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# A moving block having mass $m$, collides with another stationary block having mass $4m$. The lighter block comes to rest after the collision. When the initial velocity of the lighter block is $v$, then the value of the coefficient of restitution ($e$) will be:(A) $0.5$(B) $0.25$(C) $0.4$(D) $0.8$

Last updated date: 15th Sep 2024
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Hint: Here we will use the conservation of linear momentum since the collision is non-sticky. The conservation of linear momentum will give the solution $m \times v = 4m \times u$. Using this equation, we will find the coefficient of restitution as $e = \dfrac{{{v_s}}}{{{v_a}}} = \dfrac{u}{v}$, substituting $v = 4u$ from the above equation.
Formula used:
Conservation of momentum and coefficient of restitution $e = \dfrac{{{v_s}}}{{{v_a}}}$.

The coefficient of restitution is defined as the ratio of final velocity to the initial velocity between two objects after their collision. More precisely, it is the ratio of the relative velocity of the two bodies at their separation to their approach.
Using the conservation of linear momentum, we have
$m \times v + 4m \times 0 = m \times 0 + 4m \times u$,
where $u$ is the velocity of the heavier block after the collision.
We have used zero velocities for the initial velocity of the heavier block and the final velocity of the lighter block as given in the above problem.
Therefore, we have
$m \times v = 4m \times u$,
$\Rightarrow u = \dfrac{v}{4}$.
Now from the definition, we have the coefficient of restitution as
$e = \dfrac{{{v_s}}}{{{v_a}}}$,
where ${v_s}$ is the relative velocity of separation, and
${v_a}$ is the relative velocity of approach.
Substituting the value of the final velocity of the heavier block into the above equation of coefficient of restitution, we get,
$e = \dfrac{u}{v} = \dfrac{{v/4}}{v} = \dfrac{1}{4}$
$\Rightarrow e = 0.25$

Therefore, the correct answer is option (B).

Note: The coefficient of restitution is defined for the collision of two bodies only. The coefficient of restitution is unity for perfectly elastic collisions and lies between $0$ and $1$ for inelastic collisions. In the case of collisions involving the sticky mass concept, where some mass moving with an initial velocity sticks to another mass, we cannot conserve energy, since some energy is within the amalgamation of the bodies.