One particle of mass 1 kg is moving along positive x-axis with velocity 3 m/s. Another particle of mass 2 kg is moving along y-axis with 6m/s. At time t = 0, 1 kg mass is at (3m, 0) and 2 kg at (0, 9m), x-y plane is the horizontal plane. (surface is smooth for question 47 and rough for question 48 and 49)
Co-ordinates of centre of mass where it will stop finally are

Options

A. (2.0 m, 14.25 m)
B. (2.25 m, 10 m)
C. (3.75 m, 9 m)
D. (1.75 m, 12 m)

Answer

D

Explanation

Mass 1 kg will stop at, `x_(1)=3+v_(1)^(2)/(2mug) rArr 3+9/4=5.25m, x_(2)=0`
Mass 2 kg will stop at, `y_(2)=9+v_(2)^(2)/(2mug)=9+36/4=18m, y_(1)=0`
Therefore, co-ordinates of the centre of mass will be,
`x_(C)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))rArr 1.75m and y_(C)=(m_(1)y_(1)+m_(2)y_(2))/(m_(1)+m_(2))rArr 12m`
`therefore` co-ordinates of COM will be (1.75 m, 12 m)