JEE Main 2026 — Jan 23, Shift 1 — Physics solution

JEE Main2026Jan 23, Shift 1Physics

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
$\frac{1}{5}$
C
$\frac{2 + 3}{5}$
D
$\frac{2 - 3}{5}$

Solution

Step 1 — Speed of A at the lowest point. Bob A swings down from $6 0^{\circ}$ through a height $ℓ (1 - cos 6 0^{\circ})$ with $ℓ = 1 m$ :

$v_{A} = 2 g ℓ (1 - cos 6 0^{\circ}) = 2 \times 10 \times 1 \times \frac{1}{2} = 10 m/s$

Step 2 — Elastic collision with equal masses. The two bobs have equal mass, so they exchange velocities. Hence $v_{B} = 10 m/s$ and A comes to rest.

Step 3 — Minimum speed to just complete the loop. For bob B to just make it around the loop, the speed at the bottom of the loop must satisfy $v_{B} = 5 g R$ .

Step 4 — Solve for R. $10 = 5 g R \Rightarrow 10 = 5 \times 10 \times R \Rightarrow R = \frac{1}{5} m$ .

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