91影视

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,

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

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$\left(0.543\right)L$ .

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