Question 1
(a) Underline all the SI base units in the following list.
ampere coulomb current kelvin newton [1 mark]
(b) A toy car moves in a horizontal straight line. The displacement \(s\) of the car is given by the equation
\(s=\frac{v^2}{2a}\)
where \(a\) is the acceleration of the car and \(v\) is its final velocity.
State two conditions that apply to the motion of the car in order for the above equation to be valid.
1. ____________________________________________________________________________
2. ____________________________________________________________________________ [2 marks]
(c) An experiment is performed to determine the acceleration of the car in (b). The following measurements are obtained:
\(s=3.89\,\mathrm{m}\pm0.5\%\)
\(v=2.75\,\mathrm{m\,s^{-1}}\pm0.8\%\)
(i) Calculate the acceleration \(a\) of the car. [1 mark]
\(a\) = ____________________ \( \mathrm{m\,s^{-2}} \)
(ii) Determine the percentage uncertainty, to two significant figures, in \(a\). [2 marks]
percentage uncertainty = ____________________ \( \%\)
(iii) Use your answers in (c)(i) and (c)(ii) to determine the absolute uncertainty in the calculated value of \(a\). [1 mark]
absolute uncertainty = ____________________ \( \mathrm{m\,s^{-2}} \)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 1.3: Errors and uncertainties — parts (c)(ii) and (c)(iii)
• 2.1: Equations of motion — parts (b), (c)(i), (c)(ii) and (c)(iii)
▶️ Answer/Explanation
(a) SI base units [1 mark]
The SI base units in the list are ampere and kelvin.
Answer: \(\boxed{\text{ampere and kelvin}}\)
(b) Conditions for the equation to be valid [2 marks]
The equation
\(s=\frac{v^2}{2a}\)
is obtained from the equations of motion when the initial velocity is zero and the acceleration is constant.
Therefore, the two conditions are:
• The initial speed or velocity is zero.
• The acceleration is constant or uniform.
Answer: \(\boxed{\text{initial velocity is zero and acceleration is constant}}\)
(c)(i) Acceleration of the car [1 mark]
From
\(s=\frac{v^2}{2a}\)
rearrange to give
\(a=\frac{v^2}{2s}\)
Substituting the measured values:
\(a=\frac{(2.75)^2}{2(3.89)}\)
\(a=0.972\,\mathrm{m\,s^{-2}}\)
Answer: \(\boxed{0.97\,\mathrm{m\,s^{-2}}}\)
(c)(ii) Percentage uncertainty in \(a\) [2 marks]
Since
\(a=\frac{v^2}{2s}\)
the percentage uncertainty in \(v\) is multiplied by \(2\), while the percentage uncertainty in \(s\) is added directly.
Therefore,
\(\text{percentage uncertainty}=2(0.8)+0.5\)
\(\text{percentage uncertainty}=2.1\%\)
Answer: \(\boxed{2.1\%}\)
(c)(iii) Absolute uncertainty in \(a\) [1 mark]
Absolute uncertainty is calculated from
\(\text{absolute uncertainty}=\frac{\text{percentage uncertainty}}{100}\times\text{value}\)
Therefore,
\(\text{absolute uncertainty}=\frac{2.1}{100}\times0.97\)
\(\text{absolute uncertainty}=0.0204\,\mathrm{m\,s^{-2}}\)
Answer: \(\boxed{0.02\,\mathrm{m\,s^{-2}}}\)
Question 2
A motor uses a wire to raise a block, as illustrated in Fig. 2.1.
The base of the block takes a time of \(0.49\,\mathrm{s}\) to move vertically upwards from level \(X\) to level \(Y\) at a constant speed of \(0.64\,\mathrm{m\,s^{-1}}\). During this time the wire has a strain of \(0.0012\). The wire is made of metal of Young modulus \(2.2\times10^{11}\,\mathrm{Pa}\) and has a uniform cross-section.
The block has a weight of \(1.4\times10^4\,\mathrm{N}\). Assume that the weight of the wire is negligible.
(a) Calculate:
(i) the cross-sectional area \(A\) of the wire. [2 marks]
\(A\) = ____________________ \( \mathrm{m^2} \)
(ii) the increase in the gravitational potential energy of the block for the movement of its base from \(X\) to \(Y\). [3 marks]
increase in gravitational potential energy = ____________________ \( \mathrm{J} \)
(b) The motor has an efficiency of \(56\%\).
Calculate the input power to the motor as the base of the block moves from \(X\) to \(Y\). [3 marks]
input power = ____________________ \( \mathrm{W} \)
(c) The base of the block now has a uniform deceleration of magnitude \(1.3\,\mathrm{m\,s^{-2}}\) from level \(Y\) until the base of the block stops at level \(Z\).
Calculate the tension \(T\) in the wire as the base of the block moves from \(Y\) to \(Z\). [3 marks]
\(T\) = ____________________ \( \mathrm{N} \)
(d) The base of the block is at levels \(X\), \(Y\) and \(Z\) at times \(t_X\), \(t_Y\) and \(t_Z\) respectively.
On Fig. 2.2, sketch a graph to show the variation with time \(t\) of the distance \(d\) of the base of the block from level \(X\). Numerical values of \(d\) and \(t\) are not required. [2 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 5.2: Gravitational potential energy and kinetic energy — part (a)(ii)
• 5.1: Energy conservation — part (b)
• 3.1: Momentum and Newton’s laws of motion — part (c)
• 2.1: Equations of motion — part (d)
▶️ Answer/Explanation
(a)(i) Cross-sectional area of the wire [2 marks]
Young modulus is given by
\(E=\frac{\text{stress}}{\text{strain}}\)
Since \(\text{stress}=\frac{F}{A}\),
\(E=\frac{F}{A\varepsilon}\)
Rearranging,
\(A=\frac{F}{E\varepsilon}\)
Substituting \(F=1.4\times10^4\,\mathrm{N}\), \(E=2.2\times10^{11}\,\mathrm{Pa}\) and \(\varepsilon=0.0012\):
\(A=\frac{1.4\times10^4}{(2.2\times10^{11})(0.0012)}\)
\(A=5.3\times10^{-5}\,\mathrm{m^2}\)
Answer: \(\boxed{5.3\times10^{-5}\,\mathrm{m^2}}\)
(a)(ii) Increase in gravitational potential energy [3 marks]
The vertical distance moved from \(X\) to \(Y\) is
\(\Delta h=vt\)
\(\Delta h=(0.64)(0.49)\)
\(\Delta h=0.3136\,\mathrm{m}\)
The increase in gravitational potential energy is
\(\Delta E_{\mathrm{P}}=mg\Delta h\)
Since the weight is \(mg=1.4\times10^4\,\mathrm{N}\),
\(\Delta E_{\mathrm{P}}=(1.4\times10^4)(0.64)(0.49)\)
\(\Delta E_{\mathrm{P}}=4.39\times10^3\,\mathrm{J}\)
Answer: \(\boxed{4.4\times10^3\,\mathrm{J}}\)
(b) Input power to the motor [3 marks]
The useful output power is
\(P_{\mathrm{out}}=Fv\)
\(P_{\mathrm{out}}=(1.4\times10^4)(0.64)\)
\(P_{\mathrm{out}}=8960\,\mathrm{W}\)
Efficiency is
\(\eta=\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}\)
Therefore,
\(P_{\mathrm{in}}=\frac{P_{\mathrm{out}}}{\eta}\)
\(P_{\mathrm{in}}=\frac{8960}{0.56}\)
\(P_{\mathrm{in}}=1.6\times10^4\,\mathrm{W}\)
Answer: \(\boxed{1.6\times10^4\,\mathrm{W}}\)
(c) Tension in the wire [3 marks]
The mass of the block is
\(m=\frac{W}{g}\)
\(m=\frac{1.4\times10^4}{9.81}\)
\(m=1427\,\mathrm{kg}\)
The block is decelerating while moving upwards, so its acceleration is \(1.3\,\mathrm{m\,s^{-2}}\) downwards.
Taking downward as the direction of the resultant force,
\(W-T=ma\)
\(1.4\times10^4-T=(1427)(1.3)\)
\(T=1.4\times10^4-1855\)
\(T=1.21\times10^4\,\mathrm{N}\)
Answer: \(\boxed{1.2\times10^4\,\mathrm{N}}\)
(d) Distance-time graph [2 marks]
From \(t_X\) to \(t_Y\), the block moves at constant speed. Therefore, the distance increases linearly with a constant positive gradient.
From \(t_Y\) to \(t_Z\), the block continues moving upwards but decelerates. Therefore, the gradient of the distance-time graph decreases continuously until the gradient becomes zero at \(t_Z\).
Answer: \(\boxed{\text{A straight upward-sloping line from }t_X\text{ to }t_Y,\text{ followed by an upward-sloping curve with decreasing gradient that becomes horizontal at }t_Z.}\)
Question 3
A uniform beam AB is attached by a hinge to a wall at end A, as shown in Fig. 3.1.
The beam has length \(0.50\,\mathrm{m}\) and weight \(W\). A block of weight \(12\,\mathrm{N}\) rests on the beam at a distance of \(0.15\,\mathrm{m}\) from end B.
The beam is held horizontal and in equilibrium by a string attached between end B and a fixed point C. The string has a tension of \(17\,\mathrm{N}\) and is at an angle of \(50^{\circ}\) to the horizontal.
(a) State two conditions for an object to be in equilibrium.
1. ____________________________________________________________________________
2. ____________________________________________________________________________ [2 marks]
(b) Show that the vertical component of the tension in the string is \(13\,\mathrm{N}\). [1 mark]
____________________________________________________________________________
(c) By taking moments about end A, calculate the weight \(W\) of the beam. [2 marks]
\(W\) = ____________________ \( \mathrm{N} \)
(d) Calculate the magnitude of the vertical component of the force exerted on the beam by the hinge. [1 mark]
force = ____________________ \( \mathrm{N} \)
(e) The block is now moved closer to end A of the beam. Assume that the beam remains horizontal.
State whether this change will increase, decrease or have no effect on the horizontal component of the force exerted on the beam by the hinge. [1 mark]
______________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 4.2: Equilibrium of forces — parts (a), (d) and (e)
▶️ Answer/Explanation
(a) Conditions for equilibrium [2 marks]
For an object to be in equilibrium:
• The resultant force in any direction must be zero.
• The resultant moment or torque about any point must be zero.
Answer: \(\boxed{\text{resultant force = 0 and resultant moment = 0}}\)
(b) Vertical component of tension [1 mark]
The tension in the string is \(17\,\mathrm{N}\) and the string is at \(50^{\circ}\) to the horizontal.
The vertical component is
\(T_{\mathrm{vertical}}=17\sin50^{\circ}\)
\(T_{\mathrm{vertical}}=13.0\,\mathrm{N}\)
Answer: \(\boxed{13\,\mathrm{N}}\)
(c) Weight of the beam [2 marks]
Take moments about A. The horizontal component of the tension produces no moment because its line of action is along the beam.
The beam’s weight \(W\) acts at its centre, \(0.25\,\mathrm{m}\) from A.
The block is \(0.15\,\mathrm{m}\) from B, so its distance from A is
\(0.50-0.15=0.35\,\mathrm{m}\)
For rotational equilibrium,
\(W(0.25)+(12)(0.35)=(13)(0.50)\)
\(0.25W+4.2=6.5\)
\(0.25W=2.3\)
\(W=9.2\,\mathrm{N}\)
Answer: \(\boxed{9.2\,\mathrm{N}}\)
(d) Vertical component of hinge force [1 mark]
The upward vertical component of the tension is \(13\,\mathrm{N}\).
The downward forces are the beam’s weight and the block’s weight:
\(9.2+12=21.2\,\mathrm{N}\)
Therefore, the hinge must provide an upward vertical component of
\(F=21.2-13\)
\(F=8.2\,\mathrm{N}\)
Answer: \(\boxed{8\,\mathrm{N}}\)
(e) Effect on the horizontal hinge force [1 mark]
Moving the block closer to A decreases its moment about A.
Therefore, a smaller vertical component of tension is required to balance the moments. Since the tension magnitude remains fixed at \(17\,\mathrm{N}\), its horizontal component increases.
The horizontal component of the hinge force is equal and opposite to the horizontal component of the tension.
Answer: \(\boxed{\text{decrease}}\)
Question 4
Two blocks slide directly towards each other along a frictionless horizontal surface, as shown in Fig. 4.1. The blocks collide and then move as shown in Fig. 4.2.
Block \(X\) initially moves to the right with a momentum of \(0.37\,\mathrm{kg\,m\,s^{-1}}\). Block \(Y\) initially moves to the left with a momentum of \(0.65\,\mathrm{kg\,m\,s^{-1}}\). After the blocks collide, block \(X\) moves to the left back along its original path with a momentum of \(0.13\,\mathrm{kg\,m\,s^{-1}}\). Block \(Y\) also moves to the left after the collision.
(a) Block \(X\) has an initial kinetic energy of \(0.30\,\mathrm{J}\).
Calculate the mass of block \(X\). [3 marks]
mass = ____________________ \( \mathrm{kg} \)
(b) Determine the magnitude of the momentum of block \(Y\) after the collision. [1 mark]
momentum = ____________________ \( \mathrm{kg\,m\,s^{-1}} \)
(c) Block \(X\) exerts an average force of \(7.7\,\mathrm{N}\) on block \(Y\) during the collision.
Calculate the time that the blocks are in contact with each other. [2 marks]
time = ____________________ \( \mathrm{s} \)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
▶️ Answer/Explanation
(a) Mass of block \(X\) [3 marks]
The relationship between kinetic energy and momentum is
\(E_{\mathrm{K}}=\frac{1}{2}mv^2\)
and
\(p=mv\)
Therefore, \(v=\frac{p}{m}\), so
\(E_{\mathrm{K}}=\frac{1}{2}m\left(\frac{p}{m}\right)^2\)
\(E_{\mathrm{K}}=\frac{p^2}{2m}\)
Rearranging,
\(m=\frac{p^2}{2E_{\mathrm{K}}}\)
Substituting \(p=0.37\,\mathrm{kg\,m\,s^{-1}}\) and \(E_{\mathrm{K}}=0.30\,\mathrm{J}\):
\(m=\frac{(0.37)^2}{2(0.30)}\)
\(m=0.228\,\mathrm{kg}\)
Answer: \(\boxed{0.23\,\mathrm{kg}}\)
(b) Momentum of block \(Y\) after the collision [1 mark]
Take motion to the right as positive.
Before the collision, the total momentum is
\(p_{\mathrm{initial}}=0.37-0.65=-0.28\,\mathrm{kg\,m\,s^{-1}}\)
After the collision, block \(X\) has momentum \(-0.13\,\mathrm{kg\,m\,s^{-1}}\). Let the momentum of block \(Y\) be \(-p\).
Using conservation of momentum,
\(0.37-0.65=-0.13-p\)
\(-0.28=-0.13-p\)
\(p=0.15\,\mathrm{kg\,m\,s^{-1}}\)
Answer: \(\boxed{0.15\,\mathrm{kg\,m\,s^{-1}}}\)
(c) Time of contact [2 marks]
The impulse is equal to the change in momentum:
\(F\Delta t=\Delta p\)
For block \(X\), its momentum changes from \(+0.37\) to \(-0.13\,\mathrm{kg\,m\,s^{-1}}\).
Therefore, the magnitude of the change in momentum is
\(\left|\Delta p\right|=|-0.13-0.37|\)
\(\left|\Delta p\right|=0.50\,\mathrm{kg\,m\,s^{-1}}\)
Hence,
\(7.7\Delta t=0.50\)
\(\Delta t=\frac{0.50}{7.7}\)
\(\Delta t=0.0649\,\mathrm{s}\)
Answer: \(\boxed{0.065\,\mathrm{s}}\)
Question 5
(a) A microphone and cathode-ray oscilloscope (CRO) are used to analyse a sound wave of frequency \(5000\,\mathrm{Hz}\). The trace that is displayed on the screen of the CRO is shown in Fig. 5.1.
(i) Determine the time-base setting, in \(\mathrm{s\,cm^{-1}}\), of the CRO. [2 marks]
time-base setting = ____________________ \(\mathrm{s\,cm^{-1}}\)
(ii) The intensity of the sound detected by the microphone is now increased from its initial value of \(I\) to a new value of \(3I\). The frequency of the sound is unchanged. Assume that the amplitude of the trace on the CRO screen is proportional to the amplitude of the sound wave.
On Fig. 5.1, sketch the new trace shown on the screen of the CRO. [3 marks]
(b) An arrangement for demonstrating interference using light is shown in Fig. 5.2.
The wavelength of the light from the laser is \(630\,\mathrm{nm}\). The light is incident normally on the double slit. The separation of the two slits is \(3.6\times10^{-4}\,\mathrm{m}\). The perpendicular distance between the double slit and the screen is \(D\).
Coherent light waves from the slits form an interference pattern of bright and dark fringes on the screen. The distance between the centres of two adjacent bright fringes is \(4.0\times10^{-3}\,\mathrm{m}\). The central bright fringe is formed at point \(P\).
(i) Explain why a bright fringe is produced by the waves meeting at point \(P\). [1 mark]
____________________________________________________________________________
(ii) Calculate distance \(D\). [3 marks]
\(D\) = ____________________ \(\mathrm{m}\)
(c) The wavelength \(\lambda\) of the light in (b) is now varied. This causes a variation in the distance \(x\) between the centres of two adjacent bright fringes on the screen. The distance \(D\) and the separation of the two slits are unchanged.
On Fig. 5.3, sketch a graph to show the variation of \(x\) with \(\lambda\) from \(\lambda=400\,\mathrm{nm}\) to \(\lambda=700\,\mathrm{nm}\). Numerical values of \(x\) are not required. [1 mark]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 8.3: Interference — parts (b)(i), (b)(ii) and (c)
▶️ Answer/Explanation
(a)(i) Time-base setting of the CRO [2 marks]
The frequency of the sound wave is \(5000\,\mathrm{Hz}\), so its period is
\(T=\frac{1}{f}\)
\(T=\frac{1}{5000}\)
\(T=2.0\times10^{-4}\,\mathrm{s}\)
From the trace, one complete cycle occupies \(4.0\,\mathrm{cm}\) horizontally.
Therefore,
\(\text{time-base setting}=\frac{2.0\times10^{-4}}{4.0}\)
\(\text{time-base setting}=5.0\times10^{-5}\,\mathrm{s\,cm^{-1}}\)
Answer: \(\boxed{5.0\times10^{-5}\,\mathrm{s\,cm^{-1}}}\)
(a)(ii) New CRO trace [3 marks]
The intensity of a progressive wave is proportional to the square of its amplitude:
\(I\propto A^2\)
The intensity changes from \(I\) to \(3I\), so
\(\frac{3I}{I}=\frac{A_2^2}{A_1^2}\)
\(3=\frac{A_2^2}{A_1^2}\)
\(A_2=\sqrt{3}A_1\)
The original amplitude is \(1.0\,\mathrm{cm}\), so
\(A_2=\sqrt{3}(1.0)\)
\(A_2=1.7\,\mathrm{cm}\)
Therefore, the new trace should have:
• the same period and wavelength as the original trace;
• an amplitude of \(1.7\,\mathrm{cm}\).
Answer: \(\boxed{\text{same period, with amplitude }1.7\,\mathrm{cm}}\)
(b)(i) Bright fringe at \(P\) [1 mark]
Point \(P\) lies on the central bright fringe, so the distances travelled by the waves from the two slits to \(P\) are equal.
Therefore, the path difference is zero and the waves arrive in phase, producing constructive interference.
Answer: \(\boxed{\text{The path difference is zero, so the waves arrive in phase and interfere constructively.}}\)
(b)(ii) Distance \(D\) [3 marks]
For double-slit interference,
\(\lambda=\frac{ax}{D}\)
Rearranging,
\(D=\frac{ax}{\lambda}\)
Using \(a=3.6\times10^{-4}\,\mathrm{m}\), \(x=4.0\times10^{-3}\,\mathrm{m}\) and \(\lambda=630\times10^{-9}\,\mathrm{m}\):
\(D=\frac{(3.6\times10^{-4})(4.0\times10^{-3})}{630\times10^{-9}}\)
\(D=2.29\,\mathrm{m}\)
Answer: \(\boxed{2.3\,\mathrm{m}}\)
(c) Graph of \(x\) against \(\lambda\) [1 mark]
From the double-slit equation,
\(\lambda=\frac{ax}{D}\)
Therefore,
\(x=\frac{D}{a}\lambda\)
Since \(D\) and \(a\) are constant, \(x\) is directly proportional to \(\lambda\).
The graph should therefore be a straight line with a positive gradient, starting from a non-zero value of \(x\) at \(\lambda=400\,\mathrm{nm}\).
Answer: \(\boxed{\text{an upward-sloping straight line from }400\,\mathrm{nm}\text{ to }700\,\mathrm{nm}}\)
Question 6
(a) Define the potential difference across a component. [1 mark]
____________________________________________________________________________
(b) The variation with potential difference \(V\) of the current \(I\) in a semiconductor diode is shown in Fig. 6.1.
Use Fig. 6.1 to describe qualitatively:
(i) the resistance of the diode in the range \(V=0\) to \(V=0.25\,\mathrm{V}\). [1 mark]
____________________________________________________________________________
(ii) the variation, if any, in the resistance of the diode as \(V\) changes from \(V=0.75\,\mathrm{V}\) to \(V=1.0\,\mathrm{V}\). [1 mark]
____________________________________________________________________________
(c) A battery of electromotive force (e.m.f.) \(12\,\mathrm{V}\) and negligible internal resistance is connected to a uniform resistance wire \(XY\), a fixed resistor and a variable resistor, as shown in Fig. 6.2.
The fixed resistor has a resistance of \(5.0\,\Omega\). The current in the battery is \(2.7\,\mathrm{A}\) and the current in the fixed resistor is \(1.5\,\mathrm{A}\).
(i) Calculate the current in the resistance wire. [1 mark]
current = ____________________ \( \mathrm{A} \)
(ii) Determine the resistance of the variable resistor. [2 marks]
resistance = ____________________ \( \Omega \)
(iii) Wire \(XY\) has a length of \(2.0\,\mathrm{m}\). Point \(Z\) on the wire is a distance of \(1.6\,\mathrm{m}\) from point \(X\).
The fixed resistor is connected to the variable resistor at point \(W\). Determine the potential difference between points \(W\) and \(Z\). [3 marks]
potential difference = ____________________ \( \mathrm{V} \)
(iv) The resistance of the variable resistor is now increased.
By considering the currents in every part of the circuit, state and explain whether the total power produced by the battery decreases, increases or stays the same. [3 marks]
________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 9.3: Resistance and resistivity — part (b)
• 10.2: Kirchhoff’s laws — parts (c)(i) and (c)(ii)
• 10.1: Practical circuits — parts (c)(iii) and (c)(iv)
▶️ Answer/Explanation
(a) Definition of potential difference [1 mark]
Potential difference is the energy transferred per unit charge when charge passes through a component.
The relationship is
\(V=\frac{W}{Q}\)
Answer: \(\boxed{\text{energy transferred per unit charge}}\)
(b)(i) Resistance from \(0\) to \(0.25\,\mathrm{V}\) [1 mark]
From the graph, the current is approximately zero while the potential difference increases from \(0\) to \(0.25\,\mathrm{V}\).
Since
\(R=\frac{V}{I}\)
a very small current for a finite potential difference corresponds to a very large resistance.
Answer: \(\boxed{\text{The resistance is very large.}}\)
(b)(ii) Resistance from \(0.75\,\mathrm{V}\) to \(1.0\,\mathrm{V}\) [1 mark]
Over this range, the graph becomes progressively steeper. Therefore, a small increase in \(V\) produces a relatively large increase in \(I\).
Since \(R=\frac{V}{I}\), the resistance decreases as \(V\) increases.
Answer: \(\boxed{\text{The resistance decreases.}}\)
(c)(i) Current in the resistance wire [1 mark]
The current from the battery splits between the resistance wire and the lower branch containing the fixed and variable resistors.
Using Kirchhoff’s first law,
\(I_{\mathrm{battery}}=I_{\mathrm{wire}}+I_{\mathrm{fixed}}\)
\(2.7=I_{\mathrm{wire}}+1.5\)
\(I_{\mathrm{wire}}=1.2\,\mathrm{A}\)
Answer: \(\boxed{1.2\,\mathrm{A}}\)
(c)(ii) Resistance of the variable resistor [2 marks]
The fixed resistor and variable resistor are in series in the lower branch.
The p.d. across this branch is \(12\,\mathrm{V}\), and the current is \(1.5\,\mathrm{A}\).
Therefore, the total resistance of the lower branch is
\(R_{\mathrm{total}}=\frac{V}{I}\)
\(R_{\mathrm{total}}=\frac{12}{1.5}=8.0\,\Omega\)
Hence,
\(R_{\mathrm{variable}}=8.0-5.0\)
\(R_{\mathrm{variable}}=3.0\,\Omega\)
Answer: \(\boxed{3.0\,\Omega}\)
(c)(iii) Potential difference between \(W\) and \(Z\) [3 marks]
The wire \(XY\) is uniform, so the potential drop along it is proportional to length.
The current in the wire is \(1.2\,\mathrm{A}\), and the wire is connected directly across the \(12\,\mathrm{V}\) battery. Therefore, the total p.d. across \(XY\) is \(12\,\mathrm{V}\).
The p.d. across \(XZ\) is therefore
\(V_{XZ}=12\times\frac{1.6}{2.0}\)
\(V_{XZ}=9.6\,\mathrm{V}\)
The p.d. across the fixed resistor is
\(V=IR\)
\(V_{XW}=(1.5)(5.0)=7.5\,\mathrm{V}\)
Hence, the potential difference between \(W\) and \(Z\) is
\(V_{WZ}=9.6-7.5\)
\(V_{WZ}=2.1\,\mathrm{V}\)
Answer: \(\boxed{2.1\,\mathrm{V}}\)
(c)(iv) Effect of increasing the variable resistance [3 marks]
Increasing the variable resistance increases the resistance of the lower branch.
Therefore, the current in the lower branch decreases.
The resistance wire is connected directly across the ideal \(12\,\mathrm{V}\) battery, so its current remains unchanged at \(1.2\,\mathrm{A}\).
Hence, the total current supplied by the battery decreases.
Since the battery voltage remains constant,
\(P=VI\)
a decrease in total current means that the total power produced by the battery decreases.
Answer: \(\boxed{\text{The total power produced by the battery decreases.}}\)
Question 7
(a) Nuclei X and Y are different isotopes of the same element.
Nucleus X is unstable and emits a \(\beta^+\) particle to form nucleus Z.
By comparing the number of protons in each nucleus, state and explain whether the charge of nucleus X is less than, the same as or greater than the charge of:
(i) nucleus Y [1 mark]
______________________
(ii) nucleus Z [2 marks]
__________________________
(b) Hadrons can be divided into two groups (classes), P and Q. Group P is baryons.
(i) State the name of group Q. [1 mark]
_________________________
(ii) Describe, in general terms, the quark structure of hadrons that belong to group Q. [1 mark]
_________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 11.2: Fundamental particles — parts (b)(i) and (b)(ii)
▶️ Answer/Explanation
(a)(i) Charge of nucleus X compared with nucleus Y [1 mark]
Nuclei X and Y are isotopes of the same element.
Isotopes of the same element have the same number of protons.
The nuclear charge depends on the number of protons, since each proton has charge \(+e\).
Answer: \(\boxed{\text{The charge of X is the same as the charge of Y.}}\)
(a)(ii) Charge of nucleus X compared with nucleus Z [2 marks]
In \(\beta^+\) decay, a proton in the nucleus changes into a neutron and a positron is emitted.
Therefore, nucleus Z has one fewer proton than nucleus X.
Since nuclear charge is determined by the number of protons, nucleus X has a greater positive charge than nucleus Z.
Answer: \(\boxed{\text{The charge of X is greater than the charge of Z.}}\)
(b)(i) Name of group Q [1 mark]
Hadrons are divided into baryons and mesons.
Answer: \(\boxed{\text{mesons}}\)
(b)(ii) Quark structure of mesons [1 mark]
A meson consists of one quark and one antiquark.
Answer: \(\boxed{\text{one quark and one antiquark}}\)