A man (mass = 50kg ) and his son (mass = 20kg ) are standing on a...

By conservation of linear momentum,

$\overset{\to}{P}i = \overset{\to}{P}f$

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

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

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

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

$\overset{\to}{v}bm =$ velocity of boy w.r.t. man

So, $\overset{\to}{v}bm = \overset{\to}{v}b - \overset{\to}{v}m$

or $\overset{\to}{v}b = \overset{\to}{v}bm + \overset{\to}{v}m$

or $\overset{\to}{v}b = 0 . 7 + \overset{\to}{v}m$ ........(ii)

From equation (i) and (ii),

$0 = 20\left(0 . 7 + \overset{\to}{v}m\right) + 50\overset{\to}{v}m$

$\overset{\to}{v}m = - \dfrac{14}{70} = - 0 . 20m{s}^{- 1}$

Negative sign shows velocity of man is opposite to boy.