A solid sphere and a hollow sphere, both possessing the same mass...

Let $M$ be the mass and $R$ be the radius of both spheres.

The moment of inertia of the solid sphere is $I_S = \dfrac{2}{5}MR^2 and that of the hollow sphere is$I_H = \dfrac{2}{3}MR^2$.<br><br>For an object rolling down an inclined plane of height$h$, by conservation of mechanical energy:<br><br>Mgh = \dfrac{1}{2}Mv^2 + \dfrac{1}{2}I\omega^2$

Since $v = \omega R$ for pure rolling:

Mgh = \dfrac{1}{2}Mv^2 \left(1 +\dfrac{I}{MR^2}\right)

$v =$\sqrt{\dfrac{2gh}{1 + \dfrac{I}{MR^2}}}

Since $I_S < I_H$, the denominator is smaller for the solid sphere, giving $v_S > v_H$. Thus, the solid sphere arrives at the bottom with a greater speed.

The acceleration of a rolling body is $a =$\dfrac{g \sin \theta}{1 + \dfrac{I}{MR^2}}. Since $I_S < I_H$, $a_S > a_H$, meaning the solid sphere reaches the bottom first.

Linear momentum is $p = Mv$. Since $v_S > v_H$, $p_S > p_H$.

Total kinetic energy at the bottom is equal to the initial potential energy $Mgh$, which is the same for both spheres.