If force depends only on mass and acceleration, how come faster objects deal more damage?

Question

As we know from Newton's law, we have that $\mathbf{F} = m\cdot\mathbf{a}$. This means that as long as the mass stays constant, force depends solely on acceleration. But how does this agree with what we can observe in our day-to-day lives?

If I drop a coin on someone's head with my hand standing just a couple centimeters above their hair, they won't be bothered too much; but if I drop the same coin from the rooftop of a skyscraper, then it could cause very serious damage or even split their head open. And yet acceleration is pretty much constant near the surface of the earth, right? And even if we don't consider it to be constant, it definitely has the same value at $\sim1.7\text{ m}$ from the ground (where it hits the person's head) regardless of whether the motion of the coin started from $\sim1.72\text{ m}$ or from $\sim1 \text{ km}$.

So what gives? Am I somehow missing something about the true meaning of Newton's law?

Answer 1

The acceleration on the way down is the same. The acceleration when it strikes the person's head is different.

Both coins have to be stopped by the skull (we hope). The coin that is going 2m/s will not require as much acceleration to stop in the same distance as a coin that is going 10m/s. That slower acceleration will require less force.

After the coin is stopped, it will still produce a force equal to its weight. But that's only part of the force involved.

Answer 2

Yes, you are missing something! The acceleration that is relevant when the coin strikes your unfortunate victim's head is not the acceleration the coin experienced beforehand (which as you correctly point out is the constant acceleration due to gravity) but the acceleration the coin experiences as a result of the impact. Clearly a coin travelling at very high speed has to experience enormous acceleration if it is to be brought to a stop by an impact. And that is the best case- if the coin bounces elastically off the head in question, then its velocity would be reversed, which would be twice the acceleration!

Answer 3

If I drop a coin on someone's head with my hand standing just a couple centimeters above their hair, they won't be bothered too much; but if I drop the same coin from the rooftop of a skyscraper, then it could cause very serious damage or even split their head open

The damage is caused by the average impact force during the collision, not the force of gravity $mg$. The work energy theorem states that the net work done an object equals its change in kinetic energy. So if your coin is brought to a stop when contacting the head,

$$W_{net}=F_{average}d=-\frac{1}{2}mv^{2}$$

Where $F_{average}$ equals the average impact force on the head, $m$ is the mass of the coin, $v$ is its velocity upon impact with the head, and $d$ is the penetration distance into the head, assumed to be much less than the height from which the coin drops.

Note that the greater the dropping height $h$, all other things being equal, the greater the velocity and kinetic energy of the coin upon impact, since, neglecting air resistance, $v=\sqrt {2gh}$, and thus the greater the average impact force.

Note however that the actual impact velocity will be limited to the terminal velocity due to air resistance.

Hope this helps.

Answer 4

As you know, $\mathbf F=m \mathbf a$ is relevant to the falling of the coin before it hits the head. The force acting on the coin is indeed roughly constant, and so is its acceleration. But the longer the gravitational acceleration goes on the faster the coin will be moving.

When the coin collides with the head it experiences a rapid change in velocity, as it loses its downward velocity and may even acquire an upward velocity (if it bounces). Its acceleration (rate of change in velocity) will be large (depending on the time for which it has been falling) and so will be the force it experiences. The force that gives rise to this (upwards) acceleration comes from the head. The coin will exert an equal downward force on the head.

We have just applied $\mathbf F=m \mathbf a$ to the collision between coin and head. Note that this is the second time we've called on $\mathbf F=m \mathbf a$. It's important to distinguish between the gravitational acceleration of the coin while falling, and the much larger acceleration (you might prefer to call it deceleration) when it hits the head!

Answer 5

Assuming no air resistance, the coin accelerates from zero to the final impact velocity over the period of the fall. When it hits someone, the coin decelerates from the impact velocity back to zero in a fraction of a second. You have an identical change in velocity in a much shorter time, requiring a much larger force/acceleration. If the coin has gained more speed through a longer fall, it must experience a greater force to stop it, assuming that the impact is always a similar "short" duration.

It's not the fall that kills, it's the sudden stop at the end.

Answer 6

You are considering the acceleration during the free fall. But the damage does not happen during the free fall. It happens during the sudden stop at impact.

It is the acceleration (deceleration) during the impact that you should consider, not the acceleration during free-fall. When the coin hits the head it is swiftly slowed down - this requires a huge (negative) acceleration.

The head must provide the force that causes this (negative) acceleration. If it can't do that, then the coin isn't slowed down enough to fully stop and will thus penetrate the tissue and cause damage.

Answer 7

We know that $$F = ma$$

We also know that $$a = \frac{dv}{dt}$$

So the formula for force becomes $$ F = m\frac{dv}{dt}$$

Assuming mass is constant, we can just say $$ F = \frac{d(mv)}{dt}$$

Since $$momentum = mv$$

We can see that force is the rate of change of momentum. This is actually the original definition of the Second law of motion. A fast moving object has a high momentum and when it hits an object and stops in a relatively short time, the rate of change of momentum is very high meaning the force is very high.

Answer 8

Let's tackle this intuitively. More damage means more work needed to cause that damage.

The damaging work made by the coin is equal to its kinetic energy at the time of the collision (assuming an inelastic collision, with the coin coming to a complete halt), which is absorbed by the skull.

Therefore what is important in assessing the damage is the final velocity of the coin, which depends on how long it has been falling since it keeps on accelerating, and so it keeps on acquiring kinetic energy (until its limit velocity if you want to consider the atmospheric drag).

That's why a short fall is less damaging than a long fall.

Answer 9

Perhaps a numerical order of magnitude calculation will help?

Suppose the coin has a mass of $5\,\rm g$ and is dropped from a height of $2\,\rm cm$ onto a surface which deforms by $1\,\rm mm$ stopping the coin.

In falling the coin loses potential energy $mgh = (5\times 10^{-3}) \times 10 \times (2\times 10^{-2}) = 10^{-3}\,\rm J $ and this is the kinetic energy of the coin just before impact.
If after hitting the object the coin deforms the object by $1\,\rm mm$ whilst the object is applying a force of $F$ during the time the coin slows down and finally stops.
Equating the work done by the stopping force $F \times 10^{-4} = 10^{-3} \Rightarrow F= 1\,\rm N$.

Repeating the calculation but now dropping the coin $2\,\rm m$ and assuming that the stopping force has the same magnitude as before results in a deformation of $10\,\rm cm$.

Points to note.
a The acceleration whilst the coin is falling is the same $(\approx 10\,\rm m\,s^{-2})$.

b Just for illustration I have assumed that the force slowing the coin down is constant and in both case the acceleration is given by $F=ma \Rightarrow 1 = 5\times 10^{-3}\, a \Rightarrow a = 200\,\rm m\,s^{-2}$

c What the extra height whilst falling does is to increase the kinetic energy of the coin which in turn means that the coin will deform (damage) the object it hits more during the time it is slowing down.

d I accept that the retarding force will not be constant when it comes to the real life example of the coin hitting the head as there are different layers, skin and skull bone, involved but what I have tried to do is to show that although the acceleration whilst slowing down is important there is also the very important factor of dissipating the kinetic energy of the coin.

Answer 10

The force we care about is not the force of gravity, but the force between the coin and the persons head.

If the coin and the persons head were both rigid bodies, then the force involved would be infinite. That is clearly unrealistic though.

In reality neither the coin nor the persons head is able to remain rigid in the face of an infinite force, so one or both objects will start to deform, (in this case mostly the person's head) this will limit the force involved and hence limit the deceleration rate.

The faster the projectile is moving, the further it will travel between the initial impact and deceleration to a stop and the more deformation it will cause to the victim.

Generally materials respond differently to different levels of deformation. For small deformations they are likely to respond elastically and spring back to their previous state afterwards with no damage. Larger deformations though, are likely to result in plastic deformation, or outright breaking of the material.

Answer 11

2 main reasons.

Intuitively, the 2 objects e.g exert an electric force on eachother, due to its charge

For higher velocities, the charge 1 is going to get CLOSER to the charge 2 before being repelled away, meaning it exerts a higher force causing more work to be done on charge 2.

And also for higher relative velocities the charge 1 is also going to spend MORE time in close vicinity of the charge 2 meaning more work is being done.

Both of these make the object accelerate more for higher relative velocities.

Conservation of momentum can also show this

Answer 12

Force on the victims head is rate of change of momentum. i.e. Force=rate of change of momentum. If the time is tiny then the rate of change is huge. This is why crumple zones in cars work, they increase the time over which the passengers' momentum changes.

Answer 13

the equation for your example (Newton second law) is:

$$m\,\ddot h=m\,g$$

thus the acceleration $\ddot h~$ is the same but

with the solution

$$h=\frac{g\,t^2}{2}\quad \Rightarrow\\ v=g\,t$$ eliminate the time t you obtain that

$$v(h)=\sqrt{2\,g\,h}$$

thus the impact $~m\,v~$ is depending on the height $~h~$ from where you drop the coin

Answer 14

F=ma applies on all altitudes. The coin wll experience the same force on every height. It's the velocity the coin will get when it hits the head that counts and not the force pulling it down. You can walk around with a coin on your head without danger.

When the coin hits you with high velocity, there is an extra force you feel. In a short moment the coin decelerates because your skull exerts a force on it. The change in velocity, the time in which this happen, and the mass of the coin determine this force. If the coin is stopped in 0.0001 seconds, its velocity is 50 meters/second, and its mass is 1 gram, the force is 0.001x50/0.0001=500 Newton (a 50 kilogram coin). On a scale, the coin weighs 0.001 kilograms. The force is then 0.01 Newton.

So your skull has to give counter force to 50 kilogram effectively. Of course this is different from placing a 50 kilogram coin on your head. You would fall to the ground. It's a short impact force, but enough to cause damage. Just imagine I threw a 50 kilogram coin down.

By the way, in reality a coin can't do much damage. I have seen a coin thrown from a high hotel balcony. Near the pool down on the floor a man jumped up in full amazement...