Question

A uniform solid sphere of mass m and radius r is released from a wedge of mass 2m as shown. ABC is hemispherical position of radius R. Impulse imparted to the system consisting wedge and sphere by the vertical wall $w_1{w}_2$ till the time sphere reaches at the bottom most position of spherical portion for the first time is $m\sqrt{\frac{10g(R-r)}{\delta }}$. Friction between wedge and horizontal surface is absent and between sphere and wedge friction is sufficient to avoid slipping between them. Here $\delta$ is an integer. Find $\delta$ (Answer upto two digit after decimal places)

Answer 1

At the bottom-most point, whole potential energy is converted into kinetic energy
kinetic energy is given by
$K.E=\frac{1}{2}I{\omega }^2$
where $\omega =\frac{v}{r}$ and
$I=\frac{7}{5}m{r}^2$ as it is taken about the axis of point of contact.
$\therefore K.E=\frac{7}{10}m{v}^2=\frac{7P^2}{10m}$
$\Rightarrow mg(R-r)=\frac{7P^2}{10m}$
Since, the wedge will not move, so impulse imparted will only cause change in momentum of sphere
$\Delta P=m\sqrt{\frac{10g(R-r)}{7}}$-0 = Impulse Imparted on the system by wall
Now, comparing this by given impulse we get
$\delta =7$