QJEEM25Ph2a1a: JEE Main 2025 Physics - Class 11 JEEM25Ph2a1a Solution: As shown below, bob A of a pendulum having massless string of…

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QJEEM25Ph2a1a: JEE Main 2025 Physics - Class 11 JEEM25Ph2a1a Solution: As shown below, bob A of a pendulum having massless string of…

As shown below, bob A of a pendulum having massless string of length 'R' is released from 60° to the vertical. It hits another bob B of half the mass that is at rest on a frictionless table in the center. Assuming elastic collision, the magnitude of the velocity of bob A after the collision will be (take g as acceleration due to gravity.)

JEE Main 2025 (Online) 29th January Morning Shift Physics - Center of Mass and Collision Question 3 English

Multiple Choice Options

A $\frac{4}{3}\sqrt{Rg}$
B $\frac{1}{3}\sqrt{Rg}$
C $\sqrt{Rg}$
D $\frac{2}{3}{\sqrt{Rg}}$

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JEE Main 2025 (Online) 29th January Morning Shift Physics - Center of Mass and Collision Question 3 English Explanation

Velocity of a just before hitting :

$u = \sqrt{2 g \frac{R}{2}} = \sqrt{gR}$

Just after collision, let velocity of A and B are $\mathrm{v}_1$ and $v_2$ respectively

$∴ by COM:$

$\begin{aligned} mu = {mv}_{1} + \frac{m}{2} v_{2} \\ 2 v_{1} + v_{2} = 2 u ...... (i) \\ e = 1 = \frac{v_{2} - v_{1}}{u} \\ \Rightarrow v_{2} - v_{1} = u ...... (ii) \end{aligned}$

From (i) -(ii)

$\Rightarrow 3 v_{1} = u \Rightarrow v_{1} = \frac{u}{3} = \frac{1}{3} \sqrt{gR}$

QJEEM25Ph2a1a: JEE Main 2025 Physics - Class 11 JEEM25Ph2a1a Solution: As shown below, bob A of a pendulum having massless string of…

Multiple Choice Options

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