Monday, June 23, 2014

9702 June 2013 Paper 22 Worked Solutions | A-Level Physics

  • 9702 June 2013 Paper 22 Worked Solutions | A-Level Physics




Question 1
(a)
SI unit of power:
Power = energy/time = (force x distance/time) = kgm2s-2 / s
Power = kgm2s-3

(b)
Turbine that is used to generate electrical power from wind
Power, P = CL2ρv3
Where L is the length of each blade (3 blades) of the turbine, ρ is the density of air, v is the wind speed, C is a constant
(i)
C has no units:
C = P / L2ρv3
Units: L2 : m2              ρ = kgm-3        v3 : m3s-3
C = kgm2s-3 / (m2 kgm-3 m3s-3) = kgm2s-3 / kgm2s-3
Thus, the units of the components cancel each other, leading to C having no units.

(ii)
L = 25.0m,      ρ of air = 1.30,            C = 0.931.
Efficiency = 55%        Electric Power output, P = 3.50x105W
Wind speed = v
55% = 3.50x105W
Power available from wind, 100% = 3.50x105 x100/55 = 6.36x105W
P = CL2ρv3
So, v3 = P/ CL2ρ = 6.36x105 / (0.931 x (25)2 x 1.3)
v = 9.4ms-1

(iii)
Reasons why the electrical power output of the turbine is less than the power available from the wind:
Not all kinetic energy of the wind is converted to kinetic energy of the blades / generator. The conversion to electrical energy is also not 100% efficient since heat is produced in the generator / bearing, etc.






Question 2

{Detailed explanations for this question is available as Solution 1089 at Physics 9702 Doubts | Help Page 231 - http://physics-ref.blogspot.com/2016/01/physics-9702-doubts-help-page-231.html}





Question 3
{Detailed explanations for this question is available as Solution 864 at Physics 9702 Doubts | Help Page 173 - http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-173.html}






Question 4
(a)
Apparatus that demonstrates Brownian motion. Include diagram:
Apparatus: cell with particles e.g. smoke (container must be closed)
Diagram showing suitable arrangements with light illumination and microscope

(b)
Observations made using the apparatus:
Specks/flashes of light are seen to go in random motion.

(c)
2 conclusions about properties of molecules of a gas that follow from the observations:
choose any 2:
We cannot see what is causing the smoke to move hence, the molecules are smaller than the smoke particles.
The continuous motion of the smoke particles implies that motion of the molecules is continuous.
The random motion of the particles implies the random motion of molecules.



Question 5
(a)
Frequency, f =50Hz                transverse wave           speed, v=40ms-1

(i)
Wavelength of transverse wave on string:
v = fλ
λ = 40/50 = 0.80m

(ii)
How this arrangement may produce a stationary wave on the string:
The waves travel along the string and are reflected at Q / wall / fixed end. The incident and reflected waves interfere / superpose.

(b)
(i)
Label all nodes N and all antinodes A:
Nodes labeled at P, Q and the 2 points at zero displacement
Antinodes labeled at the 3 points of maximum displacement

(ii)
Length of string PQ:
PQ represents 1.5λ. Hence PQ = 0.8 x 1.5 = 1.2m

(iii)
Draw stationary wave at time (t+5.0ms) + explanation:
Period, T = 1/f = 1/50 = 20ms.
5ms is ¼ of cycle
Horizontal line through PQ drawn on Fig 5.2
{The reasoning as to why a horizontal line is obtained has been explained as Solution 41 (c)(iv) at Physics 9702 Doubts | Help Page 7 - http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-7.html to a question similar to this part. Read it there}



Question 6
(a)
Charge is defined as the product of current with time.

(b)
Resistance, R = 18.0Ω            Power supply of 240V            switched on for 2.60Ms
(i)
Power transformed in heater, P:
P =V2/R = (240)2 / 18 = 3200W

(ii)
Current in heater, I:
I = V/R = 240 /18 = 13.3A

(iii)
Charge passing through heater in this time, Q:
Charge, Q = It = 13.3 x 2.6x106 = 3.47 x 107C

(iv)
Number of electrons per second passing a given point in heater:
Number of electrons = 3.47x107 / 1.6x10-19 = 2.17x1026
Number of electrons per second = 2.17x1026 / 2.6x106 = 8.35x1019 s-1
  

Question 7
(a)
(i)
W = 206          X = 82
Y = 4               Z = 2

(ii)
Why mass seems not to be conserved in the reaction:
Mass – energy is conserved. Mass on the right hand side is less because energy is released.

(iii)
Meaning of spontaneous:
The reaction is spontaneous means that the reaction is not affected by external conditions / factors / environment OR 2 examples temperature and pressure.



33 comments:

  1. Here is a suggested solution for Q3(b)(iii):
    The previous part ( vertical component of T) is equal to 23 N
    Now my method of solving it is:

    Tan 50 = AC / 1.2
    AC = 1.43

    23 * 1.43 = (8.5+W) * 1.2
    W = 18.9 = 19.0 N

    Now the method in the marking scheme is a bit different, but my final answer ( 19.0 ) is still the same. Is my method acceptable or not?


    Why this method is not appropriate, even if the correct answer is obtained?
    1st mistake:
    Moment is the product of force and the perpendicular distance of the force from the pivot. Here BOTH the distance AC and the force are VERTICAL. This does not have any turning effect at A.

    2nd mistake:
    The centre of gravity does not act at 1.2m from the pivot BUT at the CENTRE.

    ReplyDelete
  2. how do we draw and describe brownian motion, please attach your suggested sample answer please thank you. :)

    ReplyDelete
    Replies
    1. The Brownian motion apparatus has been added. This is as given by Cambridge in June 2005 Paper 2 Question 2, so it should be a correct diagram.

      As for the description, look at question 5 at
      http://physics-ref.blogspot.com/2014/09/9702-november-2008-paper-2-worked.html

      let me know if you still need help

      Delete
  3. Can i draw a container (BEFORE) with smoke particles and another one clean transparent one attached but separated by a piece of wood in between. Then the wood is removed, container (AFTER) that the smoke particles diffuse and move randomly from high concentration to low concentration in the container ? Thank you :)

    ReplyDelete
    Replies
    1. I think you've got something wrong. You want to present a situation where you are causing the smoke particles to move (like by diffusion here). That's not what Brownian motion is about.

      The smoke cell SHOULD actually be closed to prevent any external forces, ... The smoke particles will be seen in random motion. This random motion is due to the smoke particles colliding with air particles which are themselves in random motion. But we cannot see air particles even with microscopes. So, the random motion of the smoke particles proves the random motion of air particles.

      Brownian motion is actually the random motion of air particles (which cannot be seen), which is inferred from the random motion of the smoke particles (which can be seen) .

      Delete
  4. in Q2 b (ii)
    how did yu use the formula v=u+at
    when both v and u is zero

    ReplyDelete
    Replies
    1. More details have been added for the explanations.

      Delete
    2. still in confusion with the last part
      yu used
      v=ut+2at
      v=0+2(3.5)
      v=7
      but
      what about the t in 2at of the equation?
      shouldn't it be 2(3.5)(2) ?

      Delete
    3. It's not v = u + 2at but v = u + at. What I'm using is the equation: Acceleration, a = change in velocity (v - u) divide by time, t. Making v the subject of formula, it becomes v = u + at.

      May be you are confusing with the equation: v^2 = u^2 + 2as where s is distance

      Delete
  5. This comment has been removed by the author.

    ReplyDelete
    Replies
    1. We can only find the weight for the rod in equilibrium with the data given in the question.

      Otherwise, the resultant torque on the rod should be given.

      Delete
  6. Can you please solve November 2012 paper 23 question 1(e), both the parts?

    ReplyDelete
    Replies
    1. See solution 658 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-131.html

      Delete
  7. Can you solve June 13 p23 q 5b iii please? Urgently, My AS examination is tomorrow

    ReplyDelete
    Replies
    1. Check solution 691 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-140.html

      Delete
  8. Can you please show the solution for june 2013 p21 q5 c

    Thanks

    ReplyDelete
    Replies
    1. Check question 25 at
      http://physics-ref.blogspot.com/2014/10/physics-9702-doubts-help-page-4.html

      Delete
    2. Thank you!!!:)

      Delete
  9. Hey,
    Could you please help me out with june 2013 paper 23, q1 b (ii) (iii), i don't get why the line is not supposed to touch the axis and also why the two graphs are curves?:)

    Thank you,

    ReplyDelete
    Replies
    1. See question 722 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-146.html

      Delete
  10. Hello sir, can you please give/tell the direction of force acting on rod A in question 3)c). Thank you it would be really helpful.

    ReplyDelete
  11. can you do may/june 13 variant 21 waves question? please?

    ReplyDelete
    Replies
    1. See solution 25 at
      http://physics-ref.blogspot.com/2014/10/physics-9702-doubts-help-page-4.html

      Delete
  12. for q2 b (ii) force * time is equal to change in momentum so if we use that for time 3 sec the momentum is supposed to be 0 since the force itself is zero? so why is there a horizontal line

    ReplyDelete
    Replies
    1. As you've said, the CHANGE in momentum is zero, i.e. momentum is not changing - it has the same value as previously. That is why there is a horizontal line.

      Delete
  13. Admin i got a problem in 2015 qp as well as 22/m/j/2013 please help ASAP please

    ReplyDelete
    Replies
    1. please help me with question Q5b 23/mj/2013

      Delete
  14. please help me with q5 23/mj/2013

    ReplyDelete
    Replies
    1. See solution 691 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-140.html

      Delete

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