JEE Advanced 2024 Paper 2 · Physics · Rotational Motion · hard

A uniform rod of mass M and length L=0.9 m hangs vertically at rest from a frictionless horizontal pivot at its upper end. A clay ball of mass m…

Numerical answer

A uniform rod of mass

M

and length

L=0.9\ \mathrm{m}

hangs vertically at rest from a frictionless horizontal pivot at its upper end. A clay ball of mass

m

moving horizontally at

6\ \mathrm{m\,s^{-1}}

strikes the lower end of the rod and sticks to it. The rod with the ball swings up and just reaches the horizontal position. Take

g=10\ \mathrm{m\,s^{-2}}

. The ratio

M/m

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Answer: 1

During the collision angular momentum about the pivot is conserved:

mvL=I\omega

with

I=\tfrac13ML^2+mL^2

. After it, mechanical energy is conserved up to the horizontal position, where the system is momentarily at rest:

\tfrac12I\omega^2=Mg\dfrac L2+mgL

. Substituting

\omega=mvL/I

\dfrac{m^2v^2L^2}{2I}=gL\left(\dfrac M2+m\right)

. Let

\mu=M/m

. Then

I=mL^2(\mu/3+1)

and the condition becomes

v^2=2gL\left(\dfrac\mu2+1\right)\left(\dfrac\mu3+1\right)=\dfrac{gL}{3}(\mu+2)(\mu+3)

. With

v^2=36

and

gL=9

(\mu+2)(\mu+3)=12

, i.e.

\mu^2+5\mu-6=0

, so

\mu=1

(the other root is negative). Hence

M/m=1

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