22
$\begingroup$

It's funny, I used to wonder this at school many years ago. For the purposes of discussion, imagine there is a big (very massive) cube that slides towards you as you're sitting on the floor. It doesn't move fast, perhaps (say) walking pace. Would it kill you on contact?

My original intuition is that it would, since presumably you would go from not moving to moving almost instantanously. Hence the acceleration would be very large. But still, the fact it's all moving so slowly still makes me want to doubt this.

EDIT: my current thinking is if I were to go a zero-momentum frame, since the cube is so massive it would basically be stationary. On the other hand, we'd be moving at ~$v$. If, as I said, $v$ was quite small, then I think we can imagine this as no different to falling with some $v$. It seems that our bodies are okay with that. So, maybe we would survive okay?

$\endgroup$
12
  • 3
    $\begingroup$ It all depends on how fast the cube is moving. What is important is your rate of change of momentum, the greater that is the greater is the potential damage done to you. $\endgroup$
    – Farcher
    Commented Feb 27 at 22:17
  • 43
    $\begingroup$ Even if the train is not compressible, your body certainly is. As with the fall, your body will deform and thus reduce the acceleration. In both cases, if the speed is low enough, you will survive intact; at somewhat higher speed (or the wrong angle) you will break some bones. Faster still, and you're history. $\endgroup$
    – hdhondt
    Commented Feb 27 at 22:31
  • 4
    $\begingroup$ I think your intuition of going to the zero-momentum frame, where the block is approximately stationary, and making an analogy to falling, is very good intuition. If you were falling, then your speed when you hit the block would depend on $t$ (the time you were falling, assuming you started falling from test) via $v=gt$, where $g=9.8{\ \rm m}\cdot{\rm s}^{-2}$ is the acceleration due to gravity on Earth. So you can think of hitting the block with small $v$ as hitting the ground after falling a short time (or small distance), or large $v$ as hitting the ground after falling a long time. $\endgroup$
    – Andrew
    Commented Feb 27 at 22:44
  • 6
    $\begingroup$ In practice, it depends on whether or not you fall over and get run over by the train. It doesn't matter how slow the train is moving if you get squished between the wheels and the tracks. $\endgroup$ Commented Feb 28 at 12:40
  • 3
    $\begingroup$ Assuming the cube is moving at walking speed, it seems like your experience should be almost exactly the same as the time I walked directly full-speed into a plate glass window, thinking it was a doorway. I was not injured, but it was very unpleasant and I strongly disrecommend the experience. (If the cube is less elastic than a large pane of glass, the force could be proportionally higher and briefer.) $\endgroup$ Commented Feb 29 at 5:01

8 Answers 8

60
$\begingroup$

One of the joys of physics is that you can often reframe the problem. In your problem, you have a moving cube and a person standing still. Thus you are thinking in the frame of the person (the person is motionless in the person's own frame). But you can also think of it in the frame of the cube. In that viewpoint, the cube is still, and the person is moving towards it at a walking pace. (the cube is motionless in the cube's own frame). Other than the minor detail of whether the persons legs are moving or if the cube is ominously floating forward, the two situations are identical, and your question doesn't concern itself with that particular detail. So let's reframe!

So thus, if you are interested in what happens in the case of that collision, you can instead ask what would happen if you walked into a steel wall. Go ahead. Try it. See what happens.

Okay, fine. Don't do it. While it wont kill you, it may hurt. What will actually happen is that your body will deform on impact, exactly as it would if you walked into a wall. The parts of our body which are most sensitive (such as the brain) are well protected by layers of flesh and bone which were evolved over millions of years to provide the critical milliseconds of protection to keep them safe.

There are times where reframing the problem will not be so helpful. Consider the same problem, but now the person is standing against a wall. You can still reframe the problem in the perspective of the cube, but now you have to deal with the pesky question of the inertia of the wall. Once the cube has contacted the person, moving them back, and then they make contact with the wall, we need to start worrying about how hard it is to accelerate the wall. Depending on stiffness, we might argue that is exactly as difficult as changing the rotation speed of the planet -- a big and hard to calculate number. Instead, we find it is easier to calculate in the wall's frame of reference, letting the cube and the person move, and then handwave away all the unnecessary nuances of how the wall moves -- simply declare that it doesn't move at all.

In principle, even here there is movement. Walls will move and deflect, often far more than you think they do. Few things actually happen instantaneously in the world of physics, although we'll often approximate them as such.

This second problem has very real real-world implications. Consider industrial robots. Industrial robots are made of very stiff steel, and have strong motors controlled by algorithms that have one goal: to move from point A to point B. If a person gets in the way, its like your cube problem. The person will get smacked by the robot, and it will hurt.

One solution is to install security barricades to keep people out of the way. This has been shown to have tremendously dangerous consequences, for now the problem is more like my second example with a cube, person, and a wall. On more than one occasion, robots have pinned people against the security barricades, crushing them not just with the power of their motors, but with the power of their motors and a decidedly well-anchored and immobile steel barricade. There's a very good reason why modern factories implement technologies like light-barriers which notify the algorithms that a human has gotten near the robot so that the algorithm can shut down the robot before something bad happens.

$\endgroup$
5
  • 11
    $\begingroup$ The nice thing about this reframing is that the mass of the train is mostly irrelevant. Most of us have walked into a wall or door, or had someone open a door into us, and survived with at worst a little bruise. $\endgroup$
    – Barmar
    Commented Feb 28 at 16:44
  • 8
    $\begingroup$ @Barmar And quite a bunch of us - me included - had actually ran full speed into a wall or door face-first and survived to tell the story just fine. $\endgroup$
    – T. Sar
    Commented Feb 29 at 11:37
  • 4
    $\begingroup$ @T.Sar my mates and I do it regularly for fun, we call it 'mountain biking' 😄 $\endgroup$ Commented Feb 29 at 23:19
  • $\begingroup$ @Barmar True, I could have drawn an analogy to a door being opened in one's face. I like that example. However, if I went with that, one's intuition about how easily a door is moved might cloud the connection to the large heavy cube. We have some instinct that the door is going to respond to being hit, and move. If the average person was used to how large safe doors move, that intuition would be directly applicable. Besides, the ability to reframe problems is really useful, and its always nice to have an example of it in use. $\endgroup$
    – Cort Ammon
    Commented Mar 2 at 14:58
  • $\begingroup$ But what if the cube hits your little toe first? $\endgroup$ Commented Mar 26 at 21:34
28
$\begingroup$

You can think of this in terms of falling from a height of say one foot (e.g accidentally rolling off your bed). You hit the surface with a low velocity and it is survivable (but not always) despite the enormous momentum of the Earth hitting you.

The key point is that it is the impulse (change of momentum imparted to you over a short period) that determines the damage done. If you have a mass of 100 kg and fall from a height of say $h = 10$ m then your momentum when you hit the floor is $m v = m \sqrt{2 g h} = 100 \times \sqrt{2 \times 9.8 \times 10} = 1400$ Kg m s$^{-1}$.

Let's estimate that the time taken to go from your impact velocity to stationary is 0.5 seconds, then the impulse received is $1400 / 0.5 = 2800$ N s

Compare this to falling from a height of 0.2 m, then the impulse force $$m v = 100 \times \sqrt{2 \times 9.8 \times 0.2} / 0.5 \approx 395 \ \textrm {Kg m s}^{-1}$$

The impulse force of falling from 10 m is approximately 7 times greater than falling from 0.2 m and the velocity is also 7 times greater, so the impulse force is proportional to the impact velocity.

If we switch to a frame where you are stationary, then it looks like the train is coming at you. If the train is moving sufficiently slowly, then the impulse imparted to you can be reasonably low and for very large masses like a train or the Earth, the mass of the larger object can be ignored.

There is a simplification in the above, in that I have assumed the impact time of the impulse is the same in both cases. The real impact time depends on factors like the size of the 'crumple zone' which is basically the distance from the COM of the falling body at impact from the the 'unmovable' object, the tensile strength of the falling object and the impact velocity. For a similar crumple zone and materials in both cases, the impulse time is shorter for the faster object, so that would make the impulse ratio even greater than above.

EDIT: my current thinking is if I were to go a zero-momentum frame, since the cube is so massive it would basically be stationary. On the other hand, we'd be moving at ~v . If, as I said, v was quite small, then I think we can imagine this as no different to falling with some v . It seems that our bodies are okay with that. So, maybe we would survive okay?

I think you have arrived at the correct solution.

$\endgroup$
5
  • 10
    $\begingroup$ After a 10m fall, your velocity is approximately 14m/s (assuming no air resitance, yada yada). It does not take half a second to stop. If we assume constant deceleration speed, and hitting feet-first, and a body length of 2m (just to be generous), it would take 0.29 seconds to use all that body length to stop (which means you're flattened into goo). If you're hitting flat and face-first, you have a lot less distance and a lot less time to stop before the corresponding deformation of your body kills you. $\endgroup$
    – Arthur
    Commented Feb 28 at 13:19
  • 3
    $\begingroup$ @Arthur: Good point. Using almost all of your body length is close to what happens if you turn the fall into a roll, which is how skilled people can handle somewhat-high falls without injury. Maybe not 10 meters onto hard ground, but maybe 5. $\endgroup$ Commented Feb 28 at 16:18
  • $\begingroup$ I'd like to see a specific citation for "it is the impulse... that determines the damage done" (that being proportional to velocity). E.g., I've seen it elsewhere argued that damage is determined by kinetic energy gained in a fall. $\endgroup$ Commented Feb 28 at 19:28
  • 2
    $\begingroup$ @DanielR.Collins not a citation but a thought experiment: if you fall from 10 meters onto a steel plate, a pool of water, and netting designed for falls, your kinetic energy will be the same in each case, but the three surfaces will dissipate the energy/momentum over different distances, with different resulting impulse. The peak force will also change. Not to say KE is irrelevant, though $\endgroup$ Commented Feb 28 at 21:47
  • $\begingroup$ @Arthur My last paragraph mentions that the impulse time is affected by the length of the 'crumple zone', the ductility of the material, the velocity at impact, the hardness of the surface etc. I do agree that my example figure of 0.5 seconds was a bit generous. $\endgroup$
    – KDP
    Commented Feb 28 at 22:31
3
$\begingroup$

There are two problems with your intuition.

  1. "presumably you would go from not moving to moving almost instantaneously" - this has nothing to do with the mass of the train! Heavy objects do not transfer momentum any faster than light ones. Rigid objects transfer momentum quickly, because they do not deform on contact. So rather than a very heavy train in your thought experiment we should change to a very rigid train (and a very rigid person).
  2. "Hence the acceleration would be very large" [and this would kill you] - if we were to collide two very rigid bodies (the masses and velocities being irrelevant), then it's true that the acceleration during the collision would be very high. But it would also be very brief, and there's no reason to suppose that brief, high acceleration is harmful.

What is known to be harmful is the total impulse of the collision - in other words, the product of the high acceleration with the time that it occurs for. A high impulse in a brief collision will hurt you, because your organs will collide with your bones and so on. But the maximum possible impulse the train can deliver to you is $2mv$ (where your mass is $m$, and the train's initial speed is $v$) - which is not very much if $v$ is around your walking speed (proof: walk into a wall).

$\endgroup$
2
$\begingroup$

for a slow collision, the inherent springiness or elasticity of your body will provide enough time in the course of the collision to bring you up to speed and to slow down the big cube by some increment. No injury results because there is no energy dissipation (crushed tissue, broken bones) in a slow collision process.

$\endgroup$
2
  • 4
    $\begingroup$ In short: "No, because you are squishy." $\endgroup$
    – ceejayoz
    Commented Feb 28 at 14:01
  • $\begingroup$ It's the acceleration that you suffer/endure/are subjected to, that kills you. The kinetic energy of the moving object colliding with you is irrelevant. Since you're squishy, as others have already stated, you deform at first ever so slightly, and then you attain the very slow velocity of the moving object. $\endgroup$
    – Dohn Joe
    Commented Feb 28 at 14:49
2
$\begingroup$

Because your body is deformable, your entire body will not have to accelerate up to the block velocity instantaneously. Only the infinitesimally tiny amount of mass adjacent to the initial contact with the cube will accelerate first, and the compression wave will travel through your body. Just think of your body as a rod being hit by the cube. The entire rod would not come up to the speed of the cube at once. This is because the rod is elastically deformable. The compression wave will travel across the rod at the speed of sound, and the extent of the compression zone grows with time. The portion of the rod inside the growing compression zone will be traveling with the rod velocity, but the portion ahead of the compression zone will not be moving yet. Ultimately, the compression zone will encompass the entire rod.

$\endgroup$
2
$\begingroup$

Imagine a treadmill which is large enough that you can sit comfortably on the band. Make it so that something solid is hovering over one end - you can simply take a big sheet of very thick/sturdy plywood or drywall and screw it on the frame where the control board is.

Set the treadmill to move at walking pace.

Sit on the treadmill as far from the board as you can.

See what happens.

I see these possibilities:

  • A very real-life treadmill has a relatively smooth band where you run on. Assuming I have long trousers, the band would just keep moving, and I would maybe have some funny feeling on my butt, or eventually it would become warm due to friction. Worst case, the material of my trousers might rip, or the flexible material of the band might rub off on it, causing black smears. For the rest of your body, the act of bumping on the wall itself (say, with your nose) would probably mostly hurt a bit without doing anything worse; after the initial bump you'd just feel the pressure against your skin.
  • The treadmill might be prepped, just for this experiment, with a surface as rough and hard as concrete. Then the result would be very unfortunate indeed - your trousers would very likely quickly be shredded to bits and pieces, and the concrete-like band would start abrading your skin; if you stay there forever (maybe because you've been a corpse or unconscious in the first place) it could eventually, maybe, pulverise you, but in practice you'd hop off the band quickly enough.
  • You might actually glue yourself to the band. Then, on impact, assuming neither the board nor the band nor the glue get ripped apart, nor the motor stalls, your trousers would be shredded even quicker than before, and the unsightly mess of trouser-cum-hardened-glue would build up before the wall; the rest would be the same as before.
  • Glue your skin to the band and repeat... very unfortunate result.

And so on and forth. The details are not so important, the point is that you can run a thought experiment like this and go from a fantastical scenario like a big moving block to something which you can intuit about from daily experience.

$\endgroup$
0
$\begingroup$

Please correct me if I'm wrong:

  • If walking speed is 3 mph that's 1.3 m/sec.
  • Gravity being 10 m/sec/sec you reach that speed after falling about 150 msec.
  • Distance travelled in that time is 10 * 0.15 * 0.15 i.e. about 25 cm

So if you fall 25 cm then you hit the planet (Earth) at about walking speed.

That's not necessarily dangerous, depending on where and how you hit -- if you soften the blow with your hands or anything -- "head first" might be a problem, I suppose, I've never tried it!

$\endgroup$
-2
$\begingroup$

The train will just kick you in the direction of its movement with the velocity double that of the train.

Explanation

Conservation of momentum:

$ m \cdot v = m \cdot u' + M \cdot v' $

where m is mass of the human, M is mass of the train, v is speed of the train and u' is human velocity after impact (0 before). v' is the speed of the train after impact, likely not much different from v.

Conservation of kinetic energy:

$ M \frac{1}{2}M \cdot v^2 = \frac{1}{2}m \cdot u'^2 + \frac{1}{2}M \cdot v'^2 $

The solution of these two equations is

$ u' = \frac{2Mv}{m + M} \approx \frac{2Mv}{M} = 2v $

Indeed, when a massive car hits a standing ball in elastic collision, the ball usually rolls away with the speed faster than the car is moving. The ball gains the kinetic energy from the car.

Due train moving slowly this is unlikely to be dangerous as humans are good at handing the low speed collisions. Usually your hand will substitute for a spring transferring the momentum over few centimeters of travel so it will not be like between two stones. While displacement to power curve of the muscle may differ from the one of the spring, the total amount of energy the muscle needs to absorb is still the same - the energy required to move the human at the speed 2*v. Skeleton will convert compression force to the extension force of the tissue of the muscle so the interaction may be still very much spring-like.

P.S. A train traveling at 100 km/h will kick you forward with the speed of 200 km/h that is another story.

$\endgroup$
6
  • 8
    $\begingroup$ Conservation of kinetic energy assumes a fully elastic collision. A ball-collision is close to that - your collision with the train is not! (Clearly total energy will be conserved, but some kinetic energy will be converted into crushing bones and heating you.) Look at what happens when a bug collides with the front of your car; does it bounce off? $\endgroup$ Commented Feb 28 at 15:38
  • $\begingroup$ No but if human collides with the train, normally the human would instinctively lift the hand to defend (assuming the slow enough motion). The further interaction would be handled via muscles very much in a way a spring does, maybe absorbing some energy rather than bouncing but anyway reproducing much of the compression step. All process would be comparable to falling down to the ground (also collision, and the Earth is awfully massive). Nothing serious would happen assuming this is from not too high. $\endgroup$
    – Nightrider
    Commented Feb 28 at 16:31
  • $\begingroup$ Sorry but this is correct. $\endgroup$
    – Nightrider
    Commented Feb 29 at 6:39
  • 2
    $\begingroup$ Muscles do not act much like springs in terms of energy transfer. When I land after a jump I don't keep bouncing. Human's coefficient of restitution is much closer to 0 than 1. $\endgroup$
    – Rick
    Commented Feb 29 at 15:53
  • $\begingroup$ Compression part is fundamentally comparable, even if force to displacement curve may differ. Extension part may be not but the muscle will only absorb as much energy as would be required for the human to move at the speed 2*v and not more. $\endgroup$
    – Nightrider
    Commented Feb 29 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.