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 as shown in figure. At distance of 1 m below point P, an identical bob B is kept at rest on a smooth horizontal surface that extends to a circular track of radius R as shown in figure. If bob B just manages to complete the circular path of radius R upto a point Q after being hit elastically by bobA, then radius R is ____ m.

Solution

Physics sir · 4 steps

Bob A drops 0.5 m and hands all its speed to B; B then needs v² = 5gR to clear the top of the loop.

1. Speed of A. Released at 60°, bob A falls through h = L(1-cos 60°) = 1 × 1/2 = 0.5 m to its lowest point. So v_A² = 2gh = 2 × 10 × 0.5 = 10 m² s⁻².
2. Elastic, equal masses. In a head-on elastic collision between equal masses, the velocities are exchanged. A stops and B moves off with v_B² = 10 m² s⁻².
3. Getting past the top. To reach Q, B must cross the top of the loop without losing contact. Just making it means vₜₒₚ² = gR there. Energy conservation from the bottom (height gain 2R) then gives the condition below.

   v_B² = vₜₒₚ² + 2g(2R) = 5gR

4. So. 10 = 5 × 10 × R, so R = 1/5 m, which is option (a). After the top, B moves down towards Q, so the normal force only grows and B stays on the track.

The idea to keep. Pendulum height is always L(1-cos θ) with θ from the vertical. Equal masses in an elastic collision swap velocities, and clearing a vertical loop needs v² = 5gR at the bottom.

Answer from the NTA final answer key. Solution by PrepWiser.