Question

Three small bodies of identical masses can move along a straight line. The central body (2) is initially at rest and bodies 1 and 3 are at a distance L and 2L from the central body respectively. Bodies 1 and 3 move towards body 2 with speeds ${v_0}$ each. The collision between 1 and 2 is perfectly elastic and the collision between body 2 and 3 is perfectly inelastic. After all the collisions are over,

A. All the bodies come to rest
B. The body 1 moves towards left, bodies 2 and 3 move towards right
C. body 2 remains at rest and other bodies 1 and 3 turn back
D. All the bodies move towards the right.

Answer 1

In case of collisions whether they are elastic or inelastic collisions there is always one thing we conserve which is momentum. If those collisions are elastic then we will conserve kinetic energy too. In case of inelastic collisions we can’t conserve the kinetic energy because it is lost in the form of heat or sound.

Formula used:
\eqalign{ & p = mv \cr & {p_i} = {p_f} \cr}
$\dfrac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}} = e$

Complete step by step answer:
According to Newton's second law as long as there is no external impulsive force momentum will always be conserved. For example if a bomb or rock explodes then we can conserve the momentum because explosion is considered as an internal reaction. So when there is no external impulsive force we find out the momentum of the rock or bomb before it gets exploded and we equate it to the momentum of the system of pieces after the rock has exploded. In this we solve many problems.
Now coming to the given question first the body 1 collides with 2 and comes to rest

We have momentum $p = mv$ where ‘m’ is the mass and ‘v’ is the velocity
We conserve the momentum for this collision ${p_i} = {p_f}$
Initial momentum before collision is ${p_i} = m \times {v_0} = m{v_0}$
Final momentum after collision will be of mass 2 only as mass 1 comes to rest ${p_f} = mv$ where ‘v’ is the velocity of mass 2 because first body comes to rest after collision and it doesn’t contribute to final momentum
${p_i} = {p_f}$
\eqalign{ & \Rightarrow {v_0}m = mv \cr & \Rightarrow {v_0} = v \cr}
Next we will solve for the collision between the second and third body. In case of inelastic collision the coefficient of restitution(e) will be zero.
$\dfrac{{{v_3} - {v_2}}}{{{u_2} - {u_3}}} = e$
Where ${u_2}$ and ${u_3}$ are the initial velocities of second and third bodies respectively while ${v_2}$ and ${v_3}$ are the final velocities of second and third bodies respectively and ‘e’ s the coefficient of restitution which will be zero for the inelastic collision.
Here ${u_2} = {v_0}$ and ${u_3} = - {v_0}$
\eqalign{ & \Rightarrow \dfrac{{{v_3} - {v_2}}}{{{u_2} - {u_3}}} = 0 \cr & \Rightarrow \dfrac{{{v_3} - {v_2}}}{{{v_0} - {v_0}}} = 0 \cr & \Rightarrow {v_3} - {v_2} = 0 \cr}
From conservation of momentum we have
\eqalign{ & {p_i} = {p_f} \cr & \Rightarrow m{v_0} - m{v_0} = m{v_2} + m{v_3} \cr & \Rightarrow 0 = {v_2} + {v_3} \cr}
From the above two equations we will get ${v_2} = 0,{v_3} = 0$.
So finally by the end of collisions all the three bodies will come to rest.

Hence option A will be the answer.

Note:
During the collision there is an external force which is gravity but still we had conserved the momentum because gravity is not the external impulsive force and collision is the very short time action and within that time gravity doesn’t create much difference to the momentum of the system.