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Suppose they have the same contact area with you.

The term "painful" is a bit ambiguous. Let me ask in another way:

Two cars with same momentum but different masses are going to hit a wall. You must stand right in front of the wall. Which car will you choose to be hit by?

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    $\begingroup$ Define painful (in physical terms). $\endgroup$ – ACuriousMind Aug 14 '14 at 20:50
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    $\begingroup$ Do experiment in many persons. Average what they tell you. $\endgroup$ – Tim Aug 14 '14 at 20:52
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    $\begingroup$ Imagine the same momentum carried by a bullet in one case, and by a multi-ton car in another. At least in this scenario and particularly for relatively low momentum values, the answer is conspicuous. $\endgroup$ – kalkanistovinko Aug 14 '14 at 20:57
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    $\begingroup$ This question appears to be off-topic because it is about pain. $\endgroup$ – ACuriousMind Aug 14 '14 at 21:18
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    $\begingroup$ @ACuriousMind your comment is rather off-topic. Just a slight modification or interpretation of terminology could render the question rather great. $\endgroup$ – kalkanistovinko Aug 14 '14 at 21:23
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This is an answer to the question version 1. Later versions invalidate the details of this answer, but some of the ideas are still valid. Will edit to current version if I get a chance.

I'll define "pain" as the change in momentum, or the energy delivered (the two are related by your velocity after the impact, provided your mass is unchanged, so unless you lose a leg or something...) of the person.

Assuming you're standing still, in the perfectly inelastic scenario, you stick to the vehicle and your final momentum is ($p$ is momentum):

$$p_{\rm person}=\frac{m_{\rm person}p_{0,\rm car}}{m_{\rm car}+m_{\rm person}}$$

and in the perfectly elastic case:

$$p_{\rm person}=\frac{2m_{\rm person}p_{0,\rm car}}{m_{\rm car}+m_{\rm person}}$$

If the momentum of the vehicle is held fixed, the mass only effects the denominator, so the momentum transferred to you will be smaller for a larger vehicle mass in both cases. This may seem counter-intuitive at first, but remember the higher mass vehicle has a lower velocity as a lower mass vehicle at the same momentum.

The energy transfer in the perfectly inelastic case is:

$$E_{\rm person}=m_{\rm person}\left(\frac{p_{0,\rm car}}{m_{\rm car}+m_{\rm person}}\right)^2$$

and similarly for the perfectly elastic case:

$$E_{\rm person}=m_{\rm person}\left(\frac{2p_{0,\rm car}}{m_{\rm car}+m_{\rm person}}\right)^2$$

Again, smaller energy transfer for higher mass vehicle at fixed momentum.

This gets more complicated if you allow the person to react (run away?) and depending on the particular values of the speeds at collision and the masses, either case may come out less "painful".

This also all assumes that the change in energy or momentum occurs instantaneously - a lower impact speed could conceivably result in a collision of longer duration, spreading out the momentum/energy delivery and potentially hurting less.

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  • $\begingroup$ Okay, that's another way of interpreting the question :) In fact, it is probably the better way. $\endgroup$ – Danu Aug 14 '14 at 22:38
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Regardless of the physical undefinability of "painfulness", I'd like to plug some numbers in a particular scenario:

Let's have a momentum of $p = 1000 $m$\cdot$kg/s,

A 0.25kg bullet would be fatal, moving at $v = 1000/0.25 = 4000$m/s,

while a 2000kg car moves at $v=0.5$m/s,

So at least in this scenario and particularly for relatively low momentum values, the answer is conspicuous.

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Let us first calculate how much energy carried by an object of mass $m$ and momentum $p$. So the velocity is $v=p/m$, and therefore, the kinetic energy is

$T=mv^2/2=p^2/2m$

Therefore, if $p$ is a constant, the heavier the object is, the less the kinetic energy $T$ it carries. This means that a heavier object hits you and you receive less energy, which probably implies that it hurts you less than a lighter object.

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Which scenario to chose?

A) I am standing with my back against a massive granite wall. A solid block of concrete of mass 100,000 kg is approaching me with a momentum of 10,000 kg m/s. It follows that the block moves at a speed of 0.1 m/s with a kinetic energy of 500 J. I stretch out my arms and when the block reaches my hands, I push with a force of 500 N. The block manages to move one more meter towards me and comes to rest an inch away from my nose. I walk away unscratched...

B) I am standing with my back against a massive granite wall. A hollow block of concrete of mass 1,000 kg is approaching me with a momentum of 10,000 kg m/s. It follows that the block moves at a speed of 10 m/s with kinetic energy of 50,000 J. I stretch out my arms and when the block reaches my hands, I push with a force of 500 N. This is to no avail. When a fraction of a second later the block reaches my nose it's kinetic energy is no less than 49,500 J. This is going to hurt...

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  • $\begingroup$ This is the right analysis. The force times distance needed to absorb the kinetic energy of the object ($p^2/2m$) tells us to chose the heavier object. $\endgroup$ – Floris Aug 16 '14 at 21:01
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In a general sense, the amount of "pain" that someone feels is a result of the pressure exerted on them (like a bed of nails vs. a single nail).

Momentum is defined as $p=mv$, and pressure is defined as $P=F/A$

Because $F=ma$ (Newton's 2nd law), and $a=\frac{\Delta v} {\Delta t}$, we can say that $P=\frac{m \Delta v} {A \Delta t}$.

Assuming that you absorb the entire energy of the car, and it stops after hitting you, then the top half of that fraction ($m \Delta v$) is equal to the momentum, which is constant. Also, because surface area of contact is constant, $A$ is constant.

Therefore, the equation for pressure becomes $P=\frac{k} {\Delta t}$, where k is some constant.

Now, what is $\Delta t$? This is the time that the car is hitting you for. Assuming that humans are a constant thickness, $\Delta t = Some Distance/v$. Therefore, $P = k*v$ so the faster car is the more painful one.

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protected by Qmechanic Aug 15 '14 at 10:44

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