Not intuitive indeed.
In fact, the energy doubles, but the recoil velocity increases by a factor of sqrt2 (1.4ish).
E=1/2mv^2.
E forwards = E backwards (conservation of energy).
1 bullet energy forwards (simplified figures for demonstration, bullet mass = 1, velocity = 2.).
E = (0.5)(1)((2)^2)) = 2
2 bullets energy forwards.
E = (0.5)(2)((2)^2)) = 4
This makes logical sense. You burnt twice as much powder and sent twice the projectile weight downrange at the same velocity.
Recoil velocity. Rifle mass = 10 for ease of calculation. E values bolded.
Rearranging the energy equation for v.
E=1/2mv^2
2E = mv^2
2E/m = v^2
Sqrt(2E/m) = v
1 bullet.
sqrt(2(2)/(10)) = v. v = 0.632
2 bullets
sqrt(2(4)/(10)) = v. v = 0.894
QED. That's what the calculator is doing also.
Unless you can magically create energy from nothing, that has to be the case. There's no way to double the recoil velocity without quadrupling the energy and that simply didn't happen. How can it? You've only burnt twice as much powder and chucked two bullets down range at the same velocity as you previously chucked one. 2x energy, not 4x.
I really tried to resist the temptation to respond and couldn't.
My apologies to the non-physics types out there.
Sorry Alistair, I have to disagree with you. Your starting assumption is flawed - this is NOT a conservation of energy problem (remember there is heat energy involved, the kinetic energy retained by the bullets flying off into the ether etc which you have not accounted for.)
As I stated before, this is a conservation of momentum problem and there is the impulse effect. Impulse is defined as being a force that is being applied for a period of time. This is in essence what you have proved - that the recoil impulse doubles with a simultaneous discharge.
Force = mass times acceleration (F= m * a)
To cut to the chase, with the simultaneous discharge, the force must be double because the mass has doubled (two bullets and two powder charges). Acceleration has stayed the same for the bullets & powder going down the barrel.
For the rifle, the mass has stayed the same but it experience the equal and opposite reaction (conservation of momentum) so it experiences twice the force of a single discharge.
Therefore the rifle must experience twice the acceleration because if force has doubled and mass is constant, then solving for the remaining variable (acceleration) tells us this.
Acceleration is the rate of change of velocity (i.e. the difference between the velocity when the trigger is pulled, which was zero. And the velocity when the bullet leaves the barrel).
We've assumed the time of this event (bullets travelling down the barrel is the same in both cases - same muzzle velocity whether 1 bullet or 2)
The starting velocity in both cases is zero.
Therefore the terminal velocity of the rifle in the double discharge case has to be double the terminal velocity of the single discharge.
If the rifle velocity has only increased by a factor of 1.41 as you've suggested, how do you get to double the force (which I believe is common cause in this scenario) seeing as your acceleration has only increased by 1.41 and not by 2?
I've tried not to go into the long-winded maths proof of this but I assure you it holds true.
