Question 1
1 (a) State Newton’s law of gravitation.
(b) A satellite is in a circular orbit around a planet. The radius of the orbit is R and the period of
the orbit is T. The planet is a uniform sphere.
Use Newton’s law of gravitation to show that R and T are related by
4π $^{2}$R $^{3}$ = GMT $^{2}$
where M is the mass of the planet and G is the gravitational constant.
(c) The Earth may be considered to be a uniform sphere of mass 5.98 × 10 $^{24}$ kg and radius 6.37 × 10 $^{6}$m.
A geostationary satellite is in orbit around the Earth.
Use the expression in (b) to determine the height of the satellite above the Earth’s surface.
height = …………………………………………….. m
(d) Another satellite is in a circular orbit around the Earth with the same orbital radius and period
as the satellite in (c).
(i) Calculate the angular speed of the satellite in this orbit. Give a unit with your answer.
angular speed = ………………………………………. unit …………….
(ii) Despite having the same orbital period, the orbit of this satellite is not geostationary.
Suggest two ways in which the orbit of this satellite could be different from the orbit of
the satellite in (c).
1 ……………………………………………………………………………………………………………………….
2 ……………………………………………………………………………………………………………………….
▶️Answer/Explanation
Ans :
1(a) (gravitational) force is (directly) proportional to product of masses
force (between point masses) is inversely proportional to the square of their separation
1(b) GMm / R$^{2}$ = mR $^{2}ω^{2}$
ω = 2 π/ T and algebra leading to $4 π^{2}R^{3}$ = GMT$^{2}$
or
GMm / R$^{2}$ = mv$^{2}$ / R
v = 2πR / T and algebra leading to 4π$^{2}$R$^{3}$ = GMT$^{2}$
1(c) 4π2 ×R$^{3}$ = 6.67 × 10$^{-11}$ ×5.98 × 10$^{24}$ × (24 × 60×60)$^{2}$
(R = 4.22 × 10$^{7}$ m)
h = R – (6.37×10$^{6}$ )
h = (4.22 ×10$^{7}$ ) – (6.37× 10$^{6}$ )
= 3.6× 10$^{7}$ m
1(d)(i) ω = 2 π / T
= 2 π/ (24×60×60)
= 7.3×10$^{-5}$ rad s$^{-1}$
1(d)(ii) orbit is from east to west
orbit is not equatorial / orbit is polar
Question 2:
2 (a) (i) State what is meant by an ideal gas.
(ii) State the temperature, in degrees Celsius, of absolute zero.
temperature = ……………………………………………. °C
(b) A sealed vessel contains a mass of 0.0424kg of an ideal gas at 227°C.
The pressure of the gas is 1.37 × 10$^{5}$Pa and the volume of the gas is 0.640m $^{3}$.
Calculate:
(i) the number of molecules of the gas in the vessel
number of molecules = …………………………………………………
(ii) the mass of one molecule of the gas
mass = ……………………………………………. kg
(iii) the root-mean-square (r.m.s.) speed v of the molecules of the gas.
v = ………………………………………… ms $^{-1}$
(c) The gas in (b) is now cooled gradually to absolute zero.
On Fig. 2.1, sketch the variation with thermodynamic temperature T of the r.m.s. speed of the molecules of the gas.
▶️Answer/Explanation
Ans :
2(a)(i) (gas that obeys) pV ∞ T (for all values of p,V and T)
where T is thermodynamic temperature
2(a)(ii) temperature =–273.15 °C
2(b)(i) pV = NkT
N = (1.37 × 10$^{5}$ × 0.640) / (1.38 × 10$^{-23}$ × (227 + 273))
= 1.27 × 10$^{25}$
2(b)(ii) mass = 0.0424 / (1.27 ×10$^{25}$)
= 3.34 × 10$^{-27}$kg
2(b)(iii) $^{\frac{1}{2}} m<c^{2}>$= (3 / 2)kT
3.34 ×10$^{-27}$ ×v$^{2}$ = 3 × 1.38× 10$^{-23}$× 500
v = 2490 m s$^{-1}$
or
pV = $^{\frac{1}{2}} Nm<c^{2}>$ and Nm = mass of gas
0.0424 ×v$^{2}$ = 3 × 1.37× 10$^{5}$ × 0.640
v = 2490 m s$^{-1}$
2(c) sketch: line from (0, 0) to (500, v)
line with decreasing positive gradient throughout.
Question 3:
3 (a) State the first law of thermodynamics. Identify the meaning of any symbols that you use.
(b) The state of an ideal gas is continuously changed according to the cycle ABCDA shown in Fig. 3.1.
(i) Complete Table 3.1 for the changes A to B and B to C by placing two ticks (3) in each row.
(ii) Use the first law of thermodynamics to describe and explain the energy transfers
associated with one complete cycle ABCDA.
▶️Answer/Explanation
Ans:
3(a) change in internal energy = work done + energy transfer by heating
increase in internal energy = work done on system + energy transferred to the system by heating
3(b)(i)
AB change in internal energy: decrease
AB work done on gas: positive
BC change in internal energy: increase
BC work done on gas: zero
3(b)(ii) more work done by gas in CD than is done on gas in AB
or
(no work done on gas in BC and DA so) (overall) gas does work
(overall) change in internal energy is zero
(must be an overall) input of thermal energy
Question 4:
A small steel sphere is oscillating vertically on the end of a spring, as shown in Fig. 4.1.
The velocity v of the sphere varies with displacement x from its equilibrium position according to
v = ± 9.7 $\sqrt{(11.6-x^{2})}$
where v is in cms $^{-1}$ and x is in cm.
(a) (i) Calculate the frequency of the oscillations.
frequency = ……………………………………………. Hz
(ii) Show that the amplitude of the oscillations is 3.4cm.
(iii) Calculate the maximum acceleration a0 of the sphere.
a$_{0}$= ………………………………………… ms$^{-2}$
(b) On Fig. 4.2, sketch the variation with x of the acceleration a of the sphere.
(c) Describe, without calculation, the interchange between the potential energy and the kinetic
energy of the oscillations.
▶️Answer/Explanation
Ans:
4(a)(i) = 2πf
f = 9.7 / 2π
= 1.5 Hz
4(a)(ii) amplitude = √(11.6) = 3.4 cm
4(a)(iii) a0 = ω2×0 = 9.72 × 3.4× 10$^{-2}$
= 3.2m s$^{-2}$
4(b) sketch: straight line through the origin with negative gradient
line with negative gradient passing through (+3.4, – a$_{0}$) and ( –3.4, + a$_{0}$)
line with ends at x = $\underline{+}$ 3.4 cm and a = $\underline{+}$a$_{0}$
4(c) sum of potential energy and kinetic energy is constant
at maximum displacement, kinetic energy is zero
or
at maximum displacement, potential energy is maximum
at zero displacement, kinetic energy is maximum
or
at zero displacement, potential energy is minimum
Question 5:
Two capacitors A and B are connected into the circuit shown in Fig. 5.1.
Capacitor A has capacitance C and capacitor B has capacitance 3C.
The electromotive force (e.m.f.) of the cell is V.
The two-way switch S is initially at position X, and capacitor B is initially uncharged.
(a) State, in terms of V and C, expressions for:
(i) the initial charge Q$_{A}$ on the plates of capacitor A
QA = …………………………………………………
(ii) the initial energy E$_{A}$ stored in capacitor A.
E$_{A}$ = ………………………………………………….
(b) The two-way switch S is now moved to position Y.
(i) State and explain what happens to the charge that was initially on the plates of capacitor A.
(ii) Show that the final potential difference (p.d.) VB across capacitor B is given by
$V_{B}=\frac{V}{4}$
Explain your reasoning.
(iii) Determine an expression, in terms of V and C, for the decrease ΔE in the total energy
that is stored in the capacitors as a result of the change of the position of the switch.
ΔE = …………………………………………………
▶️Answer/Explanation
Ans :
5(a)(i) $Q _{A}$ = CV
5(a)(ii) $E_{A}$ =$ 1⁄2CV^{2}$
5(b)(i) some of the charge transfers to (the plates of) capacitor
transfer is because the p.d.s across the capacitors are not equal
or
transfer stops when the p.d.s across the capacitors become equal
5(b)(ii) $V_{A} = V_{B}$
charge on A + charge on B = CV
CV_{B} + 3CV_{B} = CV leading to V_{B} = V / 4$
or
$C_{T} = 4C$
$Q_{T} = CV$
$V_{B} = CV / 4C = V / 4$
5(b)(iii) $\Delta E = 1⁄2CV^{2} – nCV^{2}$, where n is a multiple that is less than 1⁄2
or
$total final energy = 1⁄2 × 4 C × (V / 4)^{2} = 1⁄8CV^{2}$
$\Delta E = 1⁄2CV^{2} – 1⁄8CV^{2} = 3⁄8CV^{2}$
Question 6:
A heavy aluminium disc has a radius of 0.36m. The disc rotates with the wheels of a vehicle and
forms part of an electromagnetic braking system on the vehicle.
In order to activate the braking system, a uniform magnetic field of flux density 0.17T is switched
on. This magnetic field is perpendicular to the plane of rotation of the disc, as shown in Fig. 6.1.
(a) (i) Define magnetic flux..
(ii) Calculate the magnetic flux through the disc. Give a unit with your answer.
magnetic flux = ……………………………………….. unit ……………….
(b) The disc is rotating at a rate of 25 revolutions per second.
Calculate the magnitude of the electromotive force (e.m.f.) induced between the axle and the rim of the disc.
e.m.f. = ……………………………………………… V
(c) The axle and the rim are connected into an external circuit that enables the energy of the
rotation of the disc to be stored for future use. The direction of rotation is shown in Fig. 6.1.
Use Lenz’s law of electromagnetic induction to determine whether the current in the disc is
from the rim to the axle or from the axle to the rim. Explain your reasoning.
▶️Answer/Explanation
Ans
6(a)(i) product of (magnetic) flux density and area
area perpendicular to the (magnetic) field
6(a)(ii) flux =B×πr$^{2}$
= 0.17×π× 0.36$^{2}$
= 6.9× 10$^{-2}$ Wb
6(b) time for one revolution = 1 / 25 s
e.m.f. = rate of cutting flux or Δ$\theta$/Δt
= 0.069 × 25
= 1.7 V
6(c) current (in disc) is perpendicular to magnetic field
or
current causes force to act on disc
force opposes rotation of disc
left-hand rule indicates current is from rim to axle
Question 7:
Four diodes are used in a bridge rectifier circuit to produce rectification of a sinusoidal a.c. input
voltage V$_{IN}$.
Fig. 7.1 shows part of the circuit, but three of the diodes are missing.
The p.d. across the load resistor R is the output p.d. V$_{OUT}$ of the bridge rectifier.
(a) (i) State the name of the type of rectification produced by a bridge rectifier.
(ii) Complete Fig. 7.1 by drawing the three missing diodes, correctly connected.
(iii) On Fig. 7.1, draw an arrow to indicate the direction of the current in resistor R.
(b) V$_{IN}$ has amplitude V$_{0}$ and period T. Fig. 7.2 shows the variation with time t of V$_{IN}$.
(i) On Fig. 7.3, sketch the variation of V$_{OUT}$ with t between t = 0 and t = 2.0T.
(ii) The power dissipated in the resistor is P.
On Fig. 7.4, sketch the variation of P with t between t = 0 and t = 2.0T.
(iii) Suggest, with a reason, how the root-mean-square (r.m.s.) value of V$_{OUT}$ compares with
the r.m.s. value of V$_{IN}$.
▶️Answer/Explanation
Ans:
7(a)(i) full-wave (rectification)
7(a)(ii) lower left diode shown pointing left
lower right and upper left diodes shown pointing left
7(a)(iii) arrow indicating current direction in resistor to the right
7(b)(i) sketch: periodic line showing minimum V$_{OUT}$ = 0 and maximum V$_{OUT}$ = + V$_{0}$
line showing peak V$_{OUT}$at t = 0, 0.5 T, 1.0 T, 1.5T and 2.0 T, with
V$_{OUT}$ going to zero half-way in between each peak
line showing correct modulated sine shape
7(b)(ii) sketch: sinusoidal curve with troughs sitting on the time axis
peak power at t = 0, 0.5 T, 1.0 T, 1.5 T and 2.0
T and zero power half-way in between each peak
7(b)(iii) same power-time graph with or without rectification, so same Vrms
or
V$^{2}$-time graph is same for both V$_{OUT}$ and V$_{IN}$, so same Vrms
or
power does not depend on sign of V, so same V$_{rms}$
Question 8:
8 Fig. 8.1 shows the lowest four energy levels of an electron in an isolated atom.
Fig. 8.2 shows the lines in the emission spectrum of the atom that correspond to the transitions of
the electron from n = 3 to n = 1 and from n = 4 to n = 1.
(a) Explain, with reference to photons, why there is a single frequency of electromagnetic
radiation that corresponds to each of these transitions.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………
(b) (i) On Fig. 8.2, draw a line that corresponds to the transition of the electron from n = 2 to n = 1.
Label this line A.
(ii) On Fig. 8.2, draw a line that corresponds to the transition of the electron from n = 3 to n = 2.
Label this line B.
(c) The frequency of radiation represented by line A is f$_{A}$.
The frequency of radiation represented by line B is f$_{B}$.
The energy of the ground state (n = 1) is E$_{1}$.
Determine an expression, in terms of f$_{A}$, f$_{B}$, E$_{1}$ and the Planck constant h, for the energy E$_{3}$ of
the energy level n = 3.
E$_{3}$ = …………………………………………………
▶️Answer/Explanation
Ans:
8(a) transition (emits) (one) photon with energy equal to the difference in energy between the two levels
frequency of radiation corresponds to energy of photon
8(b)(i) line to the left of the pair in Fig. 8.2, labelled A
larger gap between line A and the nearest of the pair in Fig. 8.2 than between the lines in the pair
8(b)(ii) line to the left of both the pair in Fig. 8.2 and line A, labelled B
larger gap between line B and line A than between line A and the nearest one of the pair in Fig. 8.2
8(c) E = hf
E$_{3}$ = E$_{1}$ + h(f$_{A}$ + f$_{B}$)
Question 9:
9 (a) Define mass defect.
(b) Table 9.1 shows the mass defects of three nuclei.
The nuclear fusion process in a particular star is described by
$\frac{2}{1}H+\frac{3}{1}H\to \frac{4}{2}He+X$
where X is a particle that has no mass defect.
(i) State the name of particle X.
(ii) Show that the energy released when one nucleus of $\frac{4}{2}$ He is formed in this fusion reaction
is 2.8 × 10$^{-12}$ J.
(c) The star in (b) has a radius of 2.3 × 10$^{9}$m and a luminosity of 1.4 × 10$^{9}$W.
All the energy released from the formation of 4$\frac{4}{2}$He is radiated away from the star.
All the energy that is radiated from the star has been released in the formation of 4 2He.
Determine:
(i) the mass of $\frac{4}{2}$ He produced per unit time by the fusion process
mass per unit time = ……………………………………….. kgs$^{-1}$
(ii) the surface temperature of the star.
temperature = ……………………………………………… K
▶️Answer/Explanation
Ans:
9(a) difference between mass of nucleus and (total) mass of nucleons
when infinitely separated
9(b)(i) neutron
9(b)(ii) E =Δm c$^{2}$
Δm = (0.030377 – 0.002388 – 0.009105)u
( = 0.018884u)
energy release = (0.030377 – 0.002388 – 0.009105) × 1.66 ×10$^{-27}$ × (3.00 ×10$^{8}$)$^{2}$ = 2.8 × 10$^{-12}$ J
9(c)(i) number of atoms per unit time = (1.4 × 10$^{28}$) / (2.8 × 10$^{12}$)
( = 5.0 ×10$^{39}$ s$^{-1}$)
mass of one atom = 4 ×1.66 × 10$^{-27}$or (4 × 10$^{-3}$) / (6.02 × 10$^{23}$)
( = 6.64 × 10$^{-27}$kg)
mass per unit time = 6.64 × 10$^{-27}$ × 5.0 × 10$^{39}$
= 3.3 × 10$^{13}$kg s$^{-1}$
9(c)(ii) L = 4πσr$^{2}$T$^{4}$
1.4 × 10$^{28}$ = 4π× 5.67 × 10$^{-8}$ × (2.3 × 10$^{9}$)2 × T$^{4}$
T = 7800 K
Question 10:
(a) X-rays for use in medical diagnosis are produced in an X-ray tube. In the X-ray tube, charged
particles are accelerated towards a metal target by an applied potential difference (p.d.).
(i) State the name of the charged particles that are accelerated by the applied p.d.
(ii) Explain how X-rays are produced at the metal target.
(iii) Calculate the minimum wavelength of X-rays produced when the applied p.d. is 5.80kV.
wavelength = …………………………………………….. m
(b) X-rays pass through a medium that has an attenuation coefficient of 1.4cm$^{-1}$
Calculate the percentage of the X-ray energy that is absorbed by a 2.8cm thickness of this medium.
percentage absorbed = …………………………………………….. %
▶️Answer/Explanation
Ans:
10(a)(i) electrons
10(a)(ii) electrons are decelerated / stopped on impact with the target
(kinetic) energy lost by electrons emitted as (X-ray) photons
10(a)(iii) eV = hc/λ
λ = (6.63 ×10$^{-34}$ ×3.00 × 10$^{8}$) / (1.60 × 10$^{-19}$ × 5800)
= 2.14 × 10$^{-10}$m
10(b) I = I$_{0}$ exp (–$\mu$x)
I$_{T}$ / I$_{0}$ = exp (–(1.4 × 2.8))
= 0.020
% absorbed = (1.000 – 0.0198) × 100
= 98%