OCR A Physics AS Breadth 24/5/16 Watch

jn998
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#81
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(Original post by Adam998)
Why do that? That's so much extra hassle
Never really considered the other way and we were taught to use that method in class, I wasn't going to take any risks on the real paper with a method I hadn't tried before.
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angusfry
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#82
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Yeah I got 79 as well
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Adam998
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(Original post by jn998)
Never really considered the other way and we were taught to use that method in class, I wasn't going to take any risks on the real paper with a method I hadn't tried before.
Ah fair enough
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qwertyuiop1998
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(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 = 2.9E15 (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
For Q 27b did you have to put both the LED is in the wrong direction AND the voltage of the circuit is lower than 2.6V? Or did you have to choose one of them to explain? Also how likely is it that it would be 45 for an A?
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mahmzo
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(Original post by qwertyuiop1998)
For Q 27b did you have to put both the LED is in the wrong direction AND the voltage of the circuit is lower than 2.6V? Or did you have to choose one of them to explain? Also how likely is it that it would be 45 for an A?
I got 45 too, I didn't write about the LED being in the wrong direction, I talked about the emf being below the threshold, and refered to the graph. I would really love for 45 to be an A or even a high B so if I get an A in depth I could get an A overall. Fingers crossed
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teachercol
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#86
(Original post by qwertyuiop1998)
For Q 27b did you have to put both the LED is in the wrong direction AND the voltage of the circuit is lower than 2.6V? Or did you have to choose one of them to explain? Also how likely is it that it would be 45 for an A?
I think you'll need both to get full marks.

Tbh we're all just guessing about the boundaries since we don't have any previous papers to go on.
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mbb123mbb
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#87
(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 = 2.9E15 (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
Thank you!
For question 25 part b, I put the voltmeter across the variable resistor and said something like "we assume that all of the terminal potential difference is across the variable resistor". Would I lose marks as I didn't put the voltmeter across the cell?
Would much appreciate your help with this
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lanteacher
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#88
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I would also add that for 27 some calculations would be required for R at two different regions to show that it decreases.
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lanteacher
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#89
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#89
(Original post by mbb123mbb)
Thank you!
For question 25 part b, I put the voltmeter across the variable resistor and said something like "we assume that all of the terminal potential difference is across the variable resistor". Would I lose marks as I didn't put the voltmeter across the cell?
Would much appreciate your help with this
Voltmeter can go on either (since we are assuming there is no potential difference across wires) so you would be ok.
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Stormforge9
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Hey, i think i got like 58/70 what kind of ums i s that
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mbb123mbb
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#91
(Original post by lanteacher)
Voltmeter can go on either (since we are assuming there is no potential difference across wires) so you would be ok.
Thank you!
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Monkey XD Luffy
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(Original post by Stormforge9)
Hey, i think i got like 58/70 what kind of ums i s that
I hope that's 100 UMS
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KatieDenn
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#93
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anyone got a list of things that wasnt on the breadth paper?
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mightyned
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(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 = 2.9E15 (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
I know this is late but i just wanted to restore some hope for the second exam.for q23bii i got 7519kgm-3 using the method1.2/9.8= 0.122volume = mass/density so 0.122/1000 = 0.0001229/9.8= 0.92density= mass/volume 0.92/0.000122= 7519 would i still get all 3 marks even though it was not exactly 7500?
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mahmzo
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(Original post by mightyned)
I know this is late but i just wanted to restore some hope for the second exam.for q23bii i got 7519kgm-3 using the method1.2/9.8= 0.122volume = mass/density so 0.122/1000 = 0.0001229/9.8= 0.92density= mass/volume 0.92/0.000122= 7519 would i still get all 3 marks even though it was not exactly 7500?
No you will loose an accuracy mark as it is not to 2 sf
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2000ojw
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could you please post the questions to go with this, it would be very much appreicated
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