OCR MEI M3 Thread (Wed 24th May 2017)

OSmith99

2

Who is doing this exam?

Reply 1

TheSimpZone

1

I am, you ready? I've been getting 70s in pps except the last two were only 63 and I need 90 UMS so idrk any more...

Reply 2

JoshFarrell

1

Does anyone know of any particularly odd/difficult questions so I don't have to go looking?

Reply 3

TheSimpZone

1

June 2012 4bii is a *****... I still don't really know how they got the answer even after looking at the mark scheme and asking my teacher: I get 180-43.6 instead of 43.6 . I can see that obviously my answer is wrong because it is clearly an acute angle but no one I know can really explain the missing step.

June 2015 2 is one of those ones that seems hard but actually once you get your head round it is fairly straightforward, so might be a good one to check out.

(edited 6 years ago)

Reply 4

hugheshallettj

1

Original post by TheSimpZone

June 2012 4bii is a *****... I still don't really know how they got the answer even after looking at the mark scheme and asking my teacher: I get 180-43.6 instead of 43.6 . I can see that obviously my answer is wrong because it is clearly an acute angle but no one I know can really explain the missing step.

June 2015 2 is one of those ones that seems hard but actually once you get your head round it is fairly straightforward, so might be a good one to check out.

The problem is that sin(180-43.6)=sin43.6. So when the sine rule is used, you need to use your own judgement as to whether the angle you are looking for is acute or obtuse.

In this case we are looking for an obtuse angle (136.4). So when we get 43.6, we should technically do 180 minus your answer to get the obtuse angle, and then 180 minus that again to bring you back to the acute angle.

The mark scheme just forgot about those two steps (they probably expected you to figure that out in your head) and found the angle directly.

Does that make sense?

Reply 5

OSmith99

OP

2

Original post by TheSimpZone

June 2012 4bii is a *****... I still don't really know how they got the answer even after looking at the mark scheme and asking my teacher: I get 180-43.6 instead of 43.6 . I can see that obviously my answer is wrong because it is clearly an acute angle but no one I know can really explain the missing step.

June 2015 2 is one of those ones that seems hard but actually once you get your head round it is fairly straightforward, so might be a good one to check out.

Look up the ambiguous case for the sine rule

Reply 6

TheSimpZone

1

Yeah cheers guys, my teacher explained it again to me today and I've pretty much got the hang of it now. I got 71/72 on last years paper so feeling pretty confident now.

Reply 7

hugheshallettj

1

Original post by TheSimpZone

Yeah cheers guys, my teacher explained it again to me today and I've pretty much got the hang of it now. I got 71/72 on last years paper so feeling pretty confident now.

nice. Im surprised at how few people have posted on this thread. M3 mei isn't that popular I guess

Reply 8

MusicDrag0n

1

It was a tough one when I messed up the direction of tension for Q1, part (ii) I got 3mg(1-cos(-)) and part (iii) 4/3a.
Q2 part a is the only thing looks normal for me this exam LOL. Part B u= 0.45 something, first time I got 2.6 something after missing out a g. Two silly error in first two question cost me the time to check.
Q3 time I got 1.6 something.
Q4 angle I got 50.6

Reply 9

hugheshallettj

1

How did everyone find it? Found the COM question dodgy...

Reply 10

XenoMJ

1

Easier than last year to be sure. For the bounds of k in Q1 did people get 1<k<7 inclusive.

Reply 11

OSmith99

OP

2

I also think it was easier than last year i got 50.6 for the angle in the very last question

Reply 12

lizzieb99

1

I think it was an okay paper these are the answers I got (can't remember the question numbers)
3mg(1-costheta)
theta = 70.5 degrees (arccos 1/3)
k=7
alpha = 1, beta = 2, gamma = 1
tanalpha = 52/a so a=100
mu = 11/24 or 0.458
x= 1.2
v= 0.277
(7/3, ln2)
theta = 50.6 degrees

Reply 13

XenoMJ

1

Original post by lizzieb99

I think it was an okay paper these are the answers I got (can't remember the question numbers)
3mg(1-costheta)
theta = 70.5 degrees (arccos 1/3)
k=7
alpha = 1, beta = 2, gamma = 1
tanalpha = 52/a so a=100
mu = 11/24 or 0.458
x= 1.2
v= 0.277
(7/3, ln2)
theta = 50.6 degrees

For the second one they wanted a height not an angle, got 4a/3 for that.

Reply 14

GeorgeMartin98

Original post by XenoMJ

For the second one they wanted a height not an angle, got 4a/3 for that.

What was the minimum tension for 1)iii)?

Reply 15

yeekwan.law

9

Original post by MusicDrag0n

It was a tough one when I messed up the direction of tension for Q1, part (ii) I got 3mg(1-cos(-)) and part (iii) 4/3a.
Q2 part a is the only thing looks normal for me this exam LOL. Part B u= 0.45 something, first time I got 2.6 something after missing out a g. Two silly error in first two question cost me the time to check.
Q3 time I got 1.6 something.
Q4 angle I got 50.6

Do you remember getting sin(t) = -sqrt(5)/3 at some point for Question 3 SHM? And was second last part of Question 3 0.277 ms^-1?

(edited 6 years ago)

Reply 16

Sarnez

3

Original post by yeekwan.law

Do you remember getting sin(t) = -sqrt(5)/3 at some point for Question 3 SHM? And was second last part of Question 3 0.277 ms^-1?

i got 0.277 as well

Reply 17

OSmith99

OP

2

Answers i remember in no particular order
4a/3
K<7 (should also be greater than one too not sure if inequality should be strict or not but probably wont matter)
Alpha is 1 beta 2 gamma 1
Prove that a was 100
U was 11/24 = 0.458
Time period was 2.54
1.61 s for time to get to c moving up the plane
X bar was 7/3 y bar was ln2
Prove 39/(4+6ln2)
Angle for final question was 50.6

(edited 6 years ago)

Reply 18

hugheshallettj

1

How can theta be 50.6 or something?

xbar is 39/(4(1+6ln2)), so theta = arctan((xbar-1)/2) = 23.99

correct me if I've made a mistake, but my answer seemed pretty legitimate.

Reply 19

OSmith99

OP

2

Original post by hugheshallettj

How can theta be 50.6 or something?

xbar is 39/(4(1+6ln2)), so theta = arctan((xbar-1)/2) = 23.99

correct me if I've made a mistake, but my answer seemed pretty legitimate.

This was my working for that

OCR MEI M3 Thread (Wed 24th May 2017)

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