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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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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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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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