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Question

Two bodies with masses m1=2 kg and m2=4 kg have initial velocities v1=v2=5 m/s in the directions shown in the figure. If they collide at the origin elastically, then find the position of the center of mass 3 seconds after the collision.


A
4^i7^j
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B
4^i+7^j
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C
4^i7^j
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D
4^i+7^j
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Solution

The correct option is B 4^i+7^j

Let us calculate the velocity of centre of mass of the system in the given situation.
Let (vcom)x be the velocity of centre of mass in x direction.
(vcom)x=m1v1x+m2v2xm1+m2
Similarly,
(vcom)y=m1v1y+m2v2ym1+m2
So,
(vcom)x=2(5cos37)+(4×0)6=43 m/s
(vcom)y=2(5sin37)+(4×5)6=73 m/s

Therefore, velocity of the center of mass of the system
vcom=(43^i+73^j) m/s

Here, net force on the system in both directions x and y is zero. So, vcom will remain unchanged.
Since the collision occurs at the origin, ri=0. So position vector of center of mass 3 seconds after collision is
rcom(t=3)=ri+(vcom×t)
=0+(43^i+73^j)×3=4^i+7^j
Thus, option (b) is the correct answer.

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