Thursday, July 3, 2014

9702 November 2012 Paper 41 42 Worked Solutions | A-Level Physics

  • 9702 November 2012 Paper 41 & 42 Worked Solutions | A-Level Physics


Paper 41 and 42 are similar.


SECTION A

Question 1
(a)
Newton’s law of gravitation states that the force between two point masses is proportional to the product of the masses and inversely proportional to the square of the separation, provided that the separation is much larger than the sizes of the masses.

(b)
(i)
Considering gravitational force on satellite and centripetal force, kinetic energy Ek can be expressed as:
Gravitational force provides the centripetal force.
mv2/r = GMm/r2          and Ek = ½ mv2
Hence, Ek = GMm / 2r

(ii)
Satellite mass, m = 620kg
Initial orbit = 7.34x106m
(Resistive forces causes) New orbit = 7.30x106m

For satellite,
1.
Change in Ek
ΔEk = ½ x 4.00x1014 x 620 x ({7.30x106}-1 – {7.34x106}-1)
ΔEk = 9.26x107J (ignore any sign in answer)

2.
Change in gravitational potential energy
ΔEp = – GMm / r
ΔEp = 4.00x1014 x 620 x ({7.30x106}-1 – {7.34x106}-1)
ΔEp = 1.85x108J (ignore any sign in answer)

(iii)
Explain whether linear speed of satellite increases, decreases or remains unchanged when radius of orbit decreases:
Either ({7.30x106}-1 – {7.34x106}-1) or ΔEk is positive / Ek increased.
Speed has increased.


Question 2
Student: when an ideal heated from 100oC to 200oC, internal energy of gas is doubled
(a)
(i)
Internal energy is the sum of potential energy and kinetic energy of atoms / molecules / particles which are in random motion.

(ii)
By reference to one assumption of kinetic theory of gases, internal energy of ideal gas:
For an ideal gas, it is assumed that there are no intermolecular forces, and so, the potential energy of the molecules is zero. Therefore, internal energy of an ideal gas is the kinetic energy (from the random motion) of the molecules.

(b)
Is student’s suggestion correct?:
Kinetic energy is proportional to the thermodynamic temperature.
Either Since the temperature is given in Celsius and not in Kelvin, it is incorrect.
Or Temperature in kelvin is not doubled



Question 3
(a)
Two metal spheres are said to be in thermal equilibrium when the temperature of the two spheres is the same and there is no (net) transfer of energy between the spheres.

(b)
Water flows into tube at temperature of 18oC.
When power of heater is 3.8kW, temperature of water at outlet is 42oC.
Specific heat capacity of water, c = 4.2Jg-1K-1.
(i)
Flow rate of water through the tube, m:
Power = mcΔθ                        where m is the mass per second
3800 = m x 4.2 x (42 – 18)
m = 38gs-1

(ii)
Explain whether answer is likely to be an overestimate or underestimate of flow rate:
Some thermal energy is lost to the surroundings. So the rate is an overestimate.



Question 4
(a)
Explain features of Fig 4.2 that indicate that motion is simple harmonic:
The straight line passing through the origin shows that the acceleration is proportional to the displacement. The negative gradient shows that the gradient and displacement are in opposite directions.

(b)
For oscillation of ball:
(i)
Amplitude = 2.8cm

(ii)
Frequency:
Either gradient = ω2 and ω = 2πf       or a = - ω2x and ω = 2πf
Gradient = 13.5 / (2.8x10-2) = 482
ω = 22rads-1
Frequency = 22 / (2π) = 3.5Hz

(c)
Line AB is now vertical. Ball now oscillates. Ball now oscillates in vertical plane. One reason why oscillations may no longer be simple harmonic:
e.g. Lower spring may not be extended.
e.g. Upper spring may exceed limit of proportional / elastic limit




Question 5
{Detailed explanations for this question is available as Solution 706 at Physics 9702 Doubts | Help Page 143 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-143.html}





Question 6
(a)
(i)
Condition for charged particle to experience a force in magnetic field:
The particle must be moving with a component of velocity normal to the magnetic field.

(ii)
Expression for magnetic force F acting on charged particle in magnetic field flux density B:
F = Bqvsinθ
Where q is the charge of the particle, v is the velocity of the particle and θ is the angle between the direction of motion and magnetic field

(b)
(i)
Face where electrons tend to move due to force:
Face BCGF shaded


(ii)
Movement of electrons in magnetic field causes potential difference between 2 faces of conductor. State these faces:
Between face BCGF and face ADHE

(c)
Why potential difference causes an additional force on moving electrons in conductor:
The potential difference gives rise to an electric field.
Either FE = qE (no need to explain symbols)
Or The electric field gives rise to a force (on the electrons).


Question 7
(a)
Lenz’s law states that the induced e.m.f / current produces effects / acts in such a direction / tends to oppose the change causing it.

(b)
Why core is:
(i)
1.
Made of iron:
The core is made of iron to reduce flux losses / increase flux linkage / easily magnetized and demagnetized.

2.
Laminated:
To reduce energy / heat losses caused by eddy currents.

(ii)
Emf induced in secondary coil of transformer. Explain how current in primary coil gives rise to this induced emf:
The alternating current / voltage gives rise to a (changing) flux in the core.  This flux links the secondary coil. (By Faraday’s law), the changing flux induces an e.m.f (in secondary coil).


Question 8
(a)
A photon is a discrete quantity / packet / quantum of energy of electromagnetic radiation. Energy of photon = Planck constant x frequency

(b)
Where photoelectric emission of electrons take place, there is negligible time delay between illumination of surface and emission of electron. 3 other evidence provided by photoelectric effect for particulate nature of EM radiation:

Choose any 3:
Threshold frequency (of electromagnetic radiation required for emission of electrons.)
Rate of emission is proportional to intensity.
Maximum kinetic energy of electron dependent on frequency.
Maximum kinetic energy independent of intensity.

(c)
Work function of metal surface = 3.5eV.
Wavelength of light incident = 450nm
Determine whether electrons will be emitted by photoelectric effect from surface:

Either
E = hc /λ
λ = 450nm to give Energy = 4.4x10-19 or 2.8eV
2.8eV < 3.5eV so no emission

Or
hc /λ = eV
Work function of 3.5eV to give λ = 355nm
355nm < 450nm so no emission

Or
Work function = 3.5eV
[E=hf] Threshold frequency = 8.45x1014Hz
Wavelength of 450nm = 6.67x1014Hz
6.67x1014Hz < 8.45x1014Hz so no emission




SECTION B

Question 9
(a)
3 properties of ideal op-amp:
Choose any 3:
Zero output impedance / resistance
Infinite input impedance / resistance
Infinite (open loop) gain
Infinite bandwidth
Infinite slew rate

(b)
(i)
Comparator:  V2 constant at 1.5V. V1:sinusoidal with amplitude 2.8V,

Variation with time t of output potential Vout:
Graph: square wave
Correct cross-over points where V2 = V1
Amplitude 5V
Correct polarity (positive at t=0)

 (ii)

Draw symbols for the 2 diodes on Fig 9.1: R emits light for longer time than G:

Correct symbol for LED
Diodes connected correctly between Vout and earth
Correct polarity consistent with graph (i)
(LED R points ‘down’ if (i) correct)


Question 10
Principle of CT scanning:
X-ray images taken from different angles / X-rays directed from different angles
of one section / slice.
All the images are in the same plane
The images are combined to give an (2D) image of the section / slice
The (2D) images of successive sections / slices are combined
and a (3D) image is formed using a computer.
The image formed can be rotated / viewed from different angles.





Question 11
{Detailed explanations for this question is available as Solution 668 at Physics 9702 Doubts | Help Page 134 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-134.html}




Question 12
(a)
(i)
Cross-linking in the wire pair occurs when the signal in one wire (pair) is picked up by a neighboring wire (pair).

(ii)
Why cross-linking in coaxial cables is much less than in wire pairs:
The outer braid of coaxial cables is earthed and it shields the core from noise / external signals.

(b)
Length of wire pair = 1.4km
Constant noise power in wire pair = 3.8 x 10-8W
For input signal to wire pair of (P2=) 3.0x10-3W, signal-to-noise ratio at receiver = 25dB.
Attenuation per unit length for wire pair:

Attenuation per unit length = (1/L) x 10lg(P2/P1)
Signal power at receiver, P1 = 102.5 x 3.8x10-8 = 1.2x10-5W
Attenuation in wire pair = 10 lg({3.0x10-3}/{1.2x10-5}) = 24dB
Attenuation per unit length = (1/1.4) x 24 = 17dBkm-1



7 comments:

  1. please explain Q11 (b) 1 and 2

    ReplyDelete
  2. kindly explain 5 a ii) w12 qp 42.

    ReplyDelete
    Replies
    1. Details updated. The explanations seem quite explicit. If you are still having problems, let me know

      Delete
  3. 9702/43/O/N/12 question 9 please. part d(ii)

    ReplyDelete
    Replies
    1. Check solution 702 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-142.html

      Delete
    2. Thanks sooo much for asking that question Fatima. Exactly what I wanted.

      Delete

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