Question:
A moving block having mass $m$ collides with another stationary block of mass $5m$. After the collision, the block with mass $m$ comes to rest. If the initial velocity of the block with mass $m$ is $V$, then the value of the coefficient of restitution ($e$) is:
A moving block having mass $m$ collides with another stationary block of mass $5m$. After the collision, the block with mass $m$ comes to rest. If the initial velocity of the block with mass $m$ is $V$, then the value of the coefficient of restitution ($e$) is:
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For a collision where the incoming mass comes to a complete stop ($v_1 = 0$), the momentum equation simplifies to $m_1 u_1 = m_2 v_2 \implies v_2 = \frac{m_1}{m_2}u_1$. Substituting this directly into the definition of $e$ yields:
$$e = \frac{m_1}{m_2}$$
Here, $e = \frac{m}{5m} = \frac{1}{5} = 0.2$. This short formula works whenever the first mass stops completely after a head-on collision with a stationary target!
Updated On: Jun 10, 2026
- $0.2$
- $0.5$
- $0.7$
- $0.25$
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The Correct Option is A
Solution and Explanation
Concept:
In any collision process free from external forces, total linear momentum is conserved. The coefficient of restitution ($e$) acts as a measure of the elasticity of the collision and is defined as the ratio of the relative velocity of separation to the relative velocity of approach along the line of impact:
$$e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} = \frac{v_2 - v_1}{u_1 - u_2}$$
Step 1: Let us summarize the state variables before and after the collision:
• Before Collision: itemize
• Mass of block 1: $m_1 = m$, Initial velocity: $u_1 = V$
• Mass of block 2: $m_2 = 5m$, Initial velocity: $u_2 = 0$ (stationary)
After Collision:
• Final velocity of block 1: $v_1 = 0$ (comes to rest)
• Final velocity of block 2: $v_2 = v$
itemize Applying the Law of Conservation of Linear Momentum: $$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$ $$m(V) + 5m(0) = m(0) + 5m(v)$$ $$mV = 5mv$$ Dividing both sides by $m$ determines the final velocity of the heavier block: $$v = \frac{V}{5} = 0.2V$$
Step 2: Using the definition of the coefficient of restitution $e$: $$e = \frac{v_2 - v_1}{u_1 - u_2}$$ Substituting our known velocity terms into this ratio configuration: $$e = \frac{0.2V - 0}{V - 0}$$ $$e = \frac{0.2V}{V} = 0.2$$ Thus, the coefficient of restitution is exactly $0.2$, matching Option (A).
Step 1: Let us summarize the state variables before and after the collision:
• Before Collision: itemize
• Mass of block 1: $m_1 = m$, Initial velocity: $u_1 = V$
• Mass of block 2: $m_2 = 5m$, Initial velocity: $u_2 = 0$ (stationary)
After Collision:
• Final velocity of block 1: $v_1 = 0$ (comes to rest)
• Final velocity of block 2: $v_2 = v$
itemize Applying the Law of Conservation of Linear Momentum: $$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$ $$m(V) + 5m(0) = m(0) + 5m(v)$$ $$mV = 5mv$$ Dividing both sides by $m$ determines the final velocity of the heavier block: $$v = \frac{V}{5} = 0.2V$$
Step 2: Using the definition of the coefficient of restitution $e$: $$e = \frac{v_2 - v_1}{u_1 - u_2}$$ Substituting our known velocity terms into this ratio configuration: $$e = \frac{0.2V - 0}{V - 0}$$ $$e = \frac{0.2V}{V} = 0.2$$ Thus, the coefficient of restitution is exactly $0.2$, matching Option (A).
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