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Question

A uniform thin rod of mass, M=1 kg and length, L=2 m is hinged by a frictionless pivot at its one end O as shown in figure. A bullet of mass, m=50 g moving horizontally with a velocity v=1500 ms1 strikes the free end of the rod and gets embedded in it the angular velocity of the system about O just after the collision is-



A
98 rad s1
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B
95 rad s1
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C
100 rad s1
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D
76 rad s1
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Solution

The correct option is A 98 rad s1
As no external force involved here, hence linear and angular momentum of the system are conserved.

Using, conservation of angular momentum about the hinge point O,

Li=Lf

Lbullet+Lrod=Lbullet+rod

mvL+0=I(hinge point)ω .....(1)

Now, moment of inertia of system about hinge point O is,

I(hinge point)=13ML2+mL2

=13×1×4+0.05×4=1.53 kgm2

Using equation (1), we get

0.05×1500×2=1.53ω

ω=1501.53=98.0398 rad s1

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (A) is the correct answer.

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