Question

A man (mass$=70\text{kg}$) and his son (mass$=20\text{kg}$) are standing on a frictionless surface facing in same direction. The man pushes his son, so that he starts moving with a speed of $0.90{\text{ms}}^{-1}$ with respect to man. The speed of man with respect to the surface is

• A. $0.20{\text{ms}}^{-1}$
• B. $0.5{\text{ms}}^{-1}$
• C. $0.67{\text{ms}}^{-1}$
• D. $0.85{\text{ms}}^{-1}$

Answer 1

The correct option is A $0.20{\text{ms}}^{-1}$

There is no external force in horizontal direction on the system of (man+son), so momentum of system remains conserved in horizontal or $x-$ direction.


$\Rightarrow$ Considering man to be moving w.r.t surface with velocity $\left(v_{m,S}\right)=v^{\prime }$ along $-vex-$direction, so that final momentum of system could become $\text{zero}$, velocity of son w.r.t man is $\left(v_{s,m}\right)=0.90{\text{ms}}^{-1}$ along $+ve-x$ direction.

$\Rightarrow$velocity of son w.r.t surface is given as:
${\overrightarrow{v}}_{s,S}={\overrightarrow{v}}_{s,m}+{\overrightarrow{v}}_{m,S}$
$v_{s,S}=+\left(0.90\right)+(-v^{\prime })=0.90-v^{\prime }$
Applying momentum conservation:
$P_i=P_f=0$
$\Rightarrow P_f=(m_s\times v_{s,S})+(m_m\times v_{m,S})=0$
$\Rightarrow 20\times (0.90-v^{\prime })+70\times (-v^{\prime })=0$
$18-20v^{\prime }-70v^{\prime }=0$
$\therefore v^{\prime }=\frac{18}{90}=0.2{\text{ms}}^{-1}$