Two particles of masses m and 2 m has initial velocity `vecu_(1)=2hati+3hatj(m)/(s)` and `vecu_(2)=-4hati+3hatj(m)/(s)` Respectivley. These particles have constant acceleration `veca_(1)=4hati+3hatj((m)/(s^(2)))` and `veca_(2)=-4hati-2hatj((m)/(s^(2)))` Respectively. Path of the centre of the centre of mass of this two particle system will be:

Options

A. Straight line
B. Circular
C. Parabolic
D. Elliptical

Answer

C

Explanation

`vecu_(cm)=(mvecu_(1)+2mveru_(2))/(3m)=(vecu_(1)+2vecu_(2))/(3)`
`veca_(cm)=(mveca_(1)+2mveca_(2))/(3m)=(veca_(1)+2veca_(2))/(3)`
Since `veca_(cm)` is constant and `vecu_(cm)` & `veca_(cm)` are not parallel to path will be parabolic