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# M1 after the exam discussion watch

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1. for the moments question, was x a third of a metre?

Oh yes and what does everyone think the grade boundaries will be for this exam? Roughly?
2. EASY! BUT I F***** 7D
Could'nt get d^2 = etc...

I tried everythingggggggggggggggggg (as i had 1/2hour left lol)

even tried taking away Q in itself and P then squaring them
...tried Q^2 - P^2

but i got the quadratic for the very last part ( didnt comp it thou)

otherwise it was okay...oh btw what did ppl get for the 6 mark collision question. i got a fraction 12/14/15 i thinkkkkk
3. I think they'll be as normal
ie 80% or above is A
70 B
60 C
4. hope my the 6-7 marks i lost wont bring me dwn!
5. this thread sums up the paper so far pretty well

everything was fine except vectors lol i think i did quite well in the paper except for the "show that" question in the vector one, but hopefully it'll be method marks galore.

anyway, hows about the rest of the paper too!

heres mine, got a coefficiant to equal .751 and acceleration of that stone to = -3m/s^2 ... i think, come on people, post your answers like theres no tomorrow!
6. I put 161 mins. This question was a bit confusing. It said the time when they first see it is t = 1 (not an actual time). They wanted you to work out the other time so I worked out t

Hope its a question where you can have different answers that mean the same thing
7. (Original post by habosh)
they need the time where they dont see the light so it's between these times
Habosh, if they want the time where they cant see the light, then it is before t = 1 and at t= 2.68, not between, as between the boat is less thatn 15km away meaning they can see the light.

Yes the distance for the moments question for BC was 1/3 m.

Grade boundries will be similar to June, possibly slightly either side but around 80% A, 70% B ....
8. now you knew what p and q were if you answered correctly question C?
Now you needed to subtract Q from P (maybe the other way round)
Now you need to find the magnitude therefore its like pythagorus. i component square + j components square there you have it :P
9. 63/75 for 80 UMS i reckon
10. (Original post by ThugzMansion7)
hope my the 6-7 marks i lost wont bring me dwn!
I thought the exam went very well. Last question was okay.
You just have to say that d^2 = (q-p)^2 and then just square the coefficients of the i-vector and the j-vector.

I am literally shocked about the ocr physics exam (wave properties) today.
I can remember about 3 or 4 questions that came from the specimen paper or past papers. I couldn't believe it. They were worded exactly the same.
11. (Original post by Widowmaker)
for the moments question, was x a third of a metre?

Oh yes and what does everyone think the grade boundaries will be for this exam? Roughly?
yuuup
12. (Original post by MozMan)
this thread sums up the paper so far pretty well

everything was fine except vectors lol i think i did quite well in the paper except for the "show that" question in the vector one, but hopefully it'll be method marks galore.

anyway, hows about the rest of the paper too!

heres mine, got a coefficiant to equal .751 and acceleration of that stone to = -3m/s^2 ... i think, come on people, post your answers like theres no tomorrow!
correct too
13. (Original post by JHJH)
Habosh, if they want the time where they cant see the light, then it is before t = 1 and at t= 2.68, not between, as between the boat is less thatn 15km away meaning they can see the light.

Yes the distance for the moments question for BC was 1/3 m.

Grade boundries will be similar to June, possibly slightly either side but around 80% A, 70% B ....
,but well you cant have it all with no mistakes
14. what did u all get for the distance BC on the moments question?
15. 1a = 0.7 (deceleration)
1b = unchanged
1c = 8.25

2a = 147
2b = 1/3

3a = 162
3b = 6.2
3c = 0.56

4a = 23.0
4b = 17.6
4c = Fr > 2.5gsin theta (+explaination)

5a = 3.2
5b = 5.28
5c = 0.751
5d = constant acceleration

6a = 3 (deceleration)
b = 14.8
c = proof
d = 3.06

7a = 3i + 8j
7b = (20+3t)i + (10+8t)j and 14i + (12t-6)j
7c = proof
7d = 161 mins, 2 hours 41 mins
16. (Original post by Womble548)
1a = 0.7 (deceleration)
1b = unchanged
1c = 8.25

2a = 147
2b = 1/3

3a = 162
3b = 6.2
3c = 0.56

4a = 23.0
4b = 17.6
4c = Fr > 2.5gsin theta (+explaination)

5a = 3.2
5b = 5.28
5c = 0.751
5d = constant acceleration

6a = 3 (deceleration)
b = 14.8
c = proof
d = 3.06

7a = 3i + 8j
7b = (20+3t)i + (10+8t)j and 14i + (12t-6)j
7c = proof
7d = 161 mins, 2 hours 41 mins
yes!
wasn't sure about 4)c
but I got it.
17. (Original post by faa)
what did u all get for the distance BC on the moments question?
1/3m
18. (Original post by Womble548)
2a = 147
2b = 1/3

6a = 3 (deceleration)
b = 14.8
c = proof
d = 3.06
2a) 441N

6d) I had it rounded up to 3.07seconds (it was 3.066...)
19. (Original post by Womble548)
1a = 0.7 (deceleration)
1b = unchanged
1c = 8.25

2a = 147
2b = 1/3

3a = 162
3b = 6.2
3c = 0.56

4a = 23.0
4b = 17.6
4c = Fr > 2.5gsin theta (+explaination)

5a = 3.2
5b = 5.28
5c = 0.751
5d = constant acceleration

6a = 3 (deceleration)
b = 14.8
c = proof
d = 3.06

7a = 3i + 8j
7b = (20+3t)i + (10+8t)j and 14i + (12t-6)j
7c = proof
7d = 161 mins, 2 hours 41 mins
thx, but i thought 2a asked for 3T, T was 147, so 3T would be 441
20. (Original post by Womble548)
I put 161 mins. This question was a bit confusing. It said the time when they first see it is t = 1 (not an actual time). They wanted you to work out the other time so I worked out t

Hope its a question where you can have different answers that mean the same thing
I got that to start with, but if you put it back in it didn't work, you had to put both tension eqns together and that gave you an answer off 23/33 (to my recollection). when I put that number back in it worked.

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