Help me to humiliate my Physics Major Friend

[Xema]

To continue the analysis:

@Q.E.D.

Force is also defined as 1/2 mv[sup]2[/sup], and since the mass, m, of the shot doesn’t change measurably between the etime it leaves the barrel to the time it hits the target, and we’ve stipulated that the velocity hasn’t changed either, then the force exerted must remain constant as well.

You are right that, absent much in the way of drag, (mv[sup]2[/sup])/2 (energy, not force) is much the same at the target as it was at the gun.

Since energy = force * distance, force = energy/distance. In the gun, the energy was applied over a distance of something like 24 inches. At the target, the energy is dissipated over perhaps a tenth to a fifth of that distance, which implies a force 5 to 10 times that at the gun.

[Charlie_Tan]

I’m surprised no one has linked to INSULTINGLY STUPID MOVIE PHYSICS:

If the recoil from discharging a firearm is insufficient to throw the shooter backwards through the nearest window then the bullet also will not throw the victim backwards through the nearest window.

There is one other possible mechanism for being blown through a window: involuntary muscle contraction. The victim could be so stunned by being shot that he involuntarily jumps backwards. Since we haven’t run this experiment, and have no desire to do so, we can’t totally rule it out, but it does seem unlikely.

[vetbridge]

@Q.E.D.

F = ma is completely ignoring drag.

In Pennsylvania, game laws allow the taking of a buck even if he masquerades as a doe, making your point moot.

[matt]

Muddy waters everywhere!

Xema is correct regarding force - if we assume a constant force accelerates the shot down a 24" barrel, and the shot decelerates due to a constant drag force in the target over 6", then the force experienced by the target will be 4 times greater than that experienced by the shooter (neglecting any loss of velocity due to air drag).

BUT this force will be experienced for QUARTER THE TIME, which works out to the same change in momentum experienced by the target as by the shooter. Both time intervals are a fraction of a second in any event.

Cheesesteak’s calculation demonstrates that such a change in momentum might push an off-balance person over, just about, but won’t send them flying.

[Q.E.D]

@vetbridge

In Pennsylvania, game laws allow the taking of a buck even if he masquerades as a doe, making your point moot.

snicker

I had to read that couple times before I got it.

[treis]

@Q.E.D.

I said experience the same force, not experience the same effects.

Oops my mistake.

@Valgard

Anyhow, one of the testers sat on a cheap plastic sled in the snow and fired the 2ga. It did actually scoot him back a bit (we’re talking a foot or two).

So if that scaled-down bazooka will only budge a man who is on a relatively slippery surface, I don’t think that any firearm found in normal experience is going to send anyone flying through the air by dint of the actual impact.

I agree, the size of the gun required to send some flying would be definately classified under ‘Holy Fuck thats a big ass gun’.

@tim314

I disagree. Bracing might matter if we’re talking about only enough force to make someone lose their balance (e.g., a shove). That’s what “bracing” is – standing in a more stable way so that you’re not as easily unbalanced. But if we’re talking about enough force to send someone flying off their feet, then the only relevant question seems to be “Can the ground exert enough friction on their feet to hold them in place?”

Yes I can say from experience that you can send someone flying a foot or two and remain in the same position. The instince I am thinking of the guy was probably 20 pounds lighter than me but I got my hands ‘underneath’ his chest and pushed up then out. I would imagine it would be pretty hard to keep your balance shooting a gun of this magnitude but I would put money down that you could do it.

[Gary_Robson]

@The Gaspode

I’m surprised no one has linked to INSULTINGLY STUPID MOVIE PHYSICS:

Don’t be. See post #11 in this thread.

Very appropriate link.

[jnglmassiv]

Relevant link to short video.

[Xema]

@matt

BUT this force will be experienced for QUARTER THE TIME, which works out to the same change in momentum experienced by the target as by the shooter.

I agree. My point was simply that the forces on each end are demonstrably not the same.

[Q.E.D]

@Xema

I agree. My point was simply that the forces on each end are demonstrably not the same.

I still think you are wrong. If you put the shooter and an identically-massed target on rollers, it’s your contention that the target will roll back farther than the shooter will? I don’t think so.

[Xema]

@Q.E.D.

I still think you are wrong. If you put the shooter and an identically-massed target on rollers, it’s your contention that the target will roll back farther than the shooter will? I don’t think so.

Nor do I - because you are now talking about momentum, not force.

The discussion of why forces are not equal is presented above. Is there any part of this that you disagree with?

[Q.E.D]

I think we’re arguing from two different frames of reference here.

My argument is that since momentum is conserved, then target and shooter will be moved the same distance, and since this distance is dependant on the acceleration and F = ma or a = F/m, the the force on both the shooter and the target must be the same, too.

[Cheesesteak]

QED, I think what you’re talking about is the impulse, which is the integral of force over time, which is also equal to the change in momentum of each object. The impulse will be identical, but the actual forces involved could be vastly different if the time frame is different.

The only constraint that conservation of momentum puts on the forces is that the integrals over time wind up equal.

[Q.E.D]

Ok, that makes sense. Thanks, Cheesesteak.