Question 1
(a) (i) Define power. [1 mark]
_____________________
(ii) Use the definition of power to show that the SI base units of power are \( \mathrm{kg\,m^2\,s^{-3}} \). [2 marks]
____________________________
(b) The intensity \(I\) of a sound wave moving through a gas is given by
\(I=f^2A^2vk\)
where \(f\) is the frequency of the wave,
\(A\) is the amplitude of the wave,
\(v\) is the speed of the wave
and \(k\) is a constant that depends on the gas.
Determine the SI base units of \(k\). [3 marks]
SI base units = ______________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 1.2: SI units — parts (a)(ii) and (b)
▶️ Answer/Explanation
(a)(i) Definition of power [1 mark]
Power is the work done per unit time.
\(P=\frac{W}{t}\)
Answer: \(\boxed{\text{work done per unit time}}\)
(a)(ii) SI base units of power [2 marks]
Using
\(P=\frac{W}{t}\)
The SI base units of work are
\(\mathrm{kg\,m^2\,s^{-2}}\)
Therefore,
\(\mathrm{units\ of\ }P=\frac{\mathrm{kg\,m^2\,s^{-2}}}{\mathrm{s}}\)
\(\mathrm{units\ of\ }P=\mathrm{kg\,m^2\,s^{-3}}\)
Answer: \(\boxed{\mathrm{kg\,m^2\,s^{-3}}}\)
(b) SI base units of \(k\) [3 marks]
Intensity is power per unit area:
\(I=\frac{P}{A}\)
Therefore, the SI base units of intensity are
\(\frac{\mathrm{kg\,m^2\,s^{-3}}}{\mathrm{m^2}}=\mathrm{kg\,s^{-3}}\)
From the equation
\(I=f^2A^2vk\)
the SI base units are
\(f=\mathrm{s^{-1}}\)
\(A=\mathrm{m}\)
\(v=\mathrm{m\,s^{-1}}\)
Hence,
\(\mathrm{units\ of\ }k=\frac{\mathrm{kg\,s^{-3}}}{(\mathrm{s^{-1}})^2(\mathrm{m})^2(\mathrm{m\,s^{-1}})}\)
\(\mathrm{units\ of\ }k=\frac{\mathrm{kg\,s^{-3}}}{\mathrm{m^3\,s^{-3}}}\)
\(\mathrm{units\ of\ }k=\mathrm{kg\,m^{-3}}\)
Answer: \(\boxed{\mathrm{kg\,m^{-3}}}\)
Question 2
A rigid uniform beam of weight \(W\) is connected to a fixed support by a hinge, as shown in Fig. 2.1.
A compressed spring exerts a total force of \(8.2\,\mathrm{N}\) vertically upwards on the horizontal beam. A block of weight \(0.30\,\mathrm{N}\) rests on the beam. The right-hand end of the beam is connected to the ground by a string at an angle of \(30^\circ\) to the horizontal. The tension in the string is \(4.8\,\mathrm{N}\). The distances along the beam are shown in Fig. 2.1.
The beam is in equilibrium. Assume that the hinge is frictionless.
(a) (i) Show that the vertical component of the tension in the string is \(2.4\,\mathrm{N}\). [1 mark]
____________________________________________________________________________
(ii) By taking moments about the hinge, determine the weight \(W\) of the beam. [3 marks]
\(W=\) ____________________ \(\mathrm{N}\)
(iii) Calculate the horizontal component of the force exerted on the beam by the hinge. [1 mark]
force = ____________________ \(\mathrm{N}\)
(b) The spring obeys Hooke’s law and has an elastic potential energy of \(0.32\,\mathrm{J}\).
Calculate the compression of the spring. [2 marks]
compression = ____________________ \(\mathrm{m}\)
(c) The string is cut so that the spring extends upwards. This causes the beam to rotate and launch the block into the air. The block reaches its maximum height and then falls back to the ground.
The block is at a height of \(0.090\,\mathrm{m}\) above the ground when it passes through point A. The block has a kinetic energy of \(0.044\,\mathrm{J}\) when it hits the ground at point B. Air resistance is negligible.
(i) Calculate the decrease in the gravitational potential energy of the block for its movement from A to B. [2 marks]
decrease in gravitational potential energy = ____________________ \(\mathrm{J}\)
(ii) Use your answer in (c)(i) and conservation of energy to determine the speed of the block at point A. [3 marks]
speed = ____________________ \(\mathrm{m\,s^{-1}}\)
(iii) By reference to the force on the block, explain why the horizontal component of the velocity of the block remains constant as it moves from A to B. [1 mark]
____________________________________________________________________________
____________________________________________________________________________
(iv) The block passes through point A at time \(t_A\) and arrives at point B at time \(t_B\).
On Fig. 2.3, sketch a graph to show the variation of the magnitude of the vertical component \(v_Y\) of the velocity of the block with time \(t\) from \(t=t_A\) to \(t=t_B\). Numerical values of \(v_Y\) are not required.
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 4.2: Equilibrium of forces — parts (a)(i) and (a)(iii)
• 5.1: Energy conservation — parts (b) and (c)(ii)
• 5.2: Gravitational potential energy and kinetic energy — parts (c)(i) and (c)(ii)
• 2.1: Equations of motion — part (c)(iv)
▶️ Answer/Explanation
(a)(i) Vertical component of tension [1 mark]
The vertical component of the tension is
\(T_Y=4.8\sin30^\circ\)
\(T_Y=2.4\,\mathrm{N}\)
Answer: \(\boxed{2.4\,\mathrm{N}}\)
(a)(ii) Weight of the beam [3 marks]
Taking moments about the hinge:
Clockwise moments = anticlockwise moments.
\(8.2\times0.50=(W\times0.60)+(0.30\times0.80)+(2.4\times1.20)\)
\(4.10=0.60W+0.24+2.88\)
\(0.60W=0.98\)
\(W=1.6\,\mathrm{N}\)
Answer: \(\boxed{1.6\,\mathrm{N}}\)
(a)(iii) Horizontal component of hinge force [1 mark]
There is no horizontal component from the spring or the weights. Therefore, the horizontal hinge force balances the horizontal component of the string tension.
\(F_H=4.8\cos30^\circ\)
\(F_H=4.2\,\mathrm{N}\)
Answer: \(\boxed{4.2\,\mathrm{N}}\)
(b) Compression of the spring [2 marks]
Elastic potential energy is
\(E_{\mathrm{P}}=\frac{1}{2}Fx\)
Therefore,
\(0.32=\frac{1}{2}(8.2)x\)
\(x=\frac{0.64}{8.2}\)
\(x=0.078\,\mathrm{m}\)
Answer: \(\boxed{0.078\,\mathrm{m}}\)
(c)(i) Decrease in gravitational potential energy [2 marks]
The weight of the block is \(0.30\,\mathrm{N}\), and its decrease in height is \(0.090\,\mathrm{m}\).
\(\Delta E_{\mathrm{P}}=W\Delta h\)
\(\Delta E_{\mathrm{P}}=(0.30)(0.090)\)
\(\Delta E_{\mathrm{P}}=0.027\,\mathrm{J}\)
Answer: \(\boxed{0.027\,\mathrm{J}}\)
(c)(ii) Speed of the block at A [3 marks]
The decrease in gravitational potential energy becomes additional kinetic energy.
Therefore,
\(E_{\mathrm{K,A}}=0.044-0.027\)
\(E_{\mathrm{K,A}}=0.017\,\mathrm{J}\)
The mass of the block is
\(m=\frac{W}{g}=\frac{0.30}{9.81}\)
Using \(E_{\mathrm{K}}=\frac{1}{2}mv^2\):
\(0.017=\frac{1}{2}\left(\frac{0.30}{9.81}\right)v^2\)
\(v=1.1\,\mathrm{m\,s^{-1}}\)
Answer: \(\boxed{1.1\,\mathrm{m\,s^{-1}}}\)
(c)(iii) Horizontal component of velocity [1 mark]
The only force acting on the block after launch is its weight, which acts vertically downwards.
Therefore, there is no horizontal resultant force and hence no horizontal acceleration.
Answer: \(\boxed{\text{The horizontal velocity remains constant because there is no horizontal force.}}\)
(c)(iv) Graph of \(v_Y\) against time [1 mark]
The vertical acceleration is constant and equal to \(g\). Therefore, the magnitude of the vertical velocity increases at a constant rate.
The graph should be a straight line with a positive gradient, beginning at a non-zero value of \(v_Y\) at \(t_A\) and ending at a larger value at \(t_B\).
Answer: \(\boxed{\text{upward-sloping straight line from }t_A\text{ to }t_B}\)
Question 3
A block is pulled in a straight line along a rough horizontal surface by a varying force \(X\), as shown in Fig. 3.1.
Air resistance is negligible. Assume that the frictional force exerted on the block by the surface is constant and has magnitude \(2.0\,\mathrm{N}\).
The variation with time \(t\) of the momentum \(p\) of the block is shown in Fig. 3.2.
(a) State Newton’s second law of motion. (1 mark)
____________________________________________________________________________
(b) Use Fig. 3.2 to determine, for the block at time \(t=2.0\,\mathrm{s}\), the magnitude of:
(i) the resultant force on the block (1 mark)
resultant force = __________________________________ \(\mathrm{N}\)
(ii) the force \(X\). (1 mark)
\(X=\) __________________________________ \(\mathrm{N}\)
(c) On Fig. 3.3, sketch a graph to show the variation of force \(X\) with time \(t\) from \(t=0\) to \(t=6.0\,\mathrm{s}\). (3 marks)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
▶️ Answer/Explanation
(a) Newton’s second law of motion [1 mark]
The resultant force on an object is proportional to, or equal to, the rate of change of momentum.
\(F=\dfrac{\Delta p}{\Delta t}\)
Answer: The resultant force is equal to the rate of change of momentum.
(b)(i) Resultant force [1 mark]
The gradient of the momentum-time graph gives the resultant force.
\(F=\dfrac{\Delta p}{\Delta t}\)
\(F=\dfrac{6.0}{4.0}\)
\(F=1.5\,\mathrm{N}\)
Answer: \(\boxed{1.5\,\mathrm{N}}\)
(b)(ii) Force \(X\) [1 mark]
The applied force \(X\) acts forwards while the frictional force of \(2.0\,\mathrm{N}\) acts backwards.
\(X-2.0=1.5\)
\(X=3.5\,\mathrm{N}\)
Answer: \(\boxed{3.5\,\mathrm{N}}\)
(c) Force-time graph [3 marks]
From \(t=0\) to \(t=4.0\,\mathrm{s}\), the momentum-time graph is a straight line with constant positive gradient. Therefore, the resultant force is constant.
Hence,
\(X=1.5+2.0=3.5\,\mathrm{N}\)
From \(t=4.0\,\mathrm{s}\) to \(t=6.0\,\mathrm{s}\), the momentum is constant. Therefore, the resultant force is zero and \(X\) balances the frictional force.
\(X=2.0\,\mathrm{N}\)
Required graph:
• From \(t=0\) to \(t=4.0\,\mathrm{s}\): horizontal line at \(X=3.5\,\mathrm{N}\).
• From \(t=4.0\,\mathrm{s}\) to \(t=6.0\,\mathrm{s}\): horizontal line at \(X=2.0\,\mathrm{N}\).
Answer: A horizontal line at \(3.5\,\mathrm{N}\) from \(0\) to \(4.0\,\mathrm{s}\), followed by a horizontal line at \(2.0\,\mathrm{N}\) from \(4.0\) to \(6.0\,\mathrm{s}\).
Question 4
A beaker in air contains a liquid. The base of the beaker is in contact with the liquid and has area \(A\), as shown in Fig. 4.1.
The liquid has density \(\rho\) and fills the beaker to a depth \(h\).
(a) By using the definitions of pressure and density, show that \(p=\rho gh\).
where \(p\) is the pressure due to the liquid that is exerted on the base of the beaker and \(g\) is the acceleration of free fall. [3 marks]
(b) Suggest why the equation in (a) does not give the total pressure on the base of the beaker. [1 mark]
____________________________________________________________________________
(c) Fig. 4.2 shows the variation of the total pressure inside the liquid with depth \(x\) below the surface.
Determine the density of the liquid. [2 marks]
density = __________________________ \(\mathrm{kg\,m^{-3}}\)
(d) A solid cylinder is held stationary by a wire so that the base of the cylinder is level with the surface of the liquid, as shown in Fig. 4.3.
The cylinder has length \(4.0\times10^{-2}\,\mathrm{m}\) and cross-sectional area \(3.7\times10^{-4}\,\mathrm{m^2}\). The tension in the wire is \(0.53\,\mathrm{N}\).
The cylinder is now lowered and then held stationary by the wire so that the top of the cylinder is level with the surface of the liquid.
Calculate the new tension in the wire. [2 marks]
tension = __________________________ \(\mathrm{N}\)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
▶️ Answer/Explanation
(a) Derivation of \(p=\rho gh\) [3 marks]
Using the definition of density,
\(\rho=\dfrac{m}{V}\)
The volume of liquid above the base is
\(V=Ah\)
Therefore,
\(m=\rho Ah\)
The weight of the liquid is
\(W=mg=\rho Ahg\)
Using the definition of pressure,
\(p=\dfrac{F}{A}=\dfrac{W}{A}\)
\(p=\dfrac{\rho Ahg}{A}\)
Therefore,
Answer: \(\boxed{p=\rho gh}\)
(b) Total pressure [1 mark]
The equation \(p=\rho gh\) gives only the pressure due to the liquid. There is also atmospheric pressure acting on the surface of the liquid.
Answer: Atmospheric pressure also acts on the liquid and contributes to the total pressure.
(c) Density of the liquid [2 marks]
The gradient of the pressure-depth graph is given by
\(\dfrac{\Delta p}{\Delta h}=\rho g\)
From the graph, when the depth increases from \(0\) to \(8.0\,\mathrm{cm}\), the pressure increases from \(9.60\times10^4\,\mathrm{Pa}\) to \(9.66\times10^4\,\mathrm{Pa}\).
\(\Delta p=(9.66-9.60)\times10^4\)
\(\Delta p=6.0\times10^2\,\mathrm{Pa}\)
\(\Delta h=8.0\times10^{-2}\,\mathrm{m}\)
Therefore,
\(6.0\times10^2=\rho(9.81)(8.0\times10^{-2})\)
\(\rho\approx7.65\times10^2\,\mathrm{kg\,m^{-3}}\)
Answer: \(\boxed{7.6\times10^2\,\mathrm{kg\,m^{-3}}}\) or approximately \(\boxed{765\,\mathrm{kg\,m^{-3}}}\).
(d) New tension in the wire [2 marks]
When the cylinder is fully submerged, the upthrust is
\(F_{\mathrm{U}}=\rho gV\)
The volume of the cylinder is
\(V=Al\)
\(V=(3.7\times10^{-4})(4.0\times10^{-2})\)
\(V=1.48\times10^{-5}\,\mathrm{m^3}\)
Using \(\rho\approx760\,\mathrm{kg\,m^{-3}}\),
\(F_{\mathrm{U}}=(760)(9.81)(1.48\times10^{-5})\)
\(F_{\mathrm{U}}\approx0.11\,\mathrm{N}\)
The cylinder is stationary, so the increase in upthrust causes an equal decrease in tension.
\(T_{\mathrm{new}}=0.53-0.11\)
\(T_{\mathrm{new}}=0.42\,\mathrm{N}\)
Answer: \(\boxed{0.42\,\mathrm{N}}\)
Question 5
(a) An electromagnetic wave in a vacuum has a wavelength of \(8.4\times10^{-6}\,\mathrm{m}\).
(i) State the name of the principal region of the electromagnetic spectrum for the wave. (1 mark)
________________________________________
(ii) Calculate the frequency, in THz, of the wave. (2 marks)
frequency = ………………………………………….. \(\mathrm{THz}\)
(b) An arrangement that uses a double slit to demonstrate the interference of light from a laser is shown in Fig. 5.1.
The light from the laser has a wavelength of \(6.2\times10^{-7}\,\mathrm{m}\) and is incident normally on the slits. The separation of the two slits is \(a\). The slits and screen are parallel and separated by a distance of \(2.8\,\mathrm{m}\).
An interference pattern of bright fringes and dark fringes is formed on the screen. The distance on the screen across 8 bright fringes is \(22\,\mathrm{mm}\), as illustrated in Fig. 5.2.
(i) The light waves emerging from the two slits are coherent.
State what is meant by coherent. (1 mark)
________________________________________
(ii) Calculate the separation \(a\) of the slits. (3 marks)
\(a=\) ……………………………………………… \(\mathrm{m}\)
(c) Fringe P is the central bright fringe of the interference pattern in (b). Fringe Q and fringe R are the nearest dark fringe and the nearest bright fringe respectively to the right of fringe P, as shown in Fig. 5.2.
(i) Calculate the difference in the distances (the path difference) from each slit to the centre of fringe Q. (1 mark)
difference in the distances = ……………………………………………… \(\mathrm{m}\)
(ii) State the phase difference between the light waves meeting at the centre of fringe R. (1 mark)
phase difference = ……………………………………………….. \(^\circ\)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 7.1: Progressive waves — part (a)(ii)
• 8.3: Interference — parts (b)(i), (b)(ii), (c)(i) and (c)(ii)
▶️ Answer/Explanation
(a)(i) Principal region of the electromagnetic spectrum [1 mark]
The wavelength is \(8.4\times10^{-6}\,\mathrm{m}\), which lies in the infrared region of the electromagnetic spectrum.
Answer: \(\boxed{\text{infrared}}\)
(a)(ii) Frequency of the wave [2 marks]
For an electromagnetic wave travelling in a vacuum,
\(c=f\lambda\)
Therefore,
\(f=\frac{c}{\lambda}\)
\(f=\frac{3.0\times10^{8}}{8.4\times10^{-6}}\)
\(f=3.57\times10^{13}\,\mathrm{Hz}\)
Since \(1\,\mathrm{THz}=10^{12}\,\mathrm{Hz}\),
\(f=35.7\,\mathrm{THz}\)
Answer: \(\boxed{36\,\mathrm{THz}}\)
(b)(i) Meaning of coherent [1 mark]
Coherent waves have a constant phase difference between them.
Answer: \(\boxed{\text{constant phase difference}}\)
(b)(ii) Separation of the slits [3 marks]
For a double-slit interference pattern,
\(\lambda=\frac{ax}{D}\)
The distance across 8 bright fringes is \(22\,\mathrm{mm}\), so the fringe spacing is
\(x=\frac{22\times10^{-3}}{8}\)
\(x=2.75\times10^{-3}\,\mathrm{m}\)
Rearranging the equation,
\(a=\frac{\lambda D}{x}\)
\(a=\frac{(6.2\times10^{-7})(2.8)}{2.75\times10^{-3}}\)
\(a=6.31\times10^{-4}\,\mathrm{m}\)
Answer: \(\boxed{6.3\times10^{-4}\,\mathrm{m}}\)
(c)(i) Path difference at fringe Q [1 mark]
Q is the nearest dark fringe to the central bright fringe.
For the first dark fringe, the path difference is
\(\text{path difference}=\frac{\lambda}{2}\)
\(\text{path difference}=\frac{6.2\times10^{-7}}{2}\)
\(\text{path difference}=3.1\times10^{-7}\,\mathrm{m}\)
Answer: \(\boxed{3.1\times10^{-7}\,\mathrm{m}}\)
(c)(ii) Phase difference at fringe R [1 mark]
R is the nearest bright fringe to the central bright fringe. The path difference is one wavelength, so the waves arrive in phase.
The phase difference is therefore
\(\phi=360^\circ\)
Answer: \(\boxed{360^\circ}\)
Question 6
A metal wire in a circuit has a length of \(1.8\,\mathrm{m}\) and a cross-sectional area of \(1.5\times10^{-6}\,\mathrm{m^2}\).
The total number of free electrons (charge carriers) in the wire is \(2.3\times10^{23}\).
There is a current in the wire so that a charge of \(172\,\mathrm{C}\) moves past a fixed point in the wire in a time of \(2.5\,\mathrm{minutes}\).
(a) Show that the number density of the free electrons in the wire is \(8.5\times10^{28}\,\mathrm{m^{-3}}\). (1 mark)
________________________________________
(b) Calculate the average drift speed of the free electrons. (3 marks)
average drift speed = ………………………………………… \(\mathrm{m\,s^{-1}}\)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
▶️ Answer/Explanation
(a) Number density of the free electrons [1 mark]
The volume of the wire is
\(V=AL\)
\(V=(1.5\times10^{-6})(1.8)\)
\(V=2.7\times10^{-6}\,\mathrm{m^3}\)
Number density is the number of free electrons per unit volume:
\(n=\frac{N}{V}\)
\(n=\frac{2.3\times10^{23}}{(1.5\times10^{-6})(1.8)}\)
\(n=8.52\times10^{28}\,\mathrm{m^{-3}}\)
Answer: \(\boxed{8.5\times10^{28}\,\mathrm{m^{-3}}}\)
(b) Average drift speed of the free electrons [3 marks]
First calculate the current using
\(I=\frac{Q}{t}\)
The time is \(2.5\,\mathrm{minutes}=2.5\times60=150\,\mathrm{s}\).
Therefore,
\(I=\frac{172}{150}\)
\(I=1.15\,\mathrm{A}\)
For charge carriers moving through a conductor,
\(I=nAvq\)
where \(n\) is the number density, \(A\) is the cross-sectional area, \(v\) is the drift speed and \(q\) is the charge of one electron.
Using \(q=1.6\times10^{-19}\,\mathrm{C}\),
\(1.15=(8.5\times10^{28})(1.5\times10^{-6})v(1.6\times10^{-19})\)
Rearranging,
\(v=\frac{1.15}{(8.5\times10^{28})(1.5\times10^{-6})(1.6\times10^{-19})}\)
\(v=5.6\times10^{-5}\,\mathrm{m\,s^{-1}}\)
Answer: \(\boxed{5.6\times10^{-5}\,\mathrm{m\,s^{-1}}}\)
Question 7
A battery of electromotive force (e.m.f.) \(9.6\,\mathrm{V}\) and negligible internal resistance is connected in series with two fixed resistors and a thermistor, as shown in Fig. 7.1.
The fixed resistors have resistances of \(3400\,\Omega\) and \(5800\,\Omega\). The reading on the voltmeter in the circuit is \(6.0\,\mathrm{V}\).
(a) Calculate the current in the resistor of resistance \(5800\,\Omega\). (2 marks)
current = ………………………………………………. \(\mathrm{A}\)
(b) Calculate the resistance of the thermistor. (2 marks)
resistance = ……………………………………………… \(\Omega\)
(c) The initial energy stored in the battery is \(2.6\times10^4\,\mathrm{J}\).
Assume that the e.m.f. of the battery is constant.
Determine the final energy stored in the battery after a charge of \(330\,\mathrm{C}\) has moved through it. (2 marks)
final stored energy = ………………………………………………. \(\mathrm{J}\)
(d) The environmental conditions change causing an increase in the resistance of the thermistor.
State whether there is a decrease, increase or no change to:
(i) the temperature of the thermistor (1 mark)
________________________________________
(ii) the current in the thermistor (1 mark)
________________________________________
(iii) the potential difference across the thermistor (1 mark)
________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 9.2: Potential difference and power — parts (a), (c) and (d)(iii)
• 9.3: Resistance and resistivity — parts (b) and (d)(i), (d)(ii)
▶️ Answer/Explanation
(a) Current in the \(5800\,\Omega\) resistor [2 marks]
The \(5800\,\Omega\) resistor is in series with the rest of the circuit. The voltmeter measures \(6.0\,\mathrm{V}\) across the parallel section containing the \(3400\,\Omega\) resistor and thermistor.
Therefore, the potential difference across the \(5800\,\Omega\) resistor is
\(V=9.6-6.0=3.6\,\mathrm{V}\)
Using \(V=IR\),
\(I=\frac{V}{R}\)
\(I=\frac{3.6}{5800}\)
\(I=6.2\times10^{-4}\,\mathrm{A}\)
Answer: \(\boxed{6.2\times10^{-4}\,\mathrm{A}}\)
(b) Resistance of the thermistor [2 marks]
The \(3400\,\Omega\) resistor and thermistor are in parallel, so the potential difference across the thermistor is \(6.0\,\mathrm{V}\).
The current through the \(3400\,\Omega\) resistor is
\(I_{3400}=\frac{6.0}{3400}\)
\(I_{3400}=1.76\times10^{-3}\,\mathrm{A}\)
The total current entering the parallel section is \(6.2\times10^{-4}\,\mathrm{A}\) according to the given circuit arrangement. Using the relationship across the parallel network gives the thermistor resistance:
\(6.0=(6.2\times10^{-4})(3400+R)\)
Rearranging,
\(R=\frac{6.0}{6.2\times10^{-4}}-3400\)
\(R\approx6.3\times10^{3}\,\Omega\)
Answer: \(\boxed{6.3\times10^{3}\,\Omega}\)
(c) Final energy stored in the battery [2 marks]
The energy transferred when charge \(Q\) moves through a potential difference \(V\) is
\(\Delta E=QV\)
\(\Delta E=(330)(9.6)\)
\(\Delta E=3168\,\mathrm{J}\)
The battery loses this energy, so
\(E_{\mathrm{final}}=2.6\times10^4-3168\)
\(E_{\mathrm{final}}=2.28\times10^4\,\mathrm{J}\)
Answer: \(\boxed{2.3\times10^4\,\mathrm{J}}\)
(d)(i) Temperature of the thermistor [1 mark]
The thermistor is an NTC thermistor. For an NTC thermistor, an increase in resistance corresponds to a decrease in temperature.
Answer: \(\boxed{\text{decrease}}\)
(d)(ii) Current in the thermistor [1 mark]
The thermistor is in series with the fixed resistor arrangement, so an increase in its resistance causes the total circuit resistance to increase. With constant e.m.f., the current decreases.
Answer: \(\boxed{\text{decrease}}\)
(d)(iii) Potential difference across the thermistor [1 mark]
As the resistance of the thermistor increases, a greater proportion of the supply potential difference is across the thermistor.
Answer: \(\boxed{\text{increase}}\)
Question 8
An isolated stationary nucleus X decays by emitting an \(\alpha\)-particle to form a nucleus Y.
Nucleus Y and nucleus Z are isotopes of the same element.
(a) By comparing the number of protons in each nucleus, state and explain whether the charge of nucleus Y is less than, greater than or the same as the charge of:
(i) nucleus Z (1 mark)
________________________________________
(ii) nucleus X. (2 marks)
________________________________________
(b) Use the principle of conservation of momentum to explain why nucleus Y cannot be stationary immediately after the decay of nucleus X. (2 marks)
________________________________________
________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 3.3: Linear momentum and its conservation — part (b)
▶️ Answer/Explanation
(a)(i) Charge of nucleus Y compared with nucleus Z [1 mark]
Nuclei Y and Z are isotopes of the same element, so they have the same number of protons.
The nuclear charge depends on the number of protons. Therefore, the two nuclei have the same charge.
Answer: \(\boxed{\text{same}}\)
(a)(ii) Charge of nucleus Y compared with nucleus X [2 marks]
An \(\alpha\)-particle contains two protons. When nucleus X emits an \(\alpha\)-particle, nucleus Y has two fewer protons than nucleus X.
Therefore, nucleus Y has a smaller positive nuclear charge than nucleus X.
Answer: \(\boxed{\text{less than}}\)
(b) Conservation of momentum [2 marks]
Nucleus X is initially stationary, so its momentum before the decay is zero.
By conservation of momentum, the total momentum after the decay must also be zero.
The emitted \(\alpha\)-particle has momentum in one direction. Therefore, nucleus Y must have an equal and opposite momentum.
Hence, nucleus Y cannot be stationary and must have a velocity in the opposite direction to the \(\alpha\)-particle.
Answer: \(\boxed{\text{Nucleus Y has equal and opposite momentum to the }\alpha\text{-particle.}}\)