Question
1. (a) State what is meant by a gravitational force.
(b) A binary star system consists of two stars \(S_1\) and \(S_2\), each in a circular orbit.
The orbit of each star in the system has a period of rotation T.
Observations of the binary star from Earth are represented in Fig. 1.1.
Observed from Earth, the angular separation of the centres of \(S_1\) and \(S_2\) is \(1.2 × 10^{–5}\) rad.
The distance of the binary star system from Earth is \(1.5 × 10^{17}\)m.
Show that the separation d of the centres of \(S_1\) and \(S_2\) is \(1.8 × 10^{12}\)m.
(c) The stars \(S_1\) and \(S_2\) rotate with the same angular velocity ω about a point P, as illustrated in
Fig. 1.2.
Point P is at a distance x from the centre of star \(S_1\).
The period of rotation of the stars is 44.2 years.
(i) Calculate the angular velocity ω.
ω = ………………………………………. \(rads^{–1}\)
(ii) By considering the forces acting on the two stars, show that the ratio of the masses of the stars is given by
\(\frac{mass of S_1}{mass of S_2} = \frac{d-x}{x}\)
(iii) The mass \(M_1\) of star \(S_1\) is given by the expression
\(GM_1 = d^2 (d – x)ω^2\)
where G is the gravitational constant.
The ratio in (ii) is found to be 1.5.
Use data from (b) and your answer in (c)(i) to determine the mass \(M_1\).
\(M_1\) = ……………………………………………. kg
Answer/Explanation
Answer:
(a) force acting between two masses
or
force on mass due to another mass
or
force on mass in a gravitational field
(b) arc length = rθ
\(d = 1.5 × 10^{17} × 1.2 × 10^{–5} = 1.8 × 10^{12}\) m
(c) (i) ω = 2 π / T
= 2 π / (44.2 × 365 × 24 × 3600)
\(= 4.5 × 10^{–9} rad s^{–1}\)
(ii) gravitational forces are equal
or
centripetal force about P is the same
\(M_1xω^2 = M_2(d – x)ω^2\) so \(M_1 / M_2 = (d – x) / x\)
(iii) x = 0.4d
\(6.67 × 10^{–11} × M1 = (1.0 – 0.4) × (1.8 × 10^{12})^3 × (4.5 × 10–9)^2\)
\(M_1 = 1.1 × 10^{30}\) kg
Question
2. (a) State what is meant by the internal energy of a system.
(b) By reference to intermolecular forces, explain why the change in internal energy of an ideal
gas is equal to the change in total kinetic energy of its molecules.
(c) State and explain the change, if any, in the internal energy of a solid metal ball as it falls
under gravity in a vacuum.
Answer/Explanation
Answer:
(a) total potential energy and kinetic energy (of molecules/atoms)
reference to random motion of molecules/atoms
(b) (in ideal gas,) no intermolecular forces
no potential energy (so change in kinetic energy is change in internal energy)
(c) (random) potential energy of molecules does not change
(random) kinetic energy of molecules does not change
so internal energy does not change
or
decrease in total potential energy = gain in total kinetic energy
no external energy supplied
so internal energy does not change
or
no compression (of ball) so no work done on the ball
no resistive forces so no heating of the ball
so internal energy does not change
(c) or
no change of state so potential energy (of molecules) unchanged
no temperature rise so kinetic energy (of molecules) unchanged
so internal energy does not change
Question
3. The piston in the cylinder of a car engine moves in the cylinder with simple harmonic motion.
The piston moves between a position of maximum height in the cylinder to a position of minimum
height, as illustrated in Fig. 3.1.
The distance moved by the piston between the positions shown in Fig. 3.1 is 9.8cm.
The mass of the piston is 640g.
At one particular speed of the engine, the piston completes 2700 oscillations in 1.0 minute.
(a) For the oscillations of the piston in the cylinder, determine:
(i) the amplitude
amplitude = …………………………………………… cm
(ii) the frequency
frequency = ……………………………………………. Hz
(iii) the maximum speed
maximum speed = ………………………………………… \(ms^{–1}\)
(iv) the speed when the top of the piston is 2.3cm below its maximum height.
speed = ……………………………………….. \(ms^{–1}\)
(b) The acceleration of the piston varies.
Determine the resultant force on the piston that gives rise to its maximum acceleration.
force = …………………………………………….. N
Answer/Explanation
Answer:
(a) (i) amplitude = 4.9 cm
(ii) frequency = 2700 / 60
= 45 Hz
(iii) \(v_0 = x_0ω\) and ω = 2 πf
\(v_0 = 4.9 × 10^{–2} × 2π × 45\)
\(= 14 ms^{–1}\)
(iv) \(v = ω (x_0^2 – x^2)^{1⁄2}\)
\(= 2 π × 45 × [(4.9 × 10^{–2})^2 – (2.6 × 10^{–2})^2]^{1⁄2}\)
\(= 12 ms^{–1}\)
(b) F = ma
and
\(a_0 = v_0ω\) or \(a_0 = x_0ω^2\)
\(F = 0.64 × 13.9 × 2 π × 45\) or \(0.64 × 4.9 × (2 π × 45)^2\)
= 2500 N
Question
4. (a) (i) By reference to an ultrasound wave, explain what is meant by specific acoustic impedance.
(ii) An ultrasound wave is incident normally on the boundary between two media. The media have specific acoustic impedances \(Z_1\) and \(Z_2\).
State how the ratio
depends on the relative magnitudes of \(Z_1\) and \(Z_2\).
(b) (i) State what is meant by the attenuation of an ultrasound wave.
(ii) A parallel beam of ultrasound is passing through a medium.
The incident intensity \(I_0\) is reduced to 0.35\(I_0\) on passing through a thickness of 0.046m of the medium.
Calculate the linear attenuation coefficient μ of the ultrasound beam in the medium.
μ = ………………………………………….. \(m^{–1}\)
Answer/Explanation
Answer:
(a) (i) product of density and speed
speed of ultrasound in medium
(ii) the greater the difference between \(Z_1\) and \(Z_2\), the closer the ratio is to 1
or
if difference between
\(Z_1\) and \(Z_2\) large, ratio is close to 1
the closer together \(Z_1\) and \(Z_2\), the closer the ratio is to 0
or
if difference between
\(Z_1\) and \(Z_2\) small, ratio close to 0
(b) (i) loss of intensity/amplitude/power (of the wave)
(ii) \(I = I_0 e^{–μx}\)
\(0.35 = e^{–0.046μ}\)
\(μ = 23 m^{–1}\)
Question
5. (a) State one similarity and one difference between the fields of force produced by an isolated
point charge and by an isolated point mass.
similarity:
difference:
(b) An isolated solid metal sphere A of radius R has charge +Q, as illustrated in Fig. 5.1.
A point P is distance 2R from the surface of the sphere.
Determine an expression that includes the terms R and Q for the electric field strength E at
point P.
E = …………………………………………………
(c) A second identical solid metal sphere B is now placed near sphere A. The centres of the
spheres are separated by a distance 6R, as shown in Fig. 5.2.
Point P lies midway between spheres A and B.
Sphere B has charge –Q.
Explain why:
(i) the magnitude of the electric field strength at P is given by the sum of the magnitudes of
the field strengths due to each sphere
(ii) the electric field strength at point P due to the charged metal spheres is not, in practice,
equal to 2E, where E is the electric field strength determined in (b).
Answer/Explanation
Answer:
(a) similarity: both are radial
or
both have inverse square (variations)
difference: direction is always/only towards the mass
or
direction can be towards or away from charge
(b) field strength = \(Q / 4 π ε_0x^2\)
\(E = Q / 36 π ε_0R^2\)
(c) (i) fields (due to each sphere) are in same direction
(ii) charges on spheres attract/affect each other
or
charge distribution on each sphere distorted by the other sphere
or
charges on the surface of the spheres move
spheres are not point charges (at their centres)
Question
(a) The transmission of signals using optic fibres has, to a great extent, replaced the use of
coaxial cables.
Advantages of optic fibres include greater bandwidth and very little crosslinking.
(i) Suggest an advantage of greater bandwidth.
(ii) State what is meant by crosslinking.
(b) In telecommunications, a signal power of 1.0mW is used as a reference power.
Signal powers relative to this reference power and expressed in dB are said to be measured
in ‘dBm’.
Show that a signal power of 13dBm is equivalent to 20mW.
(c) A signal of input power 20mW is transmitted along an optic fibre for an uninterrupted distance
of 45km.
The optic fibre has an attenuation per unit length of 0.18dBk\(m^{–1}\).
Calculate the output power P from the optic fibre.
P = ………………………………………….. mW
Answer/Explanation
Answer:
(a) (i) greater information carrying capacity
(ii) power/energy is radiated
signal picked up by adjacent fibre/wire
(b) ratio / dB = 10 lg( \(P_2\) / \(P_1\))
\(13 = 10 lg [ P / (1.0 × 10^{–3})]\) and so P = 20 mW
(c) 45 × 0.18 = 10 lg (20 / P)
P = 3.1 mW
Question
The output of a microphone is processed using a non-inverting amplifier. The amplifier incorporates
an operational amplifier (op-amp).
(a) State, by reference to the input and output signals, the function of a non-inverting amplifier.
(b) The circuit for the microphone and amplifier is shown in Fig. 7.1.
The output potential difference \(V_{OUT}\) is 2.6V when the potential at point P is 84mV.
Determine:
(i) the gain of the amplifier circuit
gain = ……….
(ii) the resistance of resistor R.
resistance = …………………………………………….. Ω
(c) For the circuit of Fig. 7.1:
(i) suggest a suitable device to connect to the output such that the shape of the waveform
of the sound received by the microphone may be examined
(ii) state and explain the effect on the output potential difference \(V_{OUT}\) of increasing the
resistance of resistor R.
Answer/Explanation
Answer:
(a) output signal proportional to input signal
output signal has same sign/polarity as input signal
(b) (i) gain = \(V_{OUT} / V_{IN}\)
= 2.6 / 0.084
= 31
(ii) \(31 = 1 + (15 × 10^3) / R\)
R = 500
Ω
(c) (i) e.g. cathode-ray oscilloscope/CRO
(ii) gain is reduced
(so) \(V_{OUT}\) is smaller
Question
(a) Define the tesla.
(b) A magnet produces a uniform magnetic field of flux density B in the space between its poles.
A rigid copper wire carrying a current is balanced on a pivot. Part PQLM of the wire is between
the poles of the magnet, as illustrated in Fig. 8.1.
The wire is balanced horizontally by means of a small weight W.
The section of the wire between the poles of the magnet is shown in Fig. 8.2.
Explain why:
(i) section QL of the wire gives rise to a moment about the pivot
(ii) sections PQ and LM of the wire do not affect the equilibrium of the wire.
(c) Section QL of the wire has length 0.85cm.
The perpendicular distance of QL from the pivot is 5.6cm.
When the current in the wire is changed by 1.2 A, W is moved a distance of 2.6cm along the
wire in order to restore equilibrium. The mass of W is \(1.3 × 10^{–4}\) kg.
(i) Show that the change in moment of W about the pivot is \(3.3 × 10^{–5}\)Nm.
(ii) Use the information in (i) to determine the magnetic flux density B between the poles of the magnet.
B = ……………………………………………… T
Answer/Explanation
Answer:
(a) magnetic field normal to current
newton per ampere
newton per metre
(b) (i) current in wire QL gives rise to a force
or
wire QL is perpendicular to the magnetic field
force on wire QL is vertical
force does not act through the pivot
(ii) forces act through the same line
or
forces are horizontal
forces are equal (in magnitude) and opposite (in direction)
(c) (i) change = mg × (Δ)L
\(= 1.3 × 10^{–4} × 9.81 × 2.6 × 10^{–2} = 3.3 × 10^{–5} N m^{–1}\)
(ii) change =
B × (Δ)I × L × x
\(3.3 × 10^{–5} = B × 1.2 × 0.85 × 10^{–2} × 5.6 × 10^{–2}\)
B = 0.058 T
Question
(a) A coil of wire is situated in a uniform magnetic field of flux density B.
The coil has diameter 3.6cm and consists of 350 turns of wire, as illustrated in Fig. 9.1.
The variation with time t of B is shown in Fig. 9.2.
(i) Show that, for the time t = 0 to time t = 0.20s, the electromotive force (e.m.f.) induced in the coil is 0.080V.
(ii) On the axes of Fig. 9.3, show the variation with time t of the induced e.m.f. E for time
t = 0 to time t = 0.80s.
(b) A bar magnet is held a small distance above the surface of an aluminium disc by means of a
rod, as illustrated in Fig. 9.4.
The aluminium disc is supported horizontally and held stationary.
The magnet is rotated about a vertical axis at constant speed.
Use laws of electromagnetic induction to explain why there is a torque acting on the aluminium
disc.
Answer/Explanation
Answer:
(a) (i) e.m.f. = (Δ)B × AN / t
\(= 45 × 10^{–3 }× π × (1.8 × 10^{–2})^2 × 350 / 0.20 = 0.080 V\)
(ii) 0 to 0.2 s: straight horizontal line at 0.080 V or –0.080 V
0.2 s to 0.4 s: zero
0.4 s to 0.8 s: straight horizontal line at 0.040 V or –0.040 V
opposite polarity to 0 to 0.2 s line
(b) either disc cuts flux lines (of the magnet)
or
there is a changing flux in the disc
(by Faraday’s law) e.m.f. is induced in the disc
e.m.f. causes (eddy) currents in the disc
current in the magnetic field (of the magnet) causes force on disc
Question
(a) White light passes through a cloud of cool low-pressure gas, as illustrated in Fig. 10.1.
For light that has passed through the gas, its continuous spectrum is seen to contain a
number of darker lines.
Use the concept of discrete electron energy levels to explain the existence of these darker
lines.
(b) The uppermost electron energy bands in a solid are illustrated in Fig. 10.2.
Use band theory to explain the dependence on light intensity of the resistance of a
light-dependent resistor (LDR).
Answer/Explanation
Answer:
(a) • photon gives energy to electron (in an inner shell) or electron (in an inner shell) absorbs a photon
• electron moves (from lower) to higher energy level
• energy (of photon) is equal to difference in energy levels
• electron de-excites giving off photon (of same energy)
• photons emitted in all directions
(b) (in light) photons gives energy to electrons in VB
or
(in light) electrons in VB absorb photons
electron crosses FB/jumps to CB
(positive) holes left/created in VB
low intensity: few electrons in CB/most electrons in VB
or
high intensity: more photons so more electrons in CB
or
electron-hole pairs are charge carriers
more charge carriers results in lower resistance
Question
11. An electron, at rest, has mass \(m_e\) and charge –q.
A positron is a particle that, at rest, has mass \(m_e\) and charge +q.
A positron interacts with an electron. The electron and the positron may be considered to be at
rest.
The outcome of this interaction is that the electron and the positron become two gamma-ray (γ-ray)
photons, each having the same energy.
(a) Calculate, for one of the γ-ray photons:
(i) the photon energy, in J
energy = ……………………………………………… J
(ii) its momentum.
momentum = …………………………………………… Ns
(b) State and explain the direction, relative to each other, in which the γ-ray photons are emitted.
Answer/Explanation
Answer:
(a) (i) \(E = mc^2\)
\(= 9.11 × 10^{–31} × (3.0 × 10^8)^2\)
\(= 8.2 × 10^{–14}\) J
(ii) p = h / λ and E = hc / λ
or
E = pc
\(p = (8.2 × 10^{–14}) / (3.0 × 10^8)\)
\(= 2.7 × 10^{–22}\) Ns
(b) total momentum (before and after interaction) is zero
or
momentum must be conserved (in the interaction)
or
momentum of the photons must be equal and opposite
(photons emitted in) opposite directions
Question
(a) The decay of a sample of a radioactive isotope is said to be random and spontaneous.
Explain what is meant by the decay being:
(i) random
(ii) spontaneous.
(b) A radioactive isotope X has a half-life of 1.4 hours.
Initially, a pure sample of this isotope X has an activity of \(3.6 × 10^5\)Bq.
Determine the activity of the isotope X in the sample after a time of 2.0 hours.
activity = ……………………………………………. Bq
(c) The variation with time t of the actual activity A of the sample in (b) is shown in Fig. 12.1.
(i) The initial activity of isotope X in the sample is \(3.6 × 10^5\)Bq.
Use information from (b) to sketch, on the axes of Fig. 12.1, the variation with time t of
the activity of a pure sample of isotope X. [1]
(ii) Suggest an explanation for any difference between the actual activity of the sample
shown in Fig. 12.1 and the curve you have drawn for the activity of isotope X.
Answer/Explanation
Answer:
(a) (i) time at which a nucleus will decay cannot be predicted
or
constant probability of decay of a nucleus
(ii) decay (of a nucleus) not affected by environmental factors
(b) \(A = A_0e^{–λt}\) and \(λ = ln 2 / t^_{1⁄2}\)
\(= 3.6 × 10^5 × exp [–(2 × ln 2) / 1.4]\)
or
\(A = A_0 × 0.5^N\)
\(= 3.6 × 10^5 × 0.5\) N where N = 2 / 1.4
\(A = 1.3 × 10^5\) Bq
(c) (i) smooth curve, starting at \((0, 3.6 × 10^5)\) and passing through \((1.4, 1.8 × 10^5)\) and \((2.0, 1.3 × 10^5)\)
(ii) (activity of sample is greater than activity of X so) there must be an additional source of activity
the decay product (of isotope X) is radioactive