Three particles of equal masses are placed at the corners of an equilateral triangle as shown in the figure. Now particle A starts with a velocity v1 towards line AB, particle. B starts with a velocity v2 towards line BC and particle C starts with velocity v3 towards line CA. The displacement of CM of three particle A, B and C after time t will be (given if v1 = v2 = v3)
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a
zero
b
v1+v2+v33t
c
v1+32v2+v323t
d
v1+v2+v34t
answer is A.
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Detailed Solution
Using v→CM=i^vx+j^vy where vx=mvx1+mvx2+mvx33m=0vy=mvy1+mvy2+mvy33m=0Net velocity is zero (using vector theory) i.e., CM is at rest. So, displacement of C.M. is zero. So choices (b), (c) and (d) are wrong.