Center of Mass, Momentum & Collision

JEE Main level, numerical answer, medium

19 of 130

Question 19 of 130

A thin rod of mass 0.9kg0.9kg and length 1m1m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of move 0.1kg0.1kg moving in a straight line with velocity 80ms−180m{s}^{-1} hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in rads−1rad{s}^{-1}) of the rod immediately after the collision will be …………

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Answer: 20

Given,
Mass of the rod,
M=0.9kgM=0.9kg
Mass of particle,
m=0.1kgm=0.1kg
Length of the rod,
l=1ml=1m
Velocity of particle,
v=80ms−1v=80m{s}^{-1}
No external torque acting on the system of rod and particle, so angular momentum of system about the suspension is conserved, i.e.;
Li=Lf{L}_{i}={L}_{f}
⇒mvl=Iω...(1)\Rightarrow mvl=I\omega ...(1)
Where,
II is the momentum of inertia of rod and particle system, ω\omega is the angular velocity of system just after the collision.
⇒mvl=(Ml23+ml2)ω...(2)\Rightarrow mvl=\left(\dfrac{M{l}^{2}}{3}+m{l}^{2}\right)\omega ...(2)
Substitute the values in equation (2)

⇒0.1×80×1=(0.9×(1)23+0.1×(1)2)×ω\Rightarrow 0.1\times 80\times 1=\left(\dfrac{0.9\times (1{)}^{2}}{3}+0.1\times (1{)}^{2}\right)\times \omega
ω=20rads−1\omega =20rad{s}^{-1}

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