Two particles of equal mass have velocitiesand . First particle has an accelerationwhile the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of

Two particles of equal mass have velocitiesand . First particle has an

1.	Two particles of equal mass have velocitiesand . First particle has an accelerationwhile the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of
Answer» straight line parabola circle ellipse Solution :Here, `vecv_(1)=2hatims^(-1), vecv_(2)=2hatjms^(_1)`. `veca_(1)=(3hati+3hatj)MS^(-2), veca_(2)=0 ms^(-2)` `therefore vecv_(CM)= (m_(1)vecv_(1)+m_(2)vecv_(2))/(m_(1)+m_(2))=(vecv_(1)+vecv_(2))/(2)= (2hati+2hatj)/(2)` `(therefore m_(1)=m_(2))` `=(hati+hatj)ms^(-1)` Similarly, `veca_(CM) = (veca_(1)+veca_(2))/(2) = (3hati+3hatj+0)/(2) = 3/2(hati+hatj)ms^(-2)` SINCE, `vecV_(CM)` is parallel to `veca_(CM)`, the PATH will be straight line.