3
$\begingroup$

I am considering free falling objects that eventually hit the ground. The objects have various masses and various final (ground-impact) velocities.

What should I measure the "strength of impact" with? Momentum $p = mv_{\mathrm f}$ or perhaps kinetical energy $E_{\mathrm k} = \frac{mv^2}{2}$? Or something else?

Basically, consider two cases:

a). an apple of mass $0.1\ \mathrm{kg}$ and ground-impact velocity $v_{\mathrm f} = 200\ \mathrm{\frac{m}{s}}$

b). a box of mass $2\ \mathrm{kg}$ and ground-impact velocity $v_{\mathrm f} = 30\ \mathrm{\frac{m}{s}}$

Hence, at the time both objects hit the ground:

$p_\text{apple} < p_\text{box} \quad \land \quad E_{\mathrm k,\text{apple}} > E_{\mathrm k,\text{box}}$

Basically, my question is, which object landed "more safely"? Or: is it better to be an apple or the box?

$\endgroup$
  • 1
    $\begingroup$ Let's consider to object (a) a rubber ball and (b) a glass ball. Both have the same mass $m=1kg$ and both hit a wall with velocity $v=10m/s$. Do you think the two objects behave the same? $\endgroup$ – Semoi Jun 22 at 13:10
  • $\begingroup$ What if I am considering two very same objects, but with different final velocities. Should I measure the "strength" of impact with momentum or kinetical energy? $\endgroup$ – weno Jun 22 at 13:16
  • $\begingroup$ My point is, that there are material parameters, which have to be taken into account. First one should understand the physical reason for the damage to happen and then one is able to deduce, which physical quantity is most important. For example, most solid materials act like 3D springs. Thus their deformation is due to strain -- which is neither momentum nor energy. $\endgroup$ – Semoi Jun 22 at 13:34
  • $\begingroup$ I am calculating a task where a man is standing on top of a nearly-vertical standing ladder, the ladder starts rotating around its base falling down to the ground, and the man has a choice: jump off immedietely (free fall) or jump off just before the ladder hits the ground (I calculated the velocity of such scenario). Problem is how can I tell "what is better". By comparing momentums of both cases, or kinetical energies, or? $\endgroup$ – weno Jun 22 at 14:14
  • $\begingroup$ For your problem, the question you ask is irrelevant. As the mass of the man is the same in both cases, the higher velocity will correspond both to higher KE and higher momentum. So lower velocity will be safer. And "deformation is due to strain" is a weird formulation. Strain is a measure of deformation not the cause of deformation. $\endgroup$ – nasu Jun 22 at 22:34
2
$\begingroup$

Any answer to this conundrum: jump off or ride the ladder to the ground should take the rotational dynamics of the situation into account.

Consider the ladder as a long straight rigid rod, with mass $M$ and length $L$. The rod is originally vertical and rotates ninety degrees around the bottom end until it hits the horizontal ground. The Internet is filled with images and videos of this situations as applied to the demolition of tall chimneys (many in the "hold my beer and watch this!" genre)

The moment of inertia, $I$ of such a uniform rod is given by: $$I=\frac13 ML^2$$At the start of the fall, the kinetic energy of the rod is zero, and the potential energy is found by putting all the mass at the centre of mass:$$PE_1=mgh=Mg \frac{L}2 $$ At the end, the rod has lost all its potential energy, is rotating at an angular velocity $\omega$ and has acquired a matching kinetic energy. In terms of rotational kinetic energy:$$KE_2=\frac12 I \omega^2 =\frac12 \times \frac13 ML^2\omega^2=\frac16 ML^2\omega^2$$Equating the lost potential energy to the gained kinetic energy and rearranging, we obtain:$$\omega=\sqrt{\frac{3g}{L}}$$The linear velocity of the extreme top of the rod,$V_T$ is simply:$$V_T=L \omega = \sqrt{3gL}$$On the other hand, something leaving the top of the rod as it starts to fall will acquire the velocity for a body falling a distance $L$:$$V=\sqrt{2gL}$$ So the tip will be falling faster than gravity alone would explain. The rod transmits the force of gravity along its length to get the tip to speed up.

In the case of chimney demolition, the chimney is often not rigid enough to maintain this motion:

enter image description here

Reference: https://nobilis.nobles.edu/tcl/lib/exe/detail.php?id=courses%3Ascience%3Aphysics%3Aap_physics%3Acalendar%3Achimney_question&media=courses:science:physics:ap_physics:calendar:chim-chiminey.jpg

Note that the top part of the chimney cannot be pulled around by the bottom; the chimney breaks first.

Getting back to our individual on the tipping ladder: Staying with the ladder would take a deliberate act: the ladder would tend to accelerate away from him as it fell.

If the passenger decided to hang on to the ladder, the velocity at the ground would be greater by a factor of $sqrt{\frac32}$ So the momentum of the impact would be greater by $22.5 \%$ and the kinetic energy would by greater by $50\%$

$\endgroup$
  • $\begingroup$ Why person's mass $m$ is completely ignored in the stick-to-the-rotational-ladder case? Doesn't the person's mass somehow affect moment of inertia? $\endgroup$ – weno Jun 23 at 0:23
2
$\begingroup$

Here my answer to your ladder problem and whether or not the person should jump (a) at $t=0$ or (b) just before the impact @ $t=T$. Furthermore, I am concentrating on the "damage to the person" and not on the damage to the ladder.

In your problem energy and momentum are directly related via $E=\frac{p^2}{2m}$. Hence, if the person obtains the smaller momentum, he/she also obtains the smaller energy. Thus, the question is: In which scenario, a or b, does the person obtain the smaller momentum or kinetic energy.

Let's consider scenario a: If we are standing on the ladder at height $h$ and we jump of the ladder immediately as it drops, we obtain an additional height $dh$ due to the jump. Thus, the kinetic energy at our impact on the ground is $E = m g (h + dh)$. Thus, this is even worse than not jumping at all.

In contrast, if we jump off the ladder just before we hit the ground, we decelerate. Hence, if we assume that we add the same energy to the ladder in both cases, we effectively hit the ground with $E = m g (h - dh)$.

$\endgroup$
2
$\begingroup$

Basically, my question is, which object landed "more safely"? Or: is it better to be an apple or the box?

You haven't defined what you mean by "more safely". Also, you didn't define what you meant by "strength of impact".

I'm going to assume you want to know which object will either do more damage to the ground or to itself when it impacts. Perhaps that means how deep an impression it makes on the ground or how deformed it becomes due to the impact. Or perhaps that means how much the average impact force is on the ground or to itself. Or maybe a combination of the two.

There are other problems in comparing the box and the apple besides momentum and kinetic energy. The apple is a not a perfect sphere and will impact the ground a little differently depending on its orientation. The box, assuming it is a cube, can impact the ground on a corner, or on a flat side. Big difference in outcome there. The worst case is obviously if it impacts on a corner.

I will answer your question on the basis of concern to the damage to the ground, by substituting a sphere for the apple and the box. If you are interested in how much damage is done to the falling object, you will need to provide information on the physical properties of the objects.

The substitute sphere for the box will have the same diameter as the substitute for the apple but with a density 20 times greater to account for its mass being 20 times greater than your original apple. I will also assume the two spheres have the same rigidity. Now we can compare apples to apples (no pun intended. Oh heck, yes the pun is intended).

All other things being equal, the sphere with the greater kinetic energy upon impact should do the most damage. In order to bring the spheres to a stop, the ground has to do work on the sphere taking its kinetic energy away and transferring into the ground in the form of deformation which results in heat. The principle involved here is the Work Energy Theorem which states that the net work done on an object equals its change in kinetic energy. For the 0.1 kg sphere that is

$$W_{0.1 kg}=F_{ave}d=-\frac{(0.1)(200 m/s)^2}{2}$$

Where $F_{ave}$ is the average stopping force and $d$ is the stopping distance of the sphere after it impacts the ground.

For the 2 kg sphere we have

$$W_{2 kg}=F_{ave}d=-\frac{(2)(30 m/s)^2}{2}$$

The ground has to do more than twice the work to stop the 0.1 kg sphere than the 2 kg sphere. It should be emphasized that the above equations do not include the loss of potential energy of each sphere over the penetration distance $d$. That needs to be included to obtain the actual work required. Unless the penetration is very deep, it may not be critical in terms of comparing the two spheres since the kinetic energy of the 0.1kg sphere is more than twice as much as the 2 kg sphere. So I have not included that complication.

But we are still left with two variables, the $F_{ave}$ and $d$. The greater the average force, the shorter the stopping distance. This would be the case for a firmer ground. The lower the average force the greater the stopping distance. That would be the case for a softer ground. In any event, all other things being equal, the greater the kinetic energy on impact, the greater the relative potential damage should be.

Hope this helps.

$\endgroup$
2
$\begingroup$

First off, you need to define precisely what you mean by the "strength of impact" - depending on how you do this, you will get different answers. If you define it to be, say, the energy with which the object strikes, then of course it will be the energy that determines the strength, by definition. So we have to think a little bit about what that should "best" mean. Given that you are talking about "safety", then it stands to reason what you are after is damage - i.e. the damaging power of the impact. In that case, though, the damage depends just as much on the kind of object as on the impact itself - e.g. consider dropping a basketball versus a glass sculpture of the same mass.

That said, the most reasonable variable you will want to look at here is probably the force generated upon impact. This is because "damage", in this sense, occurs when materials are subject to forces that exceed certain strength limits, which in turn are set by the cohesive forces holding them together. In particular, denting (plastic damage) occurs when the elastic limit of the material is exceeded, and fracture (breaking into pieces) occurs when the ultimate strength is reached.

The force generated on impact is, per Newton's second law,

$$F = ma$$

related to the mass and rate of deceleration of the object. The deceleration rate, in turn, is related to the impact time $t_\mathrm{imp}$, which allows us to write this as:

$$F = It_\mathrm{imp}$$

with $I$ now the impulse, or change in momentum, delivered to the object. This will, for the case of hitting and stopping on a hard surface, equal in magnitude the object's original momentum.

So you would be tempted to say "momentum", but as you can also see, there is another factor, time, and that time depends in a complicated way upon other things like the object's velocity, the elasticity of the materials, and more and hence there isn't really any way to easily calculate it "from first principles" - you have to either do a full materials-physics simulation on a computer, or just measure the impact force empirically (usually the better and cheaper option).

Nonetheless, even with this, we can use intuition to make a rough idea that the apple will have the "worse" time, because while the needed momental change is much less, the much higher velocity means it has to stop a lot faster (a lot higher $t_\mathrm{imp}$). Indeed, from experience, we would "feel" that the apple will smash to bits. That said, depending on what your box is made of, it too could suffer damage since $30\ \mathrm{m/s}$ (= $30\ \mathrm{km/ks}$) is about highway speed.

Hence, the simplest answer would be "velocity, and the physical nature of the objects concerned".

$\endgroup$
1
$\begingroup$

It's simple. There is a quantity well reserved for this purpose called "impulse". Using this quantity you can calculate the impact of the object on the ground, which is the same as the impact of the ground on the object.

Impulse is force exerted times duration of impact.

However, the impulse is almost the same as momentum. So we can go for choosing momentum over energy. (Precisely, object's change in momentum is impulse.)

As the box has higher momentum, so the box will leave a greater impact.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.