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From time to time I see safety warning about keeping loose items in your car. The last warning used a 2kg object, and claimed that if a collision occurred at $50{km\over h}$ it would have a weight equal to 80kg. At $90{km\over h}$ it would have a weight of 256kg.

How are these "new weights" calculated?

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From $dp/dt\simeq \Delta p/\Delta t=F$. A 2 kg object at 50 km/hr has an initial momentum of $$ p=mv = 2\,{\rm kg}\cdot13.9\,{\rm m/s}=27.8\,{\rm kg\,m/s} $$ If we naively assume that the crash is over the course of a 0.05 seconds, then the force is $F=556\,{\rm N}\to m_{eff}\simeq58\,{\rm kg}$ (final momentum being 0 because the velocity is zero). Letting $\Delta t=0.03\,{\rm s}$ gives $m_{eff}=94.6\,{\rm kg}$.

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  • $\begingroup$ Did some math, and a delta-t of 0.0354s yields the correct result. However when applying the same to the 90km/h data it seems 0.02s. Hence it seems like delta-t is the time the object uses to cover a distance of 0.5m at at the speed of impact. Does that seem like a reasonable explaination? :/ $\endgroup$ – atomman Oct 14 '13 at 16:37
  • $\begingroup$ That does seem a reasonable explanation for the speed-dependence for $\Delta t$. $\endgroup$ – Kyle Kanos Oct 14 '13 at 16:41
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I'd guess it's probably some heuristic where they assume the object decelerates from the car speed to stationary in some time interval characteristic of impact times during car crashes (ie, the time it takes to decelerate when it hits you). If that's the case, the formula is $$gm_e =a m$$ where $m_e$ is the calculated "effective car-crash mass", $a$ is the deceleration encountered in the crash and $m$ is the stationary mass of the object.

For example, with $a=\frac{50\mbox{km/h}}{0.05\mbox{sec}}$ and $m=2\mbox{kg}$, this gives $m_e=56\mbox{kg}$.

Interestingly, their claimed effective mass scales nonlinearly with car speed, which may have to do with the fact that the faster the object is going when you stop, the shorter the duration it takes to travese the impact stopping distance when it strikes an object. In that case a refinement of what I gave above would probably give a more quadratic dependence. However, this is all guesswork, and the real specifics are probably in some auto safety journal.

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A change of momentum $\Delta p$ in a given amount of time $\Delta t$ translates to an applied force $F=\frac{\Delta p}{\Delta t}$.

Assuming that the change in speed puts the moving object at rest (infinite mass of the target) from an initial speed of $90\,km/h$, and assuming that this process takes $0.1\, s$ we can make an estimation of the force applied to the target.

This would give a force of $500\,N$ which would be equivalent to approximately $50\, kg$ of weight.

Probably I'm estimating the collision time too conservative to get the numbers you quote. Working backwards from them, and using the same assumption of non-recoiling target, the collision time amounts to $\Delta t\approx 0.02\, s$ which is one order of magnitude smaller.

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The matter can also be discussed in terms of some known parameters: the velocity $V$, at the moment of collision, and the crush distance, $D$, the distance the passenger compartment has to slow to a stop, as the bumpers collapse, the fenders split off, and the engine submarines. One purpose of seat belts and air-bags is to ensure that the occupants stay with the compartment as it comes to a relatively slow stop.

The relevant formula is:$$v^2=u^2+2ad$$where $v$ and $u$ are velocities, final and initial, $a$ is acceleration and $d$ is distance. In this case, $v=0$, $u=V$, and $d=D$ Substituting and rearranging, we get$$a=-\frac{V^2}{2D}$$So the acceleration of the passengers (and the so-called "weight" of a free object) would scale with the square of the velocity. In the given examples, if a speed of $50 \text{ km/hr}$ leads to a weight of $80 \text{ kg}$, then a speed of $90 \text{ km/hr}$ should lead to a weight of $80\times \frac{9^2}{5^2}$ or $259\text{ kg}$. As it does approximately...

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