Kinetic energy
Discussion
Assuming a sphere moving through a vacuum etc...
If I have a gun that fires a 1kg bullet at 100m/s, that bullet hits its target with 5000 Joules of kinetic energy. Presumably all that energy has been transferred to it from chemical energy stored in the propellant.
If I'm in a car doing 100m/s and I drop the bullet out of the window it hits with 5000 Joules of KE, gained when the car accelerated to speed.
If I shoot something in the car it hits with 5000 Joules of KE
If I fire the gun out of the back of the car the bullet is stationary with 5000-5000=0 J of KE
If I shoot out of the front of the car the bullet starts with 5000 J and gains 5000 J from the propellant, but hits the target doing 200m/s, so 20000 Joules.
Where has this "extra" energy come from? Presumably the car has lost that much energy thanks to Netwon's third law, but that was caused by 5000J of chemical energy?
Am I wrong in thinking of KE as related to relative velocity?
If I have a gun that fires a 1kg bullet at 100m/s, that bullet hits its target with 5000 Joules of kinetic energy. Presumably all that energy has been transferred to it from chemical energy stored in the propellant.
If I'm in a car doing 100m/s and I drop the bullet out of the window it hits with 5000 Joules of KE, gained when the car accelerated to speed.
If I shoot something in the car it hits with 5000 Joules of KE
If I fire the gun out of the back of the car the bullet is stationary with 5000-5000=0 J of KE
If I shoot out of the front of the car the bullet starts with 5000 J and gains 5000 J from the propellant, but hits the target doing 200m/s, so 20000 Joules.
Where has this "extra" energy come from? Presumably the car has lost that much energy thanks to Netwon's third law, but that was caused by 5000J of chemical energy?
Am I wrong in thinking of KE as related to relative velocity?
Edited by RizzoTheRat on Wednesday 12th February 10:41
now looking forward to the answer on this...
Now (me being a thicko) isn't this a similar question as to what happens if your (fictional) aircraft is flying as fast as a bullet and then fires a bullet at a target ahead of it. Does the bullet get there quicker (one would imagine so as it's leaving a platform travelling forward?), does the bullet go as far as a bullet shot from a stationary platform?
Now (me being a thicko) isn't this a similar question as to what happens if your (fictional) aircraft is flying as fast as a bullet and then fires a bullet at a target ahead of it. Does the bullet get there quicker (one would imagine so as it's leaving a platform travelling forward?), does the bullet go as far as a bullet shot from a stationary platform?
See link:
https://physics.stackexchange.com/questions/287101...
This is similar to rockets burning fuel when approaching a planet (prograde) achieves a higher difference in velocity than the other way around (as learnt from kerbal space project!).
In the case of the car the apparent difference in energy is transferred to the earth through the wheels if I understand correctly.
https://physics.stackexchange.com/questions/287101...
This is similar to rockets burning fuel when approaching a planet (prograde) achieves a higher difference in velocity than the other way around (as learnt from kerbal space project!).
In the case of the car the apparent difference in energy is transferred to the earth through the wheels if I understand correctly.
RizzoTheRat said:
If I shoot out of the front of the car the bullet starts with 5000 J and gains 5000 J from the propellant, but hits the target doing 200m/s, so 20000 Joules.
Not sure that's correct. The energy required to accelerate a 1kg mass from 0 to 100m/s is not the same as the energy required to accelerate 1kg from 100m/s to 200m/s.
If you want the mass to be travelling at 200m/s relative to the target on the ground (remember, frame of reference is important if you want to measure speed), you'll need a bigger bang.
In your example, I think the 1kg mass would hit the target at 141m/s. i.e. sqrt( (5000J+5000J) / 0.5kg)
In IrocFan's scenario the bullet travels at aircraft speed + muzzle velocity, so yes it travels further but the drag is much higher so it decelerates more quickly.
ETA: then I read Norfolk's post, that suggest the muzzle velocity is slower if fired in flight than if fired when stationary?
In the original one I think
initial Car KE + initial bullet KE + Propellant CE = Final Car KE + final bullet KE
so if the car weighs 100kg
500kJ + 5kJ + 5kJ = 499kJ + 20kJ
So the car has slowed to 99.9m/s
Does that sound right?
But that means when you fire at a target inside the car, the car is initially slowed by the recoil, but then accelerated again when the bullet strikes the target...but hitting the target will generate some noise and heat and absorb energy in deforming the bullet and the target, so the energy transfer isn't perfect, meaning the car will be slower than before you fired.
Which implies if you fitted a machine gun and a bulletproof plate inside the car it could slow the car down without any mass ever leaving the car...which doesn't seem right...
ETA: then I read Norfolk's post, that suggest the muzzle velocity is slower if fired in flight than if fired when stationary?
In the original one I think
initial Car KE + initial bullet KE + Propellant CE = Final Car KE + final bullet KE
so if the car weighs 100kg
500kJ + 5kJ + 5kJ = 499kJ + 20kJ
So the car has slowed to 99.9m/s
Does that sound right?
But that means when you fire at a target inside the car, the car is initially slowed by the recoil, but then accelerated again when the bullet strikes the target...but hitting the target will generate some noise and heat and absorb energy in deforming the bullet and the target, so the energy transfer isn't perfect, meaning the car will be slower than before you fired.
Which implies if you fitted a machine gun and a bulletproof plate inside the car it could slow the car down without any mass ever leaving the car...which doesn't seem right...
JimboM3 said:
This is similar to rockets burning fuel when approaching a planet (prograde) achieves a higher difference in velocity than the other way around (as learnt from kerbal space project!).
Funnily enough this came from trying to understand the Oberth effect in a post on the KSP forumsNorfolk B-roads said:
In your example, I think the 1kg mass would hit the target at 141m/s. i.e. sqrt( (5000J+5000J) / 0.5kg)
Edited by RizzoTheRat on Wednesday 12th February 12:05
Hang on, if Norfolk's right, then if I stood on the equator and fired a gun at mid day, the bullet would come out of the muzzle (relative to me) slower than if I fired it at midnight, because the earth's rotation means I'm travelling about 2000 mph faster at midnight. That can't be right can it?
Norfolk B-roads said:
RizzoTheRat said:
If I shoot out of the front of the car the bullet starts with 5000 J and gains 5000 J from the propellant, but hits the target doing 200m/s, so 20000 Joules.
Not sure that's correct. The energy required to accelerate a 1kg mass from 0 to 100m/s is not the same as the energy required to accelerate 1kg from 100m/s to 200m/s.
If you want the mass to be travelling at 200m/s relative to the target on the ground (remember, frame of reference is important if you want to measure speed), you'll need a bigger bang.
In your example, I think the 1kg mass would hit the target at 141m/s. i.e. sqrt( (5000J+5000J) / 0.5kg)
Norfolk B-roads said:
Relative to you, you are not travelling at all.
You can't chuck in an extra 2000mph without changing the frame of reference.
But surely that's the same in the original example?You can't chuck in an extra 2000mph without changing the frame of reference.
If the bullet were to hit the target on the ground at 141m/s then it must be doing 41m/s relative to the gun. The gun being in the car at 100m/s or on the equator at 1000mph is still a relative velocity of 0 relative to the person holding it.
Unfortunately it is even more complicated because the above all ignores relativity, which is fine at low speeds but as you approach the speed of light it gets very strange. If you are travelling at the speed of light “c” in a spaceship (not possible of course) and shone a torch out the front window the beam of light would travel at c, not 2c and similarly if you shine it out the back it would still travel at c, not zero. All very confusing (unless I assume you are a physicist).
Esceptico said:
Unfortunately it is even more complicated because the above all ignores relativity, which is fine at low speeds but as you approach the speed of light it gets very strange. If you are travelling at the speed of light “c” in a spaceship (not possible of course) and shone a torch out the front window the beam of light would travel at c, not 2c and similarly if you shine it out the back it would still travel at c, not zero. All very confusing (unless I assume you are a physicist).
Completely irrelevant to the questions asked. This reply is anti science.RizzoTheRat said:
Assuming a sphere moving through a vacuum etc...
If I have a gun that fires a 1kg bullet at 100m/s, that bullet hits its target with 5000 Joules of kinetic energy. Presumably all that energy has been transferred to it from chemical energy stored in the propellant.
If I'm in a car doing 100m/s and I drop the bullet out of the window it hits with 5000 Joules of KE, gained when the car accelerated to speed.
If I shoot something in the car it hits with 5000 Joules of KE
If I fire the gun out of the back of the car the bullet is stationary with 5000-5000=0 J of KE
If I shoot out of the front of the car the bullet starts with 5000 J and gains 5000 J from the propellant, but hits the target doing 200m/s, so 20000 Joules.
Where has this "extra" energy come from? Presumably the car has lost that much energy thanks to Netwon's third law, but that was caused by 5000J of chemical energy?
Am I wrong in thinking of KE as related to relative velocity?
Quick answer to "Am I wrong in thinking of KE as related to relative velocity?"If I have a gun that fires a 1kg bullet at 100m/s, that bullet hits its target with 5000 Joules of kinetic energy. Presumably all that energy has been transferred to it from chemical energy stored in the propellant.
If I'm in a car doing 100m/s and I drop the bullet out of the window it hits with 5000 Joules of KE, gained when the car accelerated to speed.
If I shoot something in the car it hits with 5000 Joules of KE
If I fire the gun out of the back of the car the bullet is stationary with 5000-5000=0 J of KE
If I shoot out of the front of the car the bullet starts with 5000 J and gains 5000 J from the propellant, but hits the target doing 200m/s, so 20000 Joules.
Where has this "extra" energy come from? Presumably the car has lost that much energy thanks to Netwon's third law, but that was caused by 5000J of chemical energy?
Am I wrong in thinking of KE as related to relative velocity?
Edited by RizzoTheRat on Wednesday 12th February 10:41
Answer is yes. You are wrong in thinking KE is related to a relative velocity. KE is related to velocity, which is a vector, so you can drop the relative part.
Lets leave relativity out of it until I've got my head around the Newtonian physics 
If I'm shooting a target by the roadside that velocity's obvious, if I'm shooting at another moving target it's a lot easier to think of it as a relative velocity, especially if that target is actually orbiting another planet
Gandahar said:
Quick answer to "Am I wrong in thinking of KE as related to relative velocity?"
Answer is yes. You are wrong in thinking KE is related to a relative velocity. KE is related to velocity, which is a vector, so you can drop the relative part.
That's getting in to semantics a bit though isn't it, both the bullet and the target have a velocity, I'm just using the term relative velocity to describe the vector which is the speed of the bullet in the direction of the target.Answer is yes. You are wrong in thinking KE is related to a relative velocity. KE is related to velocity, which is a vector, so you can drop the relative part.
If I'm shooting a target by the roadside that velocity's obvious, if I'm shooting at another moving target it's a lot easier to think of it as a relative velocity, especially if that target is actually orbiting another planet
You can't really forget relativity, because it's a relativity problem. As in the frame of reference is the key to the whole thing.
Where are you standing when you measure the speed of the bullet with a speedtrap gun? That will determine what result you expect to see on the speedtrap readout, which will affect the answer to the question how fast is the bullet travelling.
If you're standing on the car as the gun is fired forwards, the bullet will be doing 41m/s according to the speedtrap. If you're standing on by the roadside at the target it'll be doing -141m/s towards you. If you're standing on top of the bullet (somehow) and point the speedtrap at the bullet, its velocity will be zero.
Completely forget the 2000mph rotation of the earth, or the 67,000mph of the earth around the sun unless you're standing somewhere outside that frame of reference looking in, like from another solar system perhaps. Only then will it appear as though the bullet is hurtling through space at thousands of mph (the sum of all the vectors from the relative motions of the earth's orbit, rotation, the car, the bullet etc).
Assuming you're happy with how fast the bullet is travelling depending from where you are observing, you want to know why the sum of the two energies doesn't double the speed to 200m/s. You want to know what the relationship is between KE and velocity.
Simple, E = 1/2 mv^2, as you know.
The velocity of the bullet is proportional to the energy you fire it with (or drive it along the road with in a car), but not linearly. Non-linear means that if you double the energy you DON'T double the velocity. There's a squared term in there, so velocity increases proportional to the square root of the energy applied. So, 5000J from the car is enough to make the 1kg bullet go 100m/s, but for your next 5000J from the bullet propellant you only get 41m/s more.
Where are you standing when you measure the speed of the bullet with a speedtrap gun? That will determine what result you expect to see on the speedtrap readout, which will affect the answer to the question how fast is the bullet travelling.
If you're standing on the car as the gun is fired forwards, the bullet will be doing 41m/s according to the speedtrap. If you're standing on by the roadside at the target it'll be doing -141m/s towards you. If you're standing on top of the bullet (somehow) and point the speedtrap at the bullet, its velocity will be zero.
Completely forget the 2000mph rotation of the earth, or the 67,000mph of the earth around the sun unless you're standing somewhere outside that frame of reference looking in, like from another solar system perhaps. Only then will it appear as though the bullet is hurtling through space at thousands of mph (the sum of all the vectors from the relative motions of the earth's orbit, rotation, the car, the bullet etc).
Assuming you're happy with how fast the bullet is travelling depending from where you are observing, you want to know why the sum of the two energies doesn't double the speed to 200m/s. You want to know what the relationship is between KE and velocity.
Simple, E = 1/2 mv^2, as you know.
The velocity of the bullet is proportional to the energy you fire it with (or drive it along the road with in a car), but not linearly. Non-linear means that if you double the energy you DON'T double the velocity. There's a squared term in there, so velocity increases proportional to the square root of the energy applied. So, 5000J from the car is enough to make the 1kg bullet go 100m/s, but for your next 5000J from the bullet propellant you only get 41m/s more.
That was in response to a post about relativity, rather than relative velocities. Different things.
I completely disagree about this 41m/s thing though. If the gun fires a projectile at 100m/s when stationary, it must fire it at 100m/s when already moving at 100m/s. That was the whole point of bringing the 1000mph earth rotation in to it. If the bullet speed relative to the gun depended on the speed the gun was already travelling, it would fire it faster facing west than facing east, which is clearly b
ks. If the car was accelerating then yes it would leave the barrel more slowly, but not at a steady speed.
If a gun that fires at 100m/s when stationary fires from within the car which already doing 100m/s it will be doing 100m/s relative to the car and hit the target at 200m/s.
The "extra" energy has to come from the car being slowed down, as the propellant is imparting an equal and opposite force/impulse on both the car and the bullet. But I still can't quite get it clear in my head the difference when you fire from stationary
I completely disagree about this 41m/s thing though. If the gun fires a projectile at 100m/s when stationary, it must fire it at 100m/s when already moving at 100m/s. That was the whole point of bringing the 1000mph earth rotation in to it. If the bullet speed relative to the gun depended on the speed the gun was already travelling, it would fire it faster facing west than facing east, which is clearly b
ks. If the car was accelerating then yes it would leave the barrel more slowly, but not at a steady speed.If a gun that fires at 100m/s when stationary fires from within the car which already doing 100m/s it will be doing 100m/s relative to the car and hit the target at 200m/s.
The "extra" energy has to come from the car being slowed down, as the propellant is imparting an equal and opposite force/impulse on both the car and the bullet. But I still can't quite get it clear in my head the difference when you fire from stationary
Bill said:
Are we not into E=mc^2 territory.
That's about turning matter in energy, useful if we want to vaporise the car, bullet, and target with a nuclear explosion but a bit OTT for the moment Bill said:
The bullet fired out of the front of the car has 2x the speed, so 4x the energy.
Spot onRizzoTheRat said:
That was in response to a post about relativity, rather than relative velocities. Different things.
I completely disagree about this 41m/s thing though. If the gun fires a projectile at 100m/s when stationary, it must fire it at 100m/s when already moving at 100m/s. That was the whole point of bringing the 1000mph earth rotation in to it. If the bullet speed relative to the gun depended on the speed the gun was already travelling, it would fire it faster facing west than facing east, which is clearly bks. If the car was accelerating then yes it would leave the barrel more slowly, but not at a steady speed.
If a gun that fires at 100m/s when stationary fires from within the car which already doing 100m/s it will be doing 100m/s relative to the car and hit the target at 200m/s.
The "extra" energy has to come from the car being slowed down, as the propellant is imparting an equal and opposite force/impulse on both the car and the bullet. But I still can't quite get it clear in my head the difference when you fire from stationary
Kinetic energy is a relative quantity so you can only directly apply conservation of energy within a single reference frame. In your example you are jumping between difference reference frames, so the results make no sense.I completely disagree about this 41m/s thing though. If the gun fires a projectile at 100m/s when stationary, it must fire it at 100m/s when already moving at 100m/s. That was the whole point of bringing the 1000mph earth rotation in to it. If the bullet speed relative to the gun depended on the speed the gun was already travelling, it would fire it faster facing west than facing east, which is clearly b
ks. If the car was accelerating then yes it would leave the barrel more slowly, but not at a steady speed.If a gun that fires at 100m/s when stationary fires from within the car which already doing 100m/s it will be doing 100m/s relative to the car and hit the target at 200m/s.
The "extra" energy has to come from the car being slowed down, as the propellant is imparting an equal and opposite force/impulse on both the car and the bullet. But I still can't quite get it clear in my head the difference when you fire from stationary
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