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Bullet Fired From Moving Train Scenario

  1. Apr 10, 2010 #1
    Hello everyone. I am new here and am quite excited to start learning about physics which intrigues me greatly. I"m not studying physics or anything like that but just very interested in it and wish to work my way up to asking more advanced questions but first feel I need to be able to grasp some more primitive concepts.

    So here is the first one:

    if I'm standing on top of a train moving 500mph in the west direction, and I fire a bullet from a gun, what happens if the bullet travels at 500mph?

    If the train is going west and I fire the bullet west, does the bullet now travel 500mph from my frame of reference (since I'm already on a 500mph train) and thus 1000mph from an observer that's standing on the ground?

    And what happens if I fire the bullet in the east direction? Does an observer on the ground just see the bullet spin in place and drop on the ground near him?


    Thank you all.
     
  2. jcsd
  3. Apr 10, 2010 #2

    Doc Al

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    Staff: Mentor

    Yes. (Ignoring the effects of air resistance, of course.)

    Yes.
     
  4. Apr 10, 2010 #3
    Wow...thank you. But since a bullet can't just hover above the ground for an intermittent period of time..does that mean that basically as soon as the bullet is fired in the east direction, it falls to the ground right next to our observer who is by the side of the railroad tracks?
     
  5. Apr 10, 2010 #4

    Doc Al

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    Yes. Whether fired east or west, as soon as the bullet leaves the barrel it begins to fall. (Again, we ignore the significant effect of air resistance.)
     
  6. Apr 10, 2010 #5
    Thank you.

    Now to take it slightly to the next level, I basically used this example in an attempt to grasp the fundamentals in order to then try to grasp the fundamental of high light works from different frames of reference.

    So let's say I'm on that same train and I fire a laser pointer, light is now going in both directions at the same exact roughly 180k mps (miles per second) speed?

    I'm just trying to grasp how this concept works.

    For example let's say that the speed of light is exactly 180,000 miles per second. So let's say I am on a rocket that is going 175,000 mps and I fire a laser pointer in the same direction as the rocket is going. Obviously the light cannot possibly be now going 355,000mps because that would be super luminal so it remains the same speed 180k.
    However from an outside observer's perspective who is 'stationary' in relation to us ...what does he see?
     
  7. Apr 10, 2010 #6

    Doc Al

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    Every observer sees the light traveling at its usual speed. When dealing with speeds close to (or equal to) the speed of light, you must add the speeds relativistically. The usual addition of speeds which gave you Vbullet/ground = Vbullet/train + Vtrain/ground is only an approximation good for low speeds. (Low compared to the speed of light, that is.)
     
  8. Apr 10, 2010 #7
    What is the explanation for light always traveling at a constant? Is it the fact that light has no mass?

    For example I can understand why the bullet travels 1000mph when it is fired at 500mph and is already on a 500mph train. Because the bullet has mass and thus being loaded in the gun it is already accelerated to 500mph by the train since I am holding the gun and am rooted to the train by standing on it. So it can be said that that bullet is already traveling 500mph and then is being accelerated by the gun by an additional 500mph when it is fired for a total of 1000mph.

    But light I'm assuming has no mass and is therefore not 'accelerated' by whatever object it is on. Not to mention that light doesn't "sit in the chamber" like a bullet but rather is generated at the spot at the moment of its creation and then leaves at its constant speed.
    But my question is, is it the mass that is responsible for it, or is it the fact that like I said it doesn't sit 'cocked in the chamber' and is generated in that one moment only?
     
  9. Apr 10, 2010 #8

    Doc Al

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    I'd say it's better to realize that space & time do not work as you might expect. Even for things with mass, speeds don't simply add (as I'm sure you realize, since the speed of light is the universal "speed limit"). For example, let your train move relativistically at 0.9c. Then shoot your rocket from the train at 0.9c. Using pre-relativistic kinematics (galilean relativity), the speed of the rocket with respect to the ground would be 0.9c + 0.9c = 1.8c. But those speeds really add to (0.9c + 0.9c)(1 + 0.9*0.9) = 0.994c. Which is less than the speed of light.

    But it's true that being massless allows light to travel at its speed. And anything moving at that speed--the speed of light--in one frame will move at that same speed in any frame.
     
  10. Apr 10, 2010 #9
    thanks for clearing that up.

    Can you tell me please why if light is massless, does it exert minute force/pressure on things?

    Is it because light has mass but is very tiny and negligible? Or does no one really know the answer to that.
     
  11. Apr 10, 2010 #10

    Doc Al

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    Despite being massless, it still has momentum and energy.
     
  12. Apr 10, 2010 #11
    But

    1. I thought that momentum is a product of mass and velocity. We know the velocity of light, that means it must have mass also, no?

    2. you said that it has energy. well according to e=mc2 whatever has energy must have mass and vice versa, no? Or am I understanding that wrong.
     
  13. Apr 10, 2010 #12

    Doc Al

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    No. For massive particles, momentum is no longer simply mass X velocity, but [itex]mv/\sqrt{(1 - v^2/c^2)}[/itex]. For everyday slow speeds (compared to light speed), that is closely approximated by the familiar mass x velocity.

    For massless particles, momentum can be shown to equal energy/c.

    E = mc² refers to the "rest energy" something has by virtue of its rest mass. Light has no rest energy.
     
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