A boy of mass $20\text{kg}$ is standing on a $80\text{kg}$ free to move long cart. There is negligible friction between cart and ground. Initially, the boy is standing $25\text{m}$ from a wall. If he walks $10\text{m}$ on the cart towards the wall, then the final distance of the boy from the wall will be

The correct option is D $17\text{m}$
Given:
Mass of boy $m_1=20\text{kg}$
Mass of cart $m_2=80\text{kg}$
Initial distance between boy and cart is$25\text{m}$
Distance moved by boy towards wall is $10\text{m}$

As there is no external force, so displacement of centre of mass of the (cart + boy) system parallel to the surface is zero.

$\therefore x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}=0$

Let when the boy moves $10\text{m}$ towards the wall, the cart moves away from the wall a distance $x$
So, displacement of man w.r.t. ground towards the wall is
$x_1=10-x$
And the displacement of cart w.r.t. ground towards the wall is
$x_2=-x$

$20\times (10-x)+(80\times (-x\left)\right)=0$

$\Rightarrow x=2\text{m}$

i.e. Final distance between boy and wall,
$=25-10+2=17\text{m}$

Hence, option $\left(D\right)$ is correct.