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Monday, September 22, 2014

9702 November 2007 Paper 4 Worked Solutions | A-Level Physics

  • 9702 November 2007 Paper 4 Worked Solutions | A-Level Physics

Paper 4


SECTION A

Question 1
(a)
(i)
A radian is the angle subtended at the centre of a circle such that the arc is equal in length to the radius.

(ii)
Why 1 complete revolution is equivalent to angular displacement of 2π rad:
Length of arc = rθ and for one revolution, the length of arc formed (= circumference) = 2πr. So, θ = 2πr / r = 2π

(b)
Elastic cord has unextended length of 13.0cm. One end of cord attached to fixed point C. Small mass of weight 5.0N is hung from free end of cord. Cord extends to length of 14.8cm, as shown.
Cord and mass are now made to rotate at constant angular speed ω in vertical plane about point C. When cord is vertical and above C, its length is the unextended length of 13.0cm, as shown.
(i)
Show that angular speed ω of cord and mass = 8.7rads-1:
EITHER The weight provides / equals the centripetal force OR The acceleration of free fall is the centripetal acceleration.
(a = rω2)
9.8 = 0.13ω2
ω = 8.7 rads-1

(ii)
Cord and mass rotate so that cord is vertically below C, as shown. Length L of cord, assuming it obeys Hooke’s law:

{When the cord and mass is vertically below C, the centripetal force {FC} is provided by the resultant force from the force in the cord [tension in the cord {T} – which is upwards] and the weight [W - which is downwards]. (T should be greater than W so that the mass remains attached to that cord) So, FC = T – W giving T = W + FC}
Force in the cord = Weight + Centripetal force

{To know the force in the cord, the spring constant, which is always constant should be found first. Hooke’s law: Force, F = kx. Extension, x = L – 13 and spring constant, k = F /x. When F = 5.0N, x = 14.8 – 13 = 1.8cm, k = spring constant [this always constant] = 5.0 / 1.8} 
Force in the cord = (L – 13) x 5/1.8    or force constant = 5.0 / 1.8


{Force constant mentioned above = spring constant}

{Weight = 5.0N, Centripetal force = m rω2. Since only the weight is given, we need to find the mass to calculate the centripetal force. Mass, m = weight / g = 5.0 / 9.8}

{Force in the cord = Weight + Centripetal force}
(L – 13) x 5/1.8 = 5.0 + (5/9.8) (L x 10-2) (8.72)
Length of cord, L = 17.2cm



Question 2
(a)
Amount of 1.00mol of Helium-4 gas contained in cylinder at pressure of 1.02x105Pa and temperature of 27oC.
(i)
Volume of gas in cylinder:
pV = nRT
(T = 273 + 27 = 300)
V = (nRT/p =) (8.31 x 300) / (1.02x105) = 0.0244m3

(ii)
Show that average separation of gas atoms in cylinder is approximately 3.4x10-9m:
Volume occupied by 1 atom = 0.0244 / (6.02x1023) = 4.06x10-26m3
Separation ≈ (4.06x10-26)1/3 = 3.44x10-9m

(b)
(i)
Gravitational force between 2 Helium-4 atoms that are separated by distance of 3.4x10-9m:
Force = GMm / r2 = [(6.67x10-11) (4{1.66x10-27})2] / (3.44x10-9)2 = 2.49x10-46N

(ii)
Ratio of weight of a Helium-4 atom to gravitational force between 2 Helium-4 atoms that with separation 3.4x10-9m:
Ratio = [4(1.66x10-27) x 9.8] / (2.49x10-46) = 2.6x1020

(c)
Comment on answer to (b)(ii) with reference to 1of the assumptions of kinetic theory of gases:
One assumption of the kinetic theory of gases is that the forces between the atoms are negligible.
Comment: e.g
The ratio shows the gravitational force to be very small
The force is very much less than the weight
If there are forces, they are not gravitational forces.





Question 3
{Detailed explanations for this question is available as Solution 1125 at Physics 9702 Doubts | Help Page 243 - http://physics-ref.blogspot.com/2016/10/physics-9702-doubts-help-page-243.html}

 



Question 4
{Detailed explanations for this question is available as Solution 714 at Physics 9702 Doubts | Help Page 144 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-144.html}
 

 



Question 5
A capacitor is charged to a potential difference of 15 V and then connected in series with a switch, a resistor of resistance 12 kΩ and a sensitive ammeter, as shown in Fig. 5.1.





Question 6
A straight conductor carrying a current I is at an angle θ to a uniform magnetic field of flux density B, as shown in Fig.6.1. ... 



  
Question 7
{Detailed explanations for this question is available as Solution 1029 at Physics 9702 Doubts | Help Page 215 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-215.html}






SECTION B

Question 8
Fig. 8.1 shows a circuit incorporating an ideal operational amplifier (op-amp).
 




Question 9
(a)
Acoustic impedance is the product of density (of the medium) and the speed of sound (in the medium).

(b)
Why acoustic impedance is important when considering reflection of ultrasound at boundary between 2 media:
The difference in acoustic impedance determines the fraction of incident intensity that is reflected / amount of reflection.

(c)
Principles behind use of ultrasound to obtain diagnostic information about structures within body:
A pulse of ultrasound (which is directed into the body) is reflected at the boundary (between the tissues). (The reflected pulse is) detected and processed. The time for return of the echo gives (information on the) depth and the amount of reflection gives information on the tissue structures.




Question 10
Fig. 10.1 shows the variation with frequency f of the power P of a radio signal.


  

Question 11
In cellular phone network, country is divided into a number of cells, each with its own base station. Fig shows a number of these base stations and their connection to cellular exchange.
(a)
Explain why country is divided into a number of cells:
The carrier frequencies can be re-used (simultaneously without interference) so that the number of handsets possible is increased.
OR
Any sensible e.g. UHF used, so ‘line of sight’

(b)
What happens at base station and cellular exchange when mobile phone handset is switched on, before call is made?:
The handset sends out an (identifying) signal which is communicated by the base stations to (the computer at) the exchange. The computer selects a base station with the strongest signal and allocates a (carrier) frequency.


27 comments:

  1. asalamualaikum.
    Can you please explain Q3 part b(ii)? How do you find the amplitude using the graph and/or otherwise?

    ReplyDelete
    Replies
    1. Wa`alaikum-us-salaam.
      I have added some details. See it you understand now

      Delete
  2. 04/m/j/07 q4c(ii) pls

    ReplyDelete
  3. s06 qp 4, question 8 b ii) v(velocity)= electric potential/ magnetic force, as the velocity increases shouldn't the magnetic field decrease ;since, they are inversely proportional? meaning the E should increase as the velocity increases but why is that not the case?

    ReplyDelete
    Replies
    1. It's explained as solution 654 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-130.html

      Delete
  4. M/J 07 Q3(b) and 4(b) and (c) please. Thank you :)

    ReplyDelete
    Replies
    1. For question 3, see solution 26 at
      http://physics-ref.blogspot.com/2014/10/physics-9702-doubts-help-page-4.html

      Delete
    2. For question 4, see solution 699 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-141.html

      Delete
  5. Can you explain why is the answer for question 4(d.) ? MJ 2007

    ReplyDelete
    Replies
    1. See question 699 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-141.html

      Delete
  6. Can you explain ON2007 q4(d)? Thank you.**

    ReplyDelete
  7. MJ 2007 7(b) ,can you explain what's the function of resistor and how to draw the graph? Thank you :) thank you

    ReplyDelete
    Replies
    1. Check question 106 at
      http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-20.html

      Delete
  8. May Allah bless you for your kindness!!

    ReplyDelete
  9. Can you explain Q5 b2) because im getting a different answer and i dont understand how to calculate the area like explained in marking scheme. I solved it using trapezium rule, and by calculating the area of trapeziums normally.

    ReplyDelete
    Replies
    1. Explanation updated.

      It can work with trapezia. Make sure you follow what each cm represents as explained.

      Errors may occur due to overestimation or underestimation

      Delete
  10. PLZ EXPLAIN THE REASONS FOR THE TICKS AND CROSSES IN 0N/07 Q.8(part B)

    ReplyDelete
  11. PLEASE MAKE A COMPLETE ILLUSTRATED DIAGRAM FOR ON/07 Q.8 P(part C).. THANKSSS

    ReplyDelete
  12. Can you explain Q 10 part c by giving a graphical diagram

    ReplyDelete
  13. can you give a graphical diagram of solution of Question 10 part c

    ReplyDelete
  14. May Allah (SWT) give u success in this life and the Hereafter...Ameen!

    ReplyDelete

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