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B Mass of an impact

  1. Jul 23, 2016 #1

    imagine the following scenario: a persons sits and stands up. At mid- distance the person's hits the head on a shelf that is right above the head. A Human Head weighs about four Kilogramm. the question I have now is, what is the mass of the impact? is it exactly four kilogram or more, because the whole body is moving upwards, not only the head.
  2. jcsd
  3. Jul 23, 2016 #2
    I don't think it makes much sense to say the 'mass' of an impact. Maybe you mean the momentum of the whole person, that would have to be summed up of each small 'part' of the man as I=mv.
    And a human head weighs about 4.7-5.1 kg, I think.

  4. Jul 23, 2016 #3


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    Yes. The whole body should be used for any calculation. Perhaps an easier model to discuss, initially, would be if you throw a 5kg ball upwards and it hits the shelf.
  5. Jul 23, 2016 #4
    I am happy that you both understood what I meant. i am not looking for an exact calculation which is impossible anyway.

    So if a person weighs 50 Kilogramm, the mass that hits the shelf must be between 5(head) and 50(whole body) Kilogramm, is that correct?

    What I also do not fully understand is what l=mv means.
  6. Jul 23, 2016 #5


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    It seems that you want for there to be a single "equivalent mass" that describes the body in sufficient detail to compute the results of the collision. But there is not.
  7. Jul 23, 2016 #6
    No Jbriggs. I think you misunderstood what I was trying to say. imagine that the shelf is a special scale. The hit would show us how many kg the impact had. This is the mass I would be looking for.
  8. Jul 23, 2016 #7
    Okay I just noticed that my last post was nonsense, this would demonstrate as the force behind the impact not the actual 'mass' that hits Shelf. i'm sorry but I have huge problems explaining what i mean, I am not good at physics.
  9. Jul 23, 2016 #8


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    It's a good start that you figured that out. But even still, the force of an impact is still a very complicated thing. Not only is it not constant, but it depends on the very specific properties of the materials and structures involved. And it doesn't generally tell you anything very useful. For example, what if I told you the average force was 5,000 N. What does that information do for you?
  10. Jul 23, 2016 #9
    Nothing. My question was more theoretically. replusz and sophiecentaur have pretty much answered my question.
  11. Jul 23, 2016 #10


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    It is a rare skill and there is no way round the hard learning process that's required to become 'good' at it. The jumbly way that physics terms are used in everyday conversation only serves to make things worse for the uninitiated. "The Force of Impact" is a phrase that's used very frequently on PF and it is not an easy quantity to predict. It is a useless concept when discussing who's fault it is when cars collide and most collision questions involve traffic collisions (and legal matters).
    There is a quantity called Momentum, which is absolutely essential for describing collisions. Momentum (classical mechanics) is mass times velocity (mv) and it has direction (it is a vector quantity). Momentum is always described in a particular frame of reference - often the frame where Earth is stationary. When there is a collision, momentum is transferred between two objects. If an object is slowed to zero, it loses its momentum in the collision and this can be achieved by a combination of a Force and the time that the force acts.
    Your head hitting the shelf (and perhaps stopping) will lose some of or all its momentum. That could be very quick and with a large force involved (granite shelf, for instance) or take a longer time (padded shelf or hat) and the force will be much less.
    Your original question involves the body pushing up as well and that adds a load of extra complications with momentum change as the neck joint pushes against the rising body and the leg muscles etc. keep supplying their force contribution.
  12. Jul 23, 2016 #11
    Wow, great answer Sophie. You couldn't have explained it better! Thanks.
  13. Jul 24, 2016 #12
    One last question:

    Regarding the formula f=m*a, when we are looking for the peak force that affected the shelf,
    the peak mass that hits the shelf, m must be somewhere between 5(head) and 50(whole body)kg

    Is that correct?

    Or is the only way we could calculate the force
    through l =v*m?
    Last edited: Jul 24, 2016
  14. Jul 24, 2016 #13
    Yes, but you wouldn't know a (acceleration) as it depends on gravity AND the hitting of the shelf AND on the force in your legs.
    I=mv isn't force but momentum (https://en.wikipedia.org/wiki/Momentum).
    And if you use I=mv and F=I/t you get F=mv/t so F=ma, which isnt't so helpful.
    So all we can say of the peak force F= (some value 5-50)*(an unknown number)
  15. Jul 24, 2016 #14


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    The only way would involve knowing / measuring / inventing values for all the variables involved. Bodies, which are far from perfect (nothing personal here lol) are particularly hard to characterise, which is why we start with billiard balls (pretty near perfect models) and work upwards.
  16. Jul 29, 2016 #15
    There are still many things I don't understand but I would really like to.

    let's take a different example: person A weighs 50 kilograms and is sitting. person B stands behind and stares downwards at the top of person A's head. both have, instead of a human head, a billiard ball head that weighs 5 kilograms. the sitting person stands up with an acceleration of 15 m/s^2 and after 25 centimeters both heads collide. the moment of the impact the velocity of person A's head is 2 m/s.

    To calculate the force of the impact and the acceleration of person B's head I would do the following:


    F= (a value between 5 and 50, I take 20) 20*15




    Is that correct?
  17. Jul 29, 2016 #16


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    If you want to understand more then you need to approach it from the 'right' way. The term "Force of impact' has very little meaning (none, really). The effect of any impact will depend on the length of time the force is applied or the distance over which it acts.
    The value of acceleration that you have arrived would only apply for an infinitesimal length of time so it is 'just a number' and no more. It doesn't tell you what will actually happen.
    The theory of impacts is taught at A level (UK) at a simple level and would certainly not involve the sort of involved scenario that you have brought up. In Physics, you have to walk before you run. I don't think there is a quick way to leap half way in from zero.
    This link could help, perhaps.
  18. Jul 29, 2016 #17
    so if we knew these two things we could calculate my example right?

    let's say the impact lasted 0.05s or the force was applied for 0.5mm.
  19. Jul 29, 2016 #18


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    You are just guessing at both figures. The calculated results will only be as accurate as your guesses.

    So we have a 5kg head moving at 2 meters per second colliding with a stationary head. The collision lasts 0.05 seconds. We assume that the collision is brief enough that the heads alone are involved and not the bodies. We assume an inelastic collision. By conservation of momentum, both heads end up moving at 1 meter/second.

    That is a change in velocity of 1 m/s of a head mass of 5 kg. That's 5 kg m/sec. It took place over a duration of 0.05 seconds. So that's an average of 100 kg m/sec2. 100 Newtons. Which is just about twice the weight of the head.

    One can sanity-check that result. The acceleration of gravity is 10 m/sec2. So every tenth of a second, a freely falling head would change its velocity by 1 meter/sec. We're talking about doing that in one twentieth of a second instead. That's 2 g's. The calculation checks out.

    But let's do another sanity check... Over this 0.05 second interval the head is decelerating from 1 m/sec (relative to the collision plane between the two heads) to 0 m/sec. That's an average speed of 0.5 m/sec. Multiplied by 0.05 seconds that makes 0.025 meters. 2.5 centimeters. That's a pretty crushing head blow. It does not match up at all with the half millimeter that you had guessed.

    So your guesswork does not check out. We have to chunk this calculation into the trash.
  20. Jul 29, 2016 #19
    I am glad that you have taken the time to show me how this could be calculated. again I was not looking for a concrete result, the only thing I wanted is to show a demonstration how it could be calculated by using fictional numbers.
  21. Jul 29, 2016 #20
    Another question: is it correct that the harder the surface of the two balls is the less long the impact would last and therefore the 'force of the impact' and the acceleratuion of the second ball would be higher?

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