# 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 E

_{k}can be expressed as:
Gravitational force provides the
centripetal force.

mv

^{2}/r = GMm/r^{2}and E_{k}= ½ mv^{2}
Hence, E

_{k}= GMm / 2r
(ii)

Satellite mass, m = 620kg

Initial orbit = 7.34x10

^{6}m
(Resistive forces causes) New orbit
= 7.30x10

^{6}m
For satellite,

1.

Change in E

_{k}
ΔE

_{k}= ½ x 4.00x10^{14}x 620 x ({7.30x10^{6}}^{-1}– {7.34x10^{6}}^{-1})
ΔE

_{k}= 9.26x10^{7}J (ignore any sign in answer)
2.

Change in gravitational potential
energy

ΔE

_{p}= – GMm / r
ΔE

_{p}= 4.00x10^{14}x 620 x ({7.30x10^{6}}^{-1}– {7.34x10^{6}}^{-1})
ΔE

_{p}= 1.85x10^{8}J (ignore any sign in answer)
(iii)

Explain whether linear speed of
satellite increases, decreases or remains unchanged when radius of orbit decreases:

Either ({7.30x10

^{6}}^{-1}– {7.34x10^{6}}^{-1}) or ΔE_{k}is positive / E_{k}increased.
Speed has increased.

__Question 2__
Student: when an ideal heated from
100

^{o}C to 200^{o}C, 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 18

^{o}C.
When power of heater is 3.8kW, temperature
of water at outlet is 42

^{o}C.
Specific heat capacity of water, c =
4.2Jg

^{-1}K^{-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 = ω

^{2}and ω = 2πf or a = - ω^{2}x 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 F

_{E}= 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 = 8.45x10

^{14}Hz
Wavelength of 450nm = 6.67x10

^{14}Hz
6.67x10

^{14}Hz < 8.45x10^{14}Hz 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__: V

_{2}constant at 1.5V. V

_{1}:sinusoidal with amplitude 2.8V,

Variation with time t of output
potential V

_{out}:
Graph: square wave

Correct cross-over points where V

_{2}= V_{1}
Amplitude 5V

(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 V

_{out}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

^{-8}W
For input signal to wire pair of (P

_{2}=) 3.0x10^{-3}W, signal-to-noise ratio at receiver = 25dB.
Attenuation per unit length for wire
pair:

Attenuation per unit length = (1/L)
x 10lg(P

_{2}/P_{1})
Signal power at receiver, P

_{1}= 10^{2.5}x 3.8x10^{-8}= 1.2x10^{-5}W
Attenuation in wire pair = 10 lg({3.0x10

^{-3}}/{1.2x10^{-5}}) = 24dB
Attenuation per unit length = (1/1.4)
x 24 = 17dBkm

^{-1}
please explain Q11 (b) 1 and 2

ReplyDeleteDetails have been updated

Deletekindly explain 5 a ii) w12 qp 42.

ReplyDeleteDetails updated. The explanations seem quite explicit. If you are still having problems, let me know

Delete9702/43/O/N/12 question 9 please. part d(ii)

ReplyDeleteCheck solution 702 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-142.html

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

DeleteWhere is 42 paper..need it

ReplyDeletePaper 41 and 42 are the same. SO, they are both here.

DeleteGod bless you!!!!!!

ReplyDelete