Question

A block of mass \( m \) collides with another stationary block of mass \( 2m \). The lighter block comes to rest after collision. If the velocity of the first block is \( u \), then the value of the coefficient of restitution is:

• A. 0.4
• B. 0.8
• C. 0.5
• D. 0.6

Hint:

The coefficient of restitution describes how elastic a collision is. A value of 0.5 indicates that the collision is partially inelastic.

Answer 1

The Correct Option is C

Step 1: Coefficient of Restitution Formula.
The coefficient of restitution \( e \) is defined as: \[ e = \frac{\text{relative velocity after collision}}{\text{relative velocity before collision}} \] For this problem, the relative velocity before the collision is \( u - 0 = u \), and the relative velocity after the collision is \( 0 - v_2 \), where \( v_2 \) is the velocity of the second block after collision. Since the lighter block comes to rest, \( v_2 = u/2 \). Hence: \[ e = \frac{v_2}{u} = \frac{1}{2} = 0.5 \] Step 2: Final Answer.
Thus, the coefficient of restitution is 0.5.