The velocities of two particles `A and B` of same mass are `vec V_A = a hat i and vec V_B = b hat j` where `a and b` are constants. The acceleration of particle `A is (2 a hat i+ 4 b hat j)` and acceleration of particle `B is (a hat i- b hat j)` (in m//s^2)`. The centre of mass of two particle will move in :

Options

A. Straight line
B. Parabola
C. Ellipse
D. Circle

Answer

A

Explanation

`vecV_(cm)=(m_(1)vecV_(1)+m_(2)vecV_(2))/(m_(1)+m_(2))=(vecV_(1)+vecV_(2))/(2)=(ahati+b hatj)/(2)`
`veca_(cm) = (m_(1)veca_(1) +m_(2) veca_(2))/(m_(1)+m_(2)) =(veca_(1)+veca_(2))/(2) =(3)/(2) (a hati +b hatj)`
As `vecV_(cm)` is parallel to `veca_(cm)` so path is a straight line.