Physics - Center of Mass Question with Solution | TestHub

Let $M$ be the mass of the cart and $m$ be the mass of the boy.

$M=80\mathrm{kg}$

$m=20\mathrm{kg}$

Let $x_0$ be the initial distance of the boy from the wall, $x_0=25\mathrm{m}$.

Let $d$ be the distance the boy walks on the cart, $d=10\mathrm{m}$.

Since there is no external horizontal force, the center of mass of the boy-cart system remains stationary.

Let the initial position of the boy be $x_B$ and the initial position of the center of the cart be $x_C$.

We can set the initial position of the boy at $x_B=0$.

The initial position of the center of mass of the system is:

$$X_{CM,i}=\frac{m{x}_B+M{x}_C}{m+M}$$

When the boy walks $d=10\mathrm{m}$ towards the wall, his displacement relative to the cart is $-d$.

Let the cart move a distance $\Delta x_C$ away from the wall.

The final position of the boy relative to the ground is $x_B^{\prime }=-d+\Delta x_C$.

The final position of the cart relative to the ground is $x_C^{\prime }=\Delta x_C$.

Since the center of mass does not move:

$$X_{CM,i}=X_{CM,f}$$

$$m{x}_B+M{x}_C=m{x}_B^{\prime }+M{x}_C^{\prime }$$

Assuming the initial position of the boy is at $x=0$ and the cart's center of mass is also at $x=0$ (or we consider relative displacements).

The displacement of the boy relative to the ground is $\Delta x_B$.

The displacement of the cart relative to the ground is $\Delta x_C$.

The displacement of the boy relative to the cart is $\Delta x_{B/C}=\Delta x_B-\Delta x_C$.

We are given $\Delta x_{B/C}=-10\mathrm{m}$ (towards the wall).

Since $X_{CM}$ is constant, $m\Delta x_B+M\Delta x_C=0$.

Substitute $\Delta x_B=\Delta x_{B/C}+\Delta x_C$:

$$m(\Delta x_{B/C}+\Delta x_C)+M\Delta x_C=0$$

$$m\Delta x_{B/C}+(m+M)\Delta x_C=0$$

$$\Delta x_C=-\frac{m\Delta x_{B/C}}{m+M}$$

$$\Delta x_C=-\frac{20\mathrm{kg}\times (-10\mathrm{m})}{20\mathrm{kg}+80\mathrm{kg}}=\frac{200\mathrm{kg}\cdot \mathrm{m}}{100\mathrm{kg}}=2\mathrm{m}$$

The cart moves $2\mathrm{m}$ away from the wall.

The boy's displacement relative to the ground is $\Delta x_B=\Delta x_{B/C}+\Delta x_C=-10\mathrm{m}+2\mathrm{m}=-8\mathrm{m}$.

The boy moves $8\mathrm{m}$ towards the wall.

The initial distance of the boy from the wall is $25\mathrm{m}$.

The final distance of the boy from the wall is $25\mathrm{m}-8\mathrm{m}=17\mathrm{m}$.

The final distance of the boy from the wall will be $17\mathrm{m}$.

The final answer is $\boxed{17\mathrm{m}}$.