Question

The sphere shown in figure lies on a rough plane when a particle of mass 'm' traveling at a speed $v_0$ collides and sticks with it. If the line of motion of the particle is at a distance h above the plane, find the angular speed of the combined system just after the collision. Assume that the mass M of the sphere is large compared to the mass of the particle so that the center of mass of the combined system is not appreciably shifted from the center of the sphere.

• A. $\frac{m{v}_0(h-R)}{(\frac{2}{5}M+m)R^2}$
• B. $\frac{m{v}_0(h-R)}{(m+\frac{2}{5}M)R^2}$
• C. $\frac{m{v}_0(h-R)}{(\frac{2}{5}M+m)R^2}$
• D. $\frac{m{v}_0(h-R)}{(\frac{2}{5}m+M)R^2}$

Answer 1

The correct option is A

$\frac{m{v}_0(h-R)}{(\frac{2}{5}M+m)R^2}$

we shall use conservation of angular momentum about the centre of mass, which is to be taken at the centre of the sphere $(M>>m)$. Angular momentum of the particle before collision is $m{v}_0(h-R)$. If the system rotates with angular speed $\omega$ after collision, the angular momentum of the system becomes

$(\frac{2}{5}M{R}^2+m{R}^2)\omega$.

hence,

$m{v}_0(h-R)=(\frac{2}{5}M+m)R^2\omega$

or, $\omega =\frac{m{v}_0(h-R)}{(\frac{2}{5}M+m)R^2}$