Falling objects: remedial education desperately needed

[usererror]

I apologize many times in advance for asking what I suspect to be that which is the most repeated question on this site. Last night over dinner with friends, somebody whose judgment was impaired dared me to prove that two falling objects released simultaneously (the usual suspects like friction etc. ignored) of mass m and mass 2m would move side by side. My friend played a two year old asking ”why, why why?” no matter what I said. Not a problem: I was a philosophy student, so I was – am - one of those kids. Anyway, I knew from Middle School the classical laws of motion and the law of gravity, so I dived in. First, I used Galileo’s (I think he said this) thought experiment of two falling masses chained together. But, that seemed to beg the question and so I had to buy another round. Then, I resorted to deriving (via algebra) from f = ma that the acceleration for 2m is 1/2 that for 1m, thus showing that the acceleration of m is twice of 2m. But so what? I needed to prove, but I failed to prove, that, at each instant, m and 2m are side by side. In fact, my showing of the differences in acceleration seemed to falsify my point. I had managed to confuse myself. I had to buy another round. There was laughter. I don’t speak Calculus. Is there an algebraic manipulation of Galileo’s laws that gives the answer? Better yet, is there a thought experiment that does this without begging the question? thank you, physics folk.

 

[bugatti79]

Check out this link

http://www.physicsclassroom.com/mmedia/newtlaws/efff.cfm

Should help you.

 

[Doc Al]

First, I used Galileo’s (I think he said this) thought experiment of two falling masses chained together. But, that seemed to beg the question and so I had to buy another round.

Galileo's clever argument should have been all you needed.

Then, I resorted to deriving (via algebra) from f = ma that the acceleration for 2m is 1/2 that for 1m, thus showing that the acceleration of m is twice of 2m.

The problem here is that you assumed that the force is constant, but it's not. The force of gravity is proportional to the mass. Twice the mass then twice the gravitational force. Just right to keep the acceleration constant.

 

[usererror]

Thanks to both responders. Now I get it. I am proud to have demonstrated that using the correct formula is not the same thing as understanding that formula. Thanks for the help.

 

[usererror]

"both" should be "all". (Is there anything else that I can screw up?)