Two particles are kept on a smooth surface move towards
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Two particles are kept on a smooth surface, move towards each other due to mutual attraction. If the position and velocities of the particles at t=0 is as shown in the figure. Then position of the center of mass at t=2 sec is
A.
x=7 m
B.
x=5 m
C.
x=3 m
D.
x=2 m
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Initial position of centre of mass
xCM=m1x1+m2x2m1+m2
xCM=1×0+1×102=5 m
Velocity of centre of mass
vCM=m1v1+m2v2m1+m2
vCM=1×5−1×32=1 m/s
Displacement of CM in 2 sec=vCM×t=2 m
So location of CM =5+2=7 m