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OCR A Physics AS Breadth 24/5/16

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 multi-choice 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 = 5E-5m 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 = 4.8E-19 - 3.2E-19 = 1.6E-19. Stick this equal to 12 mv^2 and get v = 5.93E5 ms-1 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

(edited 7 years ago)

Scroll to see replies

Reply 1

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 = (680-12) (45 - 10) = 2a so a=8.0 ms-2 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 m-3 (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 1.7E-27 = - 420 x 1.7E-27 + 2.0E-26v
v = 78 ms-1 (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 / (1.2E-6 x 6.0E-3) = 1.33E24
I - nAvq so v = 3.0E-3 / (1.33E25 x 1.2E-6 x 1.6E-19 ) = 1.12E-3 ms-1 (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 = 3.0E8/480E-9 = 6.25E14Hz
E = hf = 4.14E-19J
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

Col

(edited 7 years ago)

Reply 2

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 :smile:

Reply 3

teachercol

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 :smile:

Yes - but I'm not allowed by the exam board to post them.

Reply 4

TheHobbit

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 = (680-12) (45 - 10) = 2a so a=8.0 ms-2 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 m-3 (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 1.7E-27 = - 420 x 1.7E-27 + 2.0E-26v
v = 78 ms-1 (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 / (1.2E-6 x 6.0E-3) = 1.33E24
I - nAvq so v = 3.0E-3 / (1.33E25 x 1.2E-6 x 1.6E-19 ) = 1.12E-3 ms-1 (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 = 3.0E8/480E-9 = 6.25E14Hz
E = hf = 4.14E-19J
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

Do you think I'd lose marks on the internal resistance question because in my circuit diagram I just put the cell and didn't draw a resistor in a dotted circle to represent the internal resistance, or would it not matter?

Reply 5

teachercol

Probably doesn't matter

THANKS!!

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?

Reply 8

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

Reply 9

dodojonny

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

Reply 10

TheHobbit

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

Reply 11

Ramoose

I believe that E/3 is the correct answer as well.

Reply 12

kuashdfkus

Fairly sure quesiton 8 was E/3.

Also for question 20 i got D (like 49 degrees or)

Reply 13

julesaquilina

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

Reply 14

kuashdfkus

also for question 11 did people not get current clockwise?

Reply 15

teachercol

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

Reply 16

edwowe

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?

Yh for question 8 I got E/3 because E = 1/2 Fx and it said the same force is applied to each of the springs and if you use E = kx^2 then you have to account for the change in k surely? Thanks for the mark scheme!

Reply 17

kuashdfkus

are you sure about 11 + 20

Reply 18

Parhomus

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

Yes, that's exactly what they wanted people to think, at least in my opinion; because clearly if the same force is applied and extension is different then the value of the force constant will be different.