A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

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) 0.8

(d) 0.4

Detailed Answer:

The coefficient of restitution (\(e\)) is given by the formula: \[ e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}}. \] For this problem: - Block A (with mass \(m\)) moves with velocity \(v\). - Block B (with mass \(4m\)) is initially at rest. After the collision: - Block A comes to rest, so its final velocity is \(0\). - Block B moves with some velocity, say \(v_B\). From the conservation of momentum, we know: \[ \text{Initial momentum} = \text{Final momentum}. \] This gives: \[ mv = 4m v_B. \] Simplifying: \[ v_B = \frac{v}{4}. \] The relative velocity of approach is the velocity of block A before the collision, which is \(v\), and the relative velocity of separation is the velocity of block B after the collision, which is \(v_B = \frac{v}{4}\). Now, applying the coefficient of restitution formula: \[ e = \frac{v_B}{v} = \frac{\frac{v}{4}}{v} = \frac{1}{4}. \] Therefore, the value of the coefficient of restitution is: \[ \boxed{0.25} \ (\text{Option (b)}). \]