Question

A small bob A of mass m is attached to a massless rigid rod of length 1 m pivoted at point P and kept at an angle of 60° with vertical. At 1 m below P, bob B is kept on a smooth surface. If bob B just manages to complete the circular path of radius R after being hit elastically by A, then radius R is_______ m :}
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• A. 3/5
• B. (2 - √3)/5
• C. 1/5
• D. (2 + √3)/5

Hint:

In elastic collisions between two equal masses, if one is at rest, they simply swap velocities. This is a huge time-saver in JEE!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem combines conservation of energy (for bob A's swing) and circular motion requirements (for bob B). To "just complete" a circle of radius \(R\), the velocity at the bottom must be \(v = \sqrt{5gR}\).
Step 2: Key Formula or Approach:
1. Velocity of A at bottom: \(v_A = \sqrt{2gh}\).
2. Height \(h = L(1 - \cos \theta)\).
3. Elastic collision of equal masses: Velocities are exchanged.
Step 3: Detailed Explanation:
Height of A: \(h = 1(1 - \cos 60^\circ) = 1(1 - 0.5) = 0.5\,\text{m}\).
Velocity of A just before collision: \(v_A = \sqrt{2 \times g \times 0.5} = \sqrt{g}\).
Since the collision is elastic and masses are identical, bob B acquires the velocity of bob A: \(v_B = \sqrt{g}\).
For B to complete the circular track: \[ v_B \ge \sqrt{5gR} \] \[ \sqrt{g} = \sqrt{5gR} \implies g = 5gR \implies R = \frac{1}{5}\,\text{m} \]
Step 4: Final Answer:
The radius \(R\) is 1/5 m.