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    (Original post by wuhuapple001)
    Indeed R=V/I, but as the graph is plotted as current against voltage, gradient will be dI/dV, which means that it is equal to 1/R.

    It does not matter whether the line is going through the origin or anywhere else, the gradient of the graph represents the change of current with repect to voltage. As dV/dI measures resistance, dI/dV represents 1/resistance.
    Sorry but you're wrong. R is V/I NOT dV/dI.
    Its often mistaught.
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    (Original post by wuhuapple001)
    Indeed R=V/I, but as the graph is plotted as current against voltage, gradient will be dI/dV, which means that it is equal to 1/R.

    It does not matter whether the line is going through the origin or anywhere else, the gradient of the graph represents the change of current with repect to voltage. As dV/dI measures resistance, dI/dV represents 1/resistance.
    You could test with points on the graph, points further along had a lower resistance than points on the same linear section, it confused me for ages last night before I had to search it up.
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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 = 2.9E16

    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
    How many marks is 27c out of?
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    (Original post by teachercol)
    Sorry but you're wrong. R is V/I NOT dV/dI.
    Its often mistaught.
    Yeah I get it now, sorry I was wrong. But still thank you for point out my mistake!
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    (Original post by NotGambit)
    For Q1, would it not be correct to say "rate of work done" which i think was A ? :/
    That's what I thought as well but it turns out that that's the def. for power not def. of watt
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    (Original post by JN17)
    You could test with points on the graph, points further along had a lower resistance than points on the same linear section, it confused me for ages last night before I had to search it up.
    Thank you, I will test it to see what happens! But I kind of get what you mean now.
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    (Original post by teachercol)
    I don't think so. Draw lines from the origin.
    I know that the resistance isn't constant; but I thought that if the line is linear then the rate of decrease of the resistance would be constant.
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    (Original post by wuhuapple001)
    Thank you, I will test it to see what happens! But I kind of get what you mean now.
    I found that the easiest way to think of it was that if R=V/I, by the graph, if you increased V by 1 volt, I would not increase by the same amount, so R changes.
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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 = 2.9E16

    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
    Thanks for posting these, they where a great help last year and this has helped calm me a little.
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    (Original post by NotGambit)
    For Q1, would it not be correct to say "rate of work done" which i think was A ? :/
    Didn't it ask "What is a watt?"

    So you'd have to mention the other units, not what power is.
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    (Original post by AidenJK)
    How many marks is 27c out of?
    27c was 3 marks
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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 = 2.9E16 (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 so much sir/madam! Really appreciate it
    Think I got 90%; that's probably full UMS points right?
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    I don't think we had to describe how the diameter was measured since the value for the diameter was already given; but I'm not sure. :/
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    (Original post by mbb123mbb)
    Thank you so much sir/madam! Really appreciate it
    Think I got 90%; that's probably full UMS points right?
    Well done B, and yes it most definitely will be.

    (Original post by Parhomus)
    I don't think we had to describe how the diameter was measured since the value for the diameter was already given; but I'm not sure. :/
    It said the diameter was about 5cm so to get an accurate reading you should have measured it again.

    Also if OP doesn't mind, could you please redo the very last question, as I believe the answer should be to the power of 15?
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    (Original post by sanchit117)
    Well done B, and yes it most definitely will be.


    It said the diameter was about 5cm so to get an accurate reading you should have measured it again.

    Also if OP doesn't mind, could you please redo the very last question, as I believe the answer should be to the power of 15?
    Damn, bad habit of not reading questions properly :L
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    (Original post by sanchit117)
    Well done B, and yes it most definitely will be.


    It said the diameter was about 5cm so to get an accurate reading you should have measured it again.

    Also if OP doesn't mind, could you please redo the very last question, as I believe the answer should be to the power of 15?
    Yes - its a typo - I've written 15 in my notes.
    Will correct
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    Hi, just curious as to what you're basing your grade boundary predictions on? As it's a new specification I've not been able to find any indicators of what sort of grade boundaries would be imposed on the exam. Thanks
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    Thanks for letting me know i failed 3 months i advance, ****
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    (Original post by NotGambit)
    For Q1, would it not be correct to say "rate of work done" which i think was A ? :/
    Nope, cause it was specifically referencing the unit of Watts as opposed to the physical quantity of power. When I first looked at that question I just stared at it for five minutes and thought OH MY GOD THESE ARE /ALL/ THE RIGHT ANSWER HELP.
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    (Original post by WETRTDFGKHL)
    Thanks for letting me know i failed 3 months i advance, ****
    Still all to play for - theres another paper yet.
 
 
 
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