If I throw a ball into a quarter of a circular tube, what will be the final direction of the ball? 
Assuming energy is conserved of course, no friction and similar mass. looking at this as a collision I know the mass relation plays a part, but how? Also how is having 1/4 of a circle is different from 1/5 or 1/3?
edit: no gravity and ball starting velocity is $V$. Ideally, mass of ball $M_{ball}$ and mass of tube $M_{tube}$ are different, but if its easier to solve for $M_{ball} = M_{tube}$, that will do too.
 A: The thing that will simplify calculation is the assumption that the cross-sectional diameter of the tube is equal to that of the ball. Because then, at the exit point, the horizontal velocity of the ball must be equal to the horizontal velocity of the tube. Why? Because then, at any point of its journey, the ball will not be able to move radially without causing the same radial movement of the tube. At exit, the radial direction happens to be horizontal.
Let's call the final horizontal velocity of both the tube and ball at exit, $v_x (= v_{bx} = v_{tx})$, final vertical velocity of tube, $v_{ty}$, that of the ball, $v_{by}$. For brevity, let's call $M_{ball}$ $m$, and $M_{tube}$, $M$ So, we have:
$$
mV = Mv_x + mv_x
\implies v_x = \frac{m}{M+m}V \tag{1}
$$
$$
Mv_{ty} = mv_{by} \implies v_{ty} = \frac{m}{M}v_{by} \tag{2}
$$
$(1)$ and $(2)$ are by conservation of momentum. By conservation of energy we have:
$$
\begin{align}
\frac{1}{2}mV^2 &= \frac{1}{2}Mv_t^2 + \frac{1}{2}mv_b^2 \\
\implies mV^2 &= M(v_x^2 + v_{ty}^2) + m(v_x^2 + v_{by}^2) \\
&= M(v_x^2 + \frac{m^2}{M^2}v_{by}^2) + m(v_x^2 + v_{by}^2) \\
&= (m+M)v_x^2 + \frac{m}{M}(m+M) v_{by}^2 \\
\implies \frac{m}{M}v_{by}^2 &= \frac{mV^2}{M+m} - v_x^2 \\
&= \frac{mV^2}{M+m} - \frac{m^2V^2}{(M+m)^2} \\
&= \frac{mM}{(M+m)^2}V^2 \\
\implies v_{by} &= \frac{M}{(M+m)}V \tag{3}
\end{align}
$$
By $(1)$ and $(3)$,
$$
\begin{align}
\boxed{\frac{v_{by}}{v_{bx}} = \frac{M}{m}}
\end{align}
$$
This ratio gives the direction of the final velocity of the ball. If $M \gg m$, the final direction will be almost upward as expected.
$\frac{1}{4}$ of the circle maximizes the final vertical velocities, anything else than that will cause smaller final vertical velocities.
A: I am going to solve this question with the minimum amount of assumptions.
Now, one assumption I will make is that the ball is entering the tube by just touching the bottom half of the tube. I am also assuming that the size of the ball is less than that of the tube as shown in the figure. I would also assume that the moment of Inertia of the tube about its center is $I$. We can consider the rightward direction as X and Upward as Y.

Now, I would calculate the answer completely from the basics. Let's go into the frame of reference of the tube. In this frame of reference, you can clearly see that the Normal reaction on the ball will always be perpendicular to its path and hence it will only change the direction of the ball and to its speed with respect to the tube which will be $V$.
The normal reaction on the ball $N=\frac{{M_{ball}}V^2}{d}$
The distance between the centre of the ball and the centre of curvature of the tube is  $ d = R+2r-2t$
Now there will be a torque acting on the tube which you can clearly solve and about the torque as $\tau= NdSin{\phi} = {M_{ball}V^2Sin{\phi}}$.
Hence, $$\int_{0}^{\omega} Id{\omega}=\int_{-\theta}^{\theta}{M_{ball}}V^2Sin{\phi}dt$$
$$\omega = 0$$
But this could have been derived just by symmetry. Hence, the tube won't have a rotational motion at the end of motion.
Now from symmetry, we can see that after rotation the tube would end up with the same angle from the X and Y-direction.
The direction of the velocity of the ball at the end is $2\theta$ from the X-axis.
And the speed of the ball would be $V_x=V{Cos{2\theta}}+V_{x-tube}$
And $V_y=V{Sin{2\theta}}-V_{y-tube}$ considering that $V_y$ is towards +Y and $V_{y-tube}$ towards -Y
From Momentum conservation, we can see that $$M_{ball}V_y = M_{tube}V_{y-tube}$$
$$M_{ball}V=M_{ball}V_{x} + M_{tube}V_{x-tube}$$
After calculation this leads to a result of
$$V_{x}=\frac {M_{tube}VCos{2\theta}+M_{ball}V}{M_{tube}+M_{ball}}$$
$$V_{y}=\frac {M_{tube}VSin{2\theta}}{M_{tube}+M_{ball}}$$
$$V_{x-tube}=\frac {2M_{ball}V{(Sin{\theta})^2}}{M_{ball}+M_{tube}}$$
$$V_{y-tube}=\frac {M_{tube}VSin{2\theta}}{M_{ball}+M_{tube}}$$
Now, this result will only be possible in 2 cases, If the r=t or when the velocity of the ball along x-direction at every point is equal to the velocity of the tube along the x-direction, why? because that will lead to the ball colliding to the upper surface of the tube which leads to multiple collisions inside the tube due to which we won't be able to calculate velocities at final conditions. This leads to the result $\theta = 45°$
But, If r=t then the above result would be applicable at all values of $\theta$.
Of course, You can derive this by other methods such as energy conservation, etc.
If you are considering quarter of a circle then $V_{x}= \frac {M_{ball}V}{M_{ball}+M_{tube}}$ and $V_{y}= \frac {M_{tube}V}{M_{ball}+M_{tube}}$
If we consider $M_{tube} >> M_{ball}$ then $V_{x}=0$ and $V_{y}=V$ which is would be an expected result.
