Challenges in mechanics by Ronak Agarwal (Part 1)

When the small sphere is about to leave contact, the normal reaction between spheres is $0$ . Let the speed of big sphere be $v$ and that of small sphere wrt big sphere be $v_{r}$ .

With respect to the big sphere , small sphere moves in a circle of radius $R+r$ . The acceleration vector of big sphere at the instant when small sphere leaves is contact is $0$ as there is no normal reaction and hence no horizontal force. Thus,

$\dfrac{m v_{r}^2}{R+r} = mg \cos \theta$ $\dots \text{ (i)}$

Also, by using conservation of linear momentum of the system in horizontal direction ,

$m v_{r} \cos \theta = (M+m)v$ $\dots \text{ (ii)}$

Now, for small sphere,

$|\vec{v_{0}}| = \sqrt{v^2 + v_{r}^2 - 2vv_{r} \cos \theta}$

Hence, by using conservation of mechanical energy,

$\dfrac{1}{2} m (v^2 + v_{r}^2 - 2vv_{r} \cos \theta) + \dfrac{1}{2} M v^2 = mg(R+r) (1 - \cos \theta)$ $\dots \text{ (iii)}$

Put the value of $v$ in terms of $v_{r}$ from equation $\text{(ii)}$ ,and put the value of $v_{r}^2$ in terms of $\cos \theta$ from equation $\text{ (i)}$ , in the equation $\text{ (iii)}$ to get an equation in $\cos \theta$ :

$1 - \dfrac{m}{m+M} \cos^2 \theta = \dfrac{2(1 - \cos \theta)}{\cos \theta}$

Put the values of $M,m$ to obtain:

$\cos \theta = \dfrac{4}{5}$ . Hence, $\theta \approx 36.9^\circ$

Note that the value of $\theta$ doesn't depend on $R,$ and $r$