# Gun Recoil Problem, Help Please

1. May 16, 2005

### mister_okay

If one were to increase the size of his gun and the bullet in
PROPORTION...for example, 5 times the mass of the gun and 5 times
the mass of the bullet...would the recoil experienced by the shooter
be the same as the recoil experienced by the same shooter where the
masses were not increased? For example:

A normal sized gun of 20 kg, and a bullet of 2 kg (these were made
up by myself and are not actual measurements). Increase the masses
by 5, giving us a gun of 100 kg and a bullet of 10 kg. Would the
recoil experienced by the same person be exactly the same in both
cases?

Here is how I initially reasoned it:
F = ma. Assuming that an acceleration of 100 m/s² is required for
the bullet of 2 kg, the force that the bullet feels is 200 N (F =
(2kg)(100m/s²)). The equal and opposite force on the gun would be -
200 N (because every action has an equal and opposite reaction).
Therefore, the acceleration felt by the gun would be -10 m/s² (-200
N = (20kg)(acceleration)).

Increasing the masses proportionally, we now have a gun of 100 kg
and a bullet of 10 kg. If the masses increase, the force required
in order to produce the same acceleration for the bullet would also
have to increase. Therefore, 5F = 5ma...F = 1000 N. The
acceleration of the bullet will still be 100 m/s². The force that
the gun experiences will be equal and opposite: -1000 N. Therefore,
the acceleration the gun experiences will be -10 m/s² (-1000N =
100kg)(acceleration)). In both the 20 kg and the 100 kg gun, the
recoil accelerations are -10 m/s².
From these calculations I deduced that the recoil experienced by the
same shooter must be the same. However, I would like to verify if
this is correct.

2. May 16, 2005

### jdavel

mister ok,

Sometimes considering extreme cases can give you some insight.

You know those guns they have on battle ships? They fire shells that are about 15 inches in diamter. Suppose the shell weighs a ton and the gun weighs 10 tons. You want to put that baby up against your shoulder and shoot it?

3. May 16, 2005

### mister_okay

I know it sounds ridiculous, but if you do the math doesn't the recoil come out to be the same?

4. May 16, 2005

### jdavel

mr okay,

This is your chance to think like a physicist. The equations seem to be telling you one thing, but your physical intuition is screaming at you that something must be wrong with the equation!

The acceleration of the gun will be the same. But is that what matters? Suppose by the time the gun slams into your shoulder it's going 50 miles/hr. Which would you rather have hitting you at 50 miles/hour, a tiny gun or big gun? Or that huge one on the battle ship?

5. May 16, 2005

### mister_okay

Sorry I'm not a physicist, but that's why I'm asking...

The point that I do not understand is...shouldnt the acceleration component of force matter more than the force? If something were to push me with a force of 1000000000000 N, but its mass was 1000000000000 N with an acceleration of 1m/s², would it hurt me? This is NOT a rhetorical question, I am asking because I honestly do not know. Trying to use logic with an idiot like me will not work :/

6. May 16, 2005

### PeteSF

How do you quantify "recoil"?

Energy absorbed by the shooter? Impulse applied to the shooter?
I don't think that recoil acceleration alone tells the story.

7. May 16, 2005

### mister_okay

Since I'm no expert in this field, I simply defined recoil as the push that the shooter feels. Simply put, my question was would the shooter feel a harder push in a situation with a gun and bullet size increased in proportion?

8. May 16, 2005

### PeteSF

Hi mister_okay,
If something were to push you with a force of 10^12N, then you'd accelerate at a high rate (on the order of 10 billion m/s/s) and get squashed flat.

I think that the case you are thinking of is more to do with velocity than acceleration. In the case of the gun, you should consider the final speed of the gun, rather than its acceleration. What's the speed of the gun when it hits the shooter? How much will it hurt?

However, that case only applies if the shooter is holding the gun away from their shoulder, so that it accelerates before impact... but that's bad shooting form.

In the case of a weapon held properly (firmly against the shoulder), then the shooter is effectively adding their mass to that of the gun, and they gun+shooter system will accelerate as one. In this case, a heavy shooter + light gun will have low acceleration, but a light shooter + heavy gun will have high acceleration.

In this case, you can determine the force of the gun on the shooter by considering the force necessary to accelerate the shooter at that rate.

9. May 16, 2005

### Staff: Mentor

Assuming that everything increases by the same proportion--mass of bullet, mass of gun, amount of energy released by the explosion--then the recoil speed of the gun will be the same. So, as jdavel asks, what do you think would require more force to stop: an object moving at a certain speed or an object 5 times heavier moving at the same speed?

Note that the issue is the force the shooter must exert to stop the recoiling gun, not the force experienced by the gun due to the expelling bullet. Another way to look at it is from an energy point of view: the shooter must absorb the increased energy of the heavier gun.

10. May 16, 2005

### PeteSF

(considering my previous post)
Although... that's not really completely right either, because the shooter's upper body can sways a bit.

Perhaps the best quantification of "recoil" is "How fast is the shooter moving backward immediately after firing?"

That's easy to calculate by conservation of momentum. Note that the shooter+gun are effectively a single object, so you must also consider the shooter's mass when determining recoil in this way.

11. May 16, 2005

### mister_okay

Actually i was thinking more of a handgun. The reason this whole thing came up is complicated, but in short I was watching a movie where a woman was jumping sideways with quite a large gun loaded with a large bullet (Chronicles of Riddick, stupid movie...I know). When she fired, she did not seem to experience any recoil (little, but not much). I was arguing with my friends about the physics of it all. Basically what it was about was, shouldnt the woman have been thrown back by the shot while she was flying sideways in the air? My friends said that since the gun was bigger, she would. I said that if the mass of the gun and the bullet were increased proportionally the recoil would be no different from that of a proportionally smaller gun. Does that change anything?

12. May 16, 2005

### PeteSF

OK, the question is "How much would the woman have been thrown back by the shot while she was flying sideways in the air?". Note that the woman's sideways velocity will be unchanged, but she will have some rearward velocity added. We will quantify the added rearward velocity for three cases:

Case 1
Flying woman: mass = 60kg
Regular handgun: mass = 1kg
Regular bullet: mass = 0.01kg
muzzle velocity: 400m/s

Impulse = change in momentum = bullet mass x muzzle velocity = 0.01 x 400 = 4Ns
Change in woman's rearward velocity = Impulse / (mass of woman + gun) = 4 / 61 = 0.07m/s

Case 2
Flying woman: mass = 60kg
Very heavy handgun: mass = 10kg
Very heavy bullet: mass = 0.1kg
muzzle velocity: 400m/s

Impulse = change in momentum = bullet mass x muzzle velocity = 0.1 x 400 = 40Ns
Change in woman's rearward velocity = Impulse / (mass of woman + gun) = 40 / 70 = 0.6m/s

Case 3
Flying woman: mass = 60kg
Huge gun: mass = 100kg
Humungous bullet: mass = 1kg
muzzle velocity: 400m/s

Impulse = change in momentum = bullet mass x muzzle velocity = 1 x 400 = 400Ns
Change in woman's rearward velocity = Impulse / (mass of woman + gun) = 400/160 = 2.5m/s

So there you go... the impulse on the gun+woman is proportionate to the bullet mass, but the change in velocity is proportional to the mass of the woman+gun.

One thing I haven't considered here is the huge increase in powder that would be required to maintain a constant muzzle velocity. Since energy is proportional to the square of velocity, you'd need 100 times as much explosive to launch a projectile at the same speed as one 10 times lighter.

13. May 23, 2005

### quasi426

Although the recoil experienced by the gun will be the same, the shooter's mass is not changing and will therefore experience greater recoil reaction.

14. May 24, 2005

### Andrew Mason

First of all, your calculations are way off. A typical rifle bullet will exit with a muzzle velocity of over 2000 feet/sec. It acquires this speed in traveling about 2 feet ie in about 2/1000ths of a second, so the acceleration is about $a = dv/dt = 2000*1000/2 = 10^6 ft/sec^2$

Second, it is not the reaction force or momentum of the recoil, but the energy of the recoil, which is important in the determining the recoil effect. Recoil energy is proportional to the mass of the gun and the square of the recoil speed.

So, while it is true that the momentum of the recoil is always the same as the momentum of the projectile, the energy of the recoil is not. If you increase the mass of the projectile and gun proportionately to keep the same muzzle speed, the energy of recoil will increase proportionately. There will be a point at which you will be unable to stop the recoil of the gun and it will either knock you backward or rip your shoulder off.

AM