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JEE Main 202529 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Morning Shift
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2025 · 29 January · Morning Shift · Physics · Center of Mass, Momentum and Collision

As shown below, bob A of a pendulum having massless string of...

As shown below, bob AA of a pendulum having massless string of length ' RR ' is released from (60^{\circ}) to the vertical. It hits another bob BB of half the mass that is at rest on a friction less 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.) As shown below, bob A of a pendulum having massless string of length ' R ' is released from (60^ ) to the vertical. It

Options

  1. A(\frac{4}{3} \sqrt{\operatorname{Rg}})
  2. B(\frac{2}{3} \sqrt{\mathrm{Rg}})
  3. C(\sqrt{\mathrm{Rg}})
  4. D(\frac{1}{3} \sqrt{\operatorname{Rg}})

Correct answer

D. (\frac{1}{3} \sqrt{\operatorname{Rg}})

Step-by-step solution

As shown below, bob A of a pendulum having massless string of length ' R ' is released from (60^ ) to the vertical. It
Velocity of a just before hitting : u=2 gR2=gR\mathrm{u}=\sqrt{2 \mathrm{~g} \frac{\mathrm{R}}{2}}=\sqrt{\mathrm{gR}} Just after collision, let velocity of A and B are v1andv2\mathrm{v}_1 and v_2 respectively ∴ by COM: \therefore \text{ by COM: }mu=mv1+m2v22v1+v2=2u...(i)e=1=v2−v1u⇒v2−v1=u..(ii)\begin{aligned} & m u=m v_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}$From(i)−(ii)From (i) -(ii)\Rightarrow 3 \mathrm{v}_1=\mathrm{u} \Rightarrow \mathrm{v}_1=\frac{\mathrm{u}}{3}=\frac{1}{3} \sqrt{\mathrm{gR}}$
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