WAEC SSCE - Physics - 2019 (Essay)
Ibeere 1 Ìròyìn
(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.)
Awọn alaye Idahun
(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.)
Ibeere 2 Ìròyìn
(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).
Awọn alaye Idahun
(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).
Ibeere 3 Ìròyìn
(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 \(= 480\times72 = 34560\,\text{J} = mc\,\Delta\theta\):
\[ c = \frac{34560}{1.5\times60} = 384\,\text{J kg}^{-1}\,\text{K}^{-1}. \]
Awọn alaye Idahun
(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 \(= 480\times72 = 34560\,\text{J} = mc\,\Delta\theta\):
\[ c = \frac{34560}{1.5\times60} = 384\,\text{J kg}^{-1}\,\text{K}^{-1}. \]
Ibeere 4 Ìròyìn
%IMG%
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.
The photocell has the structure shown below.
(a)(i) Identification of the labelled parts
- A - the emitter (photocathode): the large curved metal plate coated with a photosensitive material.
- B - the collector (anode): the small metal rod placed in front of the cathode.
(a)(ii) Function of each part
- A emits electrons when light of sufficient frequency falls on it.
- B attracts and collects the emitted electrons, so completing the circuit and allowing a photocurrent to flow.
(a)(iii) Meaning of the terms in \(E = hf - W_0\)
- \(E\) - the maximum kinetic energy of the emitted photoelectrons.
- \(hf\) - the energy of the incident photon (\(h\) = Planck's constant, \(f\) = frequency).
- \(W_0\) - the work function of the metal, the minimum energy needed to release an electron from its surface.
(b) Frequency of the photon
The photon energy equals the maximum kinetic energy of the electron plus the work function:
\[ hf = KE_{max} + W_0 = 1.97 + 1.32 = 3.29\ \text{eV} \]Converting to joules:
\[ hf = 3.29 \times 1.6 \times 10^{-19} = 5.264 \times 10^{-19}\ \text{J} \]Therefore the frequency is:
\[ f = \frac{hf}{h} = \frac{5.264 \times 10^{-19}}{6.6 \times 10^{-34}} = 7.98 \times 10^{14}\ \text{Hz} \](c)(i) Definition of half-life
The half-life of a radioactive element is the time taken for half the nuclei (atoms) originally present in a sample to decay.
(c)(ii) Graph of \(N = N_0 e^{-\lambda t}\)
Taking an initial number \(N_0 = 800\) undecayed nuclei, the following readings of \(N\) against time \(t\) are obtained:
| Time \(t\) (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| Undecayed nuclei \(N\) | 800 | 566 | 400 | 283 | 200 | 141 | 100 | 71 | 50 |
Plotting these readings gives the exponential decay curve:
The half-life is read where the number of undecayed nuclei falls to \(N_0/2 = 400\). From the table and graph this occurs at \(t = 2\ \text{s}\); the count halves again to 200 at \(t = 4\ \text{s}\) and to 100 at \(t = 6\ \text{s}\), confirming a constant half-life:
\[ t_{1/2} = 2\ \text{s} \]This value is indicated on the graph by the dashed lines drawn from \(N = 400\) across to the curve and down to the time axis.
Awọn alaye Idahun
The photocell has the structure shown below.
(a)(i) Identification of the labelled parts
- A - the emitter (photocathode): the large curved metal plate coated with a photosensitive material.
- B - the collector (anode): the small metal rod placed in front of the cathode.
(a)(ii) Function of each part
- A emits electrons when light of sufficient frequency falls on it.
- B attracts and collects the emitted electrons, so completing the circuit and allowing a photocurrent to flow.
(a)(iii) Meaning of the terms in \(E = hf - W_0\)
- \(E\) - the maximum kinetic energy of the emitted photoelectrons.
- \(hf\) - the energy of the incident photon (\(h\) = Planck's constant, \(f\) = frequency).
- \(W_0\) - the work function of the metal, the minimum energy needed to release an electron from its surface.
(b) Frequency of the photon
The photon energy equals the maximum kinetic energy of the electron plus the work function:
\[ hf = KE_{max} + W_0 = 1.97 + 1.32 = 3.29\ \text{eV} \]Converting to joules:
\[ hf = 3.29 \times 1.6 \times 10^{-19} = 5.264 \times 10^{-19}\ \text{J} \]Therefore the frequency is:
\[ f = \frac{hf}{h} = \frac{5.264 \times 10^{-19}}{6.6 \times 10^{-34}} = 7.98 \times 10^{14}\ \text{Hz} \](c)(i) Definition of half-life
The half-life of a radioactive element is the time taken for half the nuclei (atoms) originally present in a sample to decay.
(c)(ii) Graph of \(N = N_0 e^{-\lambda t}\)
Taking an initial number \(N_0 = 800\) undecayed nuclei, the following readings of \(N\) against time \(t\) are obtained:
| Time \(t\) (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| Undecayed nuclei \(N\) | 800 | 566 | 400 | 283 | 200 | 141 | 100 | 71 | 50 |
Plotting these readings gives the exponential decay curve:
The half-life is read where the number of undecayed nuclei falls to \(N_0/2 = 400\). From the table and graph this occurs at \(t = 2\ \text{s}\); the count halves again to 200 at \(t = 4\ \text{s}\) and to 100 at \(t = 6\ \text{s}\), confirming a constant half-life:
\[ t_{1/2} = 2\ \text{s} \]This value is indicated on the graph by the dashed lines drawn from \(N = 400\) across to the curve and down to the time axis.
Ibeere 5 Ìròyìn
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 \(5\ \text{ms}^{-1}\).
Awọn alaye Idahun
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 \(5\ \text{ms}^{-1}\).
Ibeere 6 Ìròyìn
(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.
Awọn alaye Idahun
(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.
Ibeere 7 Ìròyìn
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.
Awọn alaye Idahun
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.
Ibeere 8 Ìròyìn
%IMG%
(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, \(P = \sqrt{g\,R_{max}}\).
Awọn alaye Idahun
(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, \(P = \sqrt{g\,R_{max}}\).
Ibeere 9 Ìròyìn
State three observable phenomena in which waves behave like a particle.
Waves (electromagnetic radiation) behave like particles in the following observable phenomena:
- Photoelectric effect - light falling on a metal surface ejects electrons, and the effect depends on the frequency (energy of each photon) rather than the intensity, which is only explicable if light arrives in discrete quanta (photons).
- Compton effect (Compton scattering) - when X-rays are scattered by electrons, the scattered radiation has a longer wavelength. This is explained by treating the X-ray as a particle (photon) that collides elastically with the electron, conserving momentum and energy.
- Pair production - a high-energy photon passing near a nucleus disappears and produces an electron-positron pair, showing that the photon carries a definite, particle-like quantity of energy and momentum.
(Thermionic emission and the existence of line/emission spectra are also acceptable examples.)
Awọn alaye Idahun
Waves (electromagnetic radiation) behave like particles in the following observable phenomena:
- Photoelectric effect - light falling on a metal surface ejects electrons, and the effect depends on the frequency (energy of each photon) rather than the intensity, which is only explicable if light arrives in discrete quanta (photons).
- Compton effect (Compton scattering) - when X-rays are scattered by electrons, the scattered radiation has a longer wavelength. This is explained by treating the X-ray as a particle (photon) that collides elastically with the electron, conserving momentum and energy.
- Pair production - a high-energy photon passing near a nucleus disappears and produces an electron-positron pair, showing that the photon carries a definite, particle-like quantity of energy and momentum.
(Thermionic emission and the existence of line/emission spectra are also acceptable examples.)
Ibeere 10 Ìròyìn
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).
Awọn alaye Idahun
(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).
Ibeere 11 Ìròyìn
(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 | \(\frac{1}{2}\times18\times5\) | \(45\) |
\[ \text{Total distance}=600+144+180+45=\boxed{969\ \text{m}}. \]
Awọn alaye Idahun
(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 | \(\frac{1}{2}\times18\times5\) | \(45\) |
\[ \text{Total distance}=600+144+180+45=\boxed{969\ \text{m}}. \]
Ibeere 12 Ìròyìn
(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.
Awọn alaye Idahun
(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.
