We have been discussing whether or not bigger cars are more dangerous to pedestrians. I will reduce it to a simpler question.

Imagine two cars, which are exactly identical in shape, but have different masses. One is 1000kg and the other 2000kg. One person is hit by the light car, and another is hit by the heavy car. The speed at collision is 50kmph, they both weigh 100kg, and they are identical in shape ;)

If you look at the energy itself it suggests that the heavier car will inflict more damage. But I am thinking that what will really matter for the guy's survival is how "fast" he will be accelerated (what g-force his organs must resist). I am not convinced that the 1000kg will make a big difference in how hard he is accelerated.

Could the 1000kg vs. 2000kg in some circumstances be the difference between life and death?

If the answer is that there is only little difference, it suggests that big cars with "soft" hoods or hoods designed to let the pedestrian slide over the car, could easily be less dangerous than much smaller cars with a design that did not prioritize this.

  • 2
    $\begingroup$ What do you like better: Me throwing a ball to your head, or a rock with the same shape and velocity, but ten times as heavy? $\endgroup$
    – Bernhard
    May 25, 2012 at 23:43
  • $\begingroup$ A very soft spongy ball as hard as you want. For details, see resilience on wikipedia. $\endgroup$
    – Apoorv
    May 26, 2012 at 2:50
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    $\begingroup$ @Bernhard. In that example my body is very big compared to the other object. Take it to the other extreme. If I am hit by a 50ton locomotive and you are hit by a 20,000ton freight train, do you think the forensics could tell the difference? $\endgroup$ May 26, 2012 at 22:48

1 Answer 1


Let's break the question down to its bare bones.

Let us define danger first. I think danger falls under following categories:

  1. Any changes to the body's (internal or external) structure.
  2. Mental trauma – I would attribute it to the double derivative of the velocity (i.e., jolt). But this is highly subjective. Some people would attribute it to the impact while others would attribute it to the total energy transferred to the body. This is the part I will leave out from this discussion as it is highly subjective.

So, let's focus on 1.

What can cause changes to a [human] body's structure?

One may think that structure changes with the application of force. Intuitively, larger the force applied, more changes it is likely to produce. Longer the force is applied, more changes it is likely to produce. But in concrete physical terms, structural changes happen when applied energy is sufficient to break the bonds between atoms/molecules making up the structure.

Thus, the bond strength and the amount of energy both matter. Intuitively, a ball of putty or sponge is more likely to deform itself than a hard leather or cork ball. Thus, when you are hit by a sponge ball you take less damage than when you are hit by a hard leather ball of the same mass thrown with the same velocity.

How much energy will be transferred?

In a collision, energy is transferred between the objects that collide. Some of that energy changes the velocity of the objects while rest of it is absorbed by the structure or released as heat. When no energy is absorbed by the objects colliding they are said to have collided elastically. But, as is more likely in a real world scenario, including the car collision case you described, when the final kinetic energies of the two objects do not sum up to the sum of their initial kinetic energies then it follows that some energy will be "lost" in the collision. That energy is not actually lost and is absorbed by the objects colliding. They are said to have collided inelastically in this case. It is the energy that was absorbed by the pedestrian that would contribute to the change in structure of the pedestrian's body.

Now for your actual question – what difference does the mass of the car make? So let us assume that all other factors (mass of the pedestrian, angle of the collision, initial velocities, resilience of the car and the pedestrian, co-efficient of restitution etc.) remain the same. Let us say the mass of the pedestrian is $m$ and the mass of the car is $M$. Let us also say that they collide perfectly inelastically and the initial speed of the car is $v$ (don't care about the direction, that's why I said speed). Then the energy lost is:

$$\frac{(Mv)^2}{2(m + M)}$$

Roughly thus (considering $m\ll M$), the energy lost is directly proportional to the mass of the car. In partially inelastic collisions, the formula becomes a little complicated but the essence of the discussion remains the same. So, everything same, more mass will mean more damage. You can use the formula above to calculate the energy for yourself but doubling the weight of the car will increase the energy lost substantially – and can mean life/death. Doubling the speed will cause even more damage – hence the speed limits in congested areas.

Does the acceleration matter?

You question also asks if the acceleration matters. To answer it directly, yes but not in the same way as the energy transferred. Acceleration by itself won't cause anything more sinister (other than disturbing the fluids in the pedestrian's body) as long as it is uniform. If you experience space variant acceleration (meaning different parts of the body are accelerated differently) then it can cause tearing and that is harmful. The greater the difference, more will be the tearing. Time dependent acceleration (or jolts) can cause a lot of discomfort as the body is stretched and fluids move here and there – that can cause vomiting or trauma or both. Very large accelerations start with a significant jolt (as the initial acceleration is very close to zero).

For the example case you described, the cars are at the same initial speed. The "final" velocity of the pedestrian is

$$\frac{Mv}{m + M}, \quad \text{perfectly inelastic}$$ $$\frac{2Mv}{m + M}, \quad \text{perfectly elastic}$$

Therefore, once again, more massive the car, higher will be the acceleration experienced by the pedestrian assuming collision impact times remain same. But if you can make the impact softer, you can reduce this acceleration (by increasing the time of impact) – see next section.

Other ways to make cars safer

To make cars safer for pedestrians, one should structure the car so that it absorbs most of the energy itself (softer bumpers/hoods). On formula one race tracks you can see several layers of rubber tires that are there to absorb energy when a car collides with them. One can even structure the car in the fashion you described, allowing the pedestrian to slide over the car, thereby reducing the impact of the collision. But you definitely want to avoid an elastic collision where the pedestrian is reflected by the car into something more dangerous nearby (e.g., road, traffic sign pole, railings, a food stall with knives/boiling oil etc.).

  • $\begingroup$ Regarding acceleration: Big car: 2000kg*14m/s / (100kg+2000kg) = 13.33. Small car: 1000kg*14m/s / (100kg+1000kg) = 12.72. That is a very small increase, when considering that the mass is doubled, even smaller than I expected. $\endgroup$ May 26, 2012 at 22:37
  • $\begingroup$ While it does take much force to break a coconut (or skull) it does not take much energy. I imagine the coconut in a vice. It wouldn't take many joules to turn the krank enough to crack the it? I believe that force=acceleration is a more correct measure of harm than transferred energy. $\endgroup$ May 26, 2012 at 23:01
  • $\begingroup$ Thanks for the very well written and elaborate answer, btw. Very cool :) $\endgroup$ May 26, 2012 at 23:02
  • $\begingroup$ No problems! In your first comment, you are assuming that the impact time was 1s --which is not true. It is likely to be of the order of milliseconds. Second, the force itself does not do anything. The force transfers energy from one system to another and that energy must be sufficient to break the bonds (that's what hammers do). $\endgroup$
    – Apoorv
    May 27, 2012 at 3:23
  • $\begingroup$ The ratio is the same no matter impact time, right? $\endgroup$ May 28, 2012 at 1:26

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