# Isaac Physics 'Algebraic Manipulation 5.4'; does this make sense?

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Thread starter 10 months ago
#1
https://isaacphysics.org/questions/m...7-244d29d02aa0

I'm quite confused on this question. You have four variables (c, v, Theta, Gamma), with three equations, and you have to cancel it down into an expression that consists of only one of these variables. I've tried thinking of different ways of getting around the problem, however I feel like it must be a question that pieces together all at the last steps.

For instance, if you rearranged the two equations for Gamma and equated them equal to one another, that helps to cancel one variable, but you're still left with three and only one unused equation?

I expected it to be problem where you just find both Wx and Wy in terms of either c, v, or Theta and substituted them in without the need of long expansions for the equation for W, however I have a feeling that the only way you end up getting an answer in the correct form is through the long expansions in hopes of things cancelling? Am I correct in assuming this?
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10 months ago
#2
(Original post by domm1)
https://isaacphysics.org/questions/m...7-244d29d02aa0

I expected it to be problem where you just find both Wx and Wy in terms of either c, v, or Theta and substituted them in without the need of long expansions for the equation for W, however I have a feeling that the only way you end up getting an answer in the correct form is through the long expansions in hopes of things cancelling? Am I correct in assuming this?
I wouldn't do any rearranging. Just work out expanding as necessary. Should take about 6+ lines; at least that's how long I took.
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Thread starter 10 months ago
#3
(Original post by ghostwalker)
I wouldn't do any rearranging. Just work out expanding as necessary. Should take about 6+ lines; at least that's how long I took.
I've expanded it out but it doesn't seem to look as if it's going to simplify into an expression of just one term?

Would you be able to post your workings so I can see where I've gone wrong please?
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10 months ago
#4
(Original post by domm1)
I've expanded it out but it doesn't seem to look as if it's going to simplify into an expression of just one term?

Would you be able to post your workings so I can see where I've gone wrong please?
Why don't you post *yours*? I can confirm it simplifies to a simple expression in one variable in a few lines.
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Thread starter 10 months ago
#5
(Original post by DFranklin)
Why don't you post *yours*? I can confirm it simplifies to a simple expression in one variable in a few lines.
My reply was to ghostwalker, not you...

But thanks for your input.
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10 months ago
#6
(Original post by domm1)
My reply was to ghostwalker, not you...

But thanks for your input.

My working isn't in a state to photograph, and I don't fancy typing out all the LaTex. As DFranklin said, post your working.
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1 week ago
#7
Hi- I don't suppose that you still know/have the answer to 5.4 do you?
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1 week ago
#8
(Original post by Jordan Inglis)
Hi- I don't suppose that you still know/have the answer to 5.4 do you?
Upload what you've tried / describe what problems youre having if you want some help/feedback.
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1 week ago
#9
Not the original person who asked, but also very stuck:
I expanded Wx² and Wy², and added them together to get ((c cos θ - v)² (1 - ᵛ²⁄c²) (c² sin² θ) )/ (1 - (ᵛ ᶜᵒˢ θ)⁄c) [this isn't square rooted yet], and now I just really don't know where I'm meant to head! I've tried to expand but I still can't see what I can do to simplify further.
If you can give any pointers it'd be very much appreciated.

Note: typo - it should be ((c cos θ - v)² + (1 - ᵛ²⁄c²) (c² sin² θ) )/ (1 - (ᵛ ᶜᵒˢ θ)⁄c)²
Last edited by Avogadro’s Con; 1 week ago
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1 week ago
#10
Aside:

My old browser is now unable to access isaacphysics, so I can't assist here.
Last edited by ghostwalker; 1 week ago
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1 week ago
#11
Ok no worries - the original question is here as a reference, if anybody'd like to see/help:
Last edited by Avogadro’s Con; 1 week ago
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1 week ago
#12
(Original post by Avogadro’s Con)
Not the original person who asked, but also very stuck:
I expanded Wx² and Wy², and added them together to get ((c cos θ - v)² (1 - ᵛ²⁄c²) (c² sin² θ) )/ (1 - (ᵛ ᶜᵒˢ θ)⁄c) [this isn't square rooted yet], and now I just really don't know where I'm meant to head! I've tried to expand but I still can't see what I can do to simplify further.
If you can give any pointers it'd be very much appreciated.
Dont think that expression is correct, but it may just be typos when you typed it.
It does help to upload a picture of your working if you want some help.
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1 week ago
#13
(Original post by Avogadro’s Con)
Not the original person who asked, but also very stuck:
I expanded Wx² and Wy², and added them together to get ((c cos θ - v)² (1 - ᵛ²⁄c²) (c² sin² θ) )/ (1 - (ᵛ ᶜᵒˢ θ)⁄c) [this isn't square rooted yet], and now I just really don't know where I'm meant to head! I've tried to expand but I still can't see what I can do to simplify further.
If you can give any pointers it'd be very much appreciated.
As pointed out, there are at best numerous typos there, though it looks to be on the right lines.

As a hint, you're aiming for the numerator being "something" times the denominator, so your basic trig identity is going to be involved.

If you get stuck, post your actual working - can never be sure whether something is a typo or not when you try and type maths in that fashion. LaTeX is preferred if you're unable to upload an image of your working - see here
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1 week ago
#14
I was also stuck on this question for a few days but I managed to solve it and get the answer - my advice is to write everything out literally, make sure you don't mix up the signs and it'll all cancel out nicely.
Last edited by Jail_break; 1 week ago
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1 week ago
#15
(Original post by Jail_break)
I ...
Could you pls delete and have a read of the forum sticky about not posting solutions. Thanks
https://www.thestudentroom.co.uk/sho....php?t=4919248
Last edited by mqb2766; 1 week ago
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1 week ago
#16
Oh, didn't know that it wasn't allowed lol. Anyway's I'll delete it.
1
1 week ago
#17
(Original post by mqb2766)
Dont think that expression is correct, but it may just be typos when you typed it.
It does help to upload a picture of your working if you want some help.
Oh yes, you're right! Sorry about that - here are my workings up to that point:
[Note: Workings removed]

But now I'm just struggling to see how to simplify it further - the denominator has a cosine in it, but the numerator also includes that sine which is my problem at the moment.
Last edited by Avogadro’s Con; 1 week ago
0
1 week ago
#18
Looks about right, but I'd multiply by 1 = c^2/c^2 to get normal expressions on the numerator and denominator. Then expand the numerator and use pythagorean identity on cos^2 and sin^2 to see if stuff simplifies.
1
1 week ago
#19
(Original post by mqb2766)
Looks about right, but I'd multiply by 1 = c^2/c^2 to get normal expressions on the numerator and denominator. Then expand the numerator and use pythagorean identity on cos^2 and sin^2 to see if stuff simplifies.
Oh okay thank you! By the Pythagorean identities, does that require A Level knowledge, because we haven't been taught about them yet, or is it just basic GCSE things?
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1 week ago
#20
(Original post by Avogadro’s Con)
Oh okay thank you! By the Pythagorean identities, does that require A Level knowledge, because we haven't been taught about them yet, or is it just basic GCSE things?
cos^2(x) + sin^2(x) = 1
Have you come across that? Its really just pythagoras on a unit length hypotenuse.
Last edited by mqb2766; 1 week ago
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