# Homework Help: Struck in head by rock

1. Aug 24, 2004

### pneumoped

This past Sunday I was rock climbing in Eldorado Canyon, Colorado. Another climber was on a route just left of me and well above me. When he climbed up onto the belay ledge he accidentally knocked loose several rocks. One rock hit me in the head. Fortunately, I was wearing my helmet and I was fine. The rock fell about 25 feet, it was about the size of a golf ball, and it was made of granite.

I want to express the impact that the falling rock had on my head. As a civil engineer in a field where I do not use much of my physics I had to spend some time refreshing myself with the basic equations.

Using the specific gravity of granite at about 2.6, I calculated its density at 162.24 lb/ft3 and then estimating the diameter of the rock at 1.25 inches, I calculated the mass of the rock to be about 44 grams.

I know that Force = mass * acceleration. And I know that acceleration due to gravity is 9.81 meters per second squared. So I calculated the force acting on the rock to be about 0.43 Newtons.

Based on anecdotal evidence and experience, I know that a 44 gram rock dropped on my head from a height of 1 inch is going to hurt a lot less than the same rock that fell 25 feet and hit me in the head. However, the force exerted on the rock is the same – 0.43 Newtons. And certainly the force exerted upon my head is different depending on how far the rock falls.

How do I express the force exerted upon my head? Certainly the velocity, mass, and distance are the key components in how hard the rock hits me.

Chris

2. Aug 24, 2004

### arildno

Welcome to PF!
When the rock falls onto your head, both your head and the rock needs to equalize their velocities in an extremely short period of time (if not, the rock would pass through your skull&brain and beyond..nature isn't THAT sadistic!)

If the impact velocity (actually, momentum) of the rock is greater, but the impact period remains roughly the same, clearly the forces generated (between your head and the rock) to create the needed accelerations are correspondingly larger.

3. Aug 26, 2004

### pneumoped

arildno,

So, what you are saying is that in order to calculate the force that the rock impacts with, I need to first, determine the velocity that it is traveling when it hits me, then make an assumption as to how quickly (time) it takes to decelerate. Then using the acceleration formula [ a = ( (vf - vi)/t^2) ], I can then plug that into the force formula to get the impact force.

For example:

The rock falls for 25 feet or 7.62 meters.

The time it takes to fall that far is:
t^2 = d/g = (7.62)m / (9.81)m/s^2
t = (0.881) s

Then, the velocity of the rock is:
v = a * t
v = 9.81 * 0.881 = (8.65) m/s

Then, let’s assume that my head decelerates the rock to 0.0 velocity in 0.05 seconds. The acceleration of the rock is:

a = ( (vf - vi)/t^2)
a = (0.0 – 8.65) / 0.05^2
a = - 3458.4 m/s^2

Then the force exerted by the rock on my helmet would be:
F = m * a
F = (0.04355)Kg * (-3458.4)m/s^2
F = - 151 Newtons = - 33.9 Lbf

Am I on the right track?

4. Aug 26, 2004

### arildno

You are on the right track; however, the total force on your head (the negative of the total force on the rock) is to be found using the ENTIRE time collision lasts.

Let's simplify the issue by as follows:
2. Instead of finding some "mean velocity" that both objects should attain, let's just say that at some time after the hit, the rock and your head has zero both velocity
(that is, we regard your helmet/head as sufficiently massive/undeformable that all changes is experienced by the rock)
3. The collision is by no means over; the rock must bounce off again!
That means, it must experience a force which develops a separation velocity between the roch and your head (this in contrast to getting bird droppings on your head)
4. For simplicity, assume an elastic collision, so that at the end of the collision, the rock has a velocity upwards equal in magnitude to its initial, downwards velocity.

5. The momentum change for the rock (calculated between impact and release, that is, for the whole collision period) is:
$$\begtriangle{I}=m_{rock}v_{0}-m_{rock}(-v_{0})=2m_{rock}v_{0}$$
where $$v_{0}$$ is the magnitude of initial velocity.

6.
The average force in the collision time is therefore:
$$F_{av}=\frac{2m_{rock}v_{0}}{t_{coll}}$$
where $$t_{coll}$$ is the length of the collision period.

This ought to double your estimate, if I understood you correctly.