A man (mass = 50   k g ) and his son (mass = 20   k g ) are standing on a frictionless surface…

By conservation of linear momentum,

${\overrightarrow{\mathrm{P}}}_{\mathrm{i}}={\overrightarrow{\mathrm{P}}}_{\mathrm{f}}$

Where ${\overrightarrow{\mathrm{P}}}_{\mathrm{i}}$ and ${\overrightarrow{\mathrm{P}}}_{\mathrm{f}}$ are initial and final momentum of system (boy + man)

Now,  $0={\mathrm{m}}_{\mathrm{b}}{\overrightarrow{\mathrm{v}}}_{\mathrm{b}}+{\mathrm{m}}_{\mathrm{m}}{\overrightarrow{\mathrm{v}}}_{\mathrm{m}}$......(i)

${\mathrm{v}}_{\mathrm{b}}=$ velocity of boy w.r.t. ground

${\mathrm{v}}_{\mathrm{m}}=$ velocity of man w.r.t. ground

${\overrightarrow{\mathrm{v}}}_{\mathrm{bm}}=$ velocity of boy w.r.t. man

So, ${\overrightarrow{\mathrm{v}}}_{\mathrm{bm}}={\overrightarrow{\mathrm{v}}}_{\mathrm{b}}-{\overrightarrow{\mathrm{v}}}_{\mathrm{m}}$

or ${\overrightarrow{\mathrm{v}}}_{\mathrm{b}}={\overrightarrow{\mathrm{v}}}_{\mathrm{bm}}+{\overrightarrow{\mathrm{v}}}_{\mathrm{m}}$

or ${\overrightarrow{\mathrm{v}}}_{\mathrm{b}}=0.7+{\overrightarrow{\mathrm{v}}}_{\mathrm{m}}$ ........(ii)

From equation (i) and (ii),

$0=20\left(0.7+{\overrightarrow{\mathrm{v}}}_{\mathrm{m}}\right)+50 {\overrightarrow{\mathrm{v}}}_{\mathrm{m}}$

${\overrightarrow{\mathrm{v}}}_{\mathrm{m}}=-\frac{14}{70}=-0.20 \mathrm{m} {\mathrm{s}}^{-1}$

Negative sign shows velocity of man is opposite to boy.