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: Understanding the Concept:
In a perfectly elastic collision, both momentum and total kinetic energy are conserved. No energy is lost as heat or sound or deformation.

Step 2: Check the options:
(A) Kinetic energy before equals kinetic energy after. This is the definition, so (A) is true.
(B) Kinetic energy cannot increase in an ordinary collision, so (B) is wrong.
(C) Kinetic energy decreases only in an inelastic collision, so (C) is wrong here.
(D) The coefficient of restitution is 1 for an elastic collision (zero for a perfectly inelastic one), so (D) is wrong.

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