Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
Define painful (in physical terms). –  ACuriousMind Aug 14 at 20:50
Do experiment in many persons. Average what they tell you. –  Tim Aug 14 at 20:52
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. –  kalkanistovinko Aug 14 at 20:57
This question appears to be off-topic because it is about pain. –  ACuriousMind Aug 14 at 21:18
@ACuriousMind your comment is rather off-topic. Just a slight modification or interpretation of terminology could render the question rather great. –  kalkanistovinko Aug 14 at 21:23

7 Answers 7

up vote 3 down vote accepted

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.

share|improve this answer
I assume that you assume that the car comes to a stand-still? –  Danu Aug 14 at 21:17
@Danu no, I assume the car has the same final velocity as the person (perf. inelastic) or that energy and momentum are conserved (perf. elastic). –  Kyle Aug 14 at 22:36
Okay, that's another way of interpreting the question :) In fact, it is probably the better way. –  Danu Aug 14 at 22:38
@Danu bah, if you look at the question edit history you'll see why I answered this way. –  Kyle Aug 14 at 22:39
There seems to be nothing there... –  Danu Aug 14 at 22:40

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.

share|improve this answer

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


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.

share|improve this answer

What causes damage (pain) is Energy (.../time, do you agree with Brandon?). Suppose the momentum of the bodies is 1000 kg*m/s if

A has mass 1000 Kg its KE is 500 J

If B has mass 1 kg its KE energy is 500 000 J

Car B is more destructive (painful) than A

share|improve this answer
I could accept pain being energy / time (power) but not just energy. –  Brandon Enright Aug 15 at 6:24
@BrandonEnright, so, what is the concrete difference? –  bobie Aug 15 at 6:27
The difference is falling off a cliff with or without a parachute. Roughly same energy released but vastly different outcomes. –  Brandon Enright Aug 15 at 6:29
@BrandonEnright, false,: with a parachute when you hit the ground KE is mch less than without one –  bobie Aug 15 at 6:31
As much as your sentence is true for the potential (with or without a parachute), what counts in the final analysis here is KE. And it is the KE at the moment of impact. Everything before is irrelevant. So if I get hit with a stone (that fell of a cliff) of mass m at speed v, I do not care if it was accelerating all the way down from the height of 50 meters or if it was released from 100 m, and was then somehow slowed down on the way, if the final KE of the two stones is the same. And all the pain (damage) depends only on the final v and m, nothing else. History (initial PE) does not count –  bright magus Nov 13 at 13:08

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...

share|improve this answer
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. –  Floris Aug 16 at 21:01
Let E = kinetic energy 
P = momentum 
Pr = pressure
A = contact area
E(a) = 1/2 m*v^2 => E(a) = 1/2 P * V(a)
E(b) = 1/2 m*v^2 => E(b) = 1/2 P * V(b)

if V(a) < V(b) => E(a) < E(b)

Pr = F/A = (m*a) /A = (m * v/t)/A = (p/t)/A =>
Pr = P / ( A * t)

If we assume that both vehicles have the same elasticity the time t (contact time) is the same ??? so :

Pr(a) = Pr(b)

But if we assume that the vehicles tranfer all their kinetic energy to the person, the one with the higher velocity inflicts more damage because it transfers more energy.

In reality the problem is more complex because the contact time is not the same for material with the same elasticity but with different mass. Its the same if we assume that the cars hitting the wall are not damaged at all and all their kinetic energy is transfered to the human body and the wall

share|improve this answer
Why would you make that assumption? There are exceptions. And how many wooden cars do you know of? Why would the material be different? –  HDE 226868 Aug 14 at 21:02
@HDE 226868 i corrected my answer. I wanted to illustrate that the elasticity of the material is a defining factor i.e wood,plastic,iron,carbon etc –  firephil Aug 16 at 10:53
Inelastic collision, same object in front of immovable wall. Elasticity of object hitting doesn't come into it. –  Floris Aug 17 at 11:46

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.

share|improve this answer

protected by Qmechanic Aug 15 at 10:44

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.