• 9702 June 2010 Paper 42 43 Worked Solutions | A-Level Physics

Paper 42 & 43

SECTION A

Question 1

(a)

Gravitational potential at a point is defined as the work done in moving unit mass from infinity to the point.

(b)

Earth may be considered to be isolated sphere of radius R with mass concentrated at centre. Variation of gravitational potential ϕ with distance x from centre of Earth shown. Radius R of Earth is 6.4x10⁶m.

(i)

Considering gravitational potential at Earth’s surface, Mass of Earth:

(From graph,) At R, ϕ = 6.3x10⁷Jkg^-1

ϕ = GM / R

6.3x10⁷ = (6.67x10^-11)M / (6.4x10⁶)

M = 6.0x10²⁴kg (allow 5.95 à 6.14)

(ii)

Meteorite at rest at infinity. Meteorite travels from infinity towards Earth. Speed of meteorite when it is at a distance of 2R above Earth’s surface:

(at 3R in graph,) change in potential = 2.1x107Jkg^-1

Loss in potential energy = gain in kinetic energy

½ mv² = ϕm                 or ½ mv² = GM / 3R

½ v² = 2.1x10⁷

v = 6.5x10³ms^-1

(iii)

In practice, Earth not isolated sphere because it is orbited by Moon, as illustrated. Initial path of meteorite also shown. 2 changes to motion of meteorite caused by Moon:

Example:

The speed / velocity / acceleration of the meteorite would be greater

The meteorite deviated / bends from the straight path

Question 2

Long strip of stringy steel clamped at one end so that strip is vertical. Mass of 65g attached to free end of strip, as shown. Mass is pulled to one side and then released. Variation with time t of horizontal displacement of mass shown. Mass undergoes damped simple harmonic motion.

(a)

(i)

Damping is a reduction in the energy (of the oscillations) / reduction in the amplitude / energy of oscillations due to forces that are (always) opposing the motion / resistive forces

(ii)

Whether damping is light, critical or heavy:

The amplitude is decreasing (very) gradually / the oscillations would continue (for a long time) / there are many oscillations. So, this is light damping.

(b)

(i)

Use Fig to determine frequency of vibration of mass:

Frequency (= 1 / period) = 1 / 0.3 = 3.3Hz

(ii)

Show that initial energy stored in steel strip before mass is released is approximately 3.2mJ:

Initial energy stored = ½ mv²             and      v = ωa

Energy = ½ (0.065) (2π / 0.3)² (1.5x10^-2)² = 3.2mJ

(c)

After 8 complete oscillations of mass, amplitude of vibration is reduced from 1.5cm to 1.1cm. Explain whether, after a further 8 complete oscillations, amplitude will be 0.7cm:

The amplitude reduces exponentially / does not decrease linearly. So, the amplitude will not be 0.7cm.

 

Question 3

{Detailed explanations for this question is available as Solution 500 at Physics 9702 Doubts | Help Page 97 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-97.html}

Question 4

{Detailed explanations for this question is available as Solution 1050 at Physics 9702 Doubts | Help Page 220 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-220.html}

Question 5

{Detailed explanations for this question is available as Solution 505 at Physics 9702 Doubts | Help Page 98 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-98.html}

Question 6

{Detailed explanations for this question is available as Solution 922 at Physics 9702 Doubts | Help Page 189 - http://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-189.html}

Question 7

(a)

EITHER

The root-mean-square (r.m.s.) value of an alternating voltage is the value of steady / constant voltage that produces the same power (in a resistor) as the alternating voltage.

OR

If an alternating voltage is squared and average, the r.m.s value is the square root of this averaged value.

(b)

An alternating voltage represented by            V = 220sin(120πt)       where V measured in volts and t in seconds. For alternating voltage:

(i)

Peak voltage: 220V

(ii)

r.m.s. voltage: (= 220 / √2 =)156V

(iii)

Frequency: 60Hz

(120π = ω =2πf. So, f = 120 /2 = 60Hz)

(c)

Alternating voltage in (b) applied across resistor such that mean power output from resistor is 1.5kW. Resistance of resistor:

Power, P = V_rms² / R

So, R = 156² / 1500 = 16Ω

Question 8

Americium-241 is artificially produced radioactive element that emits α-particles. Sample of americium-241 of mass 5.1μg found to have activity of 5.9x10⁵Bq.

(a)

For this sample of americium-241:

(i)

Number of nuclei:

Number of nuclei = (5.1x10^-6) (6.02x10²³) / 241 = 1.27x10¹⁶

(ii)

Decay constant:

A = λN

5.9x10⁵ = λ (1.27x10¹⁶)

Decay constant, λ = 4.65x10^-11s^-1 

(iii)

Half-life, in years:

4.65x10^-11 x t_½ = ln2

Half-life, t_½ = 1.49x10¹⁰s = 470years

(b)

Another radioactive element has half-life of approximately 4 hours. Why measurement of mass and activity of sample of this element not appropriate for determination of its half-life:

The sample / activity would decay appreciably whilst measurements are being made

SECTION B

Question 9

(a)

Negative feedback may be used in amplifier circuits.

(i)

Negative feedback is when a fraction of the output (signal) is added to the input (signal) and it is out of phase by 180^o / π rad / to the inverting input.

(ii)

2 effects of negative feedback on amplifier incorporating an operational amplifier (op-amp):

Choose any 2:

It reduces the gain

It increases the bandwidth

There is greater stability

It reduces distortion

(b)

Fig is a circuit for amplifier that is used with a microphone. Output potential difference V_OUT is 4.4V when potential at point P is 62mV.

(i)

Gain of amplifier:

Gain (= V_OUT / V_IN) = 4.4 / 0.062 = 71

(ii)

Resistance of resistor R:

71 = 1 + 120/R

Resistance of resistor R = 1.7x10³Ω

(c)

Maximum potential produced by microphone at point P on Fig is 95mV. Power supply for operational amplifier may be either +/- 5V or +/- 9V. Which power supply should be used?:

For the amplifier not to saturate, the maximum output is (71 x 95x10^-3 =) approximately 6.7V. So, the power supply that should be used is the +/- 9V one.

Question 10

{Detailed explanations for this question is available as Solution 695 at Physics 9702 Doubts | Help Page 141 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-141.html}

Question 11

Main principles behind generation of ultrasound to obtain diagnostic information about internal body structures:

EITHER A quartz OR piezo-electric crystal with the opposite faces / two sides coated (with silver) is used to act as electrodes.

EITHER the molecular structure is indicatedOR The centres of (+) and (-) charge do not coincide.

A potential difference across the crystal causes the crystal to change its shape. An alternating voltage (in US frequency range) is applied across the crystal which causes the crystal to oscillate / vibrate

(Crystal cut) so that it vibrates at resonant frequencies.

Question 12

Telephone link between 2 towns to be provided using optic fibre. Length of optic fibre between the 2 towns is 75km.

(a)

2 changes that occur in signal as it is transmitted along an optic fibre:

The signal becomes distorted / noisy

The signal loses power / energy / intensity / is attenuated

(b)

Optic fibre has attenuation per unit length of 1.6dBkm^-1. Minimum permissible signal-to-noise power ratio in fibre is 25dB. Average noise power in optic fibre is 6.1x10^-19W.

(i)

1 reason why power ratios expressed in dB:

EITHER The numbers involved are smaller / more manageable / cover a wider range

OR Calculations involve addition and subtraction rather than multiplication and division

(ii)

Signal input power to optic fibre designed to be 6.5mW. Whether repeater amplifiers are necessary in optic fibre between the 2 towns:

25 = 10 log(P_min / {6.1x10^-19})

Minimum signal power, P_min = 1.93x10^-16W

So, signal loss = 10 log ({6.5x10^-3} / {1.93x10^-16}) = 135dB

So, maximum cable length = 135 / 1.6 = 85km.

So, no repeaters are necessary.