rm{m'}} = $2m.
Therefore, the velocity of bodies after the collision is given by,
${\rm{mv}} = 2{\rm{mv'}}$
${\rm{v'}} = \dfrac{{\rm{v}}}{2} = \dfrac{{\sqrt {2{\rm{gl}}} }}{2} = \sqrt {\dfrac{{{\rm{gl}}}}{2}} {\rm{\;}}$
Initially, the tension in the string will be, ${{\rm{T}}_1} = {\rm{mg}}$
Now, the bodies may undergo circular motion. And, the required centripetal force will obtain from the tension on the string after the collision.
i.e. $\dfrac{{{\rm{m'}}{{{\rm{v'}}}^2}}}{{\rm{l}}} = {\rm{T'}} - 2{\rm{mg}}$
$\dfrac{{2{\rm{m}}}}{{\rm{l}}}\left( {\dfrac{{{\rm{gl}}}}{2}} \right) = {\rm{T'}} - 2{\rm{mg}}$
${\rm{T'}} = 3{\rm{\;mg}}$
Now, the increase in tension in the string is given by,
$ = 3{\rm{mg}} - {\rm{mg}} = 2{\rm{mg}}$

Therefore, (C) $2 mg$ is the required solution.

Note The resultant force on the string is the result of tension in the string acting in the upward direction of the string or away from the body and weight of the body acting in a downward direction.