Year 12s: what will be the first way you research your future options? >>

OCR Mei M2 Thread (17th May 2017)

Scroll to see replies

Reply 40

Sarnez

Original post by lilrabbits

And do anyone still rmb the internal forces in the framework question..

top right is -60 thrust
bottom is 100 tension
cant remember the top left one

ayyyyy

got the same

Original post by lilrabbits

Do we have to say mu is smaller than 1

There's no reason that mu is less than 1! It usually is, but in exceptional circumstances, it can be greater - e.g "Rubber against rubber results in a static coefficient of friction of 1.15"
I think they are unlikely to penalise people who say that it is less than one though...

Just realized someone has already answered this, sorry :/

(edited 6 years ago)

Reply 43

latie

Can anyone please explain how they did the question where B bounced off an angled plane? I got 60 degrees, and I can see where I went wrong...

Reply 44

Sarnez

Original post by latie

Can anyone please explain how they did the question where B bounced off an angled plane? I got 60 degrees, and I can see where I went wrong...

i got 60 degrees too

Reply 45

latie

Original post by lilrabbits

There is also one q that asks why we cannot use s=1/2(u+v)T one. Can we just say the acceleration is non-uniform?

that's what I said. By proving that the acceleration can't be constant, you can surely just state that suvat equations are only for use when the accelerations is constant?

Reply 46

Tommo791

Original post by Sarnez

i got 60 degrees too

And me

Reply 47

lilrabbits

And the magnitude of impulse is 12(1+e) right (no need to state the direction?)

Reply 48

SeanHarvey

Original post by Sarnez

i got 60 degrees too

How did you even answer it?

Reply 49

PawanAviator

Q1)
i) Va = -2/3Vb ( hence opposite due to the sign) (2)
ii) Va = 2e and Vb = 3e (5)
iii) 12(1+e)
iv) t = 18 seconds ( consider speed of seperation of A and B) (3)
v) 60 degrees (5)

Q2)
(a)
(i)W.D against Friction = 150J (3)
(ii) W.D against friction = 125/2tan(a) or something , then calculate tan(a) using part (i) (7)
(b)
(i) P = (R + 5a)V or something. (3)
(ii) F = 12.5 when v = 4 and a = 0 at P= 50 (1)
(iii) Look at january 2010 Q2 MEI M2 ==> a is not constant (3)

Q3
(i) X = 80cos(a) = 64 from looking at framework geometry and resolving the system horizontally (8)
100N , 36N , 45N ( 2 are compressions and 1 is a tension.) Can't remember
(ii) Rc = (55/169)W and F and R can be found from resolving horizontally and vertically(12)

Q4
(i) x = 4 and y = 3 (3)
(ii) let new c.o.m be (x , 2x/3) , solve x and y simultaneously to get mX = 6 (4)
(iii ) c.o.m of weird semicircular thingy = 2r/(k+pi) (4)
(iv) 6 - pi (6)

(edited 6 years ago)

Reply 50

Sarnez

Original post by SeanHarvey

How did you even answer it?

resolve parallel and perpendicular to the plane, parallel velocity is the same after collision, perpendicular velocity after collision becomes e times the original perpendicular velocity.

Reply 51

PawanAviator

Original post by Sarnez

resolve parallel and perpendicular to the plane, parallel velocity is the same after collision, perpendicular velocity after collision becomes e times the original perpendicular velocity.

was it like :

3esin(a) = 1/3 x 3ecos(a) ?

Reply 52

lilrabbits

Original post by PawanAviator

Q1)
i) Va = -2/3Vb ( hence opposite due to the sign) (2)
ii) Va = 2e and Vb = 3e (5)
iii) 12(1+e)
iv) t = 18 seconds ( consider speed of seperation of A and B) (3)
v) tan(a) = 1/3 from considering motion perp. to plane => 18. (5)

Q2)
(a)
(i)W.D against Friction = 150J (3)
(ii) W.D against friction = 125/2tan(a) or something , then calculate tan(a) using part (i) (7)
(b)
(i) P = (R + 5a)V or something. (3)
(ii) F = 12.5 when v = 4 and a = 0 at P= 50 (1)
(iii) Look at january 2010 Q2 MEI M2 ==> a is not constant (3)

Q3
(i) X = 80cos(a) = 64 from looking at framework geometry and resolving the system horizontally (8)
100N , 36N , 45N ( 2 are compressions and 1 is a tension.) Can't remember
(ii) Rc = (55/169)W and F and R can be found from resolving horizontally and vertically(12)

Q4
(i) x = 4 and y = 3 (3)
(ii) let new c.o.m be (x , 2x/3) , solve x and y simultaneously to get mX = 6 (4)
(iii ) c.o.m of weird semicircular thingy = 2r/(k+pi) (4)
(iv) 6 - pi (6)

Wow!!!! Thank u!! You are amazing

Reply 53

SeanHarvey

Original post by Sarnez

resolve parallel and perpendicular to the plane, parallel velocity is the same after collision, perpendicular velocity after collision becomes e times the original perpendicular velocity.

Came back to it at the end and made a start of doing that, hopefully won't drop all the marks in that case.

Reply 54

PawanAviator

that power question really reminds of a past paper Q from jan 2010 Q2

as.png15.3KB

Reply 55

PawanAviator

Original post by lilrabbits

Wow!!!! Thank u!! You are amazing

no worries!

Reply 56

pusha a

Anyone get 50/v - F = 5a for the equation of motion thing? Assuming F=ma is the equation of motion

Reply 57

Brux

Original post by pusha a

Anyone get 50/v - F = 5a for the equation of motion thing? Assuming F=ma is the equation of motion

Yeah that's exactly what I had

Reply 58

OSmith99

1. (a)(i) total momentum is zero so momentum after collision must be zero. This is only possible if velocities are in the opposite direction, they cant pass through eachother so both must change direction after the collision.
(ii) Va = 2e to the left and Vb=3e to the right
(Iii) 18 seconds for time between two collisions
(Iv) magintude of impulse was (12e+12) Ns
(b) angle alpha was 60 degrees

2. (A) (I) work due to friction was 150J
(II) Show work due to friction was 125/2tan(alpha)
Tan(alpha) = 5/12
(Iii) It would go past D as less work lost due to friction and gravitational potential energy gain was still the same.
(B) (i) 5a=50/v-f
(Ii) f= 12.5
(Iii) f is constant so if a doesnt equal zero then v changes which means a must change accrodingly. If a is zero then velocity remains constant and so acceleration stays zero.
Cant use the provided formula as only for constant acceleration but as velocity changes acceleration wasnt constant

3. (A) Resolve horizontally x=80cos(alpha)=64
60 N thrust
100N tension
45 N thrust

(B) show u>= 0.308 doesnt neccasarily have to be less than 1

4. (I) com at (4,3)
(Ii) show mX =6

Value of k such that Te=2Tb was k=6-pi