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Ek0Ek1 > 0by May11
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#1
Apr712, 03:32 PM

P: 8

Hi there,
I need your help to prove the following, please: Ek0Ek1>0 Thanks in advance. (: 


#2
Apr712, 03:41 PM

Mentor
P: 18,334

Try to prove that Ek1Ek0<0
Seriously, if you don't explain your notation, then we can't help obviously. 


#3
Apr712, 04:18 PM

P: 8

Ek0 is the initial kinetic energy that an object possesses.
Ek1 is the eventual kinetic energy that an object possesses. Try to prove that Ek0Ek1 (energy loss) equals a positive value. 


#4
Apr712, 04:26 PM

P: 8

Ek0Ek1 > 0
Did you get my question? need a further explanation?



#5
Apr712, 04:31 PM

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P: 7,321




#6
Apr712, 04:39 PM

Mentor
P: 11,787

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?



#7
Apr712, 04:52 PM

P: 8

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 Ek0Ek1>0 ...that is pretty much all! we are not given any further... 


#8
Apr712, 04:53 PM

P: 8

Two masses, actually.



#9
Apr712, 04:54 PM

P: 8

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


#10
Apr712, 09:29 PM

Sci Advisor
P: 2,193

[tex]E_0E_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)[/tex] Now what can you say about whether or not this is positive? 


#11
Apr812, 05:16 AM

P: 8

Mv²/2 is undoubtedly positive.
1M/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! :) 


#12
Apr812, 08:42 AM

Sci Advisor
P: 2,193

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


#13
Apr812, 03:02 PM

P: 8

I will bear that in mind !



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