11

I saw this picture on asocial networking site and it looks interesting. May I ask if this is theoretically feasible？

Emilio Pisanty
  • 132,859
  • 33
  • 351
  • 666
timeil0611
  • 141

6 Answers6

15

Yes it is possible in principle, though the bear will have to do a lot work to keep the ball moving.

Suppose the speed of the ball at its highest point when it hits the bear is $v$ and the mass of the ball is $m$, then the momentum change when it hits the bear and bounces back is:

$$ \Delta p = 2mv $$

If the ball hits the bear $n$ times a second then the total force on the bear is just the rate of change of momentum:

$$ F = 2mvn $$

And to keep the bear in the air we just need this force to be equal to the weight of the bear. So if the mass of the bear is $M$ we require:

$$ Mg = 2mvn \tag{1}$$

To get any further we need to work out the value of $n$, and for this we need the time the ball takes to travel down to the ground then back up to the bear again. If we call this time $2t$, then the value of $n$ is just $n = 1/2t$. The problem is that $t$ is given by the quadratic:

$$ h = vt + \tfrac12 g t^2 $$

where $h$ is the height of the cliff, and solving this is going to make our equation a bit messy. However let's assume the ball is travelling sufficiently fast that the change in its velocity due to gravity is small, and then we can approximate:

$$ h \approx vt $$

so:

$$ t = h/v $$

and makes $n = v/2h$ and substituting this in equation (1) gives:

$$ Mg = 2mv\frac{v}{2h} $$

And rearranging tells us how fast the bear has to throw the ball:

$$ v = \sqrt{\frac{Mgh}{m}} \tag{2} $$

To see if this is actually practical let's use some rough values. Call the mass of the ball $m = 1~\mathrm{kg}$ and the height of the cliff $h = 10~\mathrm{m}$. Wikipedia tells me that grizzly bears weight from $130$ to $360~\mathrm{kg}$ so let's choose a convenient value in this range and say the mass of the bear is $M = 200~\mathrm{kg}$. Then equation (2) tells us the speed needed is about $140~\mathrm{m/s}$.

For comparison, the fastest baseball pitch ever was about $47~\mathrm{m/s}$ so our poor bear has to throw a much heavier ball three times as fast, and catch then throw that ball repeatedly. Unless this is a truly remarkable bear that isn't going to happen, but in principle it could.

John Rennie
  • 355,118
  • Comparing your result to that of @ordptt, it seems that your answer if a really good approximation in general, but it diverges when $\frac{m}{M}$ is not small. – WordP Sep 29 '22 at 21:17
4

The only technical problem I see with this is that the ball is described in the image as having these parabolas in shape. the problem is that the mass of the bear is way more than the mass of the ball, and so to be exerting these very large momentum shifts, the ball needs to be very sharply angled where the bear makes contact with it, like it is at the ground. Also the impacts with the ground should probably be kicking up debris.

Well, that and, the bear is shown kind of walking along the ball, which might be possible if you give the ball enough energy at the outset, but really you would want to probably give it a lot of energy in the middle too, so the bear probably wants to describe a jumping motion where it gives that ball a similar absurd push to the one that it originally gave the ball such that it had enough momentum to walk on.

But yeah, other than the fact that the trajectory should be approximately a sawtooth and the bear should be jumping, and this probably requires more strength than a bear has and more integrity than a typical ball has, and the difficulty with getting a predictable bounce back from an uneven ground between the two cliffs, the remaining basic idea is not totally bananas. The problem with the image is just that it is insufficiently violent in its characterization of this motion. But that's okay because without that it's a really funny cartoon.

CR Drost
  • 37,682
2

It depends what you mean by possible. In realistic conditions on Earth, no, it is generally not possible due to the reasons given by @John Darby.

However, you can contrive an explanation for how it could work. For example, although Sean Carroll thinks time travel probably doesn't work the way it is shown in the movies, as a science advisor he might come up with an explanation for how it could:

Your job is to help them, your job is to serve the movie, tell the best, most interesting story you possibly can. And the way that I find is helpful to sort of mentally line yourself to that job, that responsibility, is to think of the screenplay as data, rather than as a theory. In other words, you’re not looking at the screenplay and going, “Uh, no, I don’t think that’s the way it would work, I don’t think that’s an accurate description of reality,” that’s a bad attitude to have to a screenplay. The screenplay is, “In this world, what actually did happen?”

Quote is from Solo: How Time Travel Could and Should Work`, which is Episode 124 of Sean Carroll's Mindscape podcast.

In this case, if the ball has sufficient mass, it is possible for a low-mass being to jump on it mid-air without altering the trajectory of the ball much. (Though the scale is much different, this is precisely what we are doing when we jump on Earth!)

Then all you have to do is allow for that the ball can bounce, and that maybe the ground is actively powered to "push" the ball up a bit for it to regain its initial height.

It would be pretty hard to time things right, but if you tried hard enough to set it up, I'm sure you could produce a setup that replicates the concept.

Jonathan Jeffrey
  • 1,359
1

No, for a real bear and a real ball. Here is an explanation without going into the mathematics of collisions.

First, the ball will not rebound up to its initial height due to energy loss when it hits the ground. (The collision with the ground is not perfectly elastic. Even if the bear throws the ball downward giving it an initial velocity, it will eventually not rebound to its initial vertical distance.)

Second, there are two forces that act on the bear: gravity downward and the impulsive force upward from the ball when the bear collides with the ball. I assume the bear is much more massive than the ball. During the collision with the ball the bear will experience a very short-duration force upward but due to the heavy mass of the bear and the light mass of the ball this is a negligible effect, and gravity continues to pull the bear downward.

John Darby
  • 9,351
  • 5
    OP asked for a theoretical reason it won't work. From a theory pov, your first point can be overcome if the bear throws the ball downwards with great enough force (and the ball and ground are stiff enough). And your second point can be overcome if the ball is heavy enough or the bear is light enough (I suppose there's a theoretical limit to how hard the bear can throw the ball downward if she is too light and the ball is too heavy). – The Photon Feb 01 '21 at 16:41
  • 1
    @The Photon. Yes, good points. I edited my answer accordingly. – John Darby Feb 01 '21 at 16:54
1

If you look at the problem from a realistic point of view, it depends on the conditions of system. With energy losses from air resistance and the energy lost during the rebound, the maximum height of the ball will decrease over time, which makes it much more likely that the bear character will not reach the other side. However, note that the bear can reach the other side in many different ways. If it is strong enough, it could traverse the canyon with a single jump, for example.

But the picture has some implicit assumptions that one can use to solve the problem quantitatively:

  • The bear goes to the other side by using many jumps along the way
  • The height at which each jump occurs is the same
  • The movement is regular and periodic, which suggests that energy losses are not taken into account.

With that in mind, it is possible to solve for the momentum the bear has to give to the ball.

Because the ball and the bear move together horizontally, the bear cannot give the ball an impulse horizontally. If it propels itself forward, the ball will be thrown in the opposite direction, and the bear will not be able to jump on it anymore. However, that is not an issue with no energy losses. If the bear and the ball start with an initial horizontal velocity, that will be maintained as long as the bear only exert force on the ball vertically. In this sense, we only need to consider the vertical motion.

At the very beginning, both the ball and the bear are mid-air with a constant horizontal velocity. To not fall, the bear jumps such that it gains a momentum $Q$ and the ball a momentum $-Q$, due to conservation of momentum. Then, the magnitude of the velocity of the bear and the ball are, respectively, $$V=\frac{Q}{M}$$ $$v=\frac{Q}{m}$$ Then, the time $t$ the bear takes to return to the same height (the height immediately after the moment it jumped) is given by the following equation of motion $$Vt-\frac{1}{2}gt^2+h=h$$ $$t=\frac{2V}{g}$$ The time $t_1$ the ball takes to reach the ground is given by $$-vt_1-\frac{1}{2}gt_1^2+h=0$$ $$t_1=\frac{-v+\sqrt{v^2+2gh}}{g}$$ where $h$ is the height of the canyon/valley. The ball reaches the ground with a velocity $$v_f=\sqrt{v^2+2gh}$$ The time $t_2$ the ball takes to return to the same height is given by $$v_ft_2-\frac{1}{2}gt_2^2=h$$ $$t_2=\frac{v_f+\sqrt{v_f^2-2gh}}{g}=\frac{\sqrt{v^2+2gh}-v}{g}$$ By using the fact that when the jump occurs the bear and ball meet, then $$t=t_1+t_2$$ $$V=\sqrt{v^2+2gh}-v$$ $$\frac{Q}{M}=\sqrt{\frac{Q^2}{m^2}+2gh}-\frac{Q}{m}$$ Solving for $Q$: $$Q=\frac{M\sqrt{2mgh}}{\sqrt{2M+m}}$$ Which has a real solution. Now, this was for the first jump, but this can be repeated as many times as necessary, because the total momentum immediately before the second jump is 0, just like immediately before the first jump, because the ball and the bear come back to the same height with the same velocities, just with opposites directions.

Using $M=200kg$, $m=1kg$, $g=9.8m/s²$, $h=10m$, then $$v\approx 140 m/s$$

ordptt
  • 300
0

There are lots of reasons why this cannot work, and there are lots of ways the bear can die.

For instance, in order to catch the ball after every bounce, both the bear and the ball must have the same velocity in x, which is already challenging to archive if there is a delay between the bear throwing the ball and jumping from the cliff (as the image suggests) -- if this condition is not met, the bear and the ball will not arrive at the same spot, and the bear will die.

Assuming this (quite restricting) condition is met, the ball must be perfect in a sense that it can bounce off the earth (and the bear) without kinetic losses, otherwise the bear will collide with the side of the other cliff and die.

Third, the whole process must happen in pure vacuum to neglect losses due to air resistance, so poor Winnie might suffocate in the process.

Fourth, both the bear and the ball must behave as point-like objects do, otherwise they might collide in a way that they start moving along the valley, which is disastrous from the bear's point of view.

If all these conditions are met, energy and momentum conservation will do the rest, as gravity is a conservative force.

Apart from that, according the image, the bear is falling to his death already, since he cannot catch the ball for the next bounce -- so he will die anyway :(

friciwolf
  • 21