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M1 OCR (Not MEI) Exam - 9/06/2015

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Original post by olliecass1
questions 5 im pretty sure is wrong as it can also accelerate upwards


But the weight of the particle was far more than 1.2N, so there would never be any upward acceleration
Reply 401
Avatar for lolkik
lolkik
yeh that's what I got . How is it 9.74?
That paper was amazing :biggrin:
My answers.

28
6.38
0.598
6
0.2
5
7
50
9.74, 35.4 deg
?
?
1.04
1.06
0
0
68
78
132
7
Proof done with calculator
1.26
-3.15
0.525
0.735
Reply 404
Avatar for xsanda
xsanda
2
Original post by olliecass1
im pretty sure your q 5s are wrong as it can also accelerate upwards


The maximum upward force was 1.2N, and the weight was 0.4g = 3.92, so weight is always greater and so it never accelerates upwards.
Original post by olliecass1
im pretty sure your q 5s are wrong as it can also accelerate upwards


I dont think so cause the resultant force was towards the surface
Original post by wat a wizard
I did that too, did you get 13.3 ish?


Yeah :smile:
Than I used tan = O/A for the angle??
I got k as 96 and for 6i) I got 32. Anyone agree?
Reply 408
Avatar for jt663
jt663
5
Original post by mechanicsonejune
My answers.

28
6.38
0.598
6
0.2
5
7
50
9.74, 35.4 deg
?
?
1.04
1.06
0
0
68
78
132
7
Proof done with calculator
1.26
-3.15
0.525
0.735


I got different answers for question 7 iirc. Think T was 1.2 for me? (Instead of 0.525)
A lot of m friends are saying that the 1.2 force was pushing down onto the block. I thought it was pulling
Original post by mechanicsonejune
My answers.

28
6.38
0.598
6
0.2
5
7
50
9.74, 35.4 deg
?
?
1.04
1.06
0
0
68
78
132
7
Proof done with calculator
1.26
-3.15
0.525
0.735

Agree with you there. The ones that stumped you got me too
Original post by Mattniemier
I dont think so cause the resultant force was towards the surface


I got an upwards acceleration of 1.2/mass because the weight and reaction cancelled out and then there was 1.2 left??
Reply 412
Avatar for c223
c223
2
Original post by olliecass1
im pretty sure your q 5s are wrong as it can also accelerate upwards


What everyone else said about the weight is correct, it wouldn't accelerate up.
Original post by smoothER
Yeah :smile:
Than I used tan = O/A for the angle??
I think iused the sine rule to get 36 or 37
Original post by xsanda
Xsanda's solutions:

Falling question
1i) v=28 [2]
1ii) s=6.384 [2]
1iii) t=0.598 [3]

Collision question
2i) u=6 [4]
2ii) m=0.2 [4]

Cycling question
3i) 20km, 5hrs [3]
3ii) 7 hrs [3]
3iii) 50km [2]

110° tick
4i) 9.74N, 35.4° [6]
4ii) 10.3N [1]
4iii) 54.6° [2]

θ force
5i) 1.04 [3]
5ii) 1.06 [3]
5iii) a=0 [3]
5iv) a=0 [2]

Coalescing
6i) v=68 [4]
6ii) s=78 [4]
6iii) k=132 [1]
6iv) v=7 [5]

Prism
7i) a=0.7 (show that), T=1.68 [5]
7ii) v=1.26 [2]
7iii) -a = 0.9 [2]
7iva) T=1.2 [3]
7ivb) Fr = 1.12 [3]


Got all of them apart from 5)iii and 5)iv). How come they were both 0 :/
Why would the acceleration = 0 for questions 5iii and 5iv??
Reply 416
Avatar for xsanda
xsanda
2
Original post by mechanicsonejune
My answers.

28
6.38
0.598
6
0.2
5
7
50
9.74, 35.4 deg
?
?
1.04
1.06
0
0
68
78
132
7
Proof done with calculator
1.26
-3.15
0.525
0.735


I disagree on the question 7, here are my solutions:

Original post by xsanda

Xsanda's solutions:
Falling question
1i) v=28 [2]
1ii) s=6.384 [2]
1iii) t=0.598 [3]

Collision question
2i) u=6 [4]
2ii) m=0.2 [4]

Cycling question
3i) 20km, 5hrs [3]
3ii) 7 hrs [3]
3iii) 50km [2]

110° tick
4i) 9.74N, 35.4° [6]
4ii) 10.3N [1]
4iii) 54.6° [2]

θ force
5i) 1.04 [3]
5ii) 1.06 [3]
5iii) a=0 [3]
5iv) a=0 [2]

Coalescing
6i) v=68 [4]
6ii) s=78 [4]
6iii) k=132 [1]
6iv) v=7 [5]

Prism
7i) a=0.7 (show that), T=1.68 [5]
7ii) v=1.26 [2]
7iii) -a = 0.9 [2]
7iva) T=1.2 [3]
7ivb) Fr = 1.12 [3]
(edited 8 years ago)
S un
U 1.26
V 0
A
T 0.4

v = u+at
v/t - u/t = a
a = -3.15

everyone seems to have gotten a different answer and i've got a feeling that i used the wrong value for time
Original post by xsanda
I disagree on the question 7, here are my solutions:


I agree on 7
Original post by aliah4666
Why would the acceleration = 0 for questions 5iii and 5iv??


^^^

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