Two particles of equal masses have velocities $\overrightarrow{{\mathrm{v}}_1}=4\widehat{\mathrm{i}}{\mathrm{ms}}^{-1}$and $\overrightarrow{{\mathrm{v}}_2}=4\widehat{\mathrm{j}}{\mathrm{ms}}^{-1}$. First particle has an acceleration  $\overrightarrow{{\mathrm{a}}_1}=(2\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}){\mathrm{ms}}^{-2}$while the acceleration of the other particle is zero.  The centre of mass of the two particles moves in a path of 

Given,
$\begin{array}{l}\overrightarrow{{\mathrm{v}}_1}=4\widehat{\mathrm{i}}{\mathrm{ms}}^{-1},\overrightarrow{{\mathrm{v}}_2}=4\widehat{\mathrm{j}}{\mathrm{ms}}^{-1}\\ \overrightarrow{{\mathrm{a}}_1}=(2\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}){\mathrm{ms}}^{-1},\overrightarrow{{\mathrm{a}}_2}=0{\mathrm{ms}}^{-2}\end{array}$
$\therefore$Velocity of centre of mass,
$\begin{array}{l}{\mathrm{v}}_{\mathrm{CM}}=\frac{{\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}=\frac{\left({\mathrm{v}}_1+{\mathrm{v}}_2\right)\mathrm{m}}{2\mathrm{m}}\\ \left[\because {\mathrm{m}}_1={\mathrm{m}}_2=\mathrm{m}\right]\\ \overrightarrow{v_{CM}}=\frac{4\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}}{2}=2(\widehat{\mathrm{i}}+\widehat{\mathrm{j}}){\mathrm{ms}}^{-1}\end{array}$
Similarly, acceleration of centre of mass,
${\mathrm{a}}_{\mathrm{CM}}=\frac{{\mathrm{a}}_1+{\mathrm{a}}_2}{2}=\frac{2\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+0}{2}=(\widehat{\mathrm{i}}+\widehat{\mathrm{j}}){\mathrm{ms}}^{-2}$
Since, from above values, it can be seen that $\overrightarrow{{\mathrm{v}}_{\mathrm{CM}}}$is parallel to   $\overrightarrow{{\mathrm{a}}_{\mathrm{CM}}}$ , so the path will be a straight line.