JEE Main 2025 Jan 29 Shift 1, Physics Q10: SHM in Pendulums

JEE Main2025Jan 29, Shift 1Physics

As shown below, bob $A$ of a pendulum having massless string of length $R$ is released from $6 0^{\circ}$ to the vertical. It hits another bob $B$ of half the mass that is at rest on a frictionless table in the centre. Assuming elastic collision, the magnitude of the velocity of bob $A$ after the collision will be (take $g$ as acceleration due to gravity)

A
$\frac{1}{3} R g$
B
$R g$
C
$\frac{4}{3} R g$
D
$\frac{2}{3} R g$

Solution

Speed of $A$ just before impact: when released from $6 0^{\circ}$ , the vertical drop is $R - R cos 6 0^{\circ} = R /2.$ Energy conservation gives $u = 2 g (R /2) = g R .$

Elastic 1D collision between masses $m$ (incoming, speed $u$ ) and $m /2$ (at rest): use the standard result $v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u .$

With $m_{1} = m, m_{2} = m /2$ : $v_{1} = \frac{m - m /2}{m + m /2} u = \frac{1/2}{3/2} u = \frac{u}{3} .$

Therefore $∣ v_{A} ∣ = \frac{1}{3} g R .$

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JEE Main 2025 Jan 29 Shift 1, Physics Q10: SHM in Pendulums

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