Consider a uniform rod of mass M = 4m and length l pivoted about its centre. A mass m moving with velocity v making angle θ = (π)/(4) to the rod’s long axis collides with one end of the rod and sticks to it. The

Correct answer: C. 3 2 7 (v)/(l) L _ oi = L _ of mV 2 × (L)/(2) = [ 4m L ^ 2 12 + m L ^ 2 4 ] × ω ω = 6V 7 2 L = 3 2 V 7L

JEE Main 20208 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 8 January
Shift: Morning Shift
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

JEE Main 2020 · 8 January · Morning Shift · Physics · Center of Mass, Momentum and Collision

Consider a uniform rod of mass M = 4m and length l pivoted about...

Consider a uniform rod of mass

M = 4m

and length

l

pivoted about its centre. A mass

m

moving with velocity

v

making angle

\theta = \dfrac{\pi}{4}

to the rod’s long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is: Consider a uniform rod of mass M = 4m and length l pivoted about its centre. A mass m moving with velocity v making

Consider a uniform rod of mass M = 4m and length l pivoted about its centre. A mass m moving with velocity v making

Options

A $\dfrac{3}{7\sqrt{2}}\dfrac{v}{l}$
B $\dfrac{3}{7}\dfrac{v}{l}$
C $\dfrac{3\sqrt{2}}{7}\dfrac{v}{l}$
D $\dfrac{4}{7}\dfrac{v}{l}$

Correct answer

C. $\dfrac{3\sqrt{2}}{7}\dfrac{v}{l}$

Step-by-step solution

{L}_{oi} = {L}_{of}

\dfrac{mV}{\sqrt{2}} \times \dfrac{L}{2} = \left[\dfrac{4m{L}^{2}}{12} + \dfrac{m{L}^{2}}{4}\right] \times \omega

\omega = \dfrac{6V}{7\sqrt{2}L} = \dfrac{3\sqrt{2}V}{7L}

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