$\text { ie., } \quad \vec{\mathrm{V}}_{\mathrm{CM}}=0$
$\frac{m_{1} \vec{v}_{1}+m_{2} \vec{v}_{2}}{m_{1}+m_{2}}=0$
or $\text { or } \quad \mathrm{m}_{1} \vec{\mathrm{v}}_{1}+\mathrm{m}_{2} \vec{\mathrm{v}}_{2}=0$
Or $\mathrm{m}_{1} \frac{\Delta \overrightarrow{\mathrm{r}}_{1}}{\Delta \mathrm{t}}+\mathrm{m}_{2} \frac{\Delta \vec{\mathrm{r}}_{2}}{\Delta \mathrm{t}}=0$
$\mathrm{or}$ $\mathrm{m}_{1} \Delta \overrightarrow{\mathrm{r}}_{1}+\mathrm{m}_{2} \Delta \overrightarrow{\mathrm{r}}_{2}=0$
Now here, in boat-man system if the man
moves towards right the boat moves towards left.
$\therefore$ $\mathrm{m}_{\mathrm{1}} \Delta \mathrm{r}_{\mathrm{1}}=\mathrm{m}_{2} \Delta \mathrm{r}_{2}$ $...(i)$
$\left(\because \Delta \mathrm{r}_{\mathrm{1}} \text { is opposite to } \Delta \mathrm{r}_{2}\right)$
If $\Delta \mathrm{r}_{2}$ is the displacement of boat relative to shore, then the displacement of man relative to
shore would be $\left(9-\Delta \mathrm{r}_{2}\right)$
$\text { i.e., } \quad \Delta r_{1}=9-\Delta r_{2}$ $...(ii)$
From $(i)$ and $(i i)$
$\mathrm{m}_{1}\left(9-\Delta \mathrm{r}_{2}\right)=\mathrm{m}_{2} \Delta \mathrm{r}_{2}$
$\text { or } \quad 100\left(9-\Delta r_{2}\right)=500 \Delta r_{2}$
$\Delta r_{2}=\frac{100 \times 9}{600}=1.5 \mathrm{m}$
ie. Boat moves $1.5 \mathrm{m}$ relative to shore in the
direction opposite to the displacement of man.
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The trajectory of particle $1$ with respect to particle $2$ will be
