The reasoning behind solving this equation

C

cherishablegirl

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Hi everyone,
your help would be appreciated so much!

I can't for the life of me get to the right answer for this question. I don't know where I'm going wrong in my algebra but I cannot get the answer. So frustrating!! And even worse because I know that it is actually really simple!!

The question is you are given two equations for special relativity:
E^2 = m^2c^4 / 1 - (v^2/c^2) and p^2 = m^2v^2 / 1 (v^2 / c^2)

It then says "if v = 1/2c, which of the following are both true?"

The answer iis E = 2mc^2/(squareroot 3) and p = mc/(squareroot 3)

I don't understand where that square root of 3 is coming from. I think my main problem is the complex fraction.

Please show me the steps to work this out - I'll be forever grateful!!!

Thank you!!
 
cherishablegirl said:
don't know where I'm going wrong in my algebra

E^2 = m^2c^4 / (1 - (v^2/c^2))

p^2 = m^2v^2 / (1 - (v^2/c^2))

v = 1/2c

which of the following are both true?

The answer iis

E = 2mc^2/(squareroot 3)

p = mc/(squareroot 3)

I don't understand where that square root of 3 is coming from
Click to expand...

Hello:

I fixed your typing above.

The sqrt(3) comes from taking the square root of both sides of the equations, after substituting v=1/2c and simplifying.

Please show us what you get, after replacing v with 1/2*c and simplifying.

Cheers :cool:
 
You told us that you think that you're going wrong in your algebra. I'm asking you to show us what you tried.

You have an equation that defines E^2.

E^2 = m^2c^4 / (1 - (v^2/c^2))

Do you see the symbol v above? (I highlighted it in red.)

We're told that symbol v is the same as (1/2)(c)

Therefore, you need to replace that v with c/2 and simplify the result.

What do you get, when you substitute c/2 for v?
 
I'm not sure how you square root 1/2c^2. I would have thought you'd want to multiply it by 2 to make it a whole number.

E^2 = m^2c^4 / (1-(1/2c^2/c^2)

square rooting

E = mc^2 / (1 - (1/2c/c))

am I even close up to this point? lol :(
 
am I meant to then multiply both sides by 2? :/
Thanks for your help I really appreciate it.
Having gaps in my math knowledge is so frustrating.
 
cherishablegirl said:
I'm not sure how you square root 1/2c^2. I would have thought you'd want to multiply it by 2 to make it a whole number.
Click to expand...
Um... If you think that one rationalizes that denominator by multiplying something (what?) by 2 to make something (what?) "a whole number", then it's very possible that you need much more help than we can here provide. :oops:
 
stapel said:
Um... If you think that one rationalizes that denominator by multiplying something (what?) by 2 to make something (what?) "a whole number", then it's very possible that you need much more help than we can here provide. :oops:
Click to expand...

You probably didn't have to write that in such a way to make me really doubt myself and feel terrible...

I think I do need to multiply in the denominator, to remove the 1/2. Multiplying by 4 would remove this.

Is this wrong?
 
I understand that you must get an influx of teenagers here wanting you to do their homework for them.
I can assure you that I'm not.

I've actually completed Calculus in my undergraduate degree and did very well (god knows how...). I have gaps in my knowledge that I'm trying to fill by reading up and trying questions.

I'm feeling pretty helpless after your comment.
 
cherishablegirl said:
I understand that you must get an influx of teenagers here wanting you to do their homework for them.
I can assure you that I'm not.

I've actually completed Calculus in my undergraduate degree and did very well (god knows how...). I have gaps in my knowledge that I'm trying to fill by reading up and trying questions.

I'm feeling pretty helpless after your comment.
Click to expand...
Cherish, I can totally feel your frustration in your text. Normally we don't show step by step how to do a problem. Instead, we give hints and suggestions and try to get the student to discover the solution themselves.

HOWEVER, I am going to make an exception and show you how they got E and you can apply the same logic to get p.

\(\displaystyle E^2=\dfrac{m^2c^4}{1-\dfrac{v^2}{c^2}}\)

Now, substitute \(\displaystyle v=\dfrac{1}{2}c=\dfrac{c}{2}\) into the above equation.

\(\displaystyle E^2=\dfrac{m^2c^4}{1-\dfrac{\left(\dfrac{c}{2}\right)^2}{c^2}}\)

\(\displaystyle E^2=\dfrac{m^2c^4}{1-\dfrac{\left(\dfrac{c^2}{4}\right)}{c^2}}\)

Now, \(\displaystyle \dfrac{\left(\dfrac{c^2}{4}\right)}{c^2} = \dfrac{c^2}{4} \cdot \dfrac{1}{c^2} = \dfrac{c^2}{4c^2} = \dfrac{1}{4}\)

\(\displaystyle E^2=\dfrac{m^2c^4}{1-\dfrac{1}{4}}\)

\(\displaystyle E^2=\dfrac{m^2c^4}{\dfrac{3}{4}} = \dfrac{m^2c^4}{1} \cdot \dfrac{4}{3} = \dfrac{4m^2c^4}{3}\)

Take the square root of both sides and you get:

\(\displaystyle E=\pm \sqrt{\dfrac{4m^2c^4}{3}} = \pm \dfrac{\sqrt{4m^2c^4}}{\sqrt{3}}\)

\(\displaystyle E=\dfrac{2mc^2}{\sqrt{3}}\) assuming E must be the positive root.
 
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srmichael said:
Cherish, I can totally feel your frustration in your text. Normally we don't show step by step how to do a problem. Instead, we give hints and suggestions and try to get the student to discover the solution themselves.

HOWEVER, I am going to make an exception and show you how they got E and you can apply the same logic to get p.

\(\displaystyle E^2=\dfrac{m^2c^2}{1-\dfrac{v^2}{c^2}}\)

Now, substitute \(\displaystyle v=\dfrac{1}{2}c=\dfrac{c}{2}\) into the above equation.

\(\displaystyle E^2=\dfrac{m^2c^2}{1-\dfrac{\left(\dfrac{c}{2}\right)^2}{c^2}}\)

\(\displaystyle E^2=\dfrac{m^2c^2}{1-\dfrac{\left(\dfrac{c^2}{4}\right)}{c^2}}\)

Now, \(\displaystyle \dfrac{\left(\dfrac{c^2}{4}\right)}{c^2} = \dfrac{c^2}{4} \cdot \dfrac{1}{c^2} = \dfrac{c^2}{4c^2} = \dfrac{1}{4}\)

\(\displaystyle E^2=\dfrac{m^2c^2}{1-\dfrac{1}{4}}\)

\(\displaystyle E^2=\dfrac{m^2c^2}{\dfrac{3}{4}} = \dfrac{m^2c^2}{1} \cdot \dfrac{4}{3} = \dfrac{4m^2c^2}{3}\)

Take the square root of both sides and you get:

\(\displaystyle E=\pm \sqrt{\dfrac{4m^2c^2}{3}} = \pm \dfrac{\sqrt{4m^2c^2}}{\sqrt{3}}\)

\(\displaystyle E=\dfrac{2mc}{\sqrt{3}}\) assuming E must be the positive root.
Click to expand...

THANKYOU!!!
this question has been driving me crazy for days and I finally get it.
Just to note though in case a student uses this for help, the E^2=m^2c^4 - you had put ^2 not ^4 - but everything else holds true and I can see it.

Thank you so much :)
 
cherishablegirl said:
THANKYOU!!!
this question has been driving me crazy for days and I finally get it.
Just to note though in case a student uses this for help, the E^2=m^2c^4 - you had put ^2 not ^4 - but everything else holds true and I can see it.

Thank you so much :)
Click to expand...

You're welcome!

I did not even catch that little typo of mine. Thanks for pointing it out. I corrected my original post.
 
cherishablegirl said:
I can assure you that I'm not [a student who is trying to get their homework done for them].

I've actually completed Calculus in my undergraduate degree and did very well
Click to expand...

You ought to explain what you're trying to accomplish, in your first post. For example, why are you doing this exercise? Is it self-review?

Volunteer tutors cannot know what your situation is, until after you tell them.

Please read this summary page of our guidelines, before posting.

Thank you! :cool:
 
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Denis said:
found that to be a "weird" problem
Click to expand...

Sometimes, you seem too smart for your own good, Denis. ;)

This near-beginning algebra exercise involves nothing more than substitution, simplification, and order-of-operation concepts.


no "SQRT(3)" within reach
Click to expand...

I think that you would get zero credit, on this exercise, lol.

See srmichael's explanation...
 
cherishablegirl said:
I finally get it.
Click to expand...

I'm glad. (So, you're okay with the second part, now?)

Though, I'm a bit puzzled why you could not substitute c/2 for v, and thus show srmichael's first step (last night)...

Check out the Read Before Posting info. I think that will save you some time, in future requests for tutoring here.
 
Quaid said:
Sometimes, you seem too smart for your own good, Denis. ;) ..................Yeah - go to the corner now - hockey-puck!!

This near-beginning algebra exercise involves nothing more than substitution, simplification, and order-of-operation concepts.




I think that you would get zero credit, on this exercise, lol.

See srmichael's explanation...
Click to expand...
.
 
Quaid said:
I'm glad. (So, you're okay with the second part, now?)

Though, I'm a bit puzzled why you could not substitute c/2 for v, and thus show srmichael's first step (last night)...

Check out the Read Before Posting info. I think that will save you some time, in future requests for tutoring here.
Click to expand...

Yeah the way he had written it made complete sense. For some reason I was putting 1/2 c into the equation, instead of c/2, so I was totally confusing myself.

I know it's just an easy substitution question, I obviously need to brush up on my algebra rules, it's been way too long! :)
 
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