Dropped off of an airplane, into a wall.

[DavidC]

Ok, the next question is a little funnier. I was curious what scenario would create more impact force.

A) An airplane drops an object from 500 feet, which impacts into the ground (assume just solid, level granite).

B) An airplane drops an object from 500 feet, which falls 200 feet before hitting a wall (solid granite). The object then falls the remaining 300 feet to the ground and hits the ground.

I'm just curious if the total force hitting this object would be the same or different in A and B.

[Fels krone]

The 500 foot drop would be more. The kinetic energy of the object is defined by velocity squared, so the higher velocity would be much more impact force. It could be limited by its terminal velocity, but if that velocity was more than 150 mph it wouldnt have time to reach it.

[Sabrazack]

I believe acceleration slows as falling speed increases. This might have an effect here? But then again, maybe not on such a short fall.

[Matt R.]

500 ft = 152.4m

From a height of 152.4m, with initial velocity of zero, the object has final velocity
vf = sqrt(2g*152.4) = 54.7 m/s
F = dp/dt = m dv/dt
Assuming the object decelerates at a constant rate upon impact,
F = (54.7 m/s)*m/dt
where m is mass of object, and dt is amount of time it takes to stop the object (pretend that says delta t).

From 300 ft = 91.44 m
vf = sqrt(2g*91.44) = 42.3 m/s

And from 200 ft = 60.96 m
vf = 34.6 m/s

F(total) = 42.3*m/dt + 34.6*m/dt = 76.9*m/dt

76.9 > 54.7, so the total force impacting the object that "stops" once at 200 ft is greater than the total force impacting the object falling the full 500 ft without stopping. This is assuming the amount of time to decelerate the object is the same each time (which wouldn't quite be true, but I think this is a good estimate).

[Silent A]

You have to fall about a kilometer or so to reach terminal velocity so 500 feet is not enough.

While it's true that the energy is proportional to the velocity squared, the impact force is proportional to the square root of the kinetic energy (assuming the ground acts like a spring), so the impact force is proportional to the speed.

Unfortunately, at these heights, air resistence is enough to significantly slow down the object but not enough to reach terminal velocity so it's impossible to calculate the impact velocity accurately without doing some calculus (which I don't have time for).

But if you ignore air resistence you'll find that in (b) the two impact forces together will be greater than the single impact force in (a), and if you think about the physics a little you'll realize that air resistence will make this difference even bigger because it will have a bigger effect for a 500 ft drop than on a 200 or 300 ft drop.

So more total force in case (b).

However, adding forces in this way doesn't have any physical meaning. You can only add forces when they occur at the same time.

[Silent A]

[ QUOTE ]
F(total) = 42.3*m/dt + 34.6*m/dt = 76.9*m/dt

76.9 > 54.7, so the total force impacting the object that "stops" once at 200 ft is greater than the total force impacting the object falling the full 500 ft without stopping. This is assuming the amount of time to decelerate the object is the same each time (which wouldn't quite be true, but I think this is a good estimate).

[/ QUOTE ]

dt for 500ft >= dt for 300 ft >= dt for 200 ft

If you think about it, this would only make the difference even greater.

But again, adding the forces in case (b) has no physical meaning.

[Matt R.]

Yep, I initially thought the sum of the forces in (b) would be equal to the force from the "full drop", and after seeing they weren't equal I thought the differences in dt would account for this. But then I realized it only makes the difference greater.

Also, yeah, I agree summing the forces in (b) doesn't tell you much about the physics except for the total force acting on the object over the two impacts. Just answering the OP's hypothetical.

[Silent A]

[ QUOTE ]
Also, yeah, I agree summing the forces in (b) doesn't tell you much about the physics except for the total force acting on the object over the two impacts. Just answering the OP's hypothetical.

[/ QUOTE ]

I expected as much. I only mentioned it for emphasis and for the benefit of others.

[DavidC]

[ QUOTE ]
[ QUOTE ]
F(total) = 42.3*m/dt + 34.6*m/dt = 76.9*m/dt

76.9 > 54.7, so the total force impacting the object that "stops" once at 200 ft is greater than the total force impacting the object falling the full 500 ft without stopping. This is assuming the amount of time to decelerate the object is the same each time (which wouldn't quite be true, but I think this is a good estimate).

[/ QUOTE ]

dt for 500ft >= dt for 300 ft >= dt for 200 ft

If you think about it, this would only make the difference even greater.

But again, adding the forces in case (b) has no physical meaning.

[/ QUOTE ]

Ok, one thing that I'm trying to reconcile is your last point against stress testing.

i.e. furniture doors slammed by a machine til breaking after thousands of reps

i.e. circuit boards going hot to cold to hot, etc. dozens of times before breaking

hockey sticks being slapshotted by a machine til breaking, etc.

While I agree that the two forces being separate has no bearing on total force, over time, wouldn't it be possible that with certain hard objects, the two-impact scenario would damage the object more than a single impact (even with less force at a given time).

I realize that this wasn't my initial question, and thanks for pointing that out.

[Silent A]

[ QUOTE ]
While I agree that the two forces being separate has no bearing on total force, over time, wouldn't it be possible that with certain hard objects, the two-impact scenario would damage the object more than a single impact (even with less force at a given time).

[/ QUOTE ]

Simple answer, no.

What you're talking about here is a phenomenon called "hysteresis". It's kind of hard to explain if you don't know the physics involved, but I'll try.

First, when an object is stressed it can deform in one of two ways: elastic and plastic deformation. For elastic deformation, when the stress is removed the object returns to its original shape (think of an elastic band). However, if you continue to apply stress there comes a point where it starts to deform plastically. In plastic deformation, at least some of the deformation becomes permanent (think of wet clay).

If an object is stressed and then unstessed many times in the elastic range it deforms and returns to its original shape. If you were to plot the deformation vs. stress you'd follow a straight line going back and forth.

In plastic deformation cycles, instead of following a straight line a plot of deformation vs. stress follows a loop (called a hysteresis loop). The area of this loop is proportional to energy being used to break down internal bonds in the atomic structure. Each cycle thus weakens the material a little more until eventually it becomes prone to sudden catastophic failure.

Now, strictly speaking, the elastic scenario above is ideal behaviour. All real materials will exhibit at least some hysteresis, although it may be negligable after even hundreds of cycles. However, eventually all materials will fail. They'll fail much quicker if you can force them into the plastic zone.

Imagine what happens when you deform a paper clip (a fairly elastic material). Initially, it will return to its original shape (elastic deformation) but if you deform it enough, it won't (plastic). If you only slightly stress the paper clip you'd have to repeat it many thousands of times before it broke. But if you deform it into the plastic zone it only takes 5 cycles (or so).

In the end, if you carefully analyze the physics involved, one big force is always worse than two significantly smaller ones. If it wasn't, falling down from a height of one foot 500 times would be worse than falling down from 500 feet once. Obviously, it's not.