Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

86 of 130

Question 86 of 130

A particle of mass mm moving with velocity vv collides with a stationary particle of mass 2m2m. After collision, they stick together and continue to move together with velocity

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Answer: (A) v3\dfrac{v}{3}

The law of conservation of momentum is given as
m1u+m2u′=(m1+m2)v′{m}_{1}u+{m}_{2}u'=\left({m}_{1}+{m}_{2}\right)v'
The given data is
u=vu′=0ms−1m1=mm2=2mu=v u'=0m{s}^{-1} {m}_{1}=m {m}_{2}=2m
Using the conservation of momentum equation,
mv+2m(0)=(m+2m)v′⇒v′=mv3m=v3mv+2m(0)=(m+2m)v' \Rightarrow v'=\dfrac{mv}{3m}=\dfrac{v}{3}

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