Ball on a slope with hollow

Question

The experiment shown in the image suggests that ball B will reach the goal faster than ball A although the balls have identical properties and they start from the same height.

The authors even suggest that this phenomenon is independent from the the depth of the hollow. Can someone explain this to effect to me especially concerning the fact that it wouldn't matter even if the hollow was 400 km deep?

Answer 1

You can distinguish from vertical and horizontal velocity. Both balls have the same horizontal velocity, the difference lies in the vertical component. Up to the second trough there is no difference, but then the second ball accelerates downwards. It can't go straight down, which means that the gravity partially accelerates it horizontally. This difference in Velocity is neutralized at the end of the trough. Now the second ball was for some amount of time faster (hotizontally) than the first one.

Answer 2

Let's call the rightward horizontal direction $+x$, and the upward vertical direction $+y$. Both balls reach point 1 at the same time, going the same speed. They both have the same $x$-component of velocity.

At the beginning of the dip in B's path, ball A remains at constant velocity, $v_1\hat{i}$, but ball B gains in $v_x$ until the bottom of the dip. It then loses in $v_x$, but remains greater than $v_1$ until the end of the dip, where it cruises along horizontally at $v_1$.

The time that ball B arrives at the $x$ location of the end of the dip is $t_B$. The time that ball A arrives at the $x$ location of the end of the dip in B's path is $t_A$.

Both balls travel the same $\Delta x$, so they both must have the same area under the $v(t)$ vs. $t$ curves.

In order to have the same area, it must be true that $t_B-t_1$ < $t_A-t_1$, or $t_B < t_A$. Ball B arrives at the $x$ position of the end of the dip before ball A, and it remains ahead of ball A.

Answer 3

If we don't consider air friction with the ball, then you can see two different effects: an increase in length of the path and an increase in velocity of the ball. Turns out that the second effect more than compensate the first.

An exact calculation isn't simple for arbitrary shapes, but we can see a simplification: let's say from (1) to (2) the length is $L$. At (1) both balls have velocity $v = \sqrt{2 g h}$ from conservation of energy. So ball A goes from (1) to (2) in time $t_A = \dfrac{L}{v} = \dfrac{L}{\sqrt{2 g h}}$.

Second assumption: instead of a smooth curve, ball B sees the path curving down at 45° with a vertical drop $d$, then an horizontal path with length $L - 2\sqrt{2} d$ and finally the path rises with an angle of 45° and goes back at the same height as before. If the ball isn't too fast this assumption is good enough to avoid jumps and recoils.

Third: to simplify even further, let's assume the balls are point-like.

How much time does B need to go from (A) to (B)? From (A) to the bottom of the hollow it's an accelerated motion with acceleration $a_1 = \frac{\sqrt{2}}{2} g$ and a length $l_1 = \sqrt{2}d$, so the time needed is $t_1 = 2\dfrac{\sqrt{h + d} - \sqrt{h}}{\sqrt{g}}$ and has a velocity $v_1 = \sqrt{2 g (h + d)}$.

Second part, this is simpler, it takes $t_2 = \dfrac{l_2}{v_1} = \dfrac{L - 2\sqrt{2}d}{\sqrt{2g(h+d)}}$, the velocity is the same.

Third part, we are almost there. Now the ball is decelerating ($a_3 = -a_1$), but $t_3 = t_1$ and the velocity is again $\sqrt{2 g h}$.

The total time needed by ball B then is $$t_B = t_1 + t_2 + t_3 = 4\dfrac{\sqrt{h + d} - \sqrt{h}}{\sqrt{g}} + \dfrac{L - 2\sqrt{2}d}{\sqrt{2g(h+d)}}$$ With a common denominator it becomes: $$t_B = \dfrac{2\sqrt{2} (2h+d) - 4 \sqrt{2h(h+d)} + L}{\sqrt{2g(h+d)}}$$ Yet again it's not so easy to show that this formula represents a smaller time than $t_A$, so let's see a few properties: if $d=0$, $t_B=t_A$, as expected. We can ask if there are other solutions to the equation $t_A = t_B$. There is one other, $d = \dfrac{L(L+4\sqrt{2}h)}{8h}$, but if we calculate the length of the second segment, it turns out it's $L-2\sqrt{2}d = - \dfrac{L(\sqrt{2}L + 4h)}{4h}$, which is negative and thus has no physical and geometrical meaning.

That means that for this particular path there is no other depth $d$ that lets ball B run the whole trajectory in the same time as ball A. So it has to be always less or always more. We can check with a random value, for example $d = h$, and we obtain $t_B(d=h) = \dfrac{L}{2 \sqrt{gh}} + (3\sqrt{2}-4) \sqrt{\dfrac{h}{g}}$. $t_B < t_A$ for $L > \dfrac{6\sqrt{2}-8}{\sqrt{2}-1}h$, which is a pretty good condition since $L$ must also be greater than $2\sqrt{2}h$, which is even bigger.

So, we show that the longest path takes less time in this particular configuration. We can imagine it's pretty much the same for any other similar configuration, but it would be less intuitive to calculate.

All the calculations are left as exercise for the reader, I don't want to spoil all the fun. I hope they are all correct.