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Edexcel A2 Mathematics: Mechanics M3 6679 01 - 16 May 2018 [Exam Discussion]

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How many marks was question 4?
Ahh makes sense,
Original post by mahclark
I recognise them all


Did u guys use suvat to work it out for the last part?? And what was question 5 about again??
Reply 43
Avatar for Axel004
Axel004
8
Thoughts on grade boundaries?
Original post by Samsamsamsam.123
Did u guys use suvat to work it out for the last part?? And what was question 5 about again??


suvat for when it was slack, shm for while taut
Original post by Barleykit
How many marks was question 4?


7 marks
Reply 46
Avatar for _Avi
_Avi
8
For that question where a particle on a string of length “l” and modulus of elasticity “2mg” slides down a slope where sin(x) = (3/5) and is instantaneously stationary at string length “kl”.
My answer for k is 2.13.
Change in GPE = Change in EPE
(edited 5 years ago)
Original post by Samsamsamsam.123
Did u guys use suvat to work it out for the last part?? And what was question 5 about again??


I used suvat for when the string was slack and x=Acoswt when the string was taut. Then added those two times.

Q5 was about the COM of the hemispheres
Reply 48
Avatar for Al_YG
Al_YG
5
Q.7 asked from D to C so ans is 0.55/2
(edited 5 years ago)
Original post by _Avi
For that question where a particle on a string of length “l” and modulus of elasticity “2mg” slides down a slope where sin(x) = (3/5) and is instantaneously stationary at string length “kl”.
My answer for k is 2.13.
Change in GPE = Change in EPE
what about work done against friction
Original post by _Avi
For that question where a particle on string slides down slope where sin(x) = (3/5) and is stationary at string length “kl”. My answer for k is 2.13.
Change in GPE = Change in EPE
Modulus of elasticity = 2mg


it was a rough slope so there was also work done against friction wasn't there
Original post by _Avi
For that question where a particle on a string of length “l” and modulus of elasticity “2mg” slides down a slope where sin(x) = (3/5) and is instantaneously stationary at string length “kl”.
My answer for k is 2.13.
Change in GPE = Change in EPE


Change in GPE = Change in EPE + Work Done against Friction, so k=1.86
Reply 52
Avatar for _Avi
_Avi
8
Original post by Al_YG
Q.5 asked from D to C so ans is 0.55/2


Q7 (last part) Time for D to C was the same time as “C to slack (x=-0.2)” slack to highest point
(edited 5 years ago)
Reply 53
Avatar for _Avi
_Avi
8
Original post by Disfuligee
Change in GPE = Change in EPE Work Done against Friction, so k=1.86


It was smooth wasn’t it??? 😬😢
(edited 5 years ago)
Original post by _Avi
It was smooth


The coefficient of friction was 1/4. Unless we did different papers, it was rough.
u=square root (5ag) , circular motion Q
k=1.86 (elastic string Q)
k=16(centre of mass Q)
time=0.55s (last SHM Q )
Reply 56
Avatar for _Avi
_Avi
8
Original post by aesthete1
it was a rough slope so there was also work done against friction wasn't there


It was smooth wasn’t it??? 😬😢
i'm not sure about your answer for question 3. The velocity at x = 2R was root(Rg/3), the velocity at x = answer was 2root(Rg/3). This gives answer = R meaning it is 0 above the earth's surface. The only way x = r/5 is if the second velocity is 2root(Rg/6). did i misread the question?
Reply 58
Avatar for _Avi
_Avi
8
Original post by Disfuligee
The coefficient of friction was 1/4. Unless we did different papers, it was rough.


Ahhhh **** 🤦🏽*♂️🤦🏽*♂️ How many marks was that question. I totally didn’t see that. How many marks do you think I would lose if I didn’t use the friction part.
Original post by _Avi
It was smooth wasn’t it??? 😬😢


nope coefficient of friction was 0.25 it was given in the question

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