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# Edexcel M2/M3 June 6th/10th 2013 Watch

1. whos doing M3 and FP2, who wants to work together?
2. (Original post by JenniS)

A light elastic string has natural length 4m and modulus of elasticity 58.8N. Particle P of mass 0.5kg attatched to one end of the string, other end attatched to a vertical point A, particle is released from rest at A and falls vertically.

a) find the distance travelled before P instantaneously comes to rest for the first time.
b)Find the speed of the particle when the string becomes slack for the first time

HELP! anyone...

GPE lost = EPE gained
3. (Original post by StUdEnTIGCSE)
Well is that a good thing?
You have students like L'Evil Fish sitting C4 at 16.

But I'm doing Further also and with no Jan exams M2 is inevitable this year, but its kinda easy.

Are you doing Further Maths?
Nothing wrong with it, it's just that most of us in yr 12 are also 17 .
Ah L'Evil Fish , earlier doesn't necessarily mean better he's still got GCSEs to focus on...

Yes also doing Further.
4. I need help immediately. D: so this is june 2002 Q3, M2

Normally we use the formula work done against resistance = energy loss- energy gain right?
Why are we using total energy loss + work done by cyclist = work done against resistance, in this question? It doesn't make any sense to me :|
Thank you so much!
5. (Original post by Exams v__v)
I need help immediately. D: so this is june 2002 Q3, M2

Normally we use the formula work done against resistance = energy loss- energy gain right?
Why are we using total energy loss + work done by cyclist = work done against resistance, in this question? It doesn't make any sense to me :|
Thank you so much!
Work Done by Cyclist is Work Done by Resistance - Energy Loss, because the cyclist is doing work therefore there's a gain in energy
6. (Original post by Boy_wonder_95)
Work Done by Cyclist is Work Done by Resistance - Energy Loss, because the cyclist is doing work therefore there's a gain in energy
The mark scheme says
Wd by cyclist = wd by resistance (20*500) - (loss in pe+ loss in ke)
I don't still understand this. Wd by cyclist and against resistance is clearly not the same? :|
7. Can someone help me with this please

A particle P receives an impulse of magnitude 9root2 in the direction of the vector i-j

How would you convert the magnitude of an impulse into a vector ?
8. (Original post by Olive123)
Can someone help me with this please

A particle P receives an impulse of magnitude 9root2 in the direction of the vector i-j

How would you convert the magnitude of an impulse into a vector ?
If you remember back to M1 you would have to find the magnitude of the direction first. Sqrt of 1^2 + 1^2 = root 2. Then you do 1 / root 2 x 9root2 x (i - j)
9. (Original post by Olive123)
Can someone help me with this please

A particle P receives an impulse of magnitude 9root2 in the direction of the vector i-j

How would you convert the magnitude of an impulse into a vector ?
I was wondering the same thing, i think it is something to do with it being parallel to i-j but i don't know what to do from then :|
10. (Original post by Exams v__v)
The mark scheme says
Wd by cyclist = wd by resistance (20*500) - (loss in pe+ loss in ke)
I don't still understand this. Wd by cyclist and against resistance is clearly not the same? :|
Yh work done by resistance would be negative since it causes an energy loss whereas work done by cyclist would be positive since it causes an energy gain so no they're clearly not the same.

Wd by R - Wd by Cyclist = Loss in Energy assuming Wd by R is greater
11. (Original post by MLogan)
It doesn't matter. If your speed turns out to be negative all it means is that it is travelling the other way round.
Actually it is important, if they do not give you direction you must work out direction as you need to work out speeds etc... if the direction is not given, the momentum in terms of m and v must be worked out and then you can assume the direction
12. (Original post by Exams v__v)
The mark scheme says
Wd by cyclist = wd by resistance (20*500) - (loss in pe+ loss in ke)
I don't still understand this. Wd by cyclist and against resistance is clearly not the same? :|
If you think about it, energy lost is energy the cyclist is not required to input into the system. If you go down a hill and then ascend a smaller hill less energy is required to go up the hill than if you started on a small hill and had to climb a bigger hill. Same goes for your initial kinetic energy. In terms of loss of energy, if there was no cyclist and it was a ball that rolled down the hill and then went back up again the loss of energy would be the work done by the friction on the ball
13. (Original post by Olive123)
Can someone help me with this please

A particle P receives an impulse of magnitude 9root2 in the direction of the vector i-j

How would you convert the magnitude of an impulse into a vector ?
(Original post by MLogan)
I was wondering the same thing, i think it is something to do with it being parallel to i-j but i don't know what to do from then :|
First find the magnitude of the direction

Then do the inverse of the magnitude x the Impulse x the parallel vector

14. (Original post by Boy_wonder_95)
Yh work done by resistance would be negative since it causes an energy loss whereas work done by cyclist would be positive since it causes an energy gain so no they're clearly not the same.

Wd by R - Wd by Cyclist = Loss in Energy assuming Wd by R is greater
How do we know wd against resistance is greater? I thought since the cyclist is doing work, it'll obviously be a lot more significant than work against friction?
So I though it should be wd by cyclist = energy loss + wd against resistance?
15. (Original post by StUdEnTIGCSE)
Well if you consider the component of weight down the slope in addition to friction as forces doing work then you do not need to consider GPE.

Loss of KE= Gain in GPE + work done by friction

is the same as

Loss of KE = (forces acting down the plane) * distance
So work done by forces down the plane is the same as KE?
16. (Original post by Boy_wonder_95)
If you remember back to M1 you would have to find the magnitude of the direction first. Sqrt of 1^2 + 1^2 = root 2. Then you do 1 / root 2 x 9root2 x (i - j)
Thanks !!
17. (Original post by kvohra)
If you think about it, energy lost is energy the cyclist is not required to input into the system. If you go down a hill and then ascend a smaller hill less energy is required to go up the hill than if you started on a small hill and had to climb a bigger hill. Same goes for your initial kinetic energy. In terms of loss of energy, if there was no cyclist and it was a ball that rolled down the hill and then went back up again the loss of energy would be the work done by the friction on the ball
Wait, are you saying loss in energy is the work done against resistance?
18. (Original post by Boy_wonder_95)

Why 1/root2 ?
19. (Original post by Exams v__v)
Wait, are you saying loss in energy is the work done against resistance?
yes but a better way to think about it is work done by the resistance on you
20. (Original post by Exams v__v)
How do we know wd against resistance is greater? I thought since the cyclist is doing work, it'll obviously be a lot more significant than work against friction?
So I though it should be wd by cyclist = energy loss + wd against resistance?
Because the question said there's an energy loss therefore whatever force is causing that loss must be greater than what's causing the gain

No no, because resistance causes energy loss but wd causes energy gain so the difference between Wd by R and energy loss is WD by cyclist

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