Two bodies with masses m1=2kg and m2=4kg have initial velocities v1=v2=5m/s in the directions shown in the figure. If they collide at the origin elastically, then find the position of the center of mass 3seconds after the collision.
A
−4^i−7^j
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B
−4^i+7^j
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C
4^i−7^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=−43m/s (vcom)y=2(−5sin37∘)+(4×5)6=73m/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 3seconds 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.