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Let there are three equal masses situated at the vertices of an equilateral triangle, as shown in figure. Now the particle A starts with a velocity v1 towards line AB, particle B starts with the velocity v2 towards line BC and particle starts with velocity v3 towards line CA. Find the displacement of the centre of mass of the three particles A, B and C after time t. What would it be the displacement of centre of mass if v1=v2=v3?


A
(v1+v2+v3)t
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B
(v1+v2+v3)t2
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C
(v1+v2+v3)t3
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D
'0'
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Solution

The correct option is D '0'
First we write the three velocities in vectorial form, taking right direction as positive x-axis and upwards as positive y-axis.
v1=12v1^i32v1^j
v2=v2^i,v3=12v3^i+32v3^j
Thus the velocity of centre of mass of the system is
vCM=v+v2+v33
=(v212v112v3)^i+32(V3V1)^j
Which can be written asvCM=vx^i+vy^j
Δr=vxt^i+vxt^j
If v1=v2=v3=v, we have vCM=0
Therefore, there is no displacement of centre of mass of the system

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