Balls bouncing on a concave surface

Let $O$ denote the center of the sphere, $L$ the lowest point on the sphere, $D$ the point the ball is dropped from, $B$ the point on the sphere the ball bounces, and $\theta := \measuredangle{LOB}$ .

Since the ball falls from a height of $R/n$ , its speed $|v|$ immediately before impact at $B$ will be given by $mg\left(\frac{R}{n}\right) = \frac{1}{2}m|v|^2 \,\,\,\,\, \Rightarrow \,\,\,\,\, |v| = \sqrt{\frac{2gR}{n}}$ and, since the collision is elastic, this will also be the speed immediately after the impact at $B$ .

Now, since $\overline{OL}\, ||\, \overline{DB}$ we know $\measuredangle{DBO} = \measuredangle{LOB} = \theta$ . Further, since the angle the ball's path makes with the normal $\overline{OB}$ is equal on both sides, we can conclude that the initial angle the ball's trajectory makes with the vertical is $2\theta$ . Letting $L$ be the point $(0,0)$ and $B = \big(R\sin\theta, R(1-\cos\theta)\big)$ , we therefore find the following equations of motion for the ball after it bounces: $x = R\sin\theta - |v|\sin(2\theta)t = R\sin\theta - \sqrt{\frac{2gR}{n}}\sin(2\theta)t \\ y = R(1-\cos\theta) + |v|\cos(2\theta)t - \frac{1}{2}gt^2 = R(1-\cos\theta) + \sqrt{\frac{2gR}{n}}\cos(2\theta)t - \frac{1}{2}gt^2$

Since the next time it collides with the sphere is at $L$ , which is $(x,y) = (0,0)$ , there will be some time $t$ such that $0 = R\sin\theta - \sqrt{\frac{2gR}{n}}\sin(2\theta)t \\ 0 = R(1-\cos\theta) + \sqrt{\frac{2gR}{n}}\cos(2\theta)t - \frac{1}{2}gt^2$ and while we could just jump right into solving this, it's useful to make a simplification: if we change variables to $u = t\sqrt{g/R}$ , then these equations become $0 = R\left(\sin\theta - \sqrt{\frac{2}{n}}\sin(2\theta)u\right) \\ 0 = R\left( (1-\cos\theta) + \sqrt{\frac{2}{n}}\cos(2\theta)u - \frac{1}{2}u^2\right)$ and since $R>0$ , we can divide both equation by $R$ to find $0 = \sin\theta - \sqrt{\frac{2}{n}}\sin(2\theta)u \\ 0 = (1-\cos\theta) + \sqrt{\frac{2}{n}}\cos(2\theta)u - \frac{1}{2}u^2$

Solving this first equation for $u$ yields $u = \sqrt{\frac{n}{8}}\sec\theta$ and then substituting this into the second equation: $0 = (1-\cos\theta) + \frac{1}{2}\cos(2\theta)\sec\theta - \frac{n}{16}\sec^2\theta$ Solving this equation for $n$ : $\begin{aligned} n &= 16\cos^2\theta\left((1-\cos\theta) + \frac{1}{2}\cos(2\theta)\sec\theta\right) \\ &= 16\cos^2\theta(1-\cos\theta) + 8\cos(2\theta)\cos\theta \\ &= 8(\cos(2\theta) + 1)(1-\cos\theta) + 8\cos(2\theta)\cos\theta \\ &= 8 + 8(\cos(2\theta) - \cos\theta)\end{aligned}$

We see that $n$ depends on $\theta$ , but since we're told that the ball starts near the vertical axis of symmetry, we may assume $\theta$ is small and therefore $\cos(2\theta) \approx 1 \approx \cos\theta$ , so that $n \approx 8 + 8(1-1) = \boxed{8}.$

Let point of collision make an angle x with centre with lim x->0. Let us assume that during collision and at the bottom most point of spherical surface its velocity was u and g is acceleration due to gravity. It can be assumed as an inclined projectile motion with inclination x with horizontal. When it strikes the bottom most point let it makes an angle θ with the incline plane.

Using condition for inclined projectile to strike and rebound vertically tan(θ)*tan(x)=1/3 .

Therefore θ-> π/2 - 3*x.

Range of projectile = S = 2 u^2/g * (sin(θ) cos(θ+x)/cos^2(x)) .

Using small angle approximation we get S-> 2 u^2/g * sin(2 x) = R*sin(x).

R/n = u^2/(2*g) = R/8.

Therefore n=8.