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Step-by-Step Solution
Step 1: Identify the System and Given Data
• We have two blocks:
– Block 1 (lighter) with mass $m$ and initial velocity $v$.
– Block 2 (heavier) with mass $4m$ and initially stationary (velocity = 0).
• After the collision, the lighter block comes to rest, i.e., its final velocity = 0.
• We need to determine the coefficient of restitution $e$.
Step 2: Apply Conservation of Linear Momentum
Before collision, total momentum = $m \cdot v + 4m \cdot 0 = m v$.
After collision, the lighter block’s velocity = 0, and let the heavier block’s velocity be $v'$.
Thus, total momentum after collision = $m \cdot 0 + 4m \cdot v' = 4m v'$.
So from conservation of momentum:
$$ m v = 4m v' \quad \Longrightarrow \quad v' = \frac{v}{4}. $$
Step 3: Use the Definition of Coefficient of Restitution
The coefficient of restitution $e$ is defined as the ratio of the relative speed of separation to the relative speed of approach. In a one-dimensional collision, when the initial velocities are $u_1$ (for the lighter block) and $u_2$ (for the heavier block) and the final velocities are $v_1$ and $v_2$ respectively,
$$ e = \frac{v_2 - v_1}{u_1 - u_2}. $$
Here,
– $u_1 = v$, $u_2 = 0$,
– $v_1 = 0$, $v_2 = v' = \frac{v}{4}.
So,
$$
e = \frac{\left(\frac{v}{4}\right) - 0}{v - 0} = \frac{\frac{v}{4}}{v} = \frac{1}{4} = 0.25.
$$
Step 4: Conclude the Result
Hence, the coefficient of restitution for this collision is $\boxed{0.25}$.
