View previous topic :: View next topic 
Author 
Message 
samthompson
Joined: 06 Feb 2012 Posts: 4 Location: California

Posted: Tue Feb 07, 2012 6:24 pm Post subject: additive velocities 


Assume that I can throw a football consistently so that it leaves my arm at 50 mph. Now imagine I get on train and it accelerates to 50 mph at which time I throw the ball.
How fast is that ball moving with respect to the train and how fast with respect to the rail road tracks? Assume that the moving air plays no role when answering the question. I know that the ball will slow down once it leaves my hand.
Thanks,
Sammy 

Back to top 


hepcj Site Admin
Joined: 23 Jun 2007 Posts: 125

Posted: Tue Feb 07, 2012 9:40 pm Post subject: 


Think about what happens when you walk down the isle of a railway carriage. What is your velocity with respect to the train?
How about your velocity as you walk with respect to someone standing at the platform when the train passes by.
The speed relative to the ground also depends on which direction you throw the ball. i.e. in the direction of travel of the train or in the opposite direction.
Incidentally, the ball will not slow down appreciably as it leaves your hand as the horizontal component of the velocity is constant for projectile motion. 

Back to top 


samthompson
Joined: 06 Feb 2012 Posts: 4 Location: California

Posted: Thu Feb 09, 2012 4:51 pm Post subject: a brief followup if you do not mind. 


Thanks for the information and I'm sorry for not indicating that the throw on the train was to be in the forward direction.
Your answer confirmed my suspuscions. It also raised a disturbing question. Naturally, when the train accelerates after I get on it, it uses energy to accomplish the task of reaching 50 mph. Of all the energy that goes into accelerating the train, a small portion of that amount actually accelerates the ball to 50 mph. I imagine that this amount is equal to the amount I apply to the ball when I throw it.
If I throw the ball (in the direction the train travels while it is moving) I use the same amount of energy used to get it up to 50 mph. This means the ball should have twice the kinetic energy but it does not, does it?
This seems to suggest there is something wrong, either with your answer or the kinetic energy formula.
Thanks,
Sam 

Back to top 


hepcj Site Admin
Joined: 23 Jun 2007 Posts: 125

Posted: Fri Feb 10, 2012 9:18 pm Post subject: 


I don't quite understand what you mean.
The kinetic energy depends on the relative velocity. When you are on the train, you and the ball are moving at 50 mph, call it V. The acceleration that occurs to get you to this speed is not important. If you throw the ball at a speed v, then the kinetic energy as measured by another person who is also on the train is 0.5 mv^2 because they are moving with the train. After all, you don't consider the velocity at which the Earth is rotating when you throw you throw the ball. But for someone standing outside the train and you throw the ball out of the window at speed v. in the direction of travel. The kinetic energy will be 0.5 m(V+v)^2 because they are not moving with the train. 

Back to top 


samthompson
Joined: 06 Feb 2012 Posts: 4 Location: California

Posted: Fri Feb 10, 2012 10:30 pm Post subject: Making my question clearer, I hope 


First things first, thank you for sticking with me. Obviously, I am not making myself clear; please allow me to rephrase using your mathematical notations. Also, I don’t think we should get into the fact that the Earth moves because that is adding unnecessary complications to a very simple issue.
If I throw the ball at v, I do work to the ball and this results in the ball having kinetic energy equal to 0.5mv^2. When I’m on the train moving at v and I throw the ball again, I have doubled the ball’s velocity with respect to the rail road track. At that point, the ball moves at 2v with respect to the rail road tracks and thus has a kinetic energy value equal to 0.5m (v+v)^2 ; just as you said. (I see that you used “V” and “v” to show the two different velocities but they are both equal to 50 mph traveling in the same direction). If you subtract the .05mv^2 from the 0.5mv(v+v)^2, you will get a value that is much larger than the 0.5mv^2 value. In other words, even though I use exactly the same amount of muscle power both times, the work done on the ball is much greater just because I happen to be moving. If we used a cannon instead of my arm, the chemical energy of the gun powder should always translate into the same amount of kinetic energy, shouldn’t it?
Is it possible that there is something wrong with the kinetic energy formula and work? Do you know where work and kinetic energy came from? What is the experimental proof for kinetic energy?
Once again thanks,
Sam 

Back to top 


hepcj Site Admin
Joined: 23 Jun 2007 Posts: 125

Posted: Sat Feb 11, 2012 1:18 am Post subject: 


The work done by the person throwing the ball is the same in each case and the extra kinetic energy comes from the train. If it is moving at velocity v and you throw the ball at a speed v at a person who is stationary then the additional K.E. comes from the moving train. 

Back to top 


samthompson
Joined: 06 Feb 2012 Posts: 4 Location: California

Posted: Sat Feb 11, 2012 12:53 pm Post subject: 


After reviewing everything you said and visiting a very unique website (physics2012.com), I have no choice but to disagree with your answer. There really is something wrong with the kinetic energy formula. To accelerate the ball to 50 mph, you have to do work equal to 0.5mv^2. If I throw it again while I'm moving at 50 mph, it means I'm doing the same work and the ball's kinetic energy should double to mv^2. Use the kinetic energy formula and it produces 0.5m(V+v)^2, which is of course is much larger.
I think that I have to go back to that website and finish going through it. I only made it to the page called "the experiment". That site said I should check everything as I went along. That is what I was doing when I came to splung.com I wanted to find out if additive velocities were fact or fiction and to see what that meant in terms of the workenergy theorem.
Thanks,
Sam 

Back to top 


