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!
Step 1: Note the values before the collision. A block of mass $m$ moves with velocity $V$ and hits a still block of mass $5m$. So before the hit, the first block has speed $V$ and the second block has speed $0$.
Step 2: Note the values after the collision. After the collision the first block stops, so its final velocity is $0$. Let the heavy block move off with velocity $v_2$.
Step 3: Apply conservation of momentum. Total momentum stays the same: \[ m V + 5m(0) = m(0) + 5m\, v_2 \] This gives \[ m V = 5m\, v_2 \]
Step 4: Find the heavy block's speed. Cancel $m$ from both sides: \[ v_2 = \frac{V}{5} \] So the heavy block moves forward at one fifth of $V$.
Step 5: Write the restitution formula. The coefficient of restitution compares how fast the blocks separate to how fast they approached: \[ e = \frac{\text{speed of separation}}{\text{speed of approach}} = \frac{v_2 - 0}{V - 0} \]
Step 6: Put in the numbers. Substitute $v_2 = \frac{V}{5}$: \[ e = \frac{V/5}{V} = \frac{1}{5} = 0.2 \] So the coefficient of restitution is $0.2$. \[ \boxed{0.2} \]
