WAEC SSCE - Physics - 2006 (Essay)
Ajụjụ 1 Ripọtì
List three observations in support of the de-Broglie's assumptions that moving particles behave like waves.
Three observations supporting de Broglie’s assumption that moving particles behave like waves are:
- Electrons produce diffraction patterns when passed through crystals or thin metal foils.
- Electrons have measurable wavelengths and frequencies associated with their motion.
- Electrons undergo reflection and refraction, like other waves.
Akọwa Nkọwa
Three observations supporting de Broglie’s assumption that moving particles behave like waves are:
- Electrons produce diffraction patterns when passed through crystals or thin metal foils.
- Electrons have measurable wavelengths and frequencies associated with their motion.
- Electrons undergo reflection and refraction, like other waves.
Ajụjụ 2 Ripọtì
(a) Distinguish between the forces of adhesion and cohension
(b) Give one example each of the forces of adhesion and cohesion.
(a) Adhesion versus cohesion
Adhesive force is the force of attraction between the molecules of two different substances (unlike molecules).
Cohesive force is the force of attraction between the molecules of the same substance (like molecules).
(b) One example each
- Adhesion: water wetting and sticking to the surface of a clean glass (or ink sticking to paper; water clinging to glass walls).
- Cohesion: mercury forming rounded droplets, or water molecules holding together to form a drop (or mercury not wetting glass).
Akọwa Nkọwa
(a) Adhesion versus cohesion
Adhesive force is the force of attraction between the molecules of two different substances (unlike molecules).
Cohesive force is the force of attraction between the molecules of the same substance (like molecules).
(b) One example each
- Adhesion: water wetting and sticking to the surface of a clean glass (or ink sticking to paper; water clinging to glass walls).
- Cohesion: mercury forming rounded droplets, or water molecules holding together to form a drop (or mercury not wetting glass).
Ajụjụ 3 Ripọtì
A particle is projected horizontally at 10 ms\(^{-1}\) from a height of 45m. Calculate the horizontal distance covered by the particle before hitting the ground. [g = 10ms\(^{-1}\)]
Horizontal projectile
Horizontal velocity \(u = 10\ \text{m s}^{-1}\), height \(h = 45\ \text{m}\), \(g = 10\ \text{m s}^{-2}\).
Time to fall (vertical motion, initial vertical velocity zero):
\(h = \tfrac{1}{2}g t^{2} \Rightarrow 45 = \tfrac{1}{2}\times10\times t^{2} = 5t^{2}\).
\(t^{2} = 9 \Rightarrow t = 3\ \text{s}\).
Horizontal distance (range): \(R = u\,t = 10\times3 = 30\ \text{m}\).
The particle covers a horizontal distance of 30 m before hitting the ground.
Akọwa Nkọwa
Horizontal projectile
Horizontal velocity \(u = 10\ \text{m s}^{-1}\), height \(h = 45\ \text{m}\), \(g = 10\ \text{m s}^{-2}\).
Time to fall (vertical motion, initial vertical velocity zero):
\(h = \tfrac{1}{2}g t^{2} \Rightarrow 45 = \tfrac{1}{2}\times10\times t^{2} = 5t^{2}\).
\(t^{2} = 9 \Rightarrow t = 3\ \text{s}\).
Horizontal distance (range): \(R = u\,t = 10\times3 = 30\ \text{m}\).
The particle covers a horizontal distance of 30 m before hitting the ground.
Ajụjụ 4 Ripọtì
A particle dropped from a vertical height and falls freely for a time interval t. Sketch and explain a graph to show how h varies with (a) t (b) t\(^{2}\).
Key idea: here \(h\) is the particle’s vertical height above the ground, not the distance it has fallen. If it is released from an initial height \(H\), then
\[h=H-\frac{1}{2}gt^2.\]
(a) Graph of \(h\) against \(t\)
The graph starts at \(h=H\) when \(t=0\). It curves downwards and becomes progressively steeper because the particle’s downward speed increases under the constant acceleration \(g\). It reaches \(h=0\) when the particle reaches the ground.
(b) Graph of \(h\) against \(t^2\)
Rearranging the equation gives
\[h=H+\left(-\frac{g}{2}\right)t^2.\]
This has the form \(y=c+mx\), so the graph is a straight line with:
- vertical intercept \(H\);
- negative gradient \(-\frac{g}{2}\).
Exam reminder: distinguish carefully between height above the ground, \(h=H-\frac12gt^2\), and distance fallen, \(s=\frac12gt^2\). Only distance fallen gives a rising graph through the origin.
Akọwa Nkọwa
Key idea: here \(h\) is the particle’s vertical height above the ground, not the distance it has fallen. If it is released from an initial height \(H\), then
\[h=H-\frac{1}{2}gt^2.\]
(a) Graph of \(h\) against \(t\)
The graph starts at \(h=H\) when \(t=0\). It curves downwards and becomes progressively steeper because the particle’s downward speed increases under the constant acceleration \(g\). It reaches \(h=0\) when the particle reaches the ground.
(b) Graph of \(h\) against \(t^2\)
Rearranging the equation gives
\[h=H+\left(-\frac{g}{2}\right)t^2.\]
This has the form \(y=c+mx\), so the graph is a straight line with:
- vertical intercept \(H\);
- negative gradient \(-\frac{g}{2}\).
Exam reminder: distinguish carefully between height above the ground, \(h=H-\frac12gt^2\), and distance fallen, \(s=\frac12gt^2\). Only distance fallen gives a rising graph through the origin.
Ajụjụ 5 Ripọtì
(a) With the aid of ray diagrams, explain total internal reflection.
(b) Describe, with the aid of a labelled diagram, the essential features of an astronomical telescope in normal adjustment.
(c) A converging lens forms a real image of a real object. If the magnification is 2 and the distance between the image and the object is 90.0 cm, determine the
(i) focal length of the lens;
(ii) object distance for which the image would be the same size as the object.
(a) Total internal reflection
Total internal reflection occurs when light travels from an optically denser medium to a less dense medium and the angle of incidence is greater than the critical angle, C. At i = C, the angle of refraction is \(90^\circ\), so that the refracted ray travels along the boundary. For \(i>C\), the ray is completely reflected back into the denser medium.
(b) Astronomical telescope in normal adjustment
An astronomical telescope consists of two converging lenses mounted on the same principal axis:
- the objective lens, of long focal length \(f_o\);
- the eyepiece lens, of short focal length \(f_e\).
Light from a distant object enters the objective as parallel rays. The objective forms a real, inverted and diminished intermediate image at its principal focus. In normal adjustment, this image is at the first focal point of the eyepiece, so that the principal foci coincide. The eyepiece then produces a final virtual image at infinity, and the emergent rays are parallel.
Hence, for normal adjustment,
\[d=f_o+f_e\]
where \(d\) is the separation of the lenses. The angular magnifying power has magnitude
\[M=\frac{f_o}{f_e}.\]
(c)
For a real image formed by a converging lens,
\[m=\frac{v}{u}=2\]
Therefore,
\[v=2u\]
Since the object and real image are on opposite sides of the lens, their separation is \(u+v\):
\[u+v=90.0\]
\[u+2u=90.0\]
\[3u=90.0\]
\[u=30.0\text{ cm},\qquad v=60.0\text{ cm}.\]
(i) Focal length
Using \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\),
\[\frac{1}{f}=\frac{1}{30.0}+\frac{1}{60.0}=\frac{3}{60.0}\]
\[\boxed{f=20.0\text{ cm}}\]
(ii) Object distance for an image of the same size
An image is the same size as the object when the object is at \(2f\).
\[u=2f=2(20.0)=\boxed{40.0\text{ cm}}\]
Akọwa Nkọwa
(a) Total internal reflection
Total internal reflection occurs when light travels from an optically denser medium to a less dense medium and the angle of incidence is greater than the critical angle, C. At i = C, the angle of refraction is \(90^\circ\), so that the refracted ray travels along the boundary. For \(i>C\), the ray is completely reflected back into the denser medium.
(b) Astronomical telescope in normal adjustment
An astronomical telescope consists of two converging lenses mounted on the same principal axis:
- the objective lens, of long focal length \(f_o\);
- the eyepiece lens, of short focal length \(f_e\).
Light from a distant object enters the objective as parallel rays. The objective forms a real, inverted and diminished intermediate image at its principal focus. In normal adjustment, this image is at the first focal point of the eyepiece, so that the principal foci coincide. The eyepiece then produces a final virtual image at infinity, and the emergent rays are parallel.
Hence, for normal adjustment,
\[d=f_o+f_e\]
where \(d\) is the separation of the lenses. The angular magnifying power has magnitude
\[M=\frac{f_o}{f_e}.\]
(c)
For a real image formed by a converging lens,
\[m=\frac{v}{u}=2\]
Therefore,
\[v=2u\]
Since the object and real image are on opposite sides of the lens, their separation is \(u+v\):
\[u+v=90.0\]
\[u+2u=90.0\]
\[3u=90.0\]
\[u=30.0\text{ cm},\qquad v=60.0\text{ cm}.\]
(i) Focal length
Using \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\),
\[\frac{1}{f}=\frac{1}{30.0}+\frac{1}{60.0}=\frac{3}{60.0}\]
\[\boxed{f=20.0\text{ cm}}\]
(ii) Object distance for an image of the same size
An image is the same size as the object when the object is at \(2f\).
\[u=2f=2(20.0)=\boxed{40.0\text{ cm}}\]
Ajụjụ 6 Ripọtì
The uncertinty in determining the duration during which an electron remains in a particular energy level before returning to the ground state is 2.0 x 10\(^{-9}\)s. Calculate the uncertainty in determining its energy at that level [Take \(\frac{h}{2\pi} = h = 1.054 \times 10^{-34}\) Js]
The Heisenberg uncertainty principle relating energy and time is:
\[ \Delta E \cdot \Delta t \geq \frac{h}{2\pi} \]Taking the equality and making \(\Delta E\) the subject:
\[ \Delta E = \frac{h/2\pi}{\Delta t} = \frac{1.054 \times 10^{-34}}{2.0 \times 10^{-9}} \] \[ \Delta E = 5.27 \times 10^{-26}\ \text{J} \]The uncertainty in the energy of the level is \(5.27 \times 10^{-26}\ \text{J}\).
In electron-volts, \(\Delta E = \dfrac{5.27 \times 10^{-26}}{1.6 \times 10^{-19}} = 3.29 \times 10^{-7}\ \text{eV}\).
Akọwa Nkọwa
The Heisenberg uncertainty principle relating energy and time is:
\[ \Delta E \cdot \Delta t \geq \frac{h}{2\pi} \]Taking the equality and making \(\Delta E\) the subject:
\[ \Delta E = \frac{h/2\pi}{\Delta t} = \frac{1.054 \times 10^{-34}}{2.0 \times 10^{-9}} \] \[ \Delta E = 5.27 \times 10^{-26}\ \text{J} \]The uncertainty in the energy of the level is \(5.27 \times 10^{-26}\ \text{J}\).
In electron-volts, \(\Delta E = \dfrac{5.27 \times 10^{-26}}{1.6 \times 10^{-19}} = 3.29 \times 10^{-7}\ \text{eV}\).
Ajụjụ 7 Ripọtì
(a) State two applications of electrolysis.
(b) Explain what is meant by the electrochemical equivalent of copper is 3.3 x 10\(^{-7}\) kgC\(^{-1}\)
(a) Two applications of electrolysis
- Electroplating (coating an object with a layer of metal, e.g. chromium or silver plating).
- Purification (refining) of metals such as copper.
(Other acceptable uses: extraction of reactive metals like aluminium; electrotyping; production of chemicals such as chlorine and caustic soda.)
(b) Meaning of "the electrochemical equivalent of copper is \(3.3\times10^{-7}\ \text{kg C}^{-1}\)"
The electrochemical equivalent (\(z\)) of copper is the mass of copper deposited (or liberated) by a charge of one coulomb during electrolysis. So the statement means that a charge of 1 coulomb passing through a copper solution deposits \(3.3\times10^{-7}\ \text{kg}\) of copper. From Faraday's first law, \(m = z\,It\), where \(I\) is the current and \(t\) the time.
Akọwa Nkọwa
(a) Two applications of electrolysis
- Electroplating (coating an object with a layer of metal, e.g. chromium or silver plating).
- Purification (refining) of metals such as copper.
(Other acceptable uses: extraction of reactive metals like aluminium; electrotyping; production of chemicals such as chlorine and caustic soda.)
(b) Meaning of "the electrochemical equivalent of copper is \(3.3\times10^{-7}\ \text{kg C}^{-1}\)"
The electrochemical equivalent (\(z\)) of copper is the mass of copper deposited (or liberated) by a charge of one coulomb during electrolysis. So the statement means that a charge of 1 coulomb passing through a copper solution deposits \(3.3\times10^{-7}\ \text{kg}\) of copper. From Faraday's first law, \(m = z\,It\), where \(I\) is the current and \(t\) the time.
Ajụjụ 8 Ripọtì
The diagram below represents the graph of the force applied in stretching a spiral spring against the corresponding extension produced within its elastic limit.
Using the notations on the graph, determine the:
(a) force constant of the spring;
(b) work done in stretching the spring from 10 x 10\(^{-2}\)m to 20 x 10\(^{-2}\)m.
Within the elastic limit a spiral spring obeys Hooke's law, so the force applied is directly proportional to the extension produced. The force-extension graph is therefore a straight line passing through the origin, as shown below.
Reading the notations on the graph, the line passes through the points: extension \(5\times10^{-2}\) m at force \(15\) N, \(10\times10^{-2}\) m at \(30\) N, \(15\times10^{-2}\) m at \(45\) N, and \(20\times10^{-2}\) m at \(60\) N.
(a) Force constant of the spring
The force constant \(k\) is the slope (gradient) of the force-extension graph:
\[ k = \frac{\Delta F}{\Delta e} = \frac{60 - 0}{(20 - 0)\times10^{-2}} = \frac{60}{20\times10^{-2}} = \frac{60}{0.20} = 300\ \text{N m}^{-1} \]Any point on the line gives the same value, e.g. \(30 / 0.10 = 300\ \text{N m}^{-1}\). Hence the force constant is 300 N m\(^{-1}\).
(b) Work done in stretching from \(10\times10^{-2}\) m to \(20\times10^{-2}\) m
The work done equals the area under the force-extension graph between the two extensions. From the graph, at \(e_1 = 10\times10^{-2}\) m the force is \(F_1 = 30\) N, and at \(e_2 = 20\times10^{-2}\) m the force is \(F_2 = 60\) N. This shaded region is a trapezium:
\[ W = \tfrac{1}{2}(F_1 + F_2)(e_2 - e_1) = \tfrac{1}{2}(30 + 60)\,(20 - 10)\times10^{-2} \]\[ W = \tfrac{1}{2}(90)(10\times10^{-2}) = \tfrac{1}{2}(90)(0.10) = 4.5\ \text{J} \]Equivalently, using \(W = \tfrac{1}{2}k(e_2^{2} - e_1^{2})\):
\[ W = \tfrac{1}{2}(300)\big[(0.20)^{2} - (0.10)^{2}\big] = 150\,(0.04 - 0.01) = 150 \times 0.03 = 4.5\ \text{J} \]The work done in stretching the spring is 4.5 J.
Akọwa Nkọwa
Within the elastic limit a spiral spring obeys Hooke's law, so the force applied is directly proportional to the extension produced. The force-extension graph is therefore a straight line passing through the origin, as shown below.
Reading the notations on the graph, the line passes through the points: extension \(5\times10^{-2}\) m at force \(15\) N, \(10\times10^{-2}\) m at \(30\) N, \(15\times10^{-2}\) m at \(45\) N, and \(20\times10^{-2}\) m at \(60\) N.
(a) Force constant of the spring
The force constant \(k\) is the slope (gradient) of the force-extension graph:
\[ k = \frac{\Delta F}{\Delta e} = \frac{60 - 0}{(20 - 0)\times10^{-2}} = \frac{60}{20\times10^{-2}} = \frac{60}{0.20} = 300\ \text{N m}^{-1} \]Any point on the line gives the same value, e.g. \(30 / 0.10 = 300\ \text{N m}^{-1}\). Hence the force constant is 300 N m\(^{-1}\).
(b) Work done in stretching from \(10\times10^{-2}\) m to \(20\times10^{-2}\) m
The work done equals the area under the force-extension graph between the two extensions. From the graph, at \(e_1 = 10\times10^{-2}\) m the force is \(F_1 = 30\) N, and at \(e_2 = 20\times10^{-2}\) m the force is \(F_2 = 60\) N. This shaded region is a trapezium:
\[ W = \tfrac{1}{2}(F_1 + F_2)(e_2 - e_1) = \tfrac{1}{2}(30 + 60)\,(20 - 10)\times10^{-2} \]\[ W = \tfrac{1}{2}(90)(10\times10^{-2}) = \tfrac{1}{2}(90)(0.10) = 4.5\ \text{J} \]Equivalently, using \(W = \tfrac{1}{2}k(e_2^{2} - e_1^{2})\):
\[ W = \tfrac{1}{2}(300)\big[(0.20)^{2} - (0.10)^{2}\big] = 150\,(0.04 - 0.01) = 150 \times 0.03 = 4.5\ \text{J} \]The work done in stretching the spring is 4.5 J.
Ajụjụ 9 Ripọtì
(a)(i) Name and explain the common defects of a primary cell.
(ii) State two advantages of a secondary cell over a primary cell.
(b) Draw a labelled diagram to show the essential parts of a dry leclanche cell.
(c)(i) Explain why six accumulators each of e.m.f 2V connected in series can be used to start the engine of a car whereas eight dry cells each of e.m.f 1.5 V connected in series cannot be used.
(ii) Name the materials used for the positive terminal, the negative terminal and the electrolyte in a
I. leclanche cell;
II. charged lead acid accumulator.
(a)(i) Common defects of a primary (simple) cell:
- Polarisation: hydrogen gas liberated at the positive (copper) plate collects there as a layer of bubbles. This layer increases the internal resistance of the cell and sets up a back e.m.f. that opposes the working e.m.f., so the current falls rapidly. It is reduced by adding a depolariser (e.g. potassium dichromate or manganese(IV) oxide) which oxidises the hydrogen to water.
- Local action: impurities (carbon, iron) in the zinc plate form many tiny short-circuited local cells with the zinc, so the zinc is slowly eaten away and current is wasted even when the cell is not in use. It is prevented by amalgamating the zinc, i.e. coating it with a thin film of mercury so that the impurities are covered and cannot act.
(a)(ii) Two advantages of a secondary cell over a primary cell:
- A secondary cell can be recharged and used again many times, whereas a primary cell cannot be recharged.
- A secondary cell has a very low internal resistance, so it can deliver a large current and supply current for a much longer working life than a primary cell.
(b) Labelled diagram of a dry Leclanche cell:
The essential parts are the outer zinc case (negative electrode), the central carbon rod (positive electrode) surrounded by a paste of manganese(IV) oxide and powdered carbon (the depolariser), the ammonium chloride paste (electrolyte), and the pitch/wax seal with a brass cap on top.
(c)(i) The eight dry cells and the six accumulators give the same total e.m.f.: \(8 \times 1.5 = 12\ \text{V}\) and \(6 \times 2 = 12\ \text{V}\). What differs is the internal resistance. A dry (Leclanche) cell has a very high internal resistance, so eight of them in series have a large total internal resistance. When such a battery is connected to the very small resistance of a starter motor, most of the e.m.f. is dropped across the cells' own internal resistance and only a small current \(\left(I = \dfrac{E}{R + r}\right)\) flows, too small to turn the engine. Each accumulator has a very low internal resistance, so six in series can force the very large current needed through the low-resistance starter motor and start the engine.
(c)(ii) Materials used:
| Positive terminal | Negative terminal | Electrolyte | |
| I. Leclanche cell | Carbon rod (surrounded by manganese(IV) oxide, \(\text{MnO}_2\), as depolariser) | Zinc (the case) | Ammonium chloride, \(\text{NH}_4\text{Cl}\) (paste/solution) |
| II. Charged lead-acid accumulator | Lead(IV) oxide, \(\text{PbO}_2\) (lead peroxide) | Spongy lead, \(\text{Pb}\) | Dilute tetraoxosulphate(VI) acid, \(\text{H}_2\text{SO}_4\) (sulphuric acid) |
Akọwa Nkọwa
(a)(i) Common defects of a primary (simple) cell:
- Polarisation: hydrogen gas liberated at the positive (copper) plate collects there as a layer of bubbles. This layer increases the internal resistance of the cell and sets up a back e.m.f. that opposes the working e.m.f., so the current falls rapidly. It is reduced by adding a depolariser (e.g. potassium dichromate or manganese(IV) oxide) which oxidises the hydrogen to water.
- Local action: impurities (carbon, iron) in the zinc plate form many tiny short-circuited local cells with the zinc, so the zinc is slowly eaten away and current is wasted even when the cell is not in use. It is prevented by amalgamating the zinc, i.e. coating it with a thin film of mercury so that the impurities are covered and cannot act.
(a)(ii) Two advantages of a secondary cell over a primary cell:
- A secondary cell can be recharged and used again many times, whereas a primary cell cannot be recharged.
- A secondary cell has a very low internal resistance, so it can deliver a large current and supply current for a much longer working life than a primary cell.
(b) Labelled diagram of a dry Leclanche cell:
The essential parts are the outer zinc case (negative electrode), the central carbon rod (positive electrode) surrounded by a paste of manganese(IV) oxide and powdered carbon (the depolariser), the ammonium chloride paste (electrolyte), and the pitch/wax seal with a brass cap on top.
(c)(i) The eight dry cells and the six accumulators give the same total e.m.f.: \(8 \times 1.5 = 12\ \text{V}\) and \(6 \times 2 = 12\ \text{V}\). What differs is the internal resistance. A dry (Leclanche) cell has a very high internal resistance, so eight of them in series have a large total internal resistance. When such a battery is connected to the very small resistance of a starter motor, most of the e.m.f. is dropped across the cells' own internal resistance and only a small current \(\left(I = \dfrac{E}{R + r}\right)\) flows, too small to turn the engine. Each accumulator has a very low internal resistance, so six in series can force the very large current needed through the low-resistance starter motor and start the engine.
(c)(ii) Materials used:
| Positive terminal | Negative terminal | Electrolyte | |
| I. Leclanche cell | Carbon rod (surrounded by manganese(IV) oxide, \(\text{MnO}_2\), as depolariser) | Zinc (the case) | Ammonium chloride, \(\text{NH}_4\text{Cl}\) (paste/solution) |
| II. Charged lead-acid accumulator | Lead(IV) oxide, \(\text{PbO}_2\) (lead peroxide) | Spongy lead, \(\text{Pb}\) | Dilute tetraoxosulphate(VI) acid, \(\text{H}_2\text{SO}_4\) (sulphuric acid) |
Ajụjụ 10 Ripọtì
(a) Define
(i) proton number;
(ii) nucleon number;
(iii) isotopes.
(b) A nuclide \(^A_ZX\) emits \(\beta\)-particle to form a daughter nuclide Y. Write a nuclear equation to illustrate the charge conservation.
(c) The radioactive nuclei \(^{210}_{84}P_o\) emits an \(\alpha\) - particle to produce \(^{206}_{82}P_b\). Calculate the energy, in MeV, released in each disintegration.
Take the masses of \(^{210}_{84}P_o\) = 209.936730 u;
\(^{206}_{82}P_b\) = 205.929421 u;
\(^{4}_{2}He\) = 4.001504 u;
and that 1u = 931 MeV
(a)(i) Proton number (atomic number): the number of protons in the nucleus of an atom.
(ii) Nucleon number (mass number): the total number of protons and neutrons in the nucleus of an atom.
(iii) Isotopes: atoms of the same element that have the same proton number but different nucleon numbers (same number of protons, different numbers of neutrons).
(b) Beta emission converts a neutron into a proton, so the proton number rises by 1 while the nucleon number is unchanged:
\[ {}^{A}_{Z}X \rightarrow {}^{A}_{Z+1}Y + {}^{0}_{-1}e \]Charge conservation: \(Z = (Z+1) + (-1)\), which balances.
(c) Mass defect in the disintegration:
\[ \Delta m = m({}^{210}_{84}Po) - \left[ m({}^{206}_{82}Pb) + m({}^{4}_{2}He) \right] \] \[ \Delta m = 209.936730 - (205.929421 + 4.001504) \] \[ \Delta m = 209.936730 - 209.930925 = 0.005805\ \text{u} \]Energy released:
\[ E = \Delta m \times 931 = 0.005805 \times 931 \] \[ E = 5.40\ \text{MeV} \]The energy released in each disintegration is about \(5.40\ \text{MeV}\).
Akọwa Nkọwa
(a)(i) Proton number (atomic number): the number of protons in the nucleus of an atom.
(ii) Nucleon number (mass number): the total number of protons and neutrons in the nucleus of an atom.
(iii) Isotopes: atoms of the same element that have the same proton number but different nucleon numbers (same number of protons, different numbers of neutrons).
(b) Beta emission converts a neutron into a proton, so the proton number rises by 1 while the nucleon number is unchanged:
\[ {}^{A}_{Z}X \rightarrow {}^{A}_{Z+1}Y + {}^{0}_{-1}e \]Charge conservation: \(Z = (Z+1) + (-1)\), which balances.
(c) Mass defect in the disintegration:
\[ \Delta m = m({}^{210}_{84}Po) - \left[ m({}^{206}_{82}Pb) + m({}^{4}_{2}He) \right] \] \[ \Delta m = 209.936730 - (205.929421 + 4.001504) \] \[ \Delta m = 209.936730 - 209.930925 = 0.005805\ \text{u} \]Energy released:
\[ E = \Delta m \times 931 = 0.005805 \times 931 \] \[ E = 5.40\ \text{MeV} \]The energy released in each disintegration is about \(5.40\ \text{MeV}\).
Ajụjụ 11 Ripọtì
(a) State the conditions for the equilibrium of a rigid body acted upon by parallel forces.
(b)(i) Describe an experiment to determine the mass of a metre rule using the principle of moments.
(ii) State two precautions necessary to obtain accurate results in the experiment described in (b)(i) above.
(c) A bullet of mass 120 g is fired horizontally into a fixed wooden block with a speed of 20 ms\(^{-1}\). If the bullet is brought to rest in the block in 0.1s by a constant resistance, calculate the (i) magnitude of the resistance; (ii) distance moved by the bullet in the wood.
(a) Conditions for equilibrium under parallel forces:
- The sum of the forces in one direction equals the sum of the forces in the opposite direction (net force = 0).
- The sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point (net moment = 0).
(b)(i) Experiment to find the mass of a metre rule: Suspend the metre rule from a knife-edge and find its centre of gravity \(G\) (the balance point with no load). Hang a known mass \(m\) at a distance \(d_1\) on one side of a chosen pivot, and adjust the position of the pivot until the rule balances horizontally. Measure the distance \(d_2\) from the pivot to \(G\) (through which the whole weight \(W = Mg\) of the rule acts). Taking moments about the pivot:
\[ m g \times d_1 = M g \times d_2 \Rightarrow M = \frac{m \, d_1}{d_2} \]which gives the mass \(M\) of the rule.
(ii) Two precautions:
- Ensure the rule is horizontal (balanced) before taking readings.
- Avoid parallax error when reading the ruler; ensure the knife-edge is sharp and the mass is firmly fixed.
(c) Bullet mass \(m = 120\ \text{g} = 0.12\ \text{kg}\), \(u = 20\ \text{ms}^{-1}\), \(v = 0\), \(t = 0.1\ \text{s}\).
(i) Resistance (retarding force):
\[ F = \frac{m(u - v)}{t} = \frac{0.12 \times (20 - 0)}{0.1} = \frac{2.4}{0.1} = 24\ \text{N} \](ii) Distance moved in the wood (average velocity method):
\[ s = \left(\frac{u + v}{2}\right) t = \left(\frac{20 + 0}{2}\right) \times 0.1 = 10 \times 0.1 = 1.0\ \text{m} \]The resistance is \(24\ \text{N}\) and the bullet travels \(1.0\ \text{m}\) into the block.
Akọwa Nkọwa
(a) Conditions for equilibrium under parallel forces:
- The sum of the forces in one direction equals the sum of the forces in the opposite direction (net force = 0).
- The sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point (net moment = 0).
(b)(i) Experiment to find the mass of a metre rule: Suspend the metre rule from a knife-edge and find its centre of gravity \(G\) (the balance point with no load). Hang a known mass \(m\) at a distance \(d_1\) on one side of a chosen pivot, and adjust the position of the pivot until the rule balances horizontally. Measure the distance \(d_2\) from the pivot to \(G\) (through which the whole weight \(W = Mg\) of the rule acts). Taking moments about the pivot:
\[ m g \times d_1 = M g \times d_2 \Rightarrow M = \frac{m \, d_1}{d_2} \]which gives the mass \(M\) of the rule.
(ii) Two precautions:
- Ensure the rule is horizontal (balanced) before taking readings.
- Avoid parallax error when reading the ruler; ensure the knife-edge is sharp and the mass is firmly fixed.
(c) Bullet mass \(m = 120\ \text{g} = 0.12\ \text{kg}\), \(u = 20\ \text{ms}^{-1}\), \(v = 0\), \(t = 0.1\ \text{s}\).
(i) Resistance (retarding force):
\[ F = \frac{m(u - v)}{t} = \frac{0.12 \times (20 - 0)}{0.1} = \frac{2.4}{0.1} = 24\ \text{N} \](ii) Distance moved in the wood (average velocity method):
\[ s = \left(\frac{u + v}{2}\right) t = \left(\frac{20 + 0}{2}\right) \times 0.1 = 10 \times 0.1 = 1.0\ \text{m} \]The resistance is \(24\ \text{N}\) and the bullet travels \(1.0\ \text{m}\) into the block.
Ajụjụ 12 Ripọtì
(a) Explain the statement the acceleration of free fall cohesion.
(b) State two factors that can affect the value of the narrow glass tube has a concave meniscus while acceleration of free fall at a place.
The OCR text of this question is partly garbled; the coherent physics content is about the acceleration of free fall, answered below.
(a) Meaning of "the acceleration of free fall at a place is \(9.8\ \text{m s}^{-2}\)"
It means that a body falling freely under gravity alone (with no air resistance) at that place has its velocity increasing at a constant rate of \(9.8\ \text{metres per second every second}\); that is, each second its downward speed increases by \(9.8\ \text{m s}^{-1}\).
(b) Two factors that affect the value of the acceleration of free fall at a place
- The latitude (distance from the equator) - \(g\) is greater at the poles than at the equator because the Earth is not a perfect sphere and because of the Earth's rotation.
- The altitude / height above sea level (distance from the Earth's centre) - \(g\) decreases as one goes higher above the Earth's surface.
Akọwa Nkọwa
The OCR text of this question is partly garbled; the coherent physics content is about the acceleration of free fall, answered below.
(a) Meaning of "the acceleration of free fall at a place is \(9.8\ \text{m s}^{-2}\)"
It means that a body falling freely under gravity alone (with no air resistance) at that place has its velocity increasing at a constant rate of \(9.8\ \text{metres per second every second}\); that is, each second its downward speed increases by \(9.8\ \text{m s}^{-1}\).
(b) Two factors that affect the value of the acceleration of free fall at a place
- The latitude (distance from the equator) - \(g\) is greater at the poles than at the equator because the Earth is not a perfect sphere and because of the Earth's rotation.
- The altitude / height above sea level (distance from the Earth's centre) - \(g\) decreases as one goes higher above the Earth's surface.
Ajụjụ 13 Ripọtì
(a) State two;
(i) laws of solid friction;
(ii) advantages of friction;
(iii) methods of reducing friction.
(b) Draw and label a diagram of a pulley system with velocity ratio of 5.
(c)(i) Show that the efficiency L, force ratio M.A. and the velocity ratio V.R. of a machine are related by the equation \(E = \frac{M.A.}{V.R}\) x 100%.
(ii) The efficiency of a machine is 80%. Calcuate the work done by a person using the machine to raise a load of 300 kg through a height of 4 m.[ g = 10 ms\(^{-2}\) ]
(a)(i) Two laws of solid friction:
- Friction opposes the relative motion (or the tendency to relative motion) between two surfaces in contact.
- The limiting (frictional) force \(F\) is directly proportional to the normal reaction \(R\) between the surfaces, so that \(F = \mu R\); it is also independent of the area of contact and depends only on the nature of the surfaces.
(a)(ii) Two advantages of friction:
- It provides the grip that prevents a person from slipping when walking and enables vehicle tyres to hold the road.
- It allows belts and pulleys to transmit power in machines and enables brakes to bring moving vehicles to rest.
(a)(iii) Two methods of reducing friction:
- Lubrication of the moving surfaces with oil or grease.
- Use of ball bearings or roller bearings between the moving parts (smoothing or polishing the surfaces also helps).
(b) Pulley system with velocity ratio of 5:
A block-and-tackle whose velocity ratio is 5 has five rope segments (strings) supporting the movable (lower) block. This is arranged with 3 pulleys in the fixed upper block and 2 pulleys in the movable lower block. The load \(W\) hangs from the lower block and the effort \(E\) is applied to the free end of the rope.
Because 5 strings support the lower block, when the effort end is pulled through a distance of 5 units the load rises through only 1 unit, giving
\[ \text{V.R.} = \frac{\text{distance moved by effort}}{\text{distance moved by load}} = \frac{5}{1} = 5. \]
(c)(i) Relationship between efficiency, mechanical advantage and velocity ratio:
Efficiency is the ratio of useful work output to work input, expressed as a percentage:
\[ E = \frac{\text{work output}}{\text{work input}} \times 100\% = \frac{\text{load} \times \text{distance moved by load}}{\text{effort} \times \text{distance moved by effort}} \times 100\%. \]
Rearranging the factors,
\[ E = \left(\frac{\text{load}}{\text{effort}}\right) \times \left(\frac{\text{distance moved by load}}{\text{distance moved by effort}}\right) \times 100\%. \]
Now \(\dfrac{\text{load}}{\text{effort}} = \text{M.A.}\) and \(\dfrac{\text{distance moved by effort}}{\text{distance moved by load}} = \text{V.R.}\), so \(\dfrac{\text{distance moved by load}}{\text{distance moved by effort}} = \dfrac{1}{\text{V.R.}}\). Therefore
\[ E = \frac{\text{M.A.}}{\text{V.R.}} \times 100\%. \]
(c)(ii) Work done by the person (work input):
Useful work output in raising the load:
\[ W_{\text{out}} = mgh = 300 \times 10 \times 4 = 12000\ \text{J}. \]
Since efficiency \(E = \dfrac{W_{\text{out}}}{W_{\text{in}}} \times 100\%\), the work input (work done by the person) is
\[ W_{\text{in}} = \frac{W_{\text{out}}}{E} = \frac{12000}{0.80} = 15000\ \text{J}. \]
The work done by the person is \(W_{\text{in}} = 15000\ \text{J} = 15\ \text{kJ}\).
Akọwa Nkọwa
(a)(i) Two laws of solid friction:
- Friction opposes the relative motion (or the tendency to relative motion) between two surfaces in contact.
- The limiting (frictional) force \(F\) is directly proportional to the normal reaction \(R\) between the surfaces, so that \(F = \mu R\); it is also independent of the area of contact and depends only on the nature of the surfaces.
(a)(ii) Two advantages of friction:
- It provides the grip that prevents a person from slipping when walking and enables vehicle tyres to hold the road.
- It allows belts and pulleys to transmit power in machines and enables brakes to bring moving vehicles to rest.
(a)(iii) Two methods of reducing friction:
- Lubrication of the moving surfaces with oil or grease.
- Use of ball bearings or roller bearings between the moving parts (smoothing or polishing the surfaces also helps).
(b) Pulley system with velocity ratio of 5:
A block-and-tackle whose velocity ratio is 5 has five rope segments (strings) supporting the movable (lower) block. This is arranged with 3 pulleys in the fixed upper block and 2 pulleys in the movable lower block. The load \(W\) hangs from the lower block and the effort \(E\) is applied to the free end of the rope.
Because 5 strings support the lower block, when the effort end is pulled through a distance of 5 units the load rises through only 1 unit, giving
\[ \text{V.R.} = \frac{\text{distance moved by effort}}{\text{distance moved by load}} = \frac{5}{1} = 5. \]
(c)(i) Relationship between efficiency, mechanical advantage and velocity ratio:
Efficiency is the ratio of useful work output to work input, expressed as a percentage:
\[ E = \frac{\text{work output}}{\text{work input}} \times 100\% = \frac{\text{load} \times \text{distance moved by load}}{\text{effort} \times \text{distance moved by effort}} \times 100\%. \]
Rearranging the factors,
\[ E = \left(\frac{\text{load}}{\text{effort}}\right) \times \left(\frac{\text{distance moved by load}}{\text{distance moved by effort}}\right) \times 100\%. \]
Now \(\dfrac{\text{load}}{\text{effort}} = \text{M.A.}\) and \(\dfrac{\text{distance moved by effort}}{\text{distance moved by load}} = \text{V.R.}\), so \(\dfrac{\text{distance moved by load}}{\text{distance moved by effort}} = \dfrac{1}{\text{V.R.}}\). Therefore
\[ E = \frac{\text{M.A.}}{\text{V.R.}} \times 100\%. \]
(c)(ii) Work done by the person (work input):
Useful work output in raising the load:
\[ W_{\text{out}} = mgh = 300 \times 10 \times 4 = 12000\ \text{J}. \]
Since efficiency \(E = \dfrac{W_{\text{out}}}{W_{\text{in}}} \times 100\%\), the work input (work done by the person) is
\[ W_{\text{in}} = \frac{W_{\text{out}}}{E} = \frac{12000}{0.80} = 15000\ \text{J}. \]
The work done by the person is \(W_{\text{in}} = 15000\ \text{J} = 15\ \text{kJ}\).
Ajụjụ 14 Ripọtì
A paralIel beam of unpolarized light is incident on a plane glass of refractive index 1.60 at an angle to the normal. If the reflected beam is completely polarized, calculate the angle of incidence of the beam.
When the reflected beam is completely (plane) polarised, the angle of incidence equals the polarising angle (Brewster's angle) \(\theta_p\), given by Brewster's law:
\[ n = \tan \theta_p \]Making \(\theta_p\) the subject:
\[ \theta_p = \tan^{-1}(n) = \tan^{-1}(1.60) \] \[ \theta_p = 57.99^{\circ} \approx 58.0^{\circ} \]The angle of incidence of the beam is about \(58^{\circ}\).
Akọwa Nkọwa
When the reflected beam is completely (plane) polarised, the angle of incidence equals the polarising angle (Brewster's angle) \(\theta_p\), given by Brewster's law:
\[ n = \tan \theta_p \]Making \(\theta_p\) the subject:
\[ \theta_p = \tan^{-1}(n) = \tan^{-1}(1.60) \] \[ \theta_p = 57.99^{\circ} \approx 58.0^{\circ} \]The angle of incidence of the beam is about \(58^{\circ}\).
Ajụjụ 15 Ripọtì
Explain why water in a narrow glass tube has a concave meniscus while mecury in the same tube, has a convex meniscus
Shape of the meniscus: water versus mercury
The shape of the meniscus depends on the relative strengths of the adhesive force (attraction between the liquid and the glass) and the cohesive force (attraction between the liquid molecules themselves).
Water (concave meniscus): for water in glass, the adhesive force between water and glass is greater than the cohesive force between the water molecules. The water therefore wets the glass and is pulled up the walls, so the surface curves upward at the edges, giving a concave (curving-down in the middle) meniscus.
Mercury (convex meniscus): for mercury in glass, the cohesive force between mercury molecules is much greater than the adhesive force between mercury and glass. The mercury does not wet the glass; its molecules are pulled together and away from the walls, so the surface is depressed at the edges and bulges up in the middle, giving a convex meniscus.
Akọwa Nkọwa
Shape of the meniscus: water versus mercury
The shape of the meniscus depends on the relative strengths of the adhesive force (attraction between the liquid and the glass) and the cohesive force (attraction between the liquid molecules themselves).
Water (concave meniscus): for water in glass, the adhesive force between water and glass is greater than the cohesive force between the water molecules. The water therefore wets the glass and is pulled up the walls, so the surface curves upward at the edges, giving a concave (curving-down in the middle) meniscus.
Mercury (convex meniscus): for mercury in glass, the cohesive force between mercury molecules is much greater than the adhesive force between mercury and glass. The mercury does not wet the glass; its molecules are pulled together and away from the walls, so the surface is depressed at the edges and bulges up in the middle, giving a convex meniscus.
