# Kinematics Thought Problem

1. Apr 3, 2008

### thewylie

I'm trying to explain to a coworker some fallacies in a thought problem he came up with that has severely confused him on some basic principles of clasical physics... but I am having some issues. He is capable of understanding the math, but he is not capable of seperating intuition from mathematical equations.

His thought problem, as presented to me, was this (Using the frame of reference of a person on the ground):

Edit ... I wanted to present the thought problem he had to see how others tore into it... instead ... I'll simply ask this question (same frame mentioned above).

If stationary person (A) shoots a bullet from a gun and we record the muzzle velocity as n meters per second, what will the muzzle velocity of a similar bullet (same starting potential energy in the accelerant) shot from a similar gun (same barrel length) if fired by a person (B) on a superfast train which is already moving at a velocity equal to that of the muzzle velocity of the first bullet (n m/s) in terms of n?

I'm pretty sure I've got the right answer but don't want to pollute the question with my own mathematically derived assumtions the way my coworker was polluting his version of this problem with his own (magically? derived) assumptions.

Looking forward to your equations and examples.

2. Apr 3, 2008

### Joker

Last edited by a moderator: Apr 23, 2017
3. Apr 3, 2008

### thewylie

Excellent! So far so good, which leads to the next question:

The train has transferred a certain amount of energy (x) to its bullet, accelerating that bullet to n meters / second. The gun on the gound spent the same amount of energy (x) accelerating its bullet to n meters / second. According to the answer implied by the link-post above the same gun fired from the front of the train would shoot it's bullet out (relative to the person standing on the ground at 2 * n meters / second. Well with some very basic physics math we can determine that the bullet moving at 2n (m/s) has had 4 times the energy (4x) transferred to it than the original bullet (1/2 m*v$$^{}2$$). The gun only has x amount of energy to accelerate it bullet. The train already used only x amount of energy accelerating its bullet to n (m/s). So we can account for 2x amount of energy. Where does the other 2x come from? or ... if it's there and being missed by this question set ... where is it being missed?

4. Apr 3, 2008

### thewylie

I really don't want to pollute this thread with my own ideas too much because I'm trying to use it as an outside source for my coworker to understand... if he thinks I "convinced" someone else to follow "my way of thinking" (read: clasical physics) ... he'll just scoff it off and I won't be any further towards his enlightenment ...

that said ... think: KE is relative to your frame of reference ...

5. Apr 3, 2008

### Joker

the way is see it is
1
if you are on ground ,u see the bullet it will be moving with a velocity 2n so its energy will be 2mn^2

2
if you are on the train ,u see the bullet moving with a velocity n so its energy will be 1/2 *mn^2

6. Apr 3, 2008

### Danger

Welcome to both of you. I just want to point out an error which is strictly a matter of terminology. The muzzle velocity is the same in both cases; that's the velocity (really only speed, since the vector is always in line with the barrel) which the propellant charge imparts to the bullet relative to the gun. The actual velocity, relative to the observer, is as you stated.
I hope that it doesn't seem overly picky of me, but proper wording can be important.

7. Apr 3, 2008

### Mapes

You're ignoring conservation of linear momentum. If the bullet acquires forward momentum, something needs to acquire momentum in the opposite direction.

It may make things easier to assume that the gun fires two bullets, one forward and one backward. This automatically conserves linear momentum. Say that each bullet has mass m. If the gun is perfectly efficient, it takes energy mv2 to shoot the two bullets. Now assume the gun (which is very light) and the bullets (whose mass is a total of 2m) is accelerated to speed v. This takes energy mv2. If the gun is fired at speed v, one of the bullets is going speed 2v and has energy 2mv2 and one of the bullets is motionless with no kinetic energy. Now the energy expended matches the energy of the final bullet, problem solved.

You can make things more complicated by assuming the backward bullet is instead combustion gases with a different mass, or assuming that the train slows down, etc., but you need to acknowledge conservation of momentum to resolve your paradox. Basically, shooting the gun expends more energy than is taken up by the bullet.

8. Apr 4, 2008

### thewylie

We don't "Know" the gun is fired at speed v on the train... What we do know is both guns fire and have the same amount of potential energy in the gunpowder with which to accelerate the guns vs. bullets in opposite directions.

If though, in some experiment where the final velocity of the forward moving bullet were measured to be 2v relative to the the ground observer, then would not the bullet going 2v have Ek1 = m(v+v)2 while the bullet going the opposite direction would now have energy Ek2 = m(v-v)2 (or 0)?

Last edited: Apr 4, 2008
9. Apr 6, 2008

### Mapes

No. Kinetic energy is 1/2 mv2. The bullet has mass m. If its speed is 2v, its kinetic energy is 2mv2.

10. Apr 6, 2008

### luben

do you mean some energy seems to have "lost" in one frame and "gained" in another? I think this is what your try to point out. But it is natural since the accelerating bullet is in non-inertial frame, and conservation of energy cannot be accounted for just in that frame. something outside that frame (in this case the train) has to participate in the conservation.