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I've been teaching myself physics and I've been wondering about the impact time in collision calculations. The scenario I've been using to learn is an object with a mass of 4000 kilograms colliding with a human being, while travelling at 17m/s. The object has a surface area the size and shape of a human elbow (which I very roughly guesstimated to be around 20cm2.

When calculating the force of this impact I need the momentum and the duration of the impact. The momentum is easy enough to calculate, but how is the duration of impact worked out? I know that it isn't referring to how long the objects are in contact, as this would mean that swords would harmlessly rub against a person if they were swung. I assume then, that the time is referring to how long it takes one object to impart the force of it's momentum into the other object.

How am I supposed to do this? The obvious way is to measure it, but given I'm an art student I can't exactly go around driving cars into people to measure how long it takes them to react to the impact. So far I've just been using .1 seconds, but I feel like this is far too slow.

Qmechanic
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Space Ostrich
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  • Not sure I understand you comment about "swords could be harmlessly swung". Yes, for the purpose of an impact calculation you are interested in the time that one object exerts a force on another; when I shoot an arrow into a target, there is a force as the arrow slows down; once it is stuck, the arrow and target are still in contact but there is no force. – Floris Apr 28 '16 at 02:33
  • Yeah, that's what I mean. I knew it wasn't "time spent in contact" because if that were the case the force of that arrow impact would be spread out indefinitely as long as you left it in the target. – Space Ostrich Apr 28 '16 at 02:44
  • The posted answers allude to the difficulty of collision problems. I wrote a similar answer awhile back that might help.. – OnStrike Sep 26 '16 at 20:27

2 Answers2

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In general you need to establish some sort of stiffness, or more importantly an natural frequency for the system of two bodies. You can hear impacts and distinguish between slow thuds with fast pings.

For example if a short duration force has a harmonic shape (with frequency $f = \frac{\omega}{2 \pi}$) and peak force $F_{\rm max}$ then the total impulse is

$$ J = \int_{-\frac{\pi}{2 \omega}}^{\frac{\pi}{2 \omega}} F_{\rm max} \cos(\omega t) = \frac{2 F_{\rm max}}{\omega}$$

This means that the peak force is

$$F_{\rm max} = \frac{J\,\omega}{2} = \pi J \,f$$

where $J$ is the total change in momentum (impulse) and $f$ is the natural frequency of the impact (in Hertz). Typically the impulse is expressed in terms of the reduced mass of the two bodies $\mu = \frac{m_1 m_2}{m_1+m_2}$ and the impact speed $v_{\rm imp}$ and the coefficient of restitution $\epsilon$ : $$ J = (1+\epsilon)\, \mu\, v_{\rm imp} $$

Combined you have

$$ \boxed{ F_{\rm max} = (1+\epsilon) \mu\, v_{\rm imp}\; \pi \,f }$$

John Alexiou
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  • This is an interesting way to approach collision problems that I wasnt aware of. Any references you can point me toward? Is it possible to apply this method without collision tests to determine the natural frequency of impact? – OnStrike Sep 26 '16 at 20:30
  • Similar answer here: http://physics.stackexchange.com/a/202927/392. In many physics based simulations contacts are considered with a "penalty method" which essentially placing springs and measuring the peak deflection of the spring to come up with the peak contact force. – John Alexiou Sep 26 '16 at 20:45
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The simplest approach to a problem like this would assume that the collision is elastic, and that you have some knowledge of the elastic constant. But a collision between car and human is not that.

Instead, let us assume that the "elbow sized object" hits the human in the mid section, and that it doesn't simply go right through him. Then the next thing that will happen is that the human will "bend in half" as the center is violently accelerated and the head and feet haven't yet caught on.

Once the human is fully bent, all parts will pick up speed. The car will barely slow down.

Very roughly, I estimate that the car has to move no more than 80 cm for the human to be folded; since it has an initial velocity of 17 m/s, that takes about 0.05 seconds. The average force to accelerate 70 kg human to 17 m/s in 0.05 s would be $$F = \frac{mv}{\Delta t} = \frac{m v^2}{\ell} = \frac{70\cdot 17^2}{0.8} = 24 ~\rm{kN}$$

That is one heck of a sucker punch. And in reality the force will not be uniformly distributed over time, so the peak force is likely to be even greater. But modeling that accurately would require a LOT more knowledge about the system.

Floris
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  • The calculations I'd wound up with was basically "4 ton human in power armor sprints directly at human at 62 km's an hour, and collides with them elbow first", I'm still new to this so the calculations I did were force = momentum / time, with time assumed to be .1 seconds. I can't quite remember how I calculated momentum but I had 68880 kg.m/s for that force calculation. I then calculated the pressure from the impact to be 344.4 megapascals. I probably should have been more concerned about acceleration than pressure. – Space Ostrich Apr 28 '16 at 02:48
  • you are saying it takes time and space to fold the human, but you use the whole mass of the human while using the time/distance to fold the human... which would be only partial mass as it is still folding and not translating the entire body axially. I don't think it makes sense? – Zero Mar 21 '20 at 19:05
  • @Zero when the human is completely folded all parts will be moving at the same speed - the “average force “ needed to change the momentum is what I computed. Only if you try to get the force profile would you have to worry about which part starts moving at which moment. – Floris Mar 21 '20 at 19:24
  • ah ic. nevermind. im out of practice – Zero Apr 13 '20 at 18:26