Edexcel M2/M3 June 6th/10th 2013

M3 help please!

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...

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?

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

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 ?

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 ?

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? :|

It doesn't matter. If your speed turns out to be negative all it means is that it is travelling the other way round.

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? :|

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 :|

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

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

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)

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

$\sqrt {1^2 + (-1)^2} = \sqrt {2}$

$\frac{1}{\sqrt {2}} \times 9\sqrt {2} \times (i - j)$

$= 18i - 18j$

Wait, are you saying loss in energy is the work done against resistance?

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?