Two particle of equal mass have velocities `vecv_(1)=2hati` m/s and `vecv_(2)=2hatj` m/s. First particle has an accelration `veca_(1)=(3hati+3hatj)m//s^(2)` while the acceleration of the other particle is zero. The centre of mass of the two particle moved on a

Options

A. circle
B. parabola
C. ellipse
D. straight line

Answer

D

Explanation

`vecv_("com")=(m_(1)vecv_(1)+m_(2)vecv_(2))/(m_(1)+m_(2))`
`(vecv_(1)+vecv_(2))/(2)=(m_(1)=m_(2))`
`=(hati+hatj)m//s`
Similarly `veca_("com")=(veca_(1)+veca_(2))/(2)=(3)/(2)(hati+hatj)m//s^(2)`
Since `vecv_("com")` is parallel to `veca_("com")` the path will be a straight line.