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At the bottom of this page says (my emphasis):

It's interesting to note the speeds in the cartoon. The propeller-beanie kid sees his tennis ball moving away from him at 30 miles per hour. So does the Sun, sitting on the stationary platform. The engineer driving the train sees the ball coming at about 80 MPH, since the train is moving 50 MPH with respect to the ground. The train and ball interact at 80 MPH. The ball rebounds from the front of the train at nearly the same 80 MPH, which can be added to the 50 MPH speed of the train, because it acquired it from the train. The result approaches a total of 130 MPH. This scenario is analogous the velocity of a spacecraft being added to the velocity of the massive speeding planet, and "rebounding" with a higher velocity still (although the spacecraft's "rebound" is a gravitational, rather than a mechanical, interaction, like in the baseball analogy).

I do not totally understand why the speed of the train has to be added (again?). It seems to me that the ball has to leave (rebound) at 80 MPH (30 plus 50).

A clearer explanation is really appreciate. It can be a silly question, but I cannot see it :(

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  • $\begingroup$ BTW, I see this link and has a more detailed (and related) explanation. Thank you all. $\endgroup$ – JuanCa Oct 10 '19 at 13:27
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If you throw a ball off an immovable wall it bounces back at approximately the same speed according to Newton’s laws. If we are in the reference frame of the train then we can treat the train like an immovable wall. So in that reference frame the ball comes in at 80 mph and bounces back at 80 mph (again, still in the reference frame of the train). In order to convert this to the speed perceived by the propeller-beanie kid in his stationary reference frame, we need to add the speed of the train. Hence the 130mph.

A flipped version of this would be if the kid was on the train and throwing the ball at a stationary wall. From the wall, the kid would be throwing the ball at 30 + 50 = 80mph, and it would rebound back at 80mph in the wall’s reference frame. But from the kid on the train’s reference frame, the ball would be bouncing away at 130mph because the train is moving 50mph away from the 80mph traveling ball.

One final alternative example that would be more in line with how you’re thinking about it. Imagine the kid is on top of the train, throwing the ball at a wall that is also on top of the train. If he throws it at 30mph, it rebounds at 30mph relative to the train. Relative to the ground it would be going 80mph because the train is traveling 50mph. The only difference between this example and the original is the kid is now on the train so we don’t have to add the trains speed at the end of the calculation to get the speed of the ball relative to the ground.

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    $\begingroup$ I think this is a really fantastic first post here. I was trying to think of a simple way to word it; but I think showing the multiple reference frames was a really good way to make it feel more intuitive. Perhaps the only thing I would add is to clarify that the equal return ball bounce only applies to an immovable stationary wall. I think that would completely tie in why you are choosing these reference frames. $\endgroup$ – JMac Oct 8 '19 at 23:07
  • $\begingroup$ Thanks, the last example is very good to me. $\endgroup$ – JuanCa Oct 9 '19 at 15:06
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This is easiest to understand if you consider it from the rest frame of the train.

In that frame the ball and the train meet at 80mph. Since the train is so much heavier than the ball, the ball will bounce off at virtually 80mph relative to the train. Does that make sense?

Ok, now convert the ball's speed to the ground frame, which is moving 50mph faster than the stationary train. Hence 130mph.

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  • $\begingroup$ Where did you get those 80 mph? $\endgroup$ – JuanCa Oct 8 '19 at 19:48
  • $\begingroup$ @JuanCa, assuming an elastic collision, the ball will rebound at the same speed that it hit the train, and it will travel in the opposite direction upon rebound. $\endgroup$ – David White Oct 8 '19 at 19:51
  • $\begingroup$ Yes, but (at least for me) the speed will be 80 MPH: 30 from the ball (this would be the rebound speed if the train was stationary) plus 50 (train speed), not 130. :( $\endgroup$ – JuanCa Oct 8 '19 at 19:57
  • $\begingroup$ I will edit my answer t clarify this. $\endgroup$ – Marco Ocram Oct 8 '19 at 19:58

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