Two particles of equal mass have velocities`vecV _(1)= 4hati ms^(-1)` and `vecV _(2)= 4hatj ms^(-1)` First particle has an acceleration `veca_(1)=(5hati+5hathatj)ms^(-1)`while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of

Options

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

Answer

(a)

Explanation

`vecv_(1)=4hatim//s`,`a_(1)=(5hatj+5hatj)m//s^(2)`
`vecv_(2)=4hatim//s`,`a_(2)=0`
` v_(cm)=(m_(1)vecv_(1)+m_(2)vecv2)/(m_(1)+m_(2))=(m(4hati)+(4hatj))/(m+n)=(4hati+4hatj)/(2)`
`a_(cm)=(m_(1)veca_(1)+m_(2)veca2)/(m_(1)+m_(2))=(m(5hati+5hatj)+m(0))/(m+n)=(veca)/(2)`
Therefore, the path of centre of mass will be a straight line.