Force of Bullet Sending Person Flying Backwards

[MG5]

I'm proving that the force of a bullet is not actually enough to send a person flying backwards like you see in the movies.

These are the variables I have so far. I'm not sure if all of them are really necessary to solve this.

distance between gun and person getting shot = 2.45 m

mass of person getting shot = 104 kg

barrel length (gun) = .5 m (not sure if I need this)

v of bullet = 385 m/s

mass of gun = 3.10 kg (Don't think I need this?)

mass of person shooting gun = 85 kg (Don't think I need this either)

mass of bullet = .04 kg

How deep the bullet penetrates = .2 m (Another one I'm not sure I need)

I know it's an inelastic collision so eventually I'll need

m1v1 + m2v2 = (m1+m2) v

and I should also use conservation of momentum

(m1v1 + m2v2)i = (m1v1 + m2v2)f

My professor left this comment on my paper:

One approach might be to use the conservation of momentum of the system to determine the velocity after the inelastic collision, but then you have to work out how to translate that into distance traveled.

I'm sure you all know this but when he says distance traveled he means how far back the person travels after being shot.

So I first started out using conservation of momentum and did...

104kg(v2) = .04kg(385m/s)

And I got v2 = .385 m/s after the collision. Doesnt sound right though. From here not quite sure what to do. Maybe use v=d/t to find the distance he traveled after getting shot? Any help is appreciated. I really need it on this on.

 

[SteamKing]

You might want to check your original calculation. I don't get v2 = .385

 

[Andrew Mason]

So I first started out using conservation of momentum and did...

104kg(v2) = .04kg(385m/s)

And I got v2 = .385 m/s after the collision. Doesnt sound right though. From here not quite sure what to do. Maybe use v=d/t to find the distance he traveled after getting shot? Any help is appreciated. I really need it on this on.

The bullet has momentum of 15.4 kg m/sec. Assuming it imparts all of its momentum into the target person (104 kg) how fast does the target recoil from the shot?

All you can determine from this is the rate of the target person's motion (ie his speed). You cannot determine how far he moves without additional assumptions about forces acting on the person that would slow the person's motion.

AM

 

[Jobrag]

The momentum of the bullet leaving the gun is the same as the momentum of the recoil on the shooter, do people fly backwards when they fire a gun? I suppose that a shot persons muscles might spasm causing them to appear to fly backwards. I think mythbusters did a show on this.

 

[CWatters]

All you can determine from this is the rate of the target person's motion (ie his speed). You cannot determine how far he moves without additional assumptions about forces acting on the person that would slow the person's motion.
AM

You could assume it kills him so he falls down. Work out time taken to fall then use that to work out the distance. You would have to assume he doesn't go skidding along the ground but that might be obvious from the velocity.

 

[MG5]

The bullet has momentum of 15.4 kg m/sec. Assuming it imparts all of its momentum into the target person (104 kg) how fast does the target recoil from the shot?

All you can determine from this is the rate of the target person's motion (ie his speed). You cannot determine how far he moves without additional assumptions about forces acting on the person that would slow the person's motion.

AM

Yes, I also got 15.4 for the momentum. And what exactly do you mean how fast does the target recoil from the shot?

You could assume it kills him so he falls down. Work out time taken to fall then use that to work out the distance. You would have to assume he doesn't go skidding along the ground but that might be obvious from the velocity.

Yeah he does not go skidding. Just gets hit by the bullet and then flies backwards landing on his back, not skidding. Wouldn't I need to find acceleration first in order to find the time taken to fall?

 

[haruspex]

Having computed the horizontal speed of bullet+target immediately after impact, you need to make an assumption about the height of centre of gravity. About 1m maybe. You can then treat the victim as a projectile of known speed at height 1m and compute the distance to landing.

 

[Andrew Mason]

Yes, I also got 15.4 for the momentum. And what exactly do you mean how fast does the target recoil from the shot?

What is the speed of the 104 kg target after stopping the bullet?

Yeah he does not go skidding. Just gets hit by the bullet and then flies backwards landing on his back, not skidding. Wouldn't I need to find acceleration first in order to find the time taken to fall?

If all you want to find is the distance he moves between the time he is hit and the time he falls, you have to find the time it takes him to fall. Use h = .5gt^2 where h is the vertical height of his centre of mass, g is acceleration due to gravity and t is the time of the fall.

AM

 

[CWatters]

Yeah he does not go skidding. Just gets hit by the bullet and then flies backwards landing on his back, not skidding. Wouldn't I need to find acceleration first in order to find the time taken to fall?

Since he's falling the acceleration in the vertical direction is that due to gravity = 9.8m/s².

Consider the vertical and horizontal motion as two separate actions. The only thing they have in common is that they both start and end at the same time.

Vertically: Apply standard equation of motion to work out the time taken to fall vertically say 1m. Initial vertical velocity is zero.

Horizontally: Use distance = horizontal velocity/time.

 

[MG5]

Having computed the horizontal speed of bullet+target immediately after impact, you need to make an assumption about the height of centre of gravity. About 1m maybe. You can then treat the victim as a projectile of known speed at height 1m and compute the distance to landing.

Ok so I guess I use v=d/t, but I first have to find the velocity after impact, right? And the time to fall.

What is the speed of the 104 kg target after stopping the bullet?


If all you want to find is the distance he moves between the time he is hit and the time he falls, you have to find the time it takes him to fall. Use h = .5gt^2 where h is the vertical height of his centre of mass, g is acceleration due to gravity and t is the time of the fall.

AM

Wouldn't I use conservation of momentum to find the speed of the 104 kg target after getting shot? I tried that earlier. I did...

104kg (v) = 15.4 kg m/s

and got v2 = .148 m/s. Does that sound right?

And for the time I did 1=.5(9.8)t^2

and got t=.452 seconds.

Since he's falling the acceleration in the vertical direction is that due to gravity = 9.8m/s².

Consider the vertical and horizontal motion as two separate actions. The only thing they have in common is that they both start and end at the same time.

Vertically: Apply standard equation of motion to work out the time taken to fall vertically say 1m. Initial vertical velocity is zero.

Horizontally: Use distance = horizontal velocity/time.

Not quite sure what you mean about the last part. I understand that problems like these involve both horizontal and vertical components, but you want me to solve for the time vertically and horizontally? Wouldn't it just be the same?

Thanks a lot, so far you guys have been a huge help.

 

[Andrew Mason]

Wouldn't I use conservation of momentum to find the speed of the 104 kg target after getting shot? I tried that earlier. I did...

104kg (v) = 15.4 kg m/s

and got v2 = .148 m/s. Does that sound right?

Ok.

And for the time I did 1=.5(9.8)t^2

and got t=.452 seconds.

Ok. Assuming his centre of mass falls one metre, which is reasonable.

Not quite sure what you mean about the last part. I understand that problems like these involve both horizontal and vertical components, but you want me to solve for the time vertically and horizontally? Wouldn't it just be the same?

How far horizontally does the target (moving at 14.8 cm/sec) travel in .452 seconds?

AM

 

[MG5]

How far horizontally does the target (moving at 14.8 cm/sec) travel in .452 seconds?

AM

I did .148 m/s = x/.452 s

and got x = .066896 m

Seems like too low of a number though.

 

[Andrew Mason]

I did .148 m/s = x/.452 s

and got x = .066896 m

Seems like too low of a number though.

Too low for Hollywood perhaps. But physics is real life.

AM

 

[MG5]

Too low for Hollywood perhaps. But physics is real life.

AM

Hahaha, very good point.

Not quite sure to go from here though. I figured out the momentum of the bullet, the time it took for the target to hit the ground after getting hit, the distance he traveled backwards after getting hit, and the speed of the target after getting hit. The acceleration is just the acceleration due to gravity so 9.8 m/s2. Kinetic energy?

 

[MG5]

bump.

So do I find the kinetic energy next? Or the force of the bullet? Something else?

 

[Andrew Mason]

bump.

So do I find the kinetic energy next? Or the force of the bullet? Something else?

What are you trying to show? I thought you were trying to determine how far a bullet will move a person when hit by it. You have done that now.

It appears that you want to determine the force of the bullet. That is relatively easy. But the force does not tell you what speed the body will attain. Force x time over which the force acts gives you that. But that is the same as the bullet momentum and we have already worked that out.

AM

 

[MG5]

What are you trying to show? I thought you were trying to determine how far a bullet will move a person when hit by it. You have done that now.

It appears that you want to determine the force of the bullet. That is relatively easy. But the force does not tell you what speed the body will attain. Force x time over which the force acts gives you that. But that is the same as the bullet momentum and we have already worked that out.

AM

What I'm trying to show is that when I person gets hit by a bullet in real life, they do not go flying backwards. So I guess I'd be trying to show how the force of a bullet is not enough to send a person flying backwards.

 

[voko]

The entire system shooter + bullet + victim must conserve momentum. It is zero before the shot. It is zero after the bullet's impact. That means the momentum of the victim must be exactly equal to that of the shooter in magnitude. Shooters, ordinarily, do not fly. So neither should victims.

The only caveat is that the victim may instinctively leap from the danger just after being hit, but that is physiology, not physics.

 

[MG5]

The entire system shooter + bullet + victim must conserve momentum. It is zero before the shot. It is zero after the bullet's impact. That means the momentum of the victim must be exactly equal to that of the shooter in magnitude. Shooters, ordinarily, do not fly. So neither should victims.

The only caveat is that the victim may instinctively leap from the danger just after being hit, but that is physiology, not physics.

Ok. I already used conservation of momentum in the beginning. Don't I need to show more just momentum is conserved for this.

 

[voko]

Energy is not conserved; angular momentum is, but, as we have a straight-line motion here, it will boil down to ordinary momentum again. I do not see what else you can do for this problem. I think you can consider it finished.

 

[MG5]

Energy is not conserved; angular momentum is, but, as we have a straight-line motion here, it will boil down to ordinary momentum again. I do not see what else you can do for this problem. I think you can consider it finished.

Seems a lot easier than I thought. Seems too easy. Thanks though.

 

[Andrew Mason]

The entire system shooter + bullet + victim must conserve momentum. It is zero before the shot. It is zero after the bullet's impact. That means the momentum of the victim must be exactly equal to that of the shooter in magnitude. Shooters, ordinarily, do not fly. So neither should victims.

The only caveat is that the victim may instinctively leap from the danger just after being hit, but that is physiology, not physics.

But the shooter is braced to absorb the impact of the rifle on his body. So you have to include the shooter and the Earth against which the shooter is braced. Of course the shooter/earth does not recoil much.

AM

 

[voko]

But the shooter is braced to absorb the impact of the rifle on his body.

That is true. However, we could assume that the victim may equally be braced. The motion of the victim then must still mirror that of the shooter, who, as you say, doesn't recoil much. A typical Hollywood victim flies violently a few meters upon impact, sometimes even smashing furniture and walls in his path; clearly no amount of bracing of a (standing) victim can absorb that kind of impulse and have the same small amount of recoil a typical shooter has.

 

[Andrew Mason]

That is true. However, we could assume that the victim may equally be braced. The motion of the victim then must still mirror that of the shooter, who, as you say, doesn't recoil much.

On the assumption that the victim is not braced, it appears that a 104 kg victim would move about 7 cm. when hit by 25 g. bullet moving at 385 m/s. before falling 1m. to the ground. That roughly corresponds to a .44 Magnum bullet

A typical Hollywood victim flies violently a few meters upon impact, sometimes even smashing furniture and walls in his path; clearly no amount of bracing of a (standing) victim can absorb that kind of impulse and have the same small amount of recoil a typical shooter has.

I agree. There would be movement but only about 15 cm/sec. if the victim was unconstrained (using the figures provided in the OP).

AM