A moving block having mass ${\text{m}}$ , collides with another stationary block having mass ${\text{4m}}$ . The lighter block comes to rest after collision. When the initial velocity of the lighter block is $v$ , find the value of the coefficient of restitution $\left( e \right)$ .
A) $0 \cdot 5$
B) $0 \cdot 25$
C) $0 \cdot 4$
D) $0 \cdot 8$

Hint:During any collision, the linear momentum of the system will be conserved. Here, the heavier block is mentioned to be stationary before the collision and so its initial velocity will be zero. The coefficient of restitution refers to the ratio of the final relative velocity of the two blocks to the initial relative velocity of the two blocks.

Formula used:
-The coefficient of restitution for two colliding objects is given by, $e = - \dfrac{{\left( {{v_1} - {v_2}} \right)}}{{\left( {{u_1} - {u_2}} \right)}}$ where ${v_1}$ , ${v_2}$ are the final velocities of the two objects and ${u_1}$ , ${u_2}$ are the initial velocities of the two objects.

Complete step by step solution:
Step 1: List the parameters given about the blocks undergoing collision.
The mass of the lighter block is given to be ${m_1} = m$ .
The mass of the heavier block is given to be ${m_2} = 4m$ .
The initial velocity of the lighter block is given to be ${u_1} = v$ .
As the lighter block comes to rest after collision, its final velocity will be ${v_1} = 0$ .
The initial velocity of the heavier block is taken to be ${u_2} = 0$ .
Let $v$’ be the final velocity of the heavier block after the collision.

Step 2: Apply the conservation of momentum principle to obtain the final velocity of the heavier block.
In the given collision, the linear momentum of the system is conserved.

$ \Rightarrow {p_{before}} = {p_{after}}$ -------- (1)

The momentum of the system before the collision is ${p_{before}} = {m_1}{u_1} + {m_2}{u_2} = mv$ --------- (2)
The momentum of the system after the collision is ${p_{after}} = {m_1}{v_1} + {m_2}{v_2} = 4mv'$ --------- (3)
Substituting equations (2) and (3) in equation (1) we get, $mv = 4mv'$

$ \Rightarrow v' = \dfrac{v}{4}$

Thus the final velocity of the heavier block is obtained to be $v' = \dfrac{v}{4}$ .

Step 3: Express the relation for the coefficient of restitution of the two blocks.
The coefficient of restitution for two colliding blocks can be expressed as

$e = - \dfrac{{\left( {{v_1} - {v_2}} \right)}}{{\left( {{u_1} - {u_2}} \right)}}$ --------- (4)

Substituting for ${u_1} = v$ , ${v_1} = 0$ , ${u_2} = 0$ and $v' = \dfrac{v}{4}$ in equation (4) we get, $e = - \dfrac{{\left( {0 - \dfrac{v}{4}} \right)}}{{\left( {v - 0} \right)}} = \dfrac{{ - \left( { - v} \right)}}{{4v}} = \dfrac{1}{4} = 0 \cdot 25$

$\therefore $ the coefficient of restitution is obtained to be $e = 0 \cdot 25$ .

Hence, the correct option is B.

Note: The coefficient of restitution determines whether the two blocks collide elastically or inelastically. For an elastic collision, the coefficient of restitution will be one while for an inelastic collision, its value lies between 0 and 1. Here the obtained coefficient of restitution is $0 \cdot 25$ . So the above-mentioned collision is inelastic in nature. For a perfectly inelastic collision, the coefficient of restitution would be zero. The final relative velocity of the two blocks is termed as the velocity of separation and the initial relative velocity of the two blocks is termed as the velocity of approach.