What if the guy on the train throws the ball backwards (u'=-.99c)?

[jdavel]

A train is moving to the right at v=.5c. Someone standing on the train throws a baseball forward (to the right) at u'=.99c (relative to him). Relative to somone on the ground, the ball appears to be traveling at a speed u which is slightly faster than .99c, but still less than c.

What if the guy on the train throws the ball backwards (u'=-.99c)? Now what speed does the guy on the ground see? Try to get the approximate answer without looking at the velocity addition formula.

 

[NateTG]

What if the guy on the train throws the ball backwards (u'=-.99c)? Now what speed does the guy on the ground see? Try to get the approximate answer without looking at the velocity addition formula.

My best guess is a speed of slightly less than .99c.

A quick thought on the topic:
Let's say we've got an explosion pushing two balls of equal mass out of a pipe. Then linear momentum is conserved if both balls go out of the pipe in opposite directions at .99c. Now, if we put this double-sided gun on the relativistic train, the outside observer will still need to have linear mometum conserved, and we know that in a absolute sense the momentum change from .5c to .99c is much much larger than the momentum from 0c to .5c. Therefore, the ball will be seen as traveling backwards at more than .5c.

 

[R0man]

If the guy on the train throws the ball backwards at u'=-.99c, then relative to someone on the ground, the ball will appear to be moving at a speed slightly less than .99c. This is because the train is already moving at v=.5c, so when the ball is thrown backwards, its overall velocity will be slightly less than .99c in the opposite direction.

To get an approximate answer without using the velocity addition formula, we can estimate that the ball will be moving at a speed of approximately .49c relative to someone on the ground. This is because the train is moving at .5c, so the ball's velocity in the opposite direction will be slightly less than .99c. And since .5c is half of c, we can estimate that the ball's velocity will also be half of .99c, which is .49c.

However, to get a more precise answer, we can use the velocity addition formula to calculate the exact velocity of the ball relative to someone on the ground. This formula is u=(u'+v)/(1+uv/c^2), where u' is the velocity of the ball relative to the train, v is the velocity of the train, and c is the speed of light. Plugging in the values, we get u=(-.99c+.5c)/(1+(-.99c)(.5c)/c^2) = -.49c/(1-.495) = -.49c/0.505 = -.97c.

Therefore, the guy on the ground will see the ball moving at a speed of approximately -.97c, which is slightly less than the estimated speed of .49c. This shows that even though the ball is thrown backwards on a train moving at a high speed, its overall velocity relative to someone on the ground will still be less than the speed of light.