This is a very well known problem, but I can't find an answer in the specific case I'm looking for.
Let's consider two balls :
- Ball 1 weighs 10 kg
- Ball 2 weighs 1 kg
- Balls have identical volumes (so Ball 1 is much more dense)
- Balls have identical shapes (perfect spheres)
Let's drop them from a rather important height, on earth, WITH air. (That's the important thing, because all the proofs that I browse take place in a vaccum).
I am arguing with a colleague. He thinks that ball 1 will fall faster in air, and that the two balls will fall at the same speed in a vacuum. I think that the identical shapes and volumes make air friction identical too and that the vaccum has no importance here. Could someone tell who's right and provide a small proof?
Answer
I am sorry to say, but your colleague is right.
Of course, air friction acts in the same way. However, the friction is, in good approximation, proportional to the square of the velocity, $F=kv^2$. At terminal velocity, this force balances gravity,
$$ m g = k v^2 $$
And thus
$$ v=\sqrt{\frac{mg}{k}}$$
So, the terminal velocity of a ball 10 times as heavy, will be approximately three times higher. In vacuum $k=0$ and there is no terminal velocity (and no friction), thus $ma=mg$ instead of $ma=mg-F$.
No comments:
Post a Comment