Question

A man $(\text{mass}=50\text{kg})$ and his son $(\text{mass}=20\text{kg})$ are standing on a frictionless surface facing each other. The man pushes his son, so that he starts moving at a speed of $0.70{\text{ms}}^{-1}$ with respect to the man. The speed of the man with respect to the surface is

• A. $0.28{\text{ms}}^{-1}$
• B. $0.20{\text{ms}}^{-1}$
• C. $0.47{\text{ms}}^{-1}$
• D. $0.14{\text{ms}}^{-1}$

Answer 1

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

The given situation can be shown as below

Using momentum conservation law,
$(\text{Total momentum}{)}_{\text{before collision}}$
$(m_1\times 0)+(m_2\times 0)=m_1{v}_1+m_2{v}_2$
$0=m_1(-v_1)\hat{i}+m_2{v}_2\hat{i}$
$\Rightarrow m_1{v}_1=m_2{v}_2$
$\Rightarrow 50v_1=20v_2$
$\Rightarrow v_2=2.5v_1....\left(i\right)$
Again, relative velocity $0.70\text{m/s}$
But from figure, relative velocity $=v_1+v_2$
$\therefore v_1+v_2=0.7.....\left(ii\right)$
From Eqs (i) and (ii), we get
$v_1+2.5v_1=0.7$
$\Rightarrow v_1\left(3.5\right)=0.7$
$v_1=\frac{0.7}{3.5}=0.20\text{m/s}$