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Newtons 2nd law?

  1. Oct 11, 2011 #1
    This morning while riding my motorcycle, a black bird smashed into my helmet.
    Considering a black bird weighs approx. 80-120 gram (according to google), I was surprised by the force of the impact.

    Anyways, now I'm wondering if anyone could calculate with what force the bird hit the helmet? How much it "weighed" at the point of impact (if you can put it like that?).

    I would say the black bird was of average size, so let's say 100 grams. When it hit me, I was riding along at 60 kph (=37.28 mph).

    The bird didn't really carry any speed into the impact, as it came in from the side, hitting the front of the helmet.

    I hope someone can help me with this calculation (I believe google suggests something about Newtons 2nd law?).

  2. jcsd
  3. Oct 11, 2011 #2


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    Poor bird!

    Now onto the physics :

    From Newtons second law we have that the force of impact is F=DP/DT. DT here is the time that the impact lasted. Typical values for DT is 0.1sec. DP is the difference of momentum of the bird after the impact minus the momentum of the bird before the impact.

    Momentum equals mass of bird times its velocity. I take mass of bird 100g.

    Momentum of bird before impact is zero (since it came sideways).

    After impact momentum of the bird is 0.1kg*60/3.6=1.6 (factor 3.6 comes into play in order to convert kph to m/s)
    So we have DP=1.6-0=1.6 and the force of impact F=DP/DT= 16Newtons.

    1Newton is the weight of a mass of 100g. So at the point of impact it is like it weighed 1600g about 16 times its normal weight of 100grams.
  4. Oct 11, 2011 #3


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    Caveat lector: This is only a very simple back of the napkin calculation and whilst it may be sufficient for your purposes it is important to note that such a calculation is unlikely to be that accurate.
  5. Oct 11, 2011 #4
    It'll take a bit more to calculate the force than simply the weight and speed. The helmet absorbs some of the impact, and passes the rest onto you.

    If an object is speeding relative to an observer, it has kinetic energy defined as (1/2)mv^2, at nonrelativistic speeds, which are basically normal everyday speeds.

    To slow an object to 0, the energy of the moving object must be removed. So if you were colliding with the bird, as it is moving relative to you, the impact removes energy. Well, I probably shouldn't say remove, since it's more of a transference. Now think of firing a bullet into clay. This is related, so don't just think of it as weird and random. If you fire a bullet into clay, the clay will absorb the kinetic energy of the bullet, and slow it down. However, different clays have different properties, so bullets will go different distances in each clay. Because of this, a different amount of force is acted upon the bullet. A long distance in the clay means a low force is applied, as it absorbs less over the same distance. A short distance in the clay means a higher force is applied, as it absorbs more kinetic energy over the same distance. However, the stiffer(shorter distance) clay, as it absorbs the kinetic energy, passes it on to you. It's like a Newton's cradle as the clay stiffens, the balls simply transfer the energy to the next ball, and absorb essentially nothing. A less stiff clay also has drawbacks, as it may not absorb enough of the kinetic energy and the bullet could pass through, still with a high amount of kinetic energy.

    Unfortunately, I can't help you calculate the force. It depends on many other things, such as how much of the impact the helmet absorbs, and how much your body resists the blow. Letting your body just go backwards from the impact will have less impact than resisting it and keeping your body straight. Some experiments may help though, such as whacking your helmet and seeing how much force/energy is passed on to a sensor where your head would normally go.
  6. Oct 11, 2011 #5
    I'm just curious, not saying you are wrong. But let's say the bird came from the side and hit the front of the helmet. Will not the speed of the driver increase the force he feels slamming on to his head?
    I mean for example imagine yourself banging your head into the wall. Do it slowly, then do it quick. The second time you will feel a lot more pain on your head due to a stronger force that was caused on your head, right? Due to a faster acceleration caused by your "faster movement" of the head, right?

    So how come you excluded that in the calculation? Or did you just thought of the bird slamming in from the side and no forward movement from the driver was exerted on to the bird?

    Sorry if I explained this in a horrible way, I'm still a newbie but I'm trying! o:)
  7. Oct 11, 2011 #6


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    If the bird was really moving entirely perpendicular to your direction of motion, then your forward speed would have no bearing on the problem, since all that matters is the relative velocity between you and the bird, and this depends only on how fast the bird is flying.

    However, it's unlikely the bird would come at you perfectly sideways, since this would mean it would have to have the same "forward" speed relative to the road as you. More likely, the bird is moving sideways relative to the road (i.e. across your path), which means that relative to you, it's actually moving towards you on a diagonal path, since the relative velocity between you and the bird has both "sideways" and "forwards" components. Therefore, in this scenario, the faster your forward speed, the larger the the relative velocity between you and the bird (due to the larger forward component), and the harder it will strike.
  8. Oct 12, 2011 #7


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    I was once nearly jousted off a motorcycle by an oversized beetle, so I would want no part of a 100 gram bird of any species.
  9. Oct 12, 2011 #8


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    In my calculation i calculated only the component of the force parallel to the motorbike velocity axis. There is also a component of force perpendicular to that axis but i consider it negligible since i consider the momentum a bird of 100g carries is negligible at least since i have no data on the (perpendicular) velocity of the bird.
  10. Oct 12, 2011 #9
    Thank you all for the help :eek:)
  11. Oct 12, 2011 #10
    Thanks! :)
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