Question
(a) State Newton’s law of gravitation. [2]
(b) Planets have been observed orbiting a star in another solar system. Measurements are made
of the orbital radius r and the time period T of each of these planets.
The variation with R3 of T2 is shown in Fig. 1.1.
The relationship between T and R is given by
\(T^{2}=\frac{4\pi ^{2}R^{3}}{GM}\)
where G is the gravitational constant and M is the mass of the star.
Determine the mass M.
M = ……………………………………………. kg [3]
(c) A rock of mass m is also in orbit around the star in (b). The radius of the orbit is r.
(i) Explain why the gravitational potential energy of the rock is negative. [3]
(ii) Show that the kinetic energy Ek of the rock is given by
\(E_{k}=\frac{GMm}{2r}\) [2]
(iii) Use the expression in (c)(ii) to derive an expression for the total energy of the rock. [2]
[Total: 12]
Answer/Explanation
Ans
(a) (gravitational) force is (directly) proportional to product of masses
force (between point masses) is inversely proportional to the square of their separation
(b) correct read offs from the graph with correct power of ten for R3
\(M=\frac{4\times \pi ^{2}\times 1.2\times 10^{34}}{6.67\times 10^{-11}\times 2.4\times \left ( 365\times 24\times 3600 \right )^{2}}\)
= × 3.0 1030 kg
(c) (i) potential energy is zero at infinity
(gravitational) forces are attractive
work must be done on the rock to move it to infinity
(c) (ii) \(\frac{GMm}{r^{2}}=\frac{mv^{2}}{r} \\) OR\(\ v^{2}=\frac{GM}{r} \ OR\ v=\sqrt{\frac{GM}{r}}\)
use of \(½ mv^2\) (e.g. multiplication by ½ m) leading to \(\frac{GMm}{r^{2}}\)
(c)(iii) \(E_p=φ m \\) and \(\ φ =\frac{-GM}{r}\) \ or\ \(E_{p}=\frac{-GMm}{r}\)
Total energy = Ek + Ep
\(Total energy=\frac{GMm}{2r}+\frac{-GMm}{r}=\frac{-GMm}{2r}\)
Question
A fixed mass of an ideal gas is at a temperature of 21°C. The pressure of the gas is 2.3 × 105 Pa and its volume is 3.5 × 10–3m3.
(a) (i) Calculate the number N of molecules in the gas.
N = ………………………………………………… [2]
(ii) The mass of one molecule of the gas is 40u.
Determine the root-mean-square (r.m.s.) speed of the gas molecules.
r.m.s. speed = ………………………………………… ms–1 [2]
(b) The temperature of the gas is increased by 84°C.
Calculate the value of the ratio
ratio = ………………………………………………… [2]
[Total: 6]
Answer/Explanation
Ans
(a) (i) pV = NkT = or pV = nRT = and N = nNA
\(N=\frac{2.3\times 10^{5}\times 3.5\times 10^{-3}}{1.38\times 10^{-23}\times 294}\)
= 2.0 × 1023
(a) (ii) \(pV=\frac{1}{3}Nmc^{2}\)
\(c^{2}=\frac{3\times 2.3\times 10^{5}\times 3.5\times 10^{-3}}{2.0\times 10^{23}\times 40\times 1.66^{-27}}\)
= 182 000
r.m.s. speed = 430 ms–1
or
\(\frac{1}{2}mc^{2}=\frac{3}{2}kT\)
\(c^{2}=\frac{3\times 1.38\times 10^{-23}\times 294}{40\times 1.66\times 10^{-27}}\)
= 183 000
r.m.s.speed = 430 ms–1
(b) \(c^{2}=\frac{3\times2.0\times10^{23}\times1.38\times 10^{-23}\times(294+84)}{2.0\times 10^{23}\times 40\times 1.66\times 10^{-27}}\)
c = 236000
c= 485
\(ratio=\left ( =\frac{485}{430} \right )=1.1\)
OR
\(v\infty \sqrt{T} \ or\ v^{2}\infty T\)
\(ratio=\sqrt{\frac{273+21+84}{273+21}} \ or\ \sqrt{\frac{378}{294}}\)
ratio = 1.1
Question
(a) Using a simple kinetic model of matter, describe the structure of a solid. [2]
(b) The specific latent heat of vaporisation is much greater than the specific latent heat of fusion for the same substance. Explain this, in terms of the spacing of molecules. [1]
(c) A heater supplies energy at a constant rate to 0.045kg of a substance. The variation with time of the temperature of the substance is shown in Fig. 3.1. The substance is perfectly insulated
from its surroundings.
(i) Determine the temperature at which the substance melts.
temperature = ……………………………………………. °C [1]
(ii) The power of the heater is 150W.
Use data from Fig. 3.1 to calculate, in \(kJkg^{–1}\), the specific latent heat of vaporisation L of
the substance.
L = ………………………………………. kJkg–1 [3]
(iii) Suggest what can be deduced from the fact that section Q on the graph is less steep
than section P. [1]
[Total: 8]
Answer/Explanation
Ans
(a) Any 2 from:
• particles / atoms / molecules / ions (very) close together / touching
• regular, repeating pattern
• vibrate about a fixed point
(b) (much) greater increase in spacing of molecules (for vaporisation compared with fusion)
(c)(i) –100 °C
(c) (ii) time = 8.5 – 3.0
= 5.5 min
Pt = mL
energy = power × time = 150 × 5.5 × 60
= 49 500 J
\(L=\frac{E}{m}\)
\(\frac{49500}{0.045}\)
=1100 kJ kg−1
(c) (iii) gas has a higher specific heat capacity (than liquid)
Question
(a) The defining equation of simple harmonic motion is
\(a = – ω ^2x\).
State the significance of the minus (–) sign in the equation. [1]
(b) A trolley rests on a bench. Two identical stretched springs are attached to the trolley as shown in Fig. 4.1. The other end of each spring is attached to a fixed support.
The unstretched length of each spring is 12.0cm. The spring constant of each spring is 8.0Nm–1. When the trolley is in equilibrium the length of each spring is 18.0cm.
The trolley is displaced 4.8cm to one side and then released. Assume that resistive forces on the trolley are negligible.
(i) Show that the resultant force on the trolley at the moment of release is 0.77N.
(ii) The mass of the trolley is 250g.
Calculate the maximum acceleration a of the trolley.
a = ………………………………………… ms–2 [1]
(iii) Use your answer in (ii) to determine the period T of the subsequent oscillation.
T = ……………………………………………… s [3]
(iv) The experiment is repeated with an initial displacement of the trolley of 2.4cm. State and explain the effect, if any, this change has on the period of the oscillation of the trolley.
[Total: 9]
Answer/Explanation
Ans
(a) acceleration and displacement are in opposite directions
(b) (i) F =kx
= 8.0 × (0.060 − 0.048) or 8.0 × (0.060 + 0.048) or
or 8.0 × 0.012 or 8.0 × 0.108
ΣF= (8.0 × 0.012) – ( 8.0 × 0.108) = 0.77
or
Σ= 0.864− 0.096 = 0.77N
(b) (ii) \(a=\frac{F}{m}\)
\(=\frac{0.77}{0.25}\)
= 3.1 ms-2
(b) (iii) a = – ω2x
\(\omega=\sqrt{\frac{3.1}{0.048}}\)
ω= 8.04
(b) (iv) (resultant) force halved and distance halved
same T
Question
(a) (i) State what is meant by the amplitude modulation (AM) of a radio wave. [2]
(ii) State two advantages of AM transmissions when compared with frequency modulation
(FM) transmissions.
1.
2. [2]
(b) The variation with frequency f of the amplitude A of a transmitted radio wave after amplitude modulation by an audio signal is shown in Fig. 5.1.
For this transmission, determine:
(i) the wavelength of the carrier wave
wavelength = …………………………………………….. m [1]
(ii) the maximum frequency of the transmitted audio signal.
frequency = ………………………………………….. kHz [1]
(c) Another audio signal with the same maximum frequency is transmitted using a different
carrier wave frequency. The lowest frequency of this modulated wave is equal to the highest
frequency of the modulated wave in (b).
Determine the frequency of this carrier wave.
frequency = ………………………………………….. kHz [1]
[Total: 7]
Answer/Explanation
Ans
5 (a) (i) amplitude of the carrier wave varies M1
in synchrony with the displacement of the (information) signal
5 (a) (ii) Any 2 from:
• fewer transmitters needed / each transmitter can cover a greater distance
• more stations can share waveband
• transmitters and receivers are cheaper
5 (b) (i) \(\lambda =\frac{v}{f}\)
\(=\frac{3.0\times 10^{8}}{1.5\times 10^{6}}=200m\)
5 (b) (ii) 10 kHz
5 (c) 1520 kHz
Question
(a) State a similarity between the gravitational field lines around a point mass and the electric field lines around a point charge. [1]
(b) The variation with radius r of the electric field strength E due to an isolated charged sphere in a vacuum is shown in Fig. 6.1.
Use data from Fig. 6.1 to:
(i) state the radius of the sphere
radius = …………………………………………… cm [1]
(ii) calculate the charge on the sphere.
charge = …………………………………………….. C [2]
(c) Using the formula for the electric potential due to an isolated point charge, determine the capacitance of the sphere in (b).
capacitance = ……………………………………………… F [3]
[Total: 7]
Answer/Explanation
Ans
(a) (both have) radial field lines
(b) (i) 2.1 cm
(b) (ii) \(E=\frac{Q}{4\pi \varepsilon _{o}r^{2}}\)
e.g. r = 2.1 cm, E = 1.30 × 105 V m–1
\(Q=4\pi \varepsilon_{o}r^{2}E\)
\(=4\times \pi \times 8.85\times 10^{-12}\times 0.021^{2}\times 1.30\times 10^{5}\)
\(=6.4\times 10^{-9}C\)
(c) \(C=\frac{Q}{V}\)
Question
(a) Fig. 7.1 shows the circuit diagram containing an operational amplifier (op-amp).
(i) State the name of this type of amplifier. [1]
(ii) Show that the gain of the amplifier is 6.0. [1]
(iii) At time t = 0 the input potential VIN is zero. VIN then gradually increases with time t as
shown in Fig. 7.2.
On Fig. 7.2 sketch a line to show the variation with time t of the output potential
VOUT from time t = 0 to time t = T. [2]
(iv) State how the circuit of Fig. 7.1 may be changed so that the gain of the amplifier is
dependent on light intensity. [1]
(b) An op-amp is to be used to switch on a high-voltage heater.
(i) State the name of the component used as the output device of the op-amp. [1]
(ii) Complete Fig. 7.3 using the device named in (i) and a diode so that the heater may be
switched on when the output of the op-amp is positive.
Answer/Explanation
Ans
(a) (i) non-inverting (amplifier)
(a) (ii) \(gain=\frac{R_{f}}{R}+1\)
\(gain=\frac{3.6}{0.72}+1=6.0\)
(a) (iii) straight line from (0,0) to (T / 2, 3)
line from origin to 3.0 V then horizontal line at 3.0 V to T
(a )(iv) ldr / light dependent resistor replaces one of the two resistors
(b) (i) relay coil
(b) (ii) relay coil between op-amp and earth B1
diode with correct polarity (pointing away from output) connected between output and device and no other connections
or diode with correct polarity (pointing towards earth) between device and earth and no other connections
switch connected to high voltage circuit
Question
(a) Two long straight wires P and Q are parallel to each other, as shown in Fig. 8.1. There is a current in each wire in the direction shown.
The pattern of the magnetic field lines in a plane normal to wire P due to the current in the wire is also shown.
(i) Draw arrows on the magnetic field lines in Fig. 8.1 around wire P to show the direction of the field. [1]
(ii) Determine the direction of the force on wire Q due to the magnetic field from wire P. [1]
(iii) The current in wire Q is less than the current in wire P.
State and explain whether the magnitude of the force on wire P is less than, equal to, or greater than the magnitude of the force on wire Q. [2]
(b) Nuclear magnetic resonance imaging (NMRI) is used to obtain diagnostic information about internal structures in the human body.
Radio waves are produced and directed towards the body. The radio waves affect the protons within the body.
(i) Explain why radio waves are used. [2]
(ii) Explain why the radio waves are applied in pulses. [2]
[Total: 8]
Answer/Explanation
Ans
(a) (i) at least one anticlockwise arrow and no clockwise arrows
(a) (ii) (force is to the) left
(a) (iii) force is the same
Newton’s third law (of motion)
or force depends on the product of the two currents
(b) (i) frequency of radio waves is equal to natural frequency of protons
resonance of protons occurs / protons absorb energy
(b) (ii) in between pulses / when pulse stops
Any 1 from:
• protons de-excite
• protons emit r.f. pulses
• emitted (r.f.) pulse (from proton) detected
Question
(a) Define magnetic flux linkage. [2]
(b) A solenoid of diameter 6.0cm and 540 turns is placed in a uniform magnetic field as shown in Fig. 9.1.
The variation with time t of the magnetic flux density is shown in Fig. 9.2.
Calculate the maximum magnitude of the induced electromotive force (e.m.f.) in the solenoid.
e.m.f. = ……………………………………………… V [3]
(c) A thin copper sheet X is supported on a rigid rod so that it hangs between the poles of a
magnet as shown in Fig. 9.3.
Sheet X is displaced to one side and then released so that it oscillates. A motion sensor is
used to record the displacement of X.
A second thin copper sheet Y replaces sheet X. Sheet Y has the same overall dimensions as
X but is cut into the shape shown in Fig. 9.4.
The motion sensor is again used to record the displacement.
The graph in Fig. 9.5 shows the variation with time t of the displacement s of each copper
sheet.
(i) State the name of the phenomenon illustrated by the gradual reduction in the amplitude
of the dashed line. [1]
(ii) Deduce which copper sheet is represented by the dashed line. Explain your answer
using the principles of electromagnetic induction. [4]
[Total: 10]
Answer/Explanation
Ans
(a) (magnetic) flux density × area × number of turns
area is perpendicular to (magnetic) field
(b) use of t = 1.2 s
\(\varepsilon=\frac{\Delta BAN}{\Delta T}\)
\(=\frac{0.250\times \pi \times 0.030^{2}\times 540}{1.2}\)
= 0.32V
(c) (i) light damping
(c) (ii) sheet cuts (magnetic) flux and causes induced emf
(induced) emf causes (eddy) currents (in sheet)
either currents (in sheet) cause resistive force
or currents (in sheet) dissipate energy
smaller currents in Y or larger currents in X, so dashed line is X
Question
The output potential difference (p.d.) of an alternating power supply is represented by
V = 320sin(100πt)
where V is the p.d. in volts and t is the time in seconds.
(a) Determine the root-mean-square (r.m.s.) p.d. of the power supply.
r.m.s. p.d. = ……………………………………………… V [1]
(b) Determine the period T of the output.
T = ……………………………………………… s [2]
(c) The power supply is connected to resistor R and a diode in the circuit shown in Fig. 10.1.
(i) State the name of the type of rectification produced by the diode in Fig. 10.1. [1]
(ii) On Fig. 10.2 sketch the variation with time t of the p.d. \(V_R\) across R from time t = 0 to
time t = 40ms.
(iii) On Fig. 10.1, draw the symbol for a component that may be connected to produce
smoothing of \(V_R\). [1]
[Total: 8]
Answer/Explanation
Ans
(a) 230 V
(b) ω = 100π
\(T=\frac{2\pi }{\omega }=\frac{2\pi }{100\pi }\)
= 0.020s
(c) (i) half-wave (rectification)
(c) (ii) sinusoidal half waves in positive V only or negative V only, peak at 320 V
line at zero for second half of cycle
two time periods shown, each of 0.020 s
(c) (iii) capacitor added in parallel with resistor
Question
(a) Electrons are accelerated through a potential difference of 15kV. The electrons collide with a metal target and a spectrum of X-rays is produced.
(i) Explain why a continuous spectrum of energies of X-ray photons is produced. [3]
(ii) Calculate the wavelength of the highest energy X-ray photon produced.
wavelength = …………………………………………….. m [3]
(b) A beam of X-rays has an initial intensity Io. The beam is directed into some body tissue. After passing through a thickness x of tissue the intensity is I. The graph in Fig. 11.1 shows the
variation with x of ln (I/Io).
(i) Determine the linear attenuation (absorption) coefficient μ for this beam of X-rays in the
tissue.
μ = ………………………………………… cm–1 [2]
(ii) Determine the thickness of tissue that the X-ray beam must pass through so that the
intensity of the beam is reduced to 5.0% of its initial value.
thickness = …………………………………………… cm [2]
[Total: 10]
Answer/Explanation
Ans
(a) (i) electrons decelerate (on hitting target) so X-ray photons produced
range of decelerations
photon energy depends on (magnitude of) deceleration
(a) (ii) \(eV=\frac{hc}{\lambda }\)
\(\lambda =\frac{6.63\times 10^{-34}\times 3.0\times 10^{8}}{1.6\times 10^{-19}\times 15000}\)
\(=8.3\times 10^{-11}m\)
0r
E = hf and c = fλ and electron energy = eV
or
E = hc / λ and electron energy = eV
electron energy = 1.6 × 10–19 × 15000
= 2.4 × 10–15
\(\lambda =\frac{6.63\times 10^{-34}\times 3.0\times 10^{8}}{2.4\times 10^{-15}}\)
λ = 8.3 × 10-11m
(b )(i) μ = – gradient or ln (I / Io) =−μx
(e.g. 2.08 / 10.0) = 0.21 cm–1
(b) (ii) ln ( 0.05) = −μx
\(x=\frac{In0.05}{-\mu x}\)
e g x = 14cm
Question
(a) Radioactive decay is both spontaneous and random.
State what is meant by:
1. spontaneous decay
2. random decay. [2]
(b) Strontium-90 
is an unstable nuclide.
The activity of a sample of 1.0 × 10–9 kg of strontium-90 is 5.2MBq.
(i) Determine the decay constant λ of strontium-90.
λ = …………………………………………… s–1 [3]
(ii) The activity of the sample after a time of 1.0 half lives is found to be greater than the expected 2.6MBq.
Suggest a possible reason for this. [1]
[Total: 6]
Answer/Explanation
Ans
(a) 1 not affected by external factors
2 cannot predict when a (particular) nucleus will decay
or cannot predict which nucleus will decay (next)
(b) (i) Number of atoms \(=\frac{1.0\times 10^{-9}}{90\times 1.66\times 10^{-27}} \ or\ \frac{1.0\times 10^{-9}\times 6.02\times 10^{23}}{90\times 10^{-3}}\)
\(=6.693\times 10^{15}\)
\(\lambda =7.8\times 10^{-10}s^{-1}\)
(b) (ii) daughter nucleus is unstable