# Newbie saying Hello

1. Feb 23, 2007

### 44HP

Newbie saying Hello and an unusual question...

Just thought I'd say Hi and ask if I can post a rather odd question.

I've not done any academic physics since leaving school at 18 (I'm a Brit) but think that knowing the basics makes life a whole lot easier.

Physics stops your shelves falling down and your car falling off the road. So it's kinda important.

I want to ask a particular question, but it might not be to every bodies taste so I'm asking permission first.

I'm a keen shooter and have always had problems about the way other shooters talk about recoil and impact on the target.

A lot of very, very experience and knowledgeable shooters say that the shooter feels the same energy (recoil) as the target (impact) due to the conservation of energy / momentum.

Personally I think that's a simplistic view and would like to get to the bottom of this once and for all.

I want to talk about the different rates of projectile acceleration and deceleration as well as the relative areas over which the force is felt.

Is this a suitable place to have this discussion?

Cheers...

Last edited: Feb 23, 2007
2. Feb 25, 2007

### Claude Bile

Of course.
First of all we need to define some basic terms.
Momentum (P) is the product of an objects mass (m) and velocity (v) (i.e. P = mv).
Force is the rate of change of momentum (i.e. F = dP/dt). Sometimes we rewrite this as F.dt = dP and label the quantity F.dt "Impulse".
Energy is the ability to do work, and can take many forms. For example kinetic energy (energy due to motion) is equal to 1/2mv^2, mechanical work is equal to Force x Distance and so on.

Now some basic laws.

Conservation of energy will not apply in this scenario, because energy is being lost to the surroundings.
Conservation of momentum however, will apply because there are no outside forces acting on the system (I am taking the "system" to be shooter + projectile + target).
Newton's Third Law states that forces act in equal and opposite pairs.

Newton's Third Law is probably the most intuitive to follow in terms of what happens in terms of the recoil and impact. When the shooter fires the projectile, they are imparting a force onto that projectile. Newton's Third law says that a force will also be imparted onto the shooter. This is the recoil. Thus the force imparted to the projectile by the shooter is equal to the recoil. Let's now examine the impact with the target. The target will impart force onto the projectile, by Newton's third law, the the projectile will also impart force onto the target (impact).Assuming the projectile comes to a complete stop and the time taken to apply both the recoil force and impact force is equal (These are important assumptions) the force required to stop the projectile will be exactly equal and opposite to the force imparted to it by the shooter. Therefore the force imparted by the target to the projectile and thus (by Newton's 3rd law) the force imparted by the projectile to the target (impact) is equal to the original recoil.

Whew. Let us however re-examine the important assumptions that were made here - firstly that the bullet comes to a complete stop. If this assumption is not true, say in the case of a deflected bullet, the recoil force will NOT be equal to the impact force.
The second assumption, that the time taken to apply the recoil and impact force is equal does not hold in general. Conservation of momentum says that the Impulse (i.e. Force x Time) of both will be equal, but if the time taken to apply the recoil force is double that of the impact force, the impact force must be double in magnitude to satisfy conservation of momentum. To reiterate in equation form;

$$F_r.t_r = F_i.t_i$$

For this particular scenario.

Keep in mind that at the end of the day, after applying conservation of momentum, all you know is what the momentum of all your masses are and (sometimes) what masses imparted what forces to whom (if you know the duration over which the force is applied). Details such as what damage was actually caused to each mass during a collision obviously requires more detailed knowledge. The example of the surface area over which the force was applied (i.e. the pressure) was a good one, it is the reason a bullet with a sharp tip is much more likely to penetrate living tissue that spherical shot, and why the recoil force from a gun will cause much less damage in human terms than the impact force from a bullet (again keeping in mind that these two forces were equal in the above case).

So I guess the take home message is this - while conservation of momentum can be applied to this type of scenario, all it will tell you is momenta and forces. Obviously in terms of shooting, momenta and forces do not tell the whole story, therefore consideration of other factors needs to be taken in order to get the all-too-elusive "whole picture".

Claude.

Last edited: Feb 26, 2007
3. Feb 26, 2007

### pallidin

Welcome to PF, 44HP, this is a great site.