Stone thrown in empty space

This is a supplementary question to What happens if object is thrown in empty space?

Via the following logic:

$$E = \frac{(mv)v}{2}\\ E = \frac{pv}{2}\\ \Delta E = \Delta p\times \Delta \frac{v}{2}\\ \Delta p = m\times \Delta v \\ \Delta p = ma\\ \Delta p = F\\ \Delta E = F \times \Delta \frac{v}{2}$$

Just when an object leaves our hand, the energy gets stored as Kinetic Energy, which can be represented as $\frac12 mv^2$. If we say that there was an initial kinetic energy and calculate the change in that, then $\Delta E$ gets related to $F$. If there is a force then there is acceleration as well, which means there should be acceleration in object.

Is the above logic correct? If not, why?

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There is another way. If you want to make a long comment, post it in this chat room, and link Wouter to it as well. To me, this question looks like a reply to that answer, hence I'm closing it. If you can make this question not look like a reply to that one (and an independant question), I can reopen it. – Manishearth Feb 20 '13 at 11:29
@Manishearth : No thanks, I am leaving this site, to one moderator it looks like a question, so he deletes this, to some one else, it looks like an answer so he closes this. Don't know what guys are up to.... There are people who down vote also, don't know why, may be they didn't understood it at that time, well some one else might understand it..... :) ... By the way thanks...... :) – viv Feb 20 '13 at 11:35
It doesn't look like an answer to me.. It looks like a long comment, directed towards Wouter. It's rather unclear. If you want to discuss with Wouter, use the chatroom I set up. If you want to ask a question then don't direct the question to Wouter. I was going to fix this question, but I couldn't tell what your intentions were :s – Manishearth Feb 20 '13 at 11:40
Well for a start it is not true mathematically that if $a = b \times c$ then $\Delta a = \Delta b \times \Delta c$. It isn't very hard to find numbers to plug in to this to give a counter-example. – Michael Brown Feb 20 '13 at 13:32
Bookkeeping aside: you can get this registered account merged with your previous unregistered account if you want. – dmckee Feb 20 '13 at 20:48

Well some how I figured it out, still don't know if it's completely correct. My Quest was if object is thrown the change in KE gets related to F , which gets related to a, that means a is not 0. So why will it not keep on accelerating or if it moves with constant speed then what this a denotes.

This has to be broken down in two steps. First :

A : When we pick the stone and throw it.
B : When it is thrown from our hand.

Since I am considering the ideal space.

ΔE b/w any point after B and B will be 0, no loss of energy.
ΔE b/w A and B will not be 0.

So this a is from A to B, and after wards object should move with constant speed.

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I'm having trouble understanding your writing, but I think you have it right. The object will only accelerate while there is a force acting on it. When there is no force on it there is no acceleration, and consequently neither the velocity or kinetic energy will change. – Michael Brown Feb 23 '13 at 9:23
hmmmmm....... . – viv Feb 23 '13 at 9:24

You first set $\Delta p = p$ (line 3), then $F = \Delta p$ (line 6), which leads to $F = p$.

Correct is that $F = \dot{p}$. Force equals the time derivative of momentum. Force does not equal momentum. So your equations are incorrect.

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