WAEC SSCE - Physics - 2019 (Essay)
Question 1 Report
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(a) The diagram above illustrates a projectile motion. Identify each of the physical quantities labeled P, β, H and R.
(b) Write an equation to show the relationship between P, g and Rmax’ where g is the acceleration due to gravity and Rmax is maximum R.
(a) Identifying the labelled quantities (standard projectile diagram):
- P is the velocity of projection (the initial speed with which the body is launched).
- β is the angle of projection (the angle the initial velocity makes with the horizontal).
- H is the maximum height reached by the projectile.
- R is the horizontal range (the horizontal distance travelled before the projectile returns to the launch level).
(b) Relationship between P, g and Rmax:
The range of a projectile launched at speed \(P\) and angle \(\beta\) is
\[ R = \frac{P^{2}\sin 2\beta}{g}. \]
The range is greatest when \(\sin 2\beta = 1\), i.e. when \(\beta = 45^{\circ}\). Then
\[ R_{max} = \frac{P^{2}}{g}. \]
Equivalently, \(P = \sqrt{g\,R_{max}}\).
Answer Details
(a) Identifying the labelled quantities (standard projectile diagram):
- P is the velocity of projection (the initial speed with which the body is launched).
- β is the angle of projection (the angle the initial velocity makes with the horizontal).
- H is the maximum height reached by the projectile.
- R is the horizontal range (the horizontal distance travelled before the projectile returns to the launch level).
(b) Relationship between P, g and Rmax:
The range of a projectile launched at speed \(P\) and angle \(\beta\) is
\[ R = \frac{P^{2}\sin 2\beta}{g}. \]
The range is greatest when \(\sin 2\beta = 1\), i.e. when \(\beta = 45^{\circ}\). Then
\[ R_{max} = \frac{P^{2}}{g}. \]
Equivalently, \(P = \sqrt{g\,R_{max}}\).
Question 2 Report
(a) (i) Explain latent heat.
(ii) State two factors that affect the rate of evaporation of a liquid
(b) Explain each of the following observations:
(i) On a dry day, water in a clay pot is cooler than water in a closed plastic container;
(ii) Food gets cooked faster in a pressure cooker than in an ordinary cooking pot.
(c) State two effects of heat on a substance.
(d) A 40 V electric heater is used to supply a current of 12 A for 1400 s to a body of mass 1.5 kg at the melting point of the body. The body melts and its temperature rises through \(60^o\)C in an extra 72 s. Determine the:
(i) latent heat of fusion of the body;
(ii) specific heat capacity of the body.
(a)(i) Latent heat: Latent heat is the quantity of heat energy absorbed or given out by a substance during a change of state (for example melting or boiling) at constant temperature. The heat is used to change the arrangement/spacing of the molecules rather than to raise the temperature.
(a)(ii) Two factors affecting the rate of evaporation: (1) Temperature of the liquid; (2) Surface area of the liquid exposed. (Draught/wind speed and humidity of the surrounding air are also acceptable.)
(b)(i) Clay pot cooler than closed plastic container: A clay pot is porous, so water seeps to the outside and evaporates. Evaporation takes latent heat of vaporisation from the remaining water, cooling it. In the sealed plastic container no evaporation (or escape of vapour) can occur, so no cooling takes place and the water stays warmer.
(b)(ii) Faster cooking in a pressure cooker: In a sealed pressure cooker the steam produced raises the pressure above the water. Increased pressure raises the boiling point of the water above 100 °C, so the food is cooked at a higher temperature, which speeds up cooking.
(c) Two effects of heat on a substance: (1) It causes expansion (increase in size); (2) It causes a rise in temperature or a change of state. (Change in electrical resistance or chemical change are also acceptable.)
(d) Power of heater \(P = VI = 40\times12 = 480\,\text{W}\).
(i) Latent heat of fusion: Heat supplied to melt the body \(= P t_1 = 480\times1400 = 672000\,\text{J}\). This equals \(mL\):
\[ L = \frac{Pt_1}{m} = \frac{672000}{1.5} = 4.48\times10^{5}\,\text{J kg}^{-1}. \]
(ii) Specific heat capacity: Heat supplied in the extra 72 s \(= 480\times72 = 34560\,\text{J} = mc\,\Delta\theta\):
\[ c = \frac{34560}{1.5\times60} = 384\,\text{J kg}^{-1}\,\text{K}^{-1}. \]
Answer Details
(a)(i) Latent heat: Latent heat is the quantity of heat energy absorbed or given out by a substance during a change of state (for example melting or boiling) at constant temperature. The heat is used to change the arrangement/spacing of the molecules rather than to raise the temperature.
(a)(ii) Two factors affecting the rate of evaporation: (1) Temperature of the liquid; (2) Surface area of the liquid exposed. (Draught/wind speed and humidity of the surrounding air are also acceptable.)
(b)(i) Clay pot cooler than closed plastic container: A clay pot is porous, so water seeps to the outside and evaporates. Evaporation takes latent heat of vaporisation from the remaining water, cooling it. In the sealed plastic container no evaporation (or escape of vapour) can occur, so no cooling takes place and the water stays warmer.
(b)(ii) Faster cooking in a pressure cooker: In a sealed pressure cooker the steam produced raises the pressure above the water. Increased pressure raises the boiling point of the water above 100 °C, so the food is cooked at a higher temperature, which speeds up cooking.
(c) Two effects of heat on a substance: (1) It causes expansion (increase in size); (2) It causes a rise in temperature or a change of state. (Change in electrical resistance or chemical change are also acceptable.)
(d) Power of heater \(P = VI = 40\times12 = 480\,\text{W}\).
(i) Latent heat of fusion: Heat supplied to melt the body \(= P t_1 = 480\times1400 = 672000\,\text{J}\). This equals \(mL\):
\[ L = \frac{Pt_1}{m} = \frac{672000}{1.5} = 4.48\times10^{5}\,\text{J kg}^{-1}. \]
(ii) Specific heat capacity: Heat supplied in the extra 72 s \(= 480\times72 = 34560\,\text{J} = mc\,\Delta\theta\):
\[ c = \frac{34560}{1.5\times60} = 384\,\text{J kg}^{-1}\,\text{K}^{-1}. \]
Question 3 Report
State three observable phenomena in which waves behave like a particle.
Observable phenomena in which electromagnetic waves behave like particles
- photoelectric effect;
- Compton effect (Compton scattering);
- pair production;
- blackbody radiation.
Answer Details
Observable phenomena in which electromagnetic waves behave like particles
- photoelectric effect;
- Compton effect (Compton scattering);
- pair production;
- blackbody radiation.
Question 4 Report
(a) (i) State Newton’s Law of Universal Gravitation.
(ii) Define gravitational field.
(b) (i) Derive the equation relating the universal gravitational constant, G, and the acceleration of free fall, g, at the surface of the earth from Newton’s law of universal gravitation.
(ii) State two assumptions for which the relationship in 8(b)(i) holds.
(c) Calculate the force of attraction between a star of mass 2.00 x 1030 kg and the earth assuming the star is located 1.50 x 108 km from the earth. [Mass of the earth = 5.98 x 1024kg; G = 6.67 x 10-11N m\(^{2}\) kg-2; g = 10 m s\(^{-2}\)
(d) (i) Define escape velocity.
(ii) State two differences between the acceleration of free fall (g) and the universal gravitational constant (G).
(a)(i) Newton's Law of Universal Gravitation: Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
\[ F = \frac{G m_1 m_2}{r^{2}} \]
(a)(ii) Gravitational field: A gravitational field is a region of space in which a body of mass experiences a force of attraction. Its strength at a point is the force per unit mass acting on a small mass placed at that point.
(b)(i) Relationship between G and g: Consider a body of mass \(m\) resting on the earth's surface. The earth (mass \(M\), radius \(R\)) attracts it with a force which, by Newton's law, is
\[ F = \frac{G M m}{R^{2}}. \]
This same force is the weight of the body, \(F = mg\). Equating the two:
\[ mg = \frac{G M m}{R^{2}} \quad\Rightarrow\quad g = \frac{G M}{R^{2}}. \]
(b)(ii) Assumptions: (1) The earth is a perfect sphere of uniform density, so its whole mass may be taken to act at its centre. (2) The body is small compared with the earth, and effects such as the earth's rotation and air resistance are neglected.
(c) Force between the star and the earth:
\[ r = 1.50\times10^{8}\,\text{km} = 1.50\times10^{11}\,\text{m} \]
\[ F = \frac{G M_{star} M_{earth}}{r^{2}} = \frac{(6.67\times10^{-11})(2.00\times10^{30})(5.98\times10^{24})}{(1.50\times10^{11})^{2}} \]
\[ F = \frac{7.98\times10^{44}}{2.25\times10^{22}} \approx 3.55\times10^{22}\,\text{N}. \]
(d)(i) Escape velocity: The minimum velocity with which a body must be projected from the surface of the earth (or a planet) so that it completely overcomes the gravitational pull and escapes without ever returning. \(v_e = \sqrt{2gR}\).
(d)(ii) Two differences between g and G:
- g (acceleration of free fall) varies from place to place on the earth and from planet to planet, whereas G (universal gravitational constant) has the same value everywhere in the universe.
- g is a vector with unit m s-2, while G is a scalar constant with unit N m2 kg-2.
Answer Details
(a)(i) Newton's Law of Universal Gravitation: Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
\[ F = \frac{G m_1 m_2}{r^{2}} \]
(a)(ii) Gravitational field: A gravitational field is a region of space in which a body of mass experiences a force of attraction. Its strength at a point is the force per unit mass acting on a small mass placed at that point.
(b)(i) Relationship between G and g: Consider a body of mass \(m\) resting on the earth's surface. The earth (mass \(M\), radius \(R\)) attracts it with a force which, by Newton's law, is
\[ F = \frac{G M m}{R^{2}}. \]
This same force is the weight of the body, \(F = mg\). Equating the two:
\[ mg = \frac{G M m}{R^{2}} \quad\Rightarrow\quad g = \frac{G M}{R^{2}}. \]
(b)(ii) Assumptions: (1) The earth is a perfect sphere of uniform density, so its whole mass may be taken to act at its centre. (2) The body is small compared with the earth, and effects such as the earth's rotation and air resistance are neglected.
(c) Force between the star and the earth:
\[ r = 1.50\times10^{8}\,\text{km} = 1.50\times10^{11}\,\text{m} \]
\[ F = \frac{G M_{star} M_{earth}}{r^{2}} = \frac{(6.67\times10^{-11})(2.00\times10^{30})(5.98\times10^{24})}{(1.50\times10^{11})^{2}} \]
\[ F = \frac{7.98\times10^{44}}{2.25\times10^{22}} \approx 3.55\times10^{22}\,\text{N}. \]
(d)(i) Escape velocity: The minimum velocity with which a body must be projected from the surface of the earth (or a planet) so that it completely overcomes the gravitational pull and escapes without ever returning. \(v_e = \sqrt{2gR}\).
(d)(ii) Two differences between g and G:
- g (acceleration of free fall) varies from place to place on the earth and from planet to planet, whereas G (universal gravitational constant) has the same value everywhere in the universe.
- g is a vector with unit m s-2, while G is a scalar constant with unit N m2 kg-2.
Question 5 Report
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The diagram above illustrates a structure of a typical photocell.
(i) Identify each of the parts labelled A and B.
(ii) State one function each of A and B.
(iii) Einstein’s photoelectric equation can be written as \(E = hf - W_o\). State what each of the terms \(E\), \(hf\) and \(W_o\) represent.
(b) A photon is incident on a metal whose work function is \(1.32\ \text{eV}\). An electron is emitted from the surface with a maximum kinetic energy of \(1.97\ \text{eV}\). Calculate the frequency of the photon. \([1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}]\)
(c)(i) Define half-life of a radioactive element.
(ii) Sketch a graph of the relation \(N = N_0e^{-\lambda t}\) and indicate the half-life.
(a)(i) Parts of the photocell (standard evacuated photocell): A is the photocathode (emitter) - a curved metal plate coated with a photosensitive material. B is the anode (collector) - a small metal rod or ring placed in front of the cathode.
(a)(ii) Functions: A (photocathode) emits electrons when light of sufficient frequency falls on it. B (anode) collects the emitted photoelectrons, so completing the circuit and allowing a photocurrent to flow.
(a)(iii) Terms in Einstein's equation \(E = hf - W_0\): \(E\) is the maximum kinetic energy of the emitted photoelectron; \(hf\) is the energy of the incident photon (h = Planck's constant, f = frequency); \(W_0\) is the work function, the minimum energy needed to release an electron from the metal surface.
(b) Frequency of the photon: Photon energy \(= KE_{max} + W_0 = 1.97 + 1.32 = 3.29\,\text{eV}\).
\[ E = 3.29\times1.6\times10^{-19} = 5.264\times10^{-19}\,\text{J}. \]
\[ f = \frac{E}{h} = \frac{5.264\times10^{-19}}{6.6\times10^{-34}} \approx 7.98\times10^{14}\,\text{Hz}. \]
(c)(i) Half-life: The half-life of a radioactive element is the time taken for half the atoms (nuclei) originally present in a sample to decay.
(c)(ii) Graph of \(N = N_0 e^{-\lambda t}\): Plot \(N\) (number of undecayed nuclei) on the vertical axis against time \(t\). The curve starts at \(N_0\) and falls exponentially towards zero. The half-life \(t_{1/2}\) is read on the time axis at the point where \(N = N_0/2\).
Answer Details
(a)(i) Parts of the photocell (standard evacuated photocell): A is the photocathode (emitter) - a curved metal plate coated with a photosensitive material. B is the anode (collector) - a small metal rod or ring placed in front of the cathode.
(a)(ii) Functions: A (photocathode) emits electrons when light of sufficient frequency falls on it. B (anode) collects the emitted photoelectrons, so completing the circuit and allowing a photocurrent to flow.
(a)(iii) Terms in Einstein's equation \(E = hf - W_0\): \(E\) is the maximum kinetic energy of the emitted photoelectron; \(hf\) is the energy of the incident photon (h = Planck's constant, f = frequency); \(W_0\) is the work function, the minimum energy needed to release an electron from the metal surface.
(b) Frequency of the photon: Photon energy \(= KE_{max} + W_0 = 1.97 + 1.32 = 3.29\,\text{eV}\).
\[ E = 3.29\times1.6\times10^{-19} = 5.264\times10^{-19}\,\text{J}. \]
\[ f = \frac{E}{h} = \frac{5.264\times10^{-19}}{6.6\times10^{-34}} \approx 7.98\times10^{14}\,\text{Hz}. \]
(c)(i) Half-life: The half-life of a radioactive element is the time taken for half the atoms (nuclei) originally present in a sample to decay.
(c)(ii) Graph of \(N = N_0 e^{-\lambda t}\): Plot \(N\) (number of undecayed nuclei) on the vertical axis against time \(t\). The curve starts at \(N_0\) and falls exponentially towards zero. The half-life \(t_{1/2}\) is read on the time axis at the point where \(N = N_0/2\).
Question 6 Report
(a) (i) Define atomic spectra.
(ii) Differentiate between emission spectra and absorption spectra.
(b)
The diagram above illustrates an electron transition from energy level \(n = 3\) to \(n = 1\). Calculate the:
(i) energy of the photon
(ii) frequency of the photon
(ii) wavelength of the photon \([h = 6.6 \times 10^{-34}\text{ J s},\ c = 3.0 \times 10^8\text{ ms}^{-1};\ 1\text{ ev} = 1.6 \times 10^{-19}\text{ J}]\)
c)(i)Differentiate between soft x-rays and hard x-rays
(ii) Draw the circuit symbol for a p-n junction diode.
(iiii) Give the reason for doping a semiconductor material
(a)(i) Atomic spectra are the characteristic discrete lines of definite wavelengths or frequencies emitted or absorbed by atoms when electrons move between quantised energy levels.
(a)(ii) An emission spectrum consists of bright lines on a dark background. It is produced when excited electrons fall from higher to lower energy levels, emitting photons. An absorption spectrum consists of dark lines on a bright continuous background. It is produced when atoms absorb photons of particular energies, causing electrons to move from lower to higher energy levels.
(b) Transition from \(n=3\) to \(n=1\)
From the energy-level diagram, \(E_3=-1.51\text{ eV}\) and \(E_1=-13.6\text{ eV}\).
(i) Energy of the photon
\[\Delta E=E_3-E_1=-1.51-(-13.6)=12.09\text{ eV}.\]
\[\Delta E=12.09\times1.6\times10^{-19}=1.934\times10^{-18}\text{ J}.\]
Energy of the photon = \(1.93\times10^{-18}\text{ J}\) \((=12.09\text{ eV})\).
(ii) Frequency of the photon
Using \(E=hf\),
\[f=\frac{E}{h}=\frac{1.934\times10^{-18}}{6.6\times10^{-34}}=2.93\times10^{15}\text{ Hz}.\]
Frequency = \(2.93\times10^{15}\text{ Hz}\).
(iii) Wavelength of the photon
\[\lambda=\frac{c}{f}=\frac{3.0\times10^8}{2.93\times10^{15}}=1.02\times10^{-7}\text{ m}.\]
Wavelength = \(1.02\times10^{-7}\text{ m}\) (about \(102\text{ nm}\)).
(c)(i) Soft X-rays have relatively longer wavelengths, lower frequencies and energies, and low penetrating power. Hard X-rays have shorter wavelengths, higher frequencies and energies, and greater penetrating power.
(c)(ii) Circuit symbol for a p-n junction diode
(c)(iii) A semiconductor is doped to increase the number of charge carriers, thereby increasing its electrical conductivity (or reducing its resistivity).
Answer Details
(a)(i) Atomic spectra are the characteristic discrete lines of definite wavelengths or frequencies emitted or absorbed by atoms when electrons move between quantised energy levels.
(a)(ii) An emission spectrum consists of bright lines on a dark background. It is produced when excited electrons fall from higher to lower energy levels, emitting photons. An absorption spectrum consists of dark lines on a bright continuous background. It is produced when atoms absorb photons of particular energies, causing electrons to move from lower to higher energy levels.
(b) Transition from \(n=3\) to \(n=1\)
From the energy-level diagram, \(E_3=-1.51\text{ eV}\) and \(E_1=-13.6\text{ eV}\).
(i) Energy of the photon
\[\Delta E=E_3-E_1=-1.51-(-13.6)=12.09\text{ eV}.\]
\[\Delta E=12.09\times1.6\times10^{-19}=1.934\times10^{-18}\text{ J}.\]
Energy of the photon = \(1.93\times10^{-18}\text{ J}\) \((=12.09\text{ eV})\).
(ii) Frequency of the photon
Using \(E=hf\),
\[f=\frac{E}{h}=\frac{1.934\times10^{-18}}{6.6\times10^{-34}}=2.93\times10^{15}\text{ Hz}.\]
Frequency = \(2.93\times10^{15}\text{ Hz}\).
(iii) Wavelength of the photon
\[\lambda=\frac{c}{f}=\frac{3.0\times10^8}{2.93\times10^{15}}=1.02\times10^{-7}\text{ m}.\]
Wavelength = \(1.02\times10^{-7}\text{ m}\) (about \(102\text{ nm}\)).
(c)(i) Soft X-rays have relatively longer wavelengths, lower frequencies and energies, and low penetrating power. Hard X-rays have shorter wavelengths, higher frequencies and energies, and greater penetrating power.
(c)(ii) Circuit symbol for a p-n junction diode
(c)(iii) A semiconductor is doped to increase the number of charge carriers, thereby increasing its electrical conductivity (or reducing its resistivity).
Question 7 Report
Explain each of the following terms as used in Electronics.
(a) free electrons;
(b) holes.
(a) Free electrons: These are the outermost (valence) electrons of the atoms of a conductor or semiconductor that are so loosely bound that they become detached from their parent atoms and are able to move about randomly throughout the material. Because they are mobile and carry negative charge, they act as the charge carriers responsible for the flow of electric current when a potential difference is applied.
(b) Holes: A hole is the vacancy (empty space) left behind in the covalent bond structure of a semiconductor when an electron leaves its position. It behaves as a mobile carrier of positive charge, equal in magnitude to the charge on an electron. When a neighbouring electron moves in to fill the vacancy, the hole appears to move in the opposite direction, so holes constitute a flow of positive charge and contribute to conduction (dominant in p-type material).
Answer Details
(a) Free electrons: These are the outermost (valence) electrons of the atoms of a conductor or semiconductor that are so loosely bound that they become detached from their parent atoms and are able to move about randomly throughout the material. Because they are mobile and carry negative charge, they act as the charge carriers responsible for the flow of electric current when a potential difference is applied.
(b) Holes: A hole is the vacancy (empty space) left behind in the covalent bond structure of a semiconductor when an electron leaves its position. It behaves as a mobile carrier of positive charge, equal in magnitude to the charge on an electron. When a neighbouring electron moves in to fill the vacancy, the hole appears to move in the opposite direction, so holes constitute a flow of positive charge and contribute to conduction (dominant in p-type material).
Question 8 Report
A stone of mass 20g is released from a catapult whose rubber is stretched through 5cm. If the force constant of the rubber is 200Nm\(^{-1}\), calculate the speed with which the stone leaves the catapult.
The stretched rubber stores elastic potential energy, which is converted into the kinetic energy of the stone when it is released.
Data: mass \(m = 20\ \text{g} = 0.02\ \text{kg}\); extension \(x = 5\ \text{cm} = 0.05\ \text{m}\); force constant \(k = 200\ \text{Nm}^{-1}\).
Elastic potential energy stored in the rubber:
\[ E = \tfrac{1}{2}kx^{2} = \tfrac{1}{2}\times200\times(0.05)^{2} = \tfrac{1}{2}\times200\times0.0025 = 0.25\ \text{J} \]By conservation of energy, this equals the kinetic energy of the stone as it leaves the catapult:
\[ \tfrac{1}{2}mv^{2} = 0.25 \] \[ v^{2} = \frac{2\times0.25}{0.02} = \frac{0.5}{0.02} = 25 \] \[ v = \sqrt{25} = 5\ \text{ms}^{-1} \]The stone leaves the catapult with a speed of \(5\ \text{ms}^{-1}\).
Answer Details
The stretched rubber stores elastic potential energy, which is converted into the kinetic energy of the stone when it is released.
Data: mass \(m = 20\ \text{g} = 0.02\ \text{kg}\); extension \(x = 5\ \text{cm} = 0.05\ \text{m}\); force constant \(k = 200\ \text{Nm}^{-1}\).
Elastic potential energy stored in the rubber:
\[ E = \tfrac{1}{2}kx^{2} = \tfrac{1}{2}\times200\times(0.05)^{2} = \tfrac{1}{2}\times200\times0.0025 = 0.25\ \text{J} \]By conservation of energy, this equals the kinetic energy of the stone as it leaves the catapult:
\[ \tfrac{1}{2}mv^{2} = 0.25 \] \[ v^{2} = \frac{2\times0.25}{0.02} = \frac{0.5}{0.02} = 25 \] \[ v = \sqrt{25} = 5\ \text{ms}^{-1} \]The stone leaves the catapult with a speed of \(5\ \text{ms}^{-1}\).
Question 9 Report
List three magnetic elements that determine the earth’s magnetic field at a point.
The three magnetic elements that fully determine the earth's magnetic field at a point are:
- The angle of declination - the angle between the geographic (true) meridian and the magnetic meridian at the point.
- The angle of dip (inclination) - the angle that the total earth's magnetic field makes with the horizontal at the point.
- The horizontal component of the earth's magnetic field intensity \((B_H)\) - the strength of the horizontal part of the field at the point.
Together these three quantities specify both the direction and the magnitude of the earth's field at any location.
Answer Details
The three magnetic elements that fully determine the earth's magnetic field at a point are:
- The angle of declination - the angle between the geographic (true) meridian and the magnetic meridian at the point.
- The angle of dip (inclination) - the angle that the total earth's magnetic field makes with the horizontal at the point.
- The horizontal component of the earth's magnetic field intensity \((B_H)\) - the strength of the horizontal part of the field at the point.
Together these three quantities specify both the direction and the magnitude of the earth's field at any location.
Question 10 Report
(a) (i) Define force and state its S.I unit.
(ii) List the two types of solid friction.
(b) A car travelling at a constant speed of \(30\ \mathrm{ms}^{-1}\) for \(20\ \mathrm{s}\) was suddenly decelerated when the driver sighted a pot-hole. It took the driver \(6\ \mathrm{s}\) to get to the pot-hole with a reduced speed of \(18\ \mathrm{ms}^{-1}\). He maintained the steady speed for another \(10\ \mathrm{s}\) to cross the pot-hole. The brakes were then applied and the car came to rest \(5\ \mathrm{s}\) later.
(i) Draw the velocity-time graph for the journey.
(ii) Calculate the deceleration during the last \(5\ \mathrm{s}\) of the journey.
(iii) Calculate the total distance covered.
(a) (i) Force is a push or pull which changes, or tends to change, the state of rest or uniform motion of a body in a straight line.
The S.I. unit of force is the newton (N).
(ii) The two types of solid friction are:
- Static friction;
- Dynamic (sliding or kinetic) friction.
(b) (i) Velocity-time graph
The successive points plotted are t time tand tvelocity tas follows: \[(0,30),\ (20,30),\ (26,18),\ (36,18),\ (41,0).\]
(ii) Deceleration during the last 5 s
For the final stage,
\[ a=\frac{v-u}{t}=\frac{0-18}{5}=-3.6\ \text{m s}^{-2}. \]
Therefore, the deceleration is \(3.6\ \text{m s}^{-2}\).
(iii) Total distance covered
The total distance is the area under the velocity-time graph.
| Part of journey | Area under graph | Distance (m) |
|---|---|---|
| \(0\) to \(20\) s | \(30\times20\) | \(600\) |
| \(20\) to \(26\) s | \(\frac{1}{2}(30+18)\times6\) | \(144\) |
| \(26\) to \(36\) s | \(18\times10\) | \(180\) |
| \(36\) to \(41\) s | \(\frac{1}{2}\times18\times5\) | \(45\) |
\[ \text{Total distance}=600+144+180+45=\boxed{969\ \text{m}}. \]
Answer Details
(a) (i) Force is a push or pull which changes, or tends to change, the state of rest or uniform motion of a body in a straight line.
The S.I. unit of force is the newton (N).
(ii) The two types of solid friction are:
- Static friction;
- Dynamic (sliding or kinetic) friction.
(b) (i) Velocity-time graph
The successive points plotted are t time tand tvelocity tas follows: \[(0,30),\ (20,30),\ (26,18),\ (36,18),\ (41,0).\]
(ii) Deceleration during the last 5 s
For the final stage,
\[ a=\frac{v-u}{t}=\frac{0-18}{5}=-3.6\ \text{m s}^{-2}. \]
Therefore, the deceleration is \(3.6\ \text{m s}^{-2}\).
(iii) Total distance covered
The total distance is the area under the velocity-time graph.
| Part of journey | Area under graph | Distance (m) |
|---|---|---|
| \(0\) to \(20\) s | \(30\times20\) | \(600\) |
| \(20\) to \(26\) s | \(\frac{1}{2}(30+18)\times6\) | \(144\) |
| \(26\) to \(36\) s | \(18\times10\) | \(180\) |
| \(36\) to \(41\) s | \(\frac{1}{2}\times18\times5\) | \(45\) |
\[ \text{Total distance}=600+144+180+45=\boxed{969\ \text{m}}. \]
Question 11 Report
(a) Define diffusion
(b) State two factors that affect the rate of diffusion
(a) Diffusion: Diffusion is the net movement of molecules of a substance from a region of higher concentration to a region of lower concentration, as a result of the random motion of the molecules, until the molecules are evenly distributed.
(b) Two factors that affect the rate of diffusion:
- Temperature - a higher temperature gives the molecules greater average kinetic energy, so they move faster and diffuse more rapidly.
- Relative molecular mass (density) of the substance - lighter (less dense) molecules move faster and therefore diffuse more quickly than heavier ones.
(The concentration gradient and the medium of diffusion, gas versus liquid, are also acceptable factors.)
Answer Details
(a) Diffusion: Diffusion is the net movement of molecules of a substance from a region of higher concentration to a region of lower concentration, as a result of the random motion of the molecules, until the molecules are evenly distributed.
(b) Two factors that affect the rate of diffusion:
- Temperature - a higher temperature gives the molecules greater average kinetic energy, so they move faster and diffuse more rapidly.
- Relative molecular mass (density) of the substance - lighter (less dense) molecules move faster and therefore diffuse more quickly than heavier ones.
(The concentration gradient and the medium of diffusion, gas versus liquid, are also acceptable factors.)
Question 12 Report
(a) State the principle of operation of fibre optics.
(b) State two applications of fibre optics in medicine.
(a) Principle of operation of fibre optics: An optical fibre works on the principle of total internal reflection. It consists of a transparent core of high refractive index surrounded by a cladding of lower refractive index. Light entering one end strikes the core-cladding boundary at an angle greater than the critical angle, so it is totally internally reflected. The light is repeatedly reflected along the fibre and is thereby guided from one end to the other with very little loss, even when the fibre is bent.
(b) Two applications of fibre optics in medicine:
- Endoscopy - fibre-optic endoscopes are used to view and examine internal organs and cavities (for example the stomach, intestines and interior of the lungs) without major surgery.
- Keyhole (minimally invasive) surgery and laser delivery - fibres are used to transmit laser light to internal tissues for cutting, cauterising or treating them, and to illuminate the operating site during such procedures.
Answer Details
(a) Principle of operation of fibre optics: An optical fibre works on the principle of total internal reflection. It consists of a transparent core of high refractive index surrounded by a cladding of lower refractive index. Light entering one end strikes the core-cladding boundary at an angle greater than the critical angle, so it is totally internally reflected. The light is repeatedly reflected along the fibre and is thereby guided from one end to the other with very little loss, even when the fibre is bent.
(b) Two applications of fibre optics in medicine:
- Endoscopy - fibre-optic endoscopes are used to view and examine internal organs and cavities (for example the stomach, intestines and interior of the lungs) without major surgery.
- Keyhole (minimally invasive) surgery and laser delivery - fibres are used to transmit laser light to internal tissues for cutting, cauterising or treating them, and to illuminate the operating site during such procedures.
