Special Relativity Moving Frames Question

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SUMMARY

The discussion centers on understanding the behavior of a bullet shot straight up from a moving cart in different reference frames, specifically the cart's frame and the ground's frame. Participants agree that in the cart's reference frame, the bullet travels straight up and down, while from the ground's perspective, it follows a parabolic trajectory due to the cart's horizontal motion. The key equation discussed is cos(θ) = v_cart / v_bullet, which relates the angle of the bullet's path to the velocities involved. The conclusion emphasizes that the bullet should be shot straight up, assuming no air resistance and no acceleration of the cart during the bullet's flight.

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  • Understanding of reference frames in physics
  • Familiarity with projectile motion concepts
  • Knowledge of basic trigonometry, specifically cosine functions
  • Concept of non-accelerating frames of reference
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  • Study the principles of special relativity and reference frames
  • Learn about projectile motion and its equations
  • Explore the implications of air resistance on projectile trajectories
  • Investigate Einstein's thought experiments related to moving frames
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Homework Statement


I am sort of reviving this thread.
https://www.physicsforums.com/showthread.php?t=182511
My question is the same.
I understand and agree with what fisix responded a few years ago.

Homework Equations


cos( θ)= v_cart / v_bullet


The Attempt at a Solution


I can gather that in the reference frame of the cart, the bullet will appear to just go straight up and straight down.
And, in the reference frame of the ground, the bullet will look following a parabolic path.
Nevertheless, I cannot discern how to obtain the angle.

It seems to me that it has to be lesser or equal than pi/2.
I am not looking for the direct answer, but I just don't know how to proceed.

Thank you.
 
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angle...

This reminds me of the Einstein thought experiment with a person standing on a moving train, tosses a ball straight up and sees it move straight up, then straight down right back into their hand.

Note, if you assume no air, the lack of a train surrounding the cart/cannon and ball makes no difference, you should shoot the cannon ball straight up, given the cart experiences no acceleration during the balls flight and there is no air resistance (as in a train, the air travels with you), the ball will naturally return from whence it was shot, the cart frame of reference is non-accelerating and therefore the ball remains in its frame at least in the direction of the carts travel, of course in the direction straight up it enters another frame, but at right angles to the carts direction and so the net effect is zero on that vector.

Now, from someone on the ground observing the cart, the cannon it carries and the ball it shoots, the situation appears very different.

Remember, the ball already has a velocity in the direction of the cart when it is 'tosses' up, and given no resistance from air, will happily keep this velocity during its flight.

So answer is, shoot the ball straight up! Don't accelerate the cart during flight, given no air resistance, get out of the way for its return as there's no terminal velocity involved to help slow the balls return :-)
 
Last edited:
Thanks.
 

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