A wooden block of mass $m$ moves with velocity $V$ and collides with another block of mass $4m$, which is at rest. After collision the block of mass $m$ comes to rest. The coefficient of restitution will be
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When an incident body comes to rest ($v_1 = 0$) after a head-on collision with a stationary target ($u_2 = 0$), the conservation of momentum combined with the definition of $e$ yields a highly direct shortcut: $e = \frac{m_1}{m_2}$. Here, $e = \frac{m}{4m} = \frac{1}{4} = 0.25$.
Step 1: Set up the collision. A block of mass $m$ moving with speed $V$ strikes a stationary block of mass $4m$. After impact the first block stops. We need the coefficient of restitution $e$. Step 2: Apply momentum conservation. Total momentum before equals total momentum after: $mV + 0 = m(0) + 4m\,v_2$. Step 3: Find the velocity of the second block. Cancel $m$: $V = 4v_2$, so $v_2 = \dfrac{V}{4}$. Step 4: Recall the restitution definition. $e = \dfrac{\text{speed of separation}}{\text{speed of approach}} = \dfrac{v_2 - v_1}{u_1 - u_2}$. Step 5: Substitute the velocities. Here $v_1 = 0$, $v_2 = \frac{V}{4}$, $u_1 = V$, $u_2 = 0$, giving $e = \dfrac{\frac{V}{4} - 0}{V - 0}$. Step 6: Simplify. $e = \dfrac{V/4}{V} = \dfrac{1}{4} = 0.25$, which is option (2). \[ \boxed{e = 0.25} \]
