JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Conservation of Momentum

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

6 0^{\circ}

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 bob A, then radius R is ______ m.

(A) $\frac{3}{5}$
(B) $\frac{1}{5}$ ✓ Correct
(C) $\frac{2 + 3}{5}$
(D) $\frac{2 - 3}{5}$

Answer: (B)

Solution

Bob

A

swings down through

6 0^{\circ}

and reaches the lowest point with speed

v_{A} = 2 g ℓ (1 - cos 6 0^{\circ}) = 2 \times 10 \times 1 \times 0.5 = 10

m/s. In an elastic collision between equal masses the velocities are exchanged, so

B

moves off with

v_{B} = 10

m/s. To just complete the vertical circle, the speed at the top

Q

satisfies

v_{top}^{2} = g R

, and energy conservation from the bottom to the top gives

v_{B}^{2} = v_{top}^{2} + 4 g R = 5 g R

. Hence

10 = 5 \times 10 \times R

, so

R = \frac{1}{5}

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Conservation of Momentum

Solution

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