Official M5 Edexcel Thread June 2016

I think I f****d up.

Anyone have any numerical answers?

Q1: -6 i + 25/2 j

Q2: cos (t) i - 2sin (2t) j

Q3: (-3, 0, 1) but this was different to the resultant x the vector they give so I probably messed up

Q4: Really messed up

Q5: Assuming this is the variable mass

v = u•ln (5) - 19.6

Q6: a) omega = 3J/2mga I think
b) T = 20pi/3 • root {g/5a}

Q7: b) Y = 7/8 mg
But I think I was supposed to take 3mg/8 away from mg/2 not add it... FFS

Posted from TSR Mobile

Pretty sure for q7 the impulse was at the bottom of the shape not in the middle (2a from axis not a) so mg/4 for last part?

What are the odds that the board will accept both cases for cylindrical shell? The answer I got for closed ends is
(M(3r^3+6hr^2+6rh^2+4h^3))/(12(r+h))

b) Resultant moment about PQ axis:

I theta double dot = amgsin (pi/6)

4/3 ma^2 theta double dot = amg/2

theta double dot = 3g/8a

c) Resultant perpendicular to PQ:

ma theta double dot = mgsin (pi/6) - F

F = mgsin (pi/6) - ma • 3g/8a

F = mg/2 - 3mg/8 = mg/8

Edit: Sorry TSR bugged out on me.

This was my method bar the last line. I did:

F = mg/2 + 3mg/8 = 7mg/8


Posted from TSR Mobile

I think I f****d up.

Anyone have any numerical answers?

Q1: -6 i + 25/2 j

Q2: cos (t) i - 2sin (2t) j

Q3: (-3, 0, 1) but this was different to the resultant x the vector they give so I probably messed up

Q4: Really messed up

Q5: Assuming this is the variable mass

v = u•ln (5) - 19.6

Q6: a) omega = 3J/2mga I think
b) T = 20pi/3 • root {g/5a}

Q7: b) Y = 7/8 mg
But I think I was supposed to take 3mg/8 away from mg/2 not add it... FFS

Posted from TSR Mobile

Yeah that's it (the first one).

His confusion is over fact that the impulse was at 2a, but in this part we only care about COM which is distance a from axis.

Q 6 is wrong: can't remember my and a but yours has units of 1/s not seconds

Any predictions on how many marks my sign error will cost me? Really annoyed I did that

Well it's A1 or A1M1 lost for wrong equation and A1 final ans so 2-3

I may have got a/5g instead of g/5a. I remember a 5 on the denominator of the root.

Edit: Yes it must have been a/5g.

I'm pretty sure there was a square root 2 under sort sign as COM was sqrt(2) from axis

CoM was root10/2 from axis not root2. CoM was at 1/2,1/2 taking the kink in the wire as the origin.
I had T=pi root(8root10a/3g) or T=2pi root(2root10a/3g)

How? Two wires of equal mass side length 2a each. So at (a,a)???

i meant combined centre of mass for the entire kinked wire. should have clarified what I meant.

Still don't understand where that number comes from? Was there only one wire and nothing else stuck on?

sum of moments= moments of sum

m(0,a)+m(a,0)=2m(x,y)

(x.y)=1/2a, 1/2a

I think 6 marks I'm not quite sure. You couldhave done it using 2 separate centres of mass on each wire but I chose to combine the CoM to make it easier.

I think 6 marks I'm not quite sure. You couldhave done it using 2 separate centres of mass on each wire but I chose to combine the CoM to make it easier.