Discovering the existence of an error should be stimulating and don't

disturb.

Maybe what I'm trying to say hasn't been understood or, maybe, I

haven't explained myself well.

And then I propose a clarifying example.

Yesterday you were walking calmly but distractedly and you bumped into

a person with the same size as you who was walking as distractedly as

you.

You exerted your force F=10 (in some unit of measurement) and he

reacted with his force F=-10.

You don't fall backwards and he doesn't fall either because neither

force prevails.

Today, again, you go back to walking distractedly and you collide with

another person who is twice your size and you end up on the ground,

why?

Certainly, the force that you exerted on him is the same as yesterday

(i.e. F=10) and if he also had only exerted a force F=-10 on you (as

the third law prescribes), you would not have fallen backwards.

And so, it means that the force he exerted on you is *greater* (and not

equal) to the one you exerted on him: it is the greater force of him

that threw you to the ground!

With your force F=10 you only slowed down the other person, while he

only used half of his strength F=-20 to stop you and had the other half

left to push you backwards and make you fall.

The third principle applies to the compressive force (the action and

reaction compressed you *exactly* as much as it compressed him) but

does not apply to the force that accelerated you backwards (the residue

of his preponderant force, to which no you were able to object!).

In my animation https://www.geogebra.org/m/snjpnvsu there is a

photograph of a schematized collision between two generic bodies A and

B at the instant of maximum compression and there is the elastic ring

whose deformation measures the intensity of the action and reaction

force, which depends exclusively on the smaller body and remains

unchanged when the larger body varies.

Luigi Fortunati

[[Mod. note --

There are several mistaken statements here, including

(a) "the force that you exerted on him is the same as yesterday",

(b) "the force he exerted on you is *greater* (and not equal) to the one

you exerted on him", and

(c) "the action and reaction compressed you *exactly* as much as it

compressed him".

To analyze a 2-body collision, it's useful to start by considering two

extreme cases.

First, let's consider the case where the collision is *perfectly elastic*,

i.e., there is no energy dissipation in the collision. This means that,

in any inertial reference frame (IRF), the final kinetic energy of the

two bodies is equal to the initial kinetic energy of the two bodies.

Another way to think of this case is that the bodies "bounce back" after

the collision.

It's particularly convenient to analyze collisions in the center-of-mass

(COM) IRF, i.e., the IRF where the COM of the two bodies is at rest

(equivalently, this is the IRF where the total linear momentum is zero).

Unless stated otherwise, EVERYTHING I write below is in the COM IRF.

Before the collision:

MA has some velocity, say VA_initial (assume this is nonzero)

MB has some velocity, say VB_initial (assume this is nonzero)

the condition that the total linear momentum is zero says

MA*VA_initial + MB*VB_initial = 0 (1)

i.e., the initial velocities must be in the ratio

VA_initial/VB_initial = - MB/MA (2)

Since the total linear momentum is conserved, *after* the collision

we also have

MA*VA_final + MB*VB_final = 0 (3)

i.e., the final velocities must be in the ratio

VA_final/VB_final = - MB/MA (4)

In order to also conserve kinetic energy, we must have

VA_final = - VA_initial (5a)

VB_final = - VB_initial (5b)

For example, if the two masses are equal, we might have

VA_initial = +10, VB_initial = -10 (6a)

VA_final = -10, VB_final = +10 (6b)

so that the net velocity *changes* between before & after the collision

are

VA_final - VA_initial = -20 (7a)

VB_final - VB_initial = +20 (7b)

i.e., the net velocity changes are equal.

In contrast, if B has twice the mass of A, we might have

VA_initial = +10, VB_initial = -5 (8a)

VA_final = -10, VB_final = +5 (8b)

so that the net velocity *changes* between before & after the collision

are

VA_final - VA_initial = -20 (9a)

VB_final - VB_initial = +10 (9b)

i.e., the net velocity change of A is twice that of B.

So far we've only considered a perfectly elastic collision, one where

the bodies recoil after the collision. Now let's consider the opposite

extreme, a perfectly *inelastic* collision, where after the collision

the bodies don't bounce back at all, but instead stick to each other

like lumps of clay.

Everything I wrote above about the situation before the collision still

applies here. Equation (3) also still holds. After the collision, the

two bodies stick to each other, i.e.,

VA_final = VB_final (10)

Combining this with (3), we see that we must have

VA_final = VB_final = 0 (11)

So, returning to our earlier two examples, if the two masses are equal,

we might have (6a) before the collision, but (11) after the collision,

so that the net velocity *changes* between before & after the collision

are

VA_final - VA_initial = -10 (12a)

VB_final - VB_initial = +10 (12b)

i.e., the net velocity changes are equal (but both are smaller than the

net velocity changes (7a) and (7b) for an elastic collision with the

same initial velocities).

If B has twice the mass of A, we might have (8a) before the collision,

but (11) after the collision, so that the net velocity *changes*

between before & after the collision are

VA_final - VA_initial = -10 (11a)

VB_final - VB_initial = +5 (11b)

i.e., the net velocity change of A is twice that of B (but both are

smaller than the net velocity changes (9a) and (9b) for an elastic

collision with the same initial velocities).

Real collisions fall somewhere between the perfectly elastic and perfectly

inelastic cases; in practice they are probably closer to the perfectly

inelastic case.

A few more final points:

1. At *every* time during the collision, the force A exerts on B is equal

to precisely -1 times the force B exerts on A (this is just Newton's 3rd

law). But if the masses are different, these same-magnitude forces result

in different accelerations, and thus different net velocity changes.

2. As I noted above, if we compare elastic and inelastic collisions with

the same initial velocities, the net velocity *changes* are smaller for

the inelastic case. This is why cars are engineered with "crumple zones"

which crush in accidents -- this absorbs kinetic energy, reducing the

velocity changes (and hence accelerations & forces) of the occupants.

(As the article I cited earlier in this thread noted, having crumple

zones also spreads out the velocity changes over longer times, further

reducing the instantaneous accelerations).

If we compare collisions with different ratios of the two masses, the

smaller mass always has a larger velocity change, and in fact a larger

instantaneous acceleration at each time during the collision. Thus,

if you're in a car accident, the very worst case for you (in terms of

how severe your injuries are likely to be) is to collide with another

vehicle that's both very rigid (it doesn't crumple much) and much more

massive than your vehicle.

3. Suppose that instead of the COM IRF, we were to use some other IRF,

say one moving at a relative velocity V_IRF with respect to the COM IRF.

This would mean that every velocity we've worked out above would have

V_IRF subtracted from it, but the net velocity *changes* would be the

same as they are in the COM IRF.

-- jt]]