Two balls of equal radius and mass, free to roll on a horizontal plane, are separated by a distance L large compared to their radius. One ball is solid, the other hollow with a thickness small compared to its radius. They are attracted by an electric force. How far will the solid ball roll before it collides with the hollow ball?

The multiplication of mass and the square of a distance of the particle from the rotation axis are known as the moment of inertia.

The moment of inertia of the solid ball is,

$I_{solid}=\frac{2}{5}M{R}^2$

Here, $M$is the of the solid ball and $R$ is the radius of the solid ball.

Let $k_1=\frac{2}{5},$ then the above equation becomes as follows.

$I_{solid}=k_{1}M{R}^2$

The moment of inertia of the hollow ball is,

$I_{hollow}=\frac{2}{3}M{R}^2$

Here,$M$is the mass of the hollow ball and$R$is the radius of the hollow ball.

Let $k_2=\frac{2}{3},$then the above equation becomes as follows.

$I_{hollow}=k_{2}M{R}^2$

The solid ball is rolling parallel to its plane and the force on each ball remains the same. Thus, the distance travelled by solid ball before collision is,

$\begin{array}{rcl}{x}_1& =& \left[f\left(k_1,k_2\right)\right]L\\ & =& \frac{L}{1+\left(\frac{k_1+1}{k_2+1}\right)}······\left(1\right)\\ & & \end{array}$

Here,$L\left(=x_1,x_2\right)$is the separation between the two balls before the collision.

The hollow ball is rolling parallel to its plane and force on each ball remains the same. Thus, the distance travelled by hollow ball before collision is,

$\begin{array}{rcl}{x}_2& =& \left[f\left(k_2,k_1\right)\right]L\\ & =& \frac{L}{1+\left(\frac{k_2+1}{k_1+1}\right)}\\ & & \end{array}$

Substitute $\frac{2}{5}$for $k_1$and $\frac{2}{3}$for $k_2$in the equation (1).

$\begin{array}{rcl}{x}_1& =& \frac{L}{1+\left(\frac{\frac{2}{5}+1}{\frac{2}{3}+1}\right)}\\ & =& \left(0.543\right)L\\ & & \end{array}$

Therefore, the distance travelled by solid ball before collision is$\left(0.543\right)L$ .

$I_{hollow}=\frac{2}{3}M{R}^2$