Q. A thick uniform bar lies on a frictionless horizontal surface and is free to move in any way on the surface. It mass is 0.16 kg and length is 1.7 m. Two particles each of mass 0.08 kg are moving on the same surface and towards the bar in the direction perpendicular to the bar, one with a velocity of 10 $ms^{- 1}$ and other with velocity $6ms^{- 1}$. If collision between particles and bar is completely inelastic, both particles strike with the bar simultaneously. The velocity of centre of mass after collision is
A) 2 $ms^{- 1}$
B) 4$\ ms^{- 1}$
C) 10$\ ms^{- 1}$
D) 167$\ ms^{- 1}$
Explanation
(b) Here, $m = 0.08$ kg
$m_{0} = 0.16$ kg
According to conservation principle of momentum
$mv_{1} + mv_{2} = \left( 2m + m_{0} \right)v_{CM}$
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$\therefore\ v_{CM} = \frac{mv_{1} + mv_{2}}{2m + m_{0}}$
$= \frac{0.08 \times 16}{0.16 + 0.16} = \frac{1.28}{1.32}$
$= \frac{128}{32} = 4ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.