Question

A mass tied to a string is whirled in a horizontal circular path with a constant angular velocity and its angular momentum is 'L'. If the length of the string is now halved, keeping angular velocity same then the angular momentum will be

• A. \(\frac{L}{4}\)
• B. \(\frac{L}{2}\)
• C. \(L\)
• D. \(2L\)

Hint:

In an elastic collision kinetic energy is conserved and the coefficient of restitution is 1.

Answer 1

The Correct Option is A

Step 1: Recall e:
Coefficient of restitution $e = \frac{\text{speed of separation}}{\text{speed of approach}}$. For a perfectly elastic collision $e = 1$.

Step 2: Energy:
With $e = 1$, the kinetic energy afterwards equals the kinetic energy before. So (A) is the correct statement.

Final Answer:
Option (A). \[ \boxed{\text{(A)}} \]