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Edexcel - M3 - 18th May 2016

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Hey, can someone kindly explain to me in the attached question + solution, why P is closer to B instead of A?
If C is in between A and B, where AC=1.4m and CB=2.6m, then surely C would be closer to the fixed point A. As a result, my extensions for each spring were the opposite to theirs (i.e. I got 0.8-x for the tension in the spring on the side of A, whilst they got 0.8+x). The problem is I get a=+w2x instead of a=-w2x, meaning that I haven't shown SHM (I think) and that I am likely doing something wrong then.

Is this a mistake on their behalf or am I identifying the extensions incorrectly? It's probably the latter. Thanks!
tsr.png72.1KB
Original post by paradoxequation
Hey, can someone kindly explain to me in the attached question + solution, why P is closer to B instead of A?
If C is in between A and B, where AC=1.4m and CB=2.6m, then surely C would be closer to the fixed point A. As a result, my extensions for each spring were the opposite to theirs (i.e. I got 0.8-x for the tension in the spring on the side of A, whilst they got 0.8+x). The problem is I get a=+w2x instead of a=-w2x, meaning that I haven't shown SHM (I think) and that I am likely doing something wrong then.

Is this a mistake on their behalf or am I identifying the extensions incorrectly? It's probably the latter. Thanks!


C is where the SHM will start not where SHM is about. SHM will be about the equilibrium hence about the middle and to prove the SHM put P on a dist x from equilibrium then work out tensions.


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Original post by physicsmaths
C is where the SHM will start not where SHM is about. SHM will be about the equilibrium hence about the middle and to prove the SHM put P on a dist x from equilibrium then work out tensions.


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Perfect, I get it now. Thanks so much!
Can someone explain when you consider displacement from the equilibrium position to be negative in those SHM time taken questions? Is it when it's moving in the opposite direction of your increasing x when you proved SHM in prior parts?
Original post by Ayman!
Can someone explain when you consider displacement from the equilibrium position to be negative in those SHM time taken questions? Is it when it's moving in the opposite direction of your increasing x when you proved SHM in prior parts?


Yep it just depends on how you modelled it firstly. So if you took x increasing as down you take a down and anything moving up is negative velocity and any position behind equilibrium point is negative displacement.


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Original post by paradoxequation
Perfect, I get it now. Thanks so much!


No problem.


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Original post by physicsmaths
Yep it just depends on how you modelled it firstly. So if you took x increasing as down you take a down and anything moving up is negative velocity and any position behind equilibrium point is negative displacement.


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And vice versa, yeah? Perfect, cheers fam!
Original post by Ayman!
And vice versa, yeah? Perfect, cheers fam!


Yep. Wlog basically.


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Could someone help me here:

Surely in reality if you have a long string, it is much harder to force it into a vertical circle than a shorter string. However on the other hand, the equation Centripetal force =MV^2/r says that if you have a longer string you need less tension and hence it makes it easier to force it into a vertical circle.

What am i missing here from my understanding?
Original post by physicsmaths


So if u pull a string down and call this pulling down positive x, then they say find the same it's moving at this speed the speed will be negative?
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wr123
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Anyone up for trying to predict the questions that will come up this May? I think there will be a F=GMm/r^2 one!
Original post by wr123
Anyone up for trying to predict the questions that will come up this May? I think there will be a F=GMm/r^2 one!


Ngl don't even know how to use that formula???
IAL Jan '14 Q5b), I've resolved the forces horizontally and vertically but where do you take moments from? Been stuck on this for a while

https://8a40d6c38bafca75cc407741c0f3e1889c8e66b2.googledrive.com/host/0B1ZiqBksUHNYWENTbktxdlk5dEU/January%202014%20(IAL)%20QP%20-%20M3%20Edexcel.pdf
Original post by BBeyond
So if u pull a string down and call this pulling down positive x, then they say find the same it's moving at this speed the speed will be negative?


Just use the form of velocity with no trig in
Original post by TheFarmerLad
IAL Jan '14 Q5b), I've resolved the forces horizontally and vertically but where do you take moments from? Been stuck on this for a while

https://8a40d6c38bafca75cc407741c0f3e1889c8e66b2.googledrive.com/host/0B1ZiqBksUHNYWENTbktxdlk5dEU/January%202014%20(IAL)%20QP%20-%20M3%20Edexcel.pdf


About the point of contact.
Original post by BBeyond
Ngl don't even know how to use that formula???


Don't you do physics, mate? :lol: I don't think they'd give it because I can't see how they'd make the maths more M3-esque rather than something out of a physics paper.

But then again, they like to throw people off at times (case: those trapezium rule questions)
Original post by BBeyond
Ngl don't even know how to use that formula???


Its a standard differential equation usually, vdv/dx = -Gmm/x^2
Original post by BBeyond
Ngl don't even know how to use that formula???


It's just your kinematics again with solving DE's in a slightly different context, it's often free marks. :tongue:
Original post by Ayman!
Don't you do physics, mate? :lol: I don't think they'd give it because I can't see how they'd make the maths more M3-esque rather than something out of a physics paper.

But then again, they like to throw people off at times (case: those trapezium rule questions)


Nah lol I do chem maths fm econ :tongue:

Original post by samb1234
Just use the form of velocity with no trig in


yh fair probs a shout
Original post by Ayman!
Don't you do physics, mate? :lol: I don't think they'd give it because I can't see how they'd make the maths more M3-esque rather than something out of a physics paper.

But then again, they like to throw people off at times (case: those trapezium rule questions)


They do give them. The first part usually involves rearranging it to get a form in terms of little g and then you have to do some de to find velocity at a given distance or similar

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