Proving Ek0-Ek1>0: A Step-by-Step Guide

[May11]

Hi there,
I need your help to prove the following, please:

Ek0-Ek1>0

Thanks in advance. (:

 

[micromass]

Try to prove that Ek1-Ek0<0

Seriously, if you don't explain your notation, then we can't help obviously.

 

[May11]

Ek0 is the initial kinetic energy that an object possesses.
Ek1 is the eventual kinetic energy that an object possesses.

Try to prove that Ek0-Ek1 (energy loss) equals a positive value.

 

[May11]

Did you get my question? need a further explanation?

 

[Integral]

Did you get my question? need a further explanation?

Sure does. This is simply not true in general. You need to tell about the system, and what it is doing. With the information given we can do nothing.

 

[jtbell]

Depending on the circumstances, an object's kinetic energy can either increase, decrease or remain constant. What are the circumstances in your case? What kind of object are we talking about? What forces act on it?

 

[May11]

It is a plastic collision, masses exert forces on each other, ending up with a joint velocity (U). We fisrt have to express the equation of the velocity at the end of the collision, that is : U= Mv/M+m
Then, express the equation of Ek0 and Ek1 :
Ek0= Mv²/2
Ek1= (m+M)u²/2 = M²v²/2(m+M)
Then, prove that Ek0-Ek1>0 ...that is pretty much all! we are not given any further...

 

[May11]

Two masses, actually.

 

[May11]

V1 (velocity of M before the collision) = v
V2 ( velocity of m before the collision) = 0

 

[Nabeshin]

It is a plastic collision, masses exert forces on each other, ending up with a joint velocity (U). We fisrt have to express the equation of the velocity at the end of the collision, that is : U= Mv/M+m
Then, express the equation of Ek0 and Ek1 :
Ek0= Mv²/2
Ek1= (m+M)u²/2 = M²v²/2(m+M)
Then, prove that Ek0-Ek1>0 ...that is pretty much all! we are not given any further...

Just write it out? It seems like you haven't even tried this.

$$E_0-E_1 = \frac{1}{2}Mv^2 - \frac{1}{2}\frac{M^2v^2}{m+M} = \frac{1}{2}Mv^2 \left( 1- \frac{M}{m+M} \right)$$

Now what can you say about whether or not this is positive?

 

[May11]

Mv²/2 is undoubtedly positive.
1-M/m+M:
1>M/m+M
M+m>M<1 = positive.
Umm, makes sense. The teacher said we have to use more formulas which are not given in the question, haven't expressed them, to branch out a little from what we are given. I will ask if your proof is valid and acceptable.
Thanks a heap! :)

 

[Nabeshin]

Umm, makes sense. The teacher said we have to use more formulas which are not given in the question, haven't expressed them, to branch out a little from what we are given. I will ask if your proof is valid and acceptable.
Thanks a heap! :)

Haha, that's an interesting stance to take. If you plan on continuing with physics, I suggest you try to get out of this mentality that there is a 'right' way of arriving at a solution. If a given derivation or proof seems logically sound to you, then it's good. I have some physics major friends who are in a similar mindset, and when doing problem sets with me they always say things like, 'can you do that?', as if there were some mystical physics police that sets the rules for how you approach physics problems! Of course there's not, and as long as you don't abuse math, everything's fine!

With re: to this problem, all the physics is essentially in solving for the final velocity (where you have to apply conservation of momentum).

 

[May11]

I will bear that in mind !