So this super scientist invents the bullet deflector, it takes the bullet in its flight, you wear it like a jacket but instead of a regular bullet proof vest, it doesn't stop the bullet (unless you want to argue for that fleeting moment of zero forward velocity) it takes all the kinetic energy in the bullet and reflects it EXACTLY back to the bullets source. Doesn't matter what angle it hits you at, the bullet goes back the way it came. Great defense. So the question is this: you get attacked by a killer, well would be killer anyway, he doesn't know about your special shield. So he is a couple hundred yards away, has an AK47 shooting 600 rounds per minute. So the bullets come flying out of the rifle and hit the reflector, it goes back to actually hit the next bullet in line. How far away from the target do the bullets meet. The bullets lose no energy in the reflection process, they go away just as fast as they come in. The bullets fly at 1000 feet per second.

Originally posted by sonhouse So this super scientist invents the bullet deflector, it takes the bullet in its flight, you wear it like a jacket but instead of a regular bullet proof vest, it doesn't stop the bullet (unless you want to argue for that fleeting moment of zero forward velocity) it takes all the kinetic energy in the bullet and reflects it EXACTLY back to the bullets source ...[text shortened]... process, they go away just as fast as they come in. The bullets fly at 1000 feet per second.

It'd be very close to the target I imagine, but I'm too lazy to do the math.

At 600 rounds per minute, the rifle is firing 10 rounds per second. At 1000 feet per second, there are 100 feet between rounds.

A "couple hundred" yards away, if taken literally to be 200 yards away, means 600 feet. So, at 1000 feet per second for the bullet, the bullet travels 600 feet to you and then heads back toward the shooter.

the bullets meet 50 feet from you, because the second bullet was 100 feet behind the first one, they converge at 50 feet, meeting at 2000 feet per second.

600 rounds/min = 1 round every .1 seconds, so at 1000 ft/sec the first two bullets are 100 ft apart when the first bullet hits the shield.

But then the first bullet reflects straight back towards the bullet #2, at the same speed. so they will meet at the "halfway point" between where they were at impact.

so they collide 50ft back from the shield. unless the AK recoils in such a way that the rounds come in at different angles/hit different places on the shield/don't travel with constant velocity due to gravity/all the real world difficulties with this particular scenario ðŸ™‚

Originally posted by sonhouse So this super scientist invents the bullet deflector, it takes the bullet in its flight, you wear it like a jacket but instead of a regular bullet proof vest, it doesn't stop the bullet (unless you want to argue for that fleeting moment of zero forward velocity) it takes all the kinetic energy in the bullet and reflects it EXACTLY back to the bullets source process, they go away just as fast as they come in. The bullets fly at 1000 feet per second.

if you take into account to kickback of the rifle they will never hit because theyre all shot at different angles and one will probably hit you in the head

Originally posted by coquette At 600 rounds per minute, the rifle is firing 10 rounds per second. At 1000 feet per second, there are 100 feet between rounds.

A "couple hundred" yards away, if taken literally to be 200 yards away, means 600 feet. So, at 1000 feet per second for the bullet, the bullet travels 600 feet to you and then heads back toward the shooter.

the bullets meet 5 ...[text shortened]... was 100 feet behind the first one, they converge at 50 feet, meeting at 2000 feet per second.

Yep, you get the gold ring! good show. I invented that puzzle out all by myselfðŸ™‚
It's interesting to note that using an AK47 is what saves the shooter also.
If it was a single shot pistol, the bullet would find its way right back to the shooter, maybe go right up the barrel of the gun. Now THAT I would like to see. What a defense against guns, eh!

Originally posted by wolfgang59 Momentum does not get converted into heat! (You are thinking of Kinetic Energy)

The two bullets have a total of zero momentum before and after the collision since momentum is a vector quantity. There is no problem.

What I meant was

WHAT HAPPENS TO THE MOMENTUM OF A SINGLE BULLET WHEN IT HITS THE REFLECTOR DEVICE?

I was thinking of the reflector device treating a bullet like a mirror treats a photon but it can't work like that, it would have to reflect through ITSELF which seems to me would destroy the bullet. So what needs to happen is like a space-time curve that would curve the flight path such that it would enter aiming one direction and then the space-time curve would force the bullet around a very short circle which happens to turn the bullet around and keeping the same momentum, forces it to go back the exact way it came from.

Originally posted by wolfgang59 WHAT HAPPENS TO THE MOMENTUM OF A SINGLE BULLET WHEN IT HITS THE REFLECTOR DEVICE?

The same thing that happens to a tennis ball that it's a wall and bounces back. Since the system isn't conservative momentum doesn't have to be a conserved quantity.

Originally posted by adam warlock The same thing that happens to a tennis ball that it's a wall and bounces back. Since the system isn't conservative momentum doesn't have to be a conserved quantity.

That tennis ball sends the wall flying backwards...but the wall can absorb the momentum of a tennis ball.

Originally posted by AThousandYoung That tennis ball sends the wall flying backwards...but the wall can absorb the momentum of a tennis ball.

A human and a bullet however...

Yeah but I was thinking about in ideal case. If a bullet proof vest has enough density to send a bullet flying backwards then it can surely absorb its momentum too.

What happens next depends on the human of course. Either he is strong enough and after the reflection stands on his feet or he falls down. I'm betting on the second case if we want to be realistic.