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)
The centre of mass of the two particles is moving in a straight line.

Options

A. y = x + 2
B. y = 4x + 2
C. y = 2x - 4
D. y = 2x + 4

Answer

B

Explanation

At time t, `x_(1)=(3+3t) and x_(2)=0, y_(2)=(9+6t) and y_(1)=0`
Thte position of their centre of mass will be
`x_(C)=(m_(2)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))=(1+t)" (1)"`
`y_(C)=(m_(1)y_(1)+m_(2)y_(2))/(m_(1)+m_(2)) rArr 6+4t" (2)"`
Eliminating from equations (1) and (2)
We have, `y_(c)=4x_(C)+2)`
So, the centre of mass is moving along the line, y = 4x + 2