Sunday, July 13, 2014

9702 June 2012 Paper 22 Worked Solutions | A-Level Physics

  • 9702 June 2012 Paper 22 Worked Solutions | A-Level Physics



Question 1
{Detailed explanations for this question is available as Solution 1022 at Physics 9702 Doubts | Help Page 213 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-213.html}





Question 2
(a)
(i)
Speed of ball as it hits the ground:
v2 = u2 +2as = (8.4)2 + 2(9.81)(5)
v = 12.99ms-1

(ii)
Show that time taken for ball to reach ground =0.47s:
s = ut + ½ at2                           or t = (v-u) / a
5 = 8.4t + ½ (9.81)t2               or t = (12.99 – 8.4) / 9.81
t = 0.468s

(b)
Reasonable shape
Suitable scale
Correctly plotted 1st and last points at (0,8.4) and (0.88 – 0.96,0) with non-vertical line at 0.47s

{Graph to show variation with time t of velocity v of ball: First point is at (0,8.4) since velocity of ball at t=0 is 8.4ms-1. Due to gravity, velocity of ball increases until v = 12.99ms-1 where the ball reaches the ground. This is at t=0.47s. The gradient between these 2 points [(0,8.4) and (0.47,12.99)] which represents the acceleration is constant and equal to 9.81ms-2. The speed after the ball rebounds is 4.2ms-1 but since the ball is now moving upwards, v = - 4.2ms-1. Additionally, since contact time is 20ms = 0.02s [non-zero], the line representing the change from the maximum +ve speed to this –ve one is not vertical. As for the final stage, the velocity of the ball at B is zero. The time taken for the velocity to be zero is t = (0 - -4.2) / (9.81) = 0.428s. So, velocity is zero at t = 0.428 + 0.47 = 0.898s. Gradient is similar to first section (=9.81). Graph drawn: on x-axis, smallest unit = 0.02s. on y-axis, smallest unit = 1ms-1.}

(c)
Mass of ball = 0.050kg. Ball moves from A and reaches B after rebounding.
(i)
1.
Change in kinetic energy:
Kinetic energy at B is zero, so ΔKE = ½ mv2   or ΔKE = ½ mu2 – ½ mv2
ΔKE = ½ (0.05)(8.4)2 = (-)1.8J

2.
Change in gravitational potential energy:
Final maximum height, h = (4.2)2 / 2(9.81) = (0.9m)
Change in PE = mgh – mgh1  
Change in PE = 0.05 x 9.81 x (0.9 – 5) = (-)2.0J

(ii)
Explain total change in energy of ball for this motion:
The total change in energy is -3.8J. Energy is lost to ground (on impact) / energy of deformation of the ball / thermal energy in ball.





Question 3
{Detailed explanations for this question is available as Solution 614 at Physics 9702 Doubts | Help Page 121 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-121.html}






Question 4
(a)
Pd across XY:
Total resistance in circuit = 8 + 12 = 20kΩ
Current = 12 / (20x103)           or potential divider formula
p.d = [12/(20x103)] x 12x103 = 7.2V

(b)
Resistance LDR = 4.0kΩ. Current in ammeter:
Parallel resistance (across XY) = [(1/4) + (1/12)]-1 = 3kΩ
Total resistance in circuit = 3 + 8 = 113kΩ
Current = 12 / (11x103) = 1.09x10-3A or 1.1x10-3A

(c)
Intensity of light on LDR increased.
(i)
Change to ammeter reading:
The resistance of the LDR decreases. So, the total resistance (of the circuit) is less. Hence, current increases.

(ii)
Change to p.d across XY:
The resistance across XY is now less. So, there is less proportion of the 12V across XY. Hence, p.d. is less.






Question 5
{Detailed explanations for this question is available as Solution 626 at Physics 9702 Doubts | Help Page 124 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-124.html}





Question 6
{Detailed explanations for this question is available as Solution 631 at Physics 9702 Doubts | Help Page 125 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-125.html}





Question 7
{Detailed explanations for this question is available as Solution 641 at Physics 9702 Doubts | Help Page 127 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-127.html}

15 comments:

  1. could you explain 3(B2) in detail please.It would be really appreciated! ^_^ in desperate need of help

    ReplyDelete
    Replies
    1. Details have been added for the question. See again

      Delete
  2. In question 5(c), why must the line pass through 1.5 on the y axis? i get that it has to pass through 0.25 on the x axis but how did you find that it has to pass through 1.5 on the y axis?

    ReplyDelete
    Replies
    1. Details have been updated for question 5

      Delete
    2. thank you :)
      can you also upload the diagram for question 6(b)?

      Delete
  3. can you please draw question 6(b)?

    ReplyDelete
  4. Hi, I am just wondering if the uncertainty is 0.09 x 10^-3,shouldnt the actual measurement be multiples of the uncertainty? or follow the decimal place of uncertainty?Confused about this part...I would really appreciate if you can answer me in time.Thanks!!!

    ReplyDelete
    Replies
    1. If you are referring to a specific question, give me the reference of the question. It's difficult to explain otherwise.

      Delete
    2. Please tell me the difference between 9702 June 2012 Paper 22 Q 1c and 9702 October 14 P23 Q2b),your help is much appreciated ^^

      Delete
    3. The explanation has been updated. See if it helps.

      Delete
  5. Can you please solve 0ctober November 2001 Paper 2 Question 7?
    Anyways great blog. Keep up the good work!

    ReplyDelete
    Replies
    1. Check solution 1093 at
      http://physics-ref.blogspot.com/2016/01/physics-9702-doubts-help-page-232.html

      Delete
  6. these have been extremely helpful,by any chance do you have one for chemistry papers 9701?

    ReplyDelete
    Replies
    1. tried to make one at
      http://chemistry-ref.blogspot.com/

      but it requires too time work/time. if anyone's willing to contribute it would help

      Delete

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