# Does hitting an obstacle hard enough to break it actually reduce self harm?

Lets say that Car-A hits barrier that is fixed to immovable posts on either side of the road. The car's momentum is such that the barrier absorbs the impact and the car comes to a near instantaneous halt and incurs some degree of smash damage.

An identical Car-B hits an identical barrier, but its momentum is such that the barrier snaps and hinges open like a gate, allowing the car to safely roll to a stop. Obviously, Car-B also incurs some degree of smash damage.

Assuming the difference in speed was about 20%, which car would you expect to incur more smash damage?

Also, given that different materials fail in different ways (snapping, shattering, warping, etc.) Does the nature of the barrier matter? Does the nature of the vehicle matter?

Can you reduce the damage to your car by pushing the accelerator instead of the brakes?

Can Superman cause MORE harm to his nemesis by throwing him more gently into a wall?

Can a stunt actor get an unexpected injury because he didn't collide with the window hard enough to punch through?

• I'm not sure about your definition of self-harm, but mine is about my body and not my car. A similar situation you might consider is hitting a wooden board with your fist and not breaking it versus hitting it just a bit harder and breaking it. Or hitting a post with a wooden bat and having it not break versus hitting it just a bit harder and having it break. But between being just short of it breaking versus just over, I think I would rather have it break. Sep 17 at 3:44
• Any action which uses energy (e.g. breaking things that aren't me) leaves less energy available to potentially injure me . Your two boldface questions at the end are rather ridiculuous. Sep 17 at 13:04
• This is a case of out of the firing pan and into the fire. A barrier protects drivers from something. The car gets through barrier with enough momentum to go over cliff! Sep 17 at 13:43

The energy required to break an object will be independent of the speed of the thing hitting it. This is known as the toughness of a material. That energy is what the striking object will need to impart and this will have the same impact energy imparted to itself. The way this is used in practice on the roadways is by having wooden posts at the end of railings where a vehicle might hit it lengthwise. (Note that some of these designs have been found to be quite faulty as the metal rail itself can become a jousting pole into the car.) Each wooden pole is relatively easy to break, giving up its energy in smaller doses, versus one big I-beam that would decelerate a vehicle suddenly. Water-filled barrels are used to the same effect.

What the idea of energy needed to break something is that, NO, you can't hit something softer by hitting it harder. And if you strike a barrier and continue through it, expect to hit something on the other side. Dumping energy into your brakes will always be a better idea than accelerating.

I will add that there are cases where breakage is better. These are cases where the object is meant to break. Karate boards come to mind. These are specific types of wood that break when you hit them right. This would not work karate-chopping a piece of dimensional lumber meant for home construction, and certainly not for the objects designed to be hit by cars.

• As a side note, the karate kid is no superman, he just knew how to exert force on a board with parallel wood grains. The holders are quite important in holding the board steady in place while it is impacted.
– r13
Sep 17 at 20:22

These scenarios are too simplified to be able to form an opinion.

But generally "unyielding barrier" will cause more damage.

Let's say the yielding barrier never breaks, only bends and moves one meter.

assume these factors just for the sake of a perspective.

• car mass M kg

• Car speed at crash V

• d, distance car moves to a full stop 2 meters for 1st case(total pancaked front), 3 meters for 2nd.

$$F_{average}d= -\frac{1}{2}MV^2, \quad F_{average}=\frac{-\frac{1}{2}MV^2}{d}$$

Just by comparing

$$V2/V1=1.2^2=1.44\ with\ d2/d1=3/2=1.5,$$ we see any deflection larger than one meter of the yielding barrier will be less force to the car.

This is not even considering the deceleration which is greater in the first case.

• As to the part of the question hitting the subject a bit harder:

That is the game plan in martial arts or karate maneuver of breaking planks or concrete blocks safely by hitting fast. Obviously, they have practiced and learned from their master what to hit and what not to.

I would not try it on a car, even though I myself in sudden emerging black ice road conditions get very close to the car in front of me to minimize the potential impact.

# Edit

I thought a short review of "hitting a bit harder to break an object could inflict less self-harm!"

Questin of impulse and hitting an object with the aim of destruction has been for centuries the subject of weaponry and armory (Andulosian swords) and using penetrating shells and resisting shields for eons. The War industry does a lot of experiments and research on alloys and speeds of shells but they keep their data well secret.

But theoretically at least in martial art, if one has a trained hand with calluses, with a strengthened skeletal-muscle system and they have the right material which will eventually break as opposed to deform. It is plausible to think if you just hit the target a bit harder to break it your hand will go through with carrying on the motion after a glance of shock and won't have to eat back all its impulse and break your bones.

Basically hitting hard and collision in the real world is a messy physics, half plastic, some elastic, a lot of heat, and in intense collisions cratering and a cascade of molten metal radiating out.

• Your statement about martial arts is incorrect. The fast-hit is done to maximize the impulse force applied, not to "magically" reduce the potential damage to your hand. Sep 17 at 13:06
• I can speak from experience that failing to break wooden boards with your hand hurts way more than succeeding. Also I never trained my hand to grow calluses. Grading night was the first time we had the chance to try it. Sep 20 at 9:53
• @LorryLaurencemcLarry and that's because a board that doesn't break returns most ofthe energy into your hand Sep 20 at 12:44

For simplicity, let's hold the speed constant, the rigidity of the walls variable. The sketch below conceptually depicts two scenarios - the car runs into a non-yielding (rigid) wall and then runs into a defective yielding wall.

Car A suffers damage due to the full impact force. The impact of car B will be absorbed partially by the crushed wall, which forces car B to decelerate, so the damage will be less severe if all other things being equal.

Approximate Comparison of Deformations:

Let $$\Delta x$$ be the deformation of the car at the collision. For case 1:

$$ma\Delta x_1 = m\dfrac{V_f^2 - V_i^2}{2}$$, $$V_f = 0$$

$$\Delta x_1 = -\dfrac{V_i^2}{2a}$$

For case 2: Assume $$V_f = 0.5V_i$$

$$ma\Delta x_2 = m\dfrac{0.25V_i^2 - V_i^2}{2} = -0.75m\dfrac{V_i^2}{2}$$

$$\Delta x_2 = -0.75\dfrac{V_i^2}{2a} < \Delta x_1$$

• all other things are NOT equal. As stated in the OP, Car-B is traveling 20% faster. Sep 17 at 2:26
• @LorryLaurencemcLarry That's why I said "if", otherwise how do you compare? Keep in mind that the relative rigidity of the walls is also unknown.
– r13
Sep 17 at 2:36
• The real question is whether or not the force imparted on the car by a failing barrier is always exactly equal to the maximum force the barrier can withstand without failure. If so, then then math becomes much more simple, and the answer to the OP is "NO". But IS this assumption actually a fact? Sep 17 at 3:30
• No damage is not directly due to ma . The damage depends heavily on the rate of energy transfer, i.e. impulse vs. slow pressure. Sep 17 at 13:07
• @CarlWitthoft I disagree. Before "transfer" the energy, there needs to be a force to induce the energy, which is ma. Both the wall and the car feels the force ma at the instant the two subjects get in touch, for infinitely rigid non-yielding wall, the result is simple.
– r13
Sep 17 at 16:22