OCR A Physics AS Breadth 24/5/16 Multiple choice
Usual disclaimers. These are just my answers and are in no sense official. I have not seen the mark scheme yet.
There may be errors and / or typos.
First impression? A lot to do in 1.5h.
The multichoice questions contain some major timewasters.
Here we go:
Q1 1W = 1J/1s so C
Q2 Work = force x distance in direction of force so =W x vert distance so B
Q3 Polarising so transverse so C
Q4 Electron diffraction so wavelength similar to spacing so D
Q5 50um = 5E5m so IR so B
Q6 acc is prop to force so graph will be same so A
Q7 Forces must make closed triangle and N is perp to slope so B
Q8 Energy equal to 1/2 Fx; F is same so E is prop to x so E/3 so B
Q9 PD across 1000ohm = 3v so PD XY = 6v. P=VI = 6 x 1mA = 6mW so A
Q10 Total force must be zero (forces are equal and opposite) so C
Q11 5v is >3v so current is anticlockwise; series so current same so D
Q12 Kirchoff1 I1+I3 = I2 +I4 so B
Q13 V has %uncert 4% and I has % uncert of2.5% so R has 6.5% so C
Q14 v = 0 at highest point so C
Q15 eV = 1/2 mv^2 so if V halvess v must x 1/sqrt2 so C
Q16 Max ke = 4.8E19  3.2E19 = 1.6E19. Stick this equal to 12 mv^2 and get v = 5.93E5 ms1 so A
Q17 4x dia so 16 x area so x is 16 times less. Same force E = 1/2 Fx so E must be 16 times more in thin wire so D
Q18 Horiz vel is vcos 30. So KE at top is v^2 cos^2 (30) so ration of KE = cos^2 (30) = 3/4 so C
Q19 Same PD. Double R has half current. P = VI so P is twice as big in small resistor so D
Q20 n = c/v = 1/5 n sin(theta) is const so 1.5 sin(10) = 1 sin (theta) so theta = 15 so B
You'll have done well to score 15+ on those.
Short answers to follow
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OCR A Physics AS Breadth 24/5/16
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Why bother with a post grad? Are they even worth it? Have your say!  26102016 

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 1
 24052016 16:43
Last edited by teachercol; 24052016 at 18:03. Reason: Edited Q8Post rating:3 
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 2
 24052016 16:43
Section B
Q21 a) Mass is a scalar; velocity is a vector
Masses add like numbers ; need to take directions into account when adding velocities (2)
b) Tension in string and weight (1)
suvat t = 0.73s (2)
Total : 5
Q22 a) Gradient of graph = 2a
If all points move left or right by same amount (systematic error) then gradient is unchanged (1)
b) Grad = (68012) (45  10) = 2a so a=8.0 ms2 so F = ma = 7360N (3)
Total: 4
Q23 a) Measure diameter in several places with micrometer or vernier callipers.
Calculate A = pi x (dia/2 )^2
Measure weight on scales. (3sf)
Calculate P = weight / area (4)
b) i) (When an object is wholly or partially immersed in a fluid) Upthrust = weight of fluid displaced (1)
ii) Nasty Upthrust = 9.0 7.8 = 1.2N
so mass of water displaced = 1.2 / 9.81
so volume of water = vol of cylinder = 1.2/9.81 / 1000
mass of cylinder = 9.0 / 9.81
so density = mass / volume = 1000 x 9.0 / 1.2 = 7500 kg m3 (3)
Total: 8
Q24 a) N2 Resultant force is equal to the rate of change of momentum of an object (and in the same direction) (1)
b) i) Conserved 2 from mass / momentum / energy / angular momentum not KE (1)
ii) forces are equal and opposite so reflect in x axis (upside down) (2)
c) Perfectly elastic so can use either KE or momentum conservation
500 x 1.7E27 =  420 x 1.7E27 + 2.0E26v
v = 78 ms1 (3)
Total: 7
Q25 a)i ) similarity both energy converted per unit charge / both measured in volts
difference emf is to electrical PD is from electrical (2)
ii) Sneaky n = N / V = 9.6E16 / (1.2E6 x 6.0E3) = 1.33E24
I  nAvq so v = 3.0E3 / (1.33E25 x 1.2E6 x 1.6E19 ) = 1.12E3 ms1 (3)
b) Circuit with cell and variable resistor. Ammeter in series; voltmeter across cell (or R)
Measure terminal PD and current. Very R repeat.
V = E  I r
Plot graph with V on y axis and I on x axis
Gradient = r so r =  gradient (4)
Total: 9
Q26 a) i) 180 out of phase / move in opposite directions (1)
ii) lambda / 2 = 40.0cm + 2.0 cm (5% error)
so v  f x lambda = 75 x 0.40 x 2 = 60 + 3 (also 5% error) (3)
b) i) waves created when release travel to ends and reflect (180 phase shift)
superposition means AN in centre (constructive interference /reinforcement) and nodes at ends (destructive interference /cancellation) (2)
ii) Node to node = lambda /2 so measure length of string
lambda = 2 x length (1)
Total: 7
Q27 a) I is zero so R is infinite
I increases as LED lights so R decreasing
I increases a lot as V increases a little bit so R continues to decrease. (4)
b) cell / LED is wrong way round . LED doesn't turn on until 2.6v+
so reverse cell / LED and add more cells. (3)
c) f = c/ lambda = 3.0E8/480E9 = 6.25E14Hz
E = hf = 4.14E19J
P = E x N so N = 2.9E15 (3)
Total : 10
and there we have it.
I think that's pretty tough. Grade boundaries will be a lot lower than any recent Mech / EWP paper.
Best guess?
A 45
B 40
C 35
D 30
E 25
Could be lower.
Good Luck
ColLast edited by teachercol; 24052016 at 19:43.Post rating:4 
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 3
 24052016 17:14
Do you have the questions with you, and for the multiple choices the other answers? It's hard to remember what I put even with your workings

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 4
 24052016 17:17
(Original post by IsaacEinstein)
Do you have the questions with you, and for the multiple choices the other answers? It's hard to remember what I put even with your workings 
 Follow
 5
 24052016 17:30
(Original post by teachercol)
Section B
Q21 a) Mass is a scalar; velocity is a vector
Masses add like numbers ; need to take directions into account when adding velocities (2)
b) Tension in string and weight (1)
suvat t = 0.73s (2)
Total : 5
Q22 a) Gradient of graph = 2a
If all points move left or right by same amount (systematic error) then gradient is unchanged (1)
b) Grad = (68012) (45  10) = 2a so a=8.0 ms2 so F = ma = 7360N (3)
Total: 4
Q23 a) Measure diameter in several places with micrometer or vernier callipers.
Calculate A = pi x (dia/2 )^2
Measure weight on scales. (3sf)
Calculate P = weight / area (4)
b) i) (When an object is wholly or partially immersed in a fluid) Upthrust = weight of fluid displaced (1)
ii) Nasty Upthrust = 9.0 7.8 = 1.2N
so mass of water displaced = 1.2 / 9.81
so volume of water = vol of cylinder = 1.2/9.81 / 1000
mass of cylinder = 9.0 / 9.81
so density = mass / volume = 1000 x 9.0 / 1.2 = 7500 kg m3 (3)
Total: 8
Q24 a) N2 Resultant force is equal to the rate of change of momentum of an object (and in the same direction) (1)
b) i) Conserved 2 from mass / momentum / energy / angular momentum not KE (1)
ii) forces are equal and opposite so reflect in x axis (upside down) (2)
c) Perfectly elastic so can use either KE or momentum conservation
500 x 1.7E27 =  420 x 1.7E27 + 2.0E26v
v = 78 ms1 (3)
Total: 7
Q25 a)i ) similarity both energy converted per unit charge / both measured in volts
difference emf is to electrical PD is from electrical (2)
ii) Sneaky n = N / V = 9.6E16 / (1.2E6 x 6.0E3) = 1.33E24
I  nAvq so v = 3.0E3 / (1.33E25 x 1.2E6 x 1.6E19 ) = 1.12E3 ms1 (3)
b) Circuit with cell and variable resistor. Ammeter in series; voltmeter across cell (or R)
Measure terminal PD and current. Very R repeat.
V = E  I r
Plot graph with V on y axis and I on x axis
Gradient = r so r =  gradient (4)
Total: 9
Q26 a) i) 180 out of phase / move in opposite directions (1)
ii) lambda / 2 = 40.0cm + 2.0 cm (5% error)
so v  f x lambda = 75 x 0.40 x 2 = 60 + 3 (also 5% error) (3)
b) i) waves created when release travel to ends and reflect (180 phase shift)
superposition means AN in centre (constructive interference /reinforcement) and nodes at ends (destructive interference /cancellation) (2)
ii) Node to node = lambda /2 so measure length of string
lambda = 2 x length (1)
Total: 7
Q27 a) I is zero so R is infinite
I increases as LED lights so R decreasing
I increases a lot as V increases a little bit so R continues to decrease. (4)
b) cell / LED is wrong way round . LED doesn't turn on until 2.6v+
so reverse cell / LED and add more cells. (3)
c) f = c/ lambda = 3.0E8/480E9 = 6.25E14Hz
E = hf = 4.14E19J
P = E x N so N = 2.9E16
and there we have it.
I think that's pretty tough. Grade boundaries will be a lot lower than any recent Mech / EWP paper.
Best guess?
A 45
B 40
C 35
D 30
E 25
Could be lower.
Good Luck
Col 
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 6
 24052016 17:31
Probably doesn't matter

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 7
 24052016 17:32
THANKS!!

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 8
 24052016 17:33
For Q8 surely the spring constant would have changed, as there was the same mass and different extensions, meaning it was only 3 times less?

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 9
 24052016 17:44
I disagree with question 8, the force constant changed so you can't use that formula, you need to use E=1/2Fx  since the force is the same, E is proportional to x and so it's E/3

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 10
 24052016 17:45
i'm with you on that, spring constant couldn't have been the same for both of them, otherwise the extension would have been the same due to it being the same mass

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 11
 24052016 17:51
(Original post by Jamvicious)
For Q8 surely the spring constant would have changed, as there was the same mass and different extensions, meaning it was only 3 times less?
Thats what I thought too... I put 1/9 and then changed it because we only knew that the force applied was the same and epe=0.5Fx 
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 12
 24052016 17:51
I believe that E/3 is the correct answer as well.

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 13
 24052016 17:56
Fairly sure quesiton 8 was E/3.
Also for question 20 i got D (like 49 degrees or) 
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 14
 24052016 17:58
for the very last question about number of photons could you have used e=pt and divided e by 1.6x10^19? Cuz they gave you the power and they said it was one second

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 15
 24052016 17:59
also for question 11 did people not get current clockwise?

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 16
 24052016 18:01
Yep  agree with correction to Q8.
E = 1/2 Fx and same F so E is prop to x so x/3 and answer is B 
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 17
 24052016 18:06
(Original post by Jamvicious)
For Q8 surely the spring constant would have changed, as there was the same mass and different extensions, meaning it was only 3 times less? 
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 18
 24052016 18:06
are you sure about 11 + 20

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 19
 24052016 18:06
(Original post by sanchit117)
I disagree with question 8, the force constant changed so you can't use that formula, you need to use E=1/2Fx  since the force is the same, E is proportional to x and so it's E/3 
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 20
 24052016 18:06
For Q1, would it not be correct to say "rate of work done" which i think was A ? :/
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Updated: June 10, 2016
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