WAEC SSCE - Physics - 2017 (Essay)
Ajụjụ 1 Ripọtì
(a) Define stable equilibrium as applied to a rigid body.
(b) Sketch a block and tackle system of pulleys with a velocity ratio of 3.
(c) At the beginning of a race, a tyre of volume \(8.0 \times 10^{-4}\) at 20°C-has a gas pressure of \(4.5 \times 10^5\) Pa. Calculate the temperature of the gas in the tyre at the end of the race if the pressure has risen to \(4.6 \times 1 0^5\) Pa.
(d)(i)
Ice point 273k |
373k Steam point |
|
Resistance/\(\Omega\) |
5.67 | 7.75 |
Pressure/Pa |
\(7.13 \times 10^4\) | \(9.74 \times 10^4\) |
The table above shows readings of the resistance and pressure of a platinum resistance thermometer and a constant-volume gas thermometer respectively, when immersed in the same liquid bath. Use this data to determine the temperature of the bath on the: (\(\alpha\)) resistance thermometer; (\(\beta\)) gas thermometer
(ii) By what percentage is the temperature measured on the platinum resistance thermometer in error?
(a) Stable equilibrium
A rigid body is in stable equilibrium if, when it is slightly displaced and released, it tends to return to its original position. Its centre of gravity rises on displacement, so that its potential energy is a minimum at the equilibrium position.
(b) Block and tackle of velocity ratio 3
The rope is fixed to the movable block. It then passes over the left fixed pulley, round the movable pulley, and over the right fixed pulley. Thus, the movable block is supported by three tensioned sections of the rope, giving:
\[\text{V.R.}=3\]
(c) Temperature of the gas at the end of the race
Since the volume of the tyre is constant,
\[\frac{P_1}{T_1}=\frac{P_2}{T_2}\]
\[P_1=4.5\times10^5\ \text{Pa},\qquad P_2=4.6\times10^5\ \text{Pa}\]
\[T_1=20^\circ\text{C}+273=293\ \text{K}\]
\[T_2=\frac{P_2T_1}{P_1}=\frac{(4.6\times10^5)(293)}{4.5\times10^5}=299.5\ \text{K}\]
\[\boxed{T_2=299.5\ \text{K}\approx 26.5^\circ\text{C}}\]
(d)(i) Temperature scales
The calibration readings are:
| Fixed point | Temperature | Resistance | Gas pressure |
|---|---|---|---|
| Ice point | 273 K | 5.67 \(\Omega\) | \(7.13\times10^4\) Pa |
| Steam point | 373 K | 7.75 \(\Omega\) | \(9.74\times10^4\) Pa |
Let \(R_\theta\) be the resistance of the platinum thermometer in the liquid bath and \(P_\theta\) be the pressure of the gas thermometer in the same bath.
(α) Platinum resistance thermometer
\[\frac{R_\theta-5.67}{7.75-5.67}=\frac{T_R-273}{373-273}\]
\[\boxed{T_R=273+\frac{100(R_\theta-5.67)}{2.08}\ \text{K}}\]
(β) Constant-volume gas thermometer
\[\frac{P_\theta-7.13\times10^4}{(9.74-7.13)\times10^4}=\frac{T_G-273}{373-273}\]
\[\boxed{T_G=273+\frac{100(P_\theta-7.13\times10^4)}{2.61\times10^4}\ \text{K}}\]
(d)(ii) Percentage error of the platinum resistance thermometer
Taking the constant-volume gas thermometer temperature as the reference temperature,
\[\boxed{\%\,\text{error}=\frac{T_R-T_G}{T_G}\times100\%}\]
where \(T_R\) and \(T_G\) are obtained from the two boxed calibration equations above.
Akọwa Nkọwa
(a) Stable equilibrium
A rigid body is in stable equilibrium if, when it is slightly displaced and released, it tends to return to its original position. Its centre of gravity rises on displacement, so that its potential energy is a minimum at the equilibrium position.
(b) Block and tackle of velocity ratio 3
The rope is fixed to the movable block. It then passes over the left fixed pulley, round the movable pulley, and over the right fixed pulley. Thus, the movable block is supported by three tensioned sections of the rope, giving:
\[\text{V.R.}=3\]
(c) Temperature of the gas at the end of the race
Since the volume of the tyre is constant,
\[\frac{P_1}{T_1}=\frac{P_2}{T_2}\]
\[P_1=4.5\times10^5\ \text{Pa},\qquad P_2=4.6\times10^5\ \text{Pa}\]
\[T_1=20^\circ\text{C}+273=293\ \text{K}\]
\[T_2=\frac{P_2T_1}{P_1}=\frac{(4.6\times10^5)(293)}{4.5\times10^5}=299.5\ \text{K}\]
\[\boxed{T_2=299.5\ \text{K}\approx 26.5^\circ\text{C}}\]
(d)(i) Temperature scales
The calibration readings are:
| Fixed point | Temperature | Resistance | Gas pressure |
|---|---|---|---|
| Ice point | 273 K | 5.67 \(\Omega\) | \(7.13\times10^4\) Pa |
| Steam point | 373 K | 7.75 \(\Omega\) | \(9.74\times10^4\) Pa |
Let \(R_\theta\) be the resistance of the platinum thermometer in the liquid bath and \(P_\theta\) be the pressure of the gas thermometer in the same bath.
(α) Platinum resistance thermometer
\[\frac{R_\theta-5.67}{7.75-5.67}=\frac{T_R-273}{373-273}\]
\[\boxed{T_R=273+\frac{100(R_\theta-5.67)}{2.08}\ \text{K}}\]
(β) Constant-volume gas thermometer
\[\frac{P_\theta-7.13\times10^4}{(9.74-7.13)\times10^4}=\frac{T_G-273}{373-273}\]
\[\boxed{T_G=273+\frac{100(P_\theta-7.13\times10^4)}{2.61\times10^4}\ \text{K}}\]
(d)(ii) Percentage error of the platinum resistance thermometer
Taking the constant-volume gas thermometer temperature as the reference temperature,
\[\boxed{\%\,\text{error}=\frac{T_R-T_G}{T_G}\times100\%}\]
where \(T_R\) and \(T_G\) are obtained from the two boxed calibration equations above.
Ajụjụ 2 Ripọtì
List three advantages of p - n junetion diode over diode valve.
Three advantages of a p-n junction (semiconductor) diode over a diode valve (vacuum-tube diode)
- Much smaller and lighter: a semiconductor diode is very compact, so circuits built with it can be made small and portable.
- No heater/filament needed: it works cold and requires no warming-up time, whereas a diode valve needs a heated cathode; the semiconductor diode therefore consumes far less power.
- More rugged, cheaper and longer-lasting: it has no fragile glass envelope or filament to break or burn out, so it is mechanically robust, inexpensive and durable.
(Other acceptable points: it operates at lower voltages and switches on instantly.)
Akọwa Nkọwa
Three advantages of a p-n junction (semiconductor) diode over a diode valve (vacuum-tube diode)
- Much smaller and lighter: a semiconductor diode is very compact, so circuits built with it can be made small and portable.
- No heater/filament needed: it works cold and requires no warming-up time, whereas a diode valve needs a heated cathode; the semiconductor diode therefore consumes far less power.
- More rugged, cheaper and longer-lasting: it has no fragile glass envelope or filament to break or burn out, so it is mechanically robust, inexpensive and durable.
(Other acceptable points: it operates at lower voltages and switches on instantly.)
Ajụjụ 3 Ripọtì
(i) What is dielectric?
(ii) A parallel plate capacitor consists of two plates each of area 9.6 x 10\(^{-2}\)m\(^2\), separated by a dielectric of thickness 2.25 x 10\(^{-3}\) and dielectric constant 900. Calculate the capacitance of the capacitor. [ \(\varepsilon_o\) = permittivity of free space = 8.85 x 10\(^{-12}\)Fm\(^{-1}\)]
(b) (i) Which of the following devices has a higher resistance; an ammeter or a voltmeter? Give a reason for your answer.
(ii)
The resistance of the voltmeter in the circuit diagram illustrated above is 800\(\Omega\), Calculate the voltmeter reading.
(c)
A battery of negligible internal resistance is connected to a set of resistors as illustrated in the circuit diagram above. Determine the equivalent resistance of the circuit.
(a)(i) Dielectric
A dielectric is an insulating material (solid, liquid or gas) placed between the two plates of a capacitor. When placed in the field it becomes polarised and increases the capacitance of the capacitor. Examples are mica, waxed paper and air.
(a)(ii) Capacitance
\[ C = \frac{\varepsilon_r\,\varepsilon_0\,A}{d} \]
with \(\varepsilon_r = 900\), \(\varepsilon_0 = 8.85\times10^{-12}\,\text{Fm}^{-1}\), \(A = 9.6\times10^{-2}\,\text{m}^2\) and \(d = 2.25\times10^{-3}\,\text{m}\):
\[ C = \frac{900 \times 8.85\times10^{-12} \times 9.6\times10^{-2}}{2.25\times10^{-3}} = 3.4\times10^{-7}\,\text{F} \]
(b)(i) Ammeter or voltmeter?
The voltmeter has the higher resistance. A voltmeter is connected in parallel with the component whose p.d. is being measured, so it must have a very high resistance to draw only a negligible current and avoid disturbing the circuit. An ammeter is connected in series and must have a very low resistance so that it does not alter the current it measures.
(b)(ii) Voltmeter reading
The two 400 \(\Omega\) resistors on the left are in parallel:
\[ \frac{1}{R_1} = \frac{1}{400} + \frac{1}{400} \Rightarrow R_1 = 200\,\Omega \]
The 800 \(\Omega\) resistor is in parallel with the 800 \(\Omega\) voltmeter:
\[ \frac{1}{R_2} = \frac{1}{800} + \frac{1}{800} \Rightarrow R_2 = 400\,\Omega \]
\(R_1\) and \(R_2\) are in series across the 6 V supply:
\[ R_T = 200 + 400 = 600\,\Omega, \qquad I = \frac{V_T}{R_T} = \frac{6}{600} = 0.01\,\text{A} \]
The voltmeter reads the p.d. across \(R_2\):
\[ V = I\,R_2 = 0.01 \times 400 = 4.0\,\text{V} \]
(c) Equivalent resistance
The two 2 \(\Omega\) resistors in parallel combine to:
\[ \frac{1}{R_p} = \frac{1}{2} + \frac{1}{2} = 1 \Rightarrow R_p = 1\,\Omega \]
This 1 \(\Omega\) is in series with the remaining 2 \(\Omega\) and 2 \(\Omega\) resistors:
\[ R_T = 2 + 1 + 2 = 5\,\Omega \]
Akọwa Nkọwa
(a)(i) Dielectric
A dielectric is an insulating material (solid, liquid or gas) placed between the two plates of a capacitor. When placed in the field it becomes polarised and increases the capacitance of the capacitor. Examples are mica, waxed paper and air.
(a)(ii) Capacitance
\[ C = \frac{\varepsilon_r\,\varepsilon_0\,A}{d} \]
with \(\varepsilon_r = 900\), \(\varepsilon_0 = 8.85\times10^{-12}\,\text{Fm}^{-1}\), \(A = 9.6\times10^{-2}\,\text{m}^2\) and \(d = 2.25\times10^{-3}\,\text{m}\):
\[ C = \frac{900 \times 8.85\times10^{-12} \times 9.6\times10^{-2}}{2.25\times10^{-3}} = 3.4\times10^{-7}\,\text{F} \]
(b)(i) Ammeter or voltmeter?
The voltmeter has the higher resistance. A voltmeter is connected in parallel with the component whose p.d. is being measured, so it must have a very high resistance to draw only a negligible current and avoid disturbing the circuit. An ammeter is connected in series and must have a very low resistance so that it does not alter the current it measures.
(b)(ii) Voltmeter reading
The two 400 \(\Omega\) resistors on the left are in parallel:
\[ \frac{1}{R_1} = \frac{1}{400} + \frac{1}{400} \Rightarrow R_1 = 200\,\Omega \]
The 800 \(\Omega\) resistor is in parallel with the 800 \(\Omega\) voltmeter:
\[ \frac{1}{R_2} = \frac{1}{800} + \frac{1}{800} \Rightarrow R_2 = 400\,\Omega \]
\(R_1\) and \(R_2\) are in series across the 6 V supply:
\[ R_T = 200 + 400 = 600\,\Omega, \qquad I = \frac{V_T}{R_T} = \frac{6}{600} = 0.01\,\text{A} \]
The voltmeter reads the p.d. across \(R_2\):
\[ V = I\,R_2 = 0.01 \times 400 = 4.0\,\text{V} \]
(c) Equivalent resistance
The two 2 \(\Omega\) resistors in parallel combine to:
\[ \frac{1}{R_p} = \frac{1}{2} + \frac{1}{2} = 1 \Rightarrow R_p = 1\,\Omega \]
This 1 \(\Omega\) is in series with the remaining 2 \(\Omega\) and 2 \(\Omega\) resistors:
\[ R_T = 2 + 1 + 2 = 5\,\Omega \]
Ajụjụ 4 Ripọtì
List three advantages of fluorescent tubes over filament bulbs.
Three advantages of fluorescent tubes over filament (incandescent) bulbs
- Higher luminous efficiency: a fluorescent tube produces much more light for the same electrical power consumed, so it is more economical to run.
- Less heat is produced: most of the energy is converted to light rather than wasted as heat, so the tube runs cool.
- Longer working life: a fluorescent tube lasts far longer than a filament bulb because it has no fragile hot filament to burn out.
(Other acceptable points: the light is more evenly spread and softer on the eyes, and the running cost is lower.)
Akọwa Nkọwa
Three advantages of fluorescent tubes over filament (incandescent) bulbs
- Higher luminous efficiency: a fluorescent tube produces much more light for the same electrical power consumed, so it is more economical to run.
- Less heat is produced: most of the energy is converted to light rather than wasted as heat, so the tube runs cool.
- Longer working life: a fluorescent tube lasts far longer than a filament bulb because it has no fragile hot filament to burn out.
(Other acceptable points: the light is more evenly spread and softer on the eyes, and the running cost is lower.)
Ajụjụ 5 Ripọtì
List three phenomena which can be explained by the molecular theory of matter.
Phenomena explained by the molecular (kinetic) theory of matter
- Diffusion - the spreading of one substance through another, due to the random motion of molecules.
- Brownian motion - the ceaseless, random, zig-zag movement of tiny particles (such as smoke or pollen grains) caused by bombardment by fast-moving air or liquid molecules.
- Evaporation - the escape of the more energetic molecules from a liquid surface, accompanied by cooling of the liquid left behind.
(Surface tension, capillarity, expansion on heating and osmosis are also acceptable.)
Akọwa Nkọwa
Phenomena explained by the molecular (kinetic) theory of matter
- Diffusion - the spreading of one substance through another, due to the random motion of molecules.
- Brownian motion - the ceaseless, random, zig-zag movement of tiny particles (such as smoke or pollen grains) caused by bombardment by fast-moving air or liquid molecules.
- Evaporation - the escape of the more energetic molecules from a liquid surface, accompanied by cooling of the liquid left behind.
(Surface tension, capillarity, expansion on heating and osmosis are also acceptable.)
Ajụjụ 6 Ripọtì
(a) State two factors on which surface tension depends.
(b) How can mosquito larvae be made to sink in stagnant water?
(a) Two factors on which surface tension depends
- Temperature of the liquid: surface tension decreases as the temperature rises (the molecules gain more energy and their mutual attraction weakens).
- Nature of the liquid and any dissolved substances (impurities): different liquids have different surface tensions, and adding impurities such as detergent or soap lowers it, while some dissolved salts raise it.
(b) Making mosquito larvae sink
Mosquito larvae hang from the underside of the water surface and breathe there, being supported by the surface tension of the water. If a thin film of oil (or kerosene, or a little detergent) is spread over the stagnant water, the surface tension is greatly reduced. The surface can then no longer support the larvae, so they lose their hold on the surface film and sink. The oil film also blocks their breathing tubes, so they are unable to take in air.
Akọwa Nkọwa
(a) Two factors on which surface tension depends
- Temperature of the liquid: surface tension decreases as the temperature rises (the molecules gain more energy and their mutual attraction weakens).
- Nature of the liquid and any dissolved substances (impurities): different liquids have different surface tensions, and adding impurities such as detergent or soap lowers it, while some dissolved salts raise it.
(b) Making mosquito larvae sink
Mosquito larvae hang from the underside of the water surface and breathe there, being supported by the surface tension of the water. If a thin film of oil (or kerosene, or a little detergent) is spread over the stagnant water, the surface tension is greatly reduced. The surface can then no longer support the larvae, so they lose their hold on the surface film and sink. The oil film also blocks their breathing tubes, so they are unable to take in air.
Ajụjụ 7 Ripọtì
(a) (i) What is nuclear fission?
(ii) State the function of each of the following ma-terials in a nuclear fision reactor:
(\(\alpha\)) graphite, (\(\beta\)) boron rods; (\(\gamma\)) liquid sodium
(b) The table below gives some of the energy levels of a hydrogen atom.
| n | 1 | 2 | 3 | 4 | 5 | \(\infty\) |
| \(E_n/\text{eV}\) | -13.60 | -3.39 | -1.51 | -0.85 | -0.54 | 0.00 |
(i) Draw the energy level diagram for the atom.
(ii) Determine the wavelength of the photon emitted when the atom goes from the energy state \(n = 3\) to the ground state. [\(h = 6.6 \times 10^{-34}\,\mathrm{Js}\), \(c = 3 \times 10^{8}\,\mathrm{ms}^{-1}\), \(e = 1.6 \times 10^{-19}\)]
(c) A piece of ancient bone from an excavation site showed \(^{14}_{6}\mathrm{C}\) activity of 9.5 disintegrations per minute per \(1.0 \times 10^{-3}\,\mathrm{kg}\). If a bone specimen from a living creature shows \(^{14}_{6}\mathrm{C}\) activity of 12.0 disintegrations per minute per \(1.0 \times 1 0^{-3}\), determine the age of the ancient bone. [Half - life of \(^{14}_{6}\mathrm{C} = 5572\) years].
(a) (i) Nuclear fission is the splitting of a heavy atomic nucleus into two lighter nuclei, with the release of a large amount of energy and neutrons.
(ii) Functions of materials in a nuclear fission reactor
| Material | Function |
|---|---|
| Graphite | It acts as a moderator, slowing down fast neutrons so that they can cause further fission. |
| Boron rods | They absorb excess neutrons and hence control the rate of the chain reaction. |
| Liquid sodium | It acts as a coolant, removing heat from the reactor core. |
(b) (i) Energy-level diagram of the hydrogen atom:
(ii) For the transition from \(n=3\) to the ground state \(n=1\),
\[\Delta E=E_3-E_1=(-1.51)-(-13.60)=12.09\ \text{eV}\]
\[\Delta E=12.09\times1.6\times10^{-19}=1.9344\times10^{-18}\ \text{J}\]
Using \(\Delta E=\dfrac{hc}{\lambda}\),
\[\lambda=\frac{hc}{\Delta E}=\frac{(6.6\times10^{-34})(3.0\times10^8)}{1.9344\times10^{-18}}\]
\[\lambda=1.02\times10^{-7}\ \text{m}\]
The wavelength of the emitted photon is \(\boxed{1.02\times10^{-7}\ \text{m}}\), or \(102\ \text{nm}\).
(c) The activity is proportional to the number of undecayed carbon-14 nuclei. Thus,
\[\frac{A}{A_0}=\frac{9.5}{12.0}=e^{-\lambda t}\]
\[\lambda=\frac{0.693}{5572}=1.244\times10^{-4}\ \text{year}^{-1}\]
\[t=\frac{1}{\lambda}\ln\left(\frac{A_0}{A}\right)=\frac{1}{1.244\times10^{-4}}\ln\left(\frac{12.0}{9.5}\right)\]
\[t=1.88\times10^3\ \text{years}\]
Therefore, the age of the ancient bone is \(\boxed{1.88\times10^3\ \text{years}}\), approximately \(\boxed{1878\ \text{years}}\).
Akọwa Nkọwa
(a) (i) Nuclear fission is the splitting of a heavy atomic nucleus into two lighter nuclei, with the release of a large amount of energy and neutrons.
(ii) Functions of materials in a nuclear fission reactor
| Material | Function |
|---|---|
| Graphite | It acts as a moderator, slowing down fast neutrons so that they can cause further fission. |
| Boron rods | They absorb excess neutrons and hence control the rate of the chain reaction. |
| Liquid sodium | It acts as a coolant, removing heat from the reactor core. |
(b) (i) Energy-level diagram of the hydrogen atom:
(ii) For the transition from \(n=3\) to the ground state \(n=1\),
\[\Delta E=E_3-E_1=(-1.51)-(-13.60)=12.09\ \text{eV}\]
\[\Delta E=12.09\times1.6\times10^{-19}=1.9344\times10^{-18}\ \text{J}\]
Using \(\Delta E=\dfrac{hc}{\lambda}\),
\[\lambda=\frac{hc}{\Delta E}=\frac{(6.6\times10^{-34})(3.0\times10^8)}{1.9344\times10^{-18}}\]
\[\lambda=1.02\times10^{-7}\ \text{m}\]
The wavelength of the emitted photon is \(\boxed{1.02\times10^{-7}\ \text{m}}\), or \(102\ \text{nm}\).
(c) The activity is proportional to the number of undecayed carbon-14 nuclei. Thus,
\[\frac{A}{A_0}=\frac{9.5}{12.0}=e^{-\lambda t}\]
\[\lambda=\frac{0.693}{5572}=1.244\times10^{-4}\ \text{year}^{-1}\]
\[t=\frac{1}{\lambda}\ln\left(\frac{A_0}{A}\right)=\frac{1}{1.244\times10^{-4}}\ln\left(\frac{12.0}{9.5}\right)\]
\[t=1.88\times10^3\ \text{years}\]
Therefore, the age of the ancient bone is \(\boxed{1.88\times10^3\ \text{years}}\), approximately \(\boxed{1878\ \text{years}}\).
Ajụjụ 8 Ripọtì
(a) What is a wavefront?
(b) (i) State two practical uses of glass prisms.
(ii) List two factors that determine the deviation of a ray of light travelling from air into a triangular glass prism.
(iii) Sketch a graph to illustrate the variation of the angle of deviation d, with that of incidence, i, for a ray of light travelling from air into a triangular glass prism. Indicate on the graph the point at which the angle of incidence equal to the angle of emergence e.
(c) (i) Draw and label a diagram of an astronomical telescope in normal adjustment.
(ii) The angular magnification of an astronomical telescope in normal adjustment is 5. If the focal length of the objective is 100cm, calculate the:
(\(\alpha\)) focal length of the eyepiece;
(\(\beta\)) length of the telescope.
(a) A wavefront is a surface or locus of points in a wave which are vibrating in the same phase.
(b)(i) Two practical uses of glass prisms are:
- to change the direction of a beam of light, for example in binoculars or periscopes;
- to disperse white light into its component colours, for example in a spectrometer.
(b)(ii) The deviation produced by a triangular glass prism depends on:
- the angle of incidence of the ray;
- the refractive index of the glass prism.
The refracting angle of the prism is also a factor.
(b)(iii) The graph of angle of deviation, d, against angle of incidence, i, is shown below. The minimum point is the point at which i = e.
(c)(i) Astronomical telescope in normal adjustment:
In normal adjustment, the real intermediate image formed by the objective lies at the first focal point of the eyepiece. The emergent rays are parallel and the final image is therefore at infinity.
(c)(ii)
For an astronomical telescope in normal adjustment,
\[M=\frac{f_o}{f_e}\]
Given \(M=5\) and \(f_o=100\text{ cm}\):
(α) Focal length of eyepiece
\[5=\frac{100}{f_e}\]
\[f_e=\frac{100}{5}=20\text{ cm}\]
(β) Length of telescope
\[L=f_o+f_e=100+20=120\text{ cm}\]
Akọwa Nkọwa
(a) A wavefront is a surface or locus of points in a wave which are vibrating in the same phase.
(b)(i) Two practical uses of glass prisms are:
- to change the direction of a beam of light, for example in binoculars or periscopes;
- to disperse white light into its component colours, for example in a spectrometer.
(b)(ii) The deviation produced by a triangular glass prism depends on:
- the angle of incidence of the ray;
- the refractive index of the glass prism.
The refracting angle of the prism is also a factor.
(b)(iii) The graph of angle of deviation, d, against angle of incidence, i, is shown below. The minimum point is the point at which i = e.
(c)(i) Astronomical telescope in normal adjustment:
In normal adjustment, the real intermediate image formed by the objective lies at the first focal point of the eyepiece. The emergent rays are parallel and the final image is therefore at infinity.
(c)(ii)
For an astronomical telescope in normal adjustment,
\[M=\frac{f_o}{f_e}\]
Given \(M=5\) and \(f_o=100\text{ cm}\):
(α) Focal length of eyepiece
\[5=\frac{100}{f_e}\]
\[f_e=\frac{100}{5}=20\text{ cm}\]
(β) Length of telescope
\[L=f_o+f_e=100+20=120\text{ cm}\]
Ajụjụ 9 Ripọtì
A spiral spring has a length of 14cm when a force of 4N is hung on it. A force of 6N extends the spring by 4cm. Calculate the unstretched length of the spring
Given: a load of 4 N makes the spring 14 cm long; a load of 6 N produces an extension of 4 cm. Let the unstretched (natural) length be \(L_0\) and the force constant be k.
By Hooke's law, extension is proportional to load, so the force constant is found from the second statement:
\[ k = \frac{F}{e} = \frac{6\ \text{N}}{4\ \text{cm}} = 1.5\ \text{N cm}^{-1} \]
Extension produced by the 4 N load:
\[ e_1 = \frac{F}{k} = \frac{4}{1.5} = 2.67\ \text{cm} \]
The length under the 4 N load is 14 cm, and this equals natural length plus extension:
\[ L_0 = 14 - e_1 = 14 - 2.67 = 11.33\ \text{cm} \]
The unstretched length of the spring is approximately 11.3 cm.
Akọwa Nkọwa
Given: a load of 4 N makes the spring 14 cm long; a load of 6 N produces an extension of 4 cm. Let the unstretched (natural) length be \(L_0\) and the force constant be k.
By Hooke's law, extension is proportional to load, so the force constant is found from the second statement:
\[ k = \frac{F}{e} = \frac{6\ \text{N}}{4\ \text{cm}} = 1.5\ \text{N cm}^{-1} \]
Extension produced by the 4 N load:
\[ e_1 = \frac{F}{k} = \frac{4}{1.5} = 2.67\ \text{cm} \]
The length under the 4 N load is 14 cm, and this equals natural length plus extension:
\[ L_0 = 14 - e_1 = 14 - 2.67 = 11.33\ \text{cm} \]
The unstretched length of the spring is approximately 11.3 cm.
Ajụjụ 10 Ripọtì
A particle is projected horizontally at 15ms\(^{-1}\) from a height of 20m. Calculate the horizontal distance covered by the particle just before hitting the ground. [g= 10ms\(^{-2}\)]
Given: horizontal velocity \(u = 15\ \text{m s}^{-1}\), height \(h = 20\ \text{m}\), \(g = 10\ \text{m s}^{-2}\).
The horizontal and vertical motions are independent. First find the time of flight from the vertical motion (initial vertical velocity is zero):
\[ h = \tfrac{1}{2}gt^2 \Rightarrow 20 = \tfrac{1}{2}\times10\times t^2 = 5t^2 \] \[ t^2 = 4 \Rightarrow t = 2\ \text{s} \]
The horizontal velocity stays constant, so the horizontal distance (range) is:
\[ R = u \times t = 15 \times 2 = 30\ \text{m} \]
The particle covers a horizontal distance of 30 m before hitting the ground.
Akọwa Nkọwa
Given: horizontal velocity \(u = 15\ \text{m s}^{-1}\), height \(h = 20\ \text{m}\), \(g = 10\ \text{m s}^{-2}\).
The horizontal and vertical motions are independent. First find the time of flight from the vertical motion (initial vertical velocity is zero):
\[ h = \tfrac{1}{2}gt^2 \Rightarrow 20 = \tfrac{1}{2}\times10\times t^2 = 5t^2 \] \[ t^2 = 4 \Rightarrow t = 2\ \text{s} \]
The horizontal velocity stays constant, so the horizontal distance (range) is:
\[ R = u \times t = 15 \times 2 = 30\ \text{m} \]
The particle covers a horizontal distance of 30 m before hitting the ground.
Ajụjụ 11 Ripọtì
A particle is dropped from a vertical height h and falls freely for a time t. With the aid of a sketch, explain how h varies with \(t^2\)
Here, \(h\) is most naturally interpreted as the particle’s height above the ground. As the particle falls freely from rest, its downward distance travelled is
\[ s=\tfrac{1}{2}gt^2. \]If its initial height is \(H\), its height remaining above the ground is therefore
\[ h=H-\tfrac{1}{2}gt^2. \]So \(h\) decreases linearly as \(t^2\) increases. A graph of \(h\) against \(t^2\) is a straight line with:
- vertical intercept \(H\), the initial height;
- negative gradient \(-\tfrac{1}{2}g\);
- \(h=0\) when the particle reaches the ground.
Important distinction: if \(h\) were instead defined as the distance fallen, then \(h=\tfrac{1}{2}gt^2\), giving a straight line through the origin with positive gradient \(\tfrac{1}{2}g\). For a question referring to the particle’s vertical height, use \(h=H-\tfrac{1}{2}gt^2\).
Akọwa Nkọwa
Here, \(h\) is most naturally interpreted as the particle’s height above the ground. As the particle falls freely from rest, its downward distance travelled is
\[ s=\tfrac{1}{2}gt^2. \]If its initial height is \(H\), its height remaining above the ground is therefore
\[ h=H-\tfrac{1}{2}gt^2. \]So \(h\) decreases linearly as \(t^2\) increases. A graph of \(h\) against \(t^2\) is a straight line with:
- vertical intercept \(H\), the initial height;
- negative gradient \(-\tfrac{1}{2}g\);
- \(h=0\) when the particle reaches the ground.
Important distinction: if \(h\) were instead defined as the distance fallen, then \(h=\tfrac{1}{2}gt^2\), giving a straight line through the origin with positive gradient \(\tfrac{1}{2}g\). For a question referring to the particle’s vertical height, use \(h=H-\tfrac{1}{2}gt^2\).
Ajụjụ 12 Ripọtì
(a) State two deductions that can be made from a displacement-time graph.
(b) If the distance beween two equal masses is doubled and their individual masses are also doubled, what would happen to the force between them? Support your answer quantitatively.
(c) State two factors that affect the maximum height attained by a bullet fired from a gun.
(d) State two practical examples of mechanical resonance. A body is released from rest at the top of a plane inclined at 30° to the horizontal and 4.0 m high. If the coefficient of friction between the body and the plane is 0.3, calculate the time the body takes to reach the bottom of the plane.
(a) Deductions from a displacement-time graph
- The velocity is obtained from the gradient of the graph. For the straight line shown,
\[v=\frac{\Delta s}{\Delta t}=\frac{10-2}{4-0}=2.0\ \text{m s}^{-1}.\]
- The initial displacement is given by the intercept on the displacement axis. From the graph, the initial displacement is \(2.0\ \text{m}\). A straight line also indicates uniform velocity.
(b) Effect on the gravitational force
The gravitational force between two masses is
\[F=\frac{Gm_1m_2}{r^2}.\]
When each mass and the distance between them are doubled,
\[F' = \frac{G(2m_1)(2m_2)}{(2r)^2}=\frac{4Gm_1m_2}{4r^2}=\frac{Gm_1m_2}{r^2}=F.\]
Therefore, the force between the masses remains unchanged.
(c) Factors affecting the maximum height attained by a fired bullet
- The initial speed, or muzzle velocity, of the bullet.
- The angle of projection of the bullet.
Other acceptable factors include acceleration due to gravity and air resistance.
(d) Practical examples of mechanical resonance
- A glass may shatter when subjected to a sound whose frequency equals the natural frequency of the glass.
- A bridge may vibrate dangerously when soldiers march across it in step.
Time taken by the body to reach the bottom of the plane
Given: \(h=4.0\ \text{m}\), \(\theta=30^\circ\), \(\mu=0.3\), \(u=0\), and \(g=10\ \text{m s}^{-2}\).
The length \(s\) of the inclined plane is
\[s=\frac{h}{\sin 30^\circ}=\frac{4.0}{0.5}=8.0\ \text{m}.\]
The acceleration down the plane is
\[a=g(\sin\theta-\mu\cos\theta)\]
\[a=10\left(\sin30^\circ-0.3\cos30^\circ\right)\]
\[a=10\left(0.5-0.3\times0.866\right)=2.40\ \text{m s}^{-2}.\]
Using \(s=ut+\tfrac12at^2\),
\[8.0=0+\frac12(2.40)t^2\]
\[t^2=\frac{16.0}{2.40}=6.67\]
\[t=2.58\ \text{s}\approx2.6\ \text{s}.\]
Hence, the body takes \(2.6\ \text{s}\) to reach the bottom of the plane.
Akọwa Nkọwa
(a) Deductions from a displacement-time graph
- The velocity is obtained from the gradient of the graph. For the straight line shown,
\[v=\frac{\Delta s}{\Delta t}=\frac{10-2}{4-0}=2.0\ \text{m s}^{-1}.\]
- The initial displacement is given by the intercept on the displacement axis. From the graph, the initial displacement is \(2.0\ \text{m}\). A straight line also indicates uniform velocity.
(b) Effect on the gravitational force
The gravitational force between two masses is
\[F=\frac{Gm_1m_2}{r^2}.\]
When each mass and the distance between them are doubled,
\[F' = \frac{G(2m_1)(2m_2)}{(2r)^2}=\frac{4Gm_1m_2}{4r^2}=\frac{Gm_1m_2}{r^2}=F.\]
Therefore, the force between the masses remains unchanged.
(c) Factors affecting the maximum height attained by a fired bullet
- The initial speed, or muzzle velocity, of the bullet.
- The angle of projection of the bullet.
Other acceptable factors include acceleration due to gravity and air resistance.
(d) Practical examples of mechanical resonance
- A glass may shatter when subjected to a sound whose frequency equals the natural frequency of the glass.
- A bridge may vibrate dangerously when soldiers march across it in step.
Time taken by the body to reach the bottom of the plane
Given: \(h=4.0\ \text{m}\), \(\theta=30^\circ\), \(\mu=0.3\), \(u=0\), and \(g=10\ \text{m s}^{-2}\).
The length \(s\) of the inclined plane is
\[s=\frac{h}{\sin 30^\circ}=\frac{4.0}{0.5}=8.0\ \text{m}.\]
The acceleration down the plane is
\[a=g(\sin\theta-\mu\cos\theta)\]
\[a=10\left(\sin30^\circ-0.3\cos30^\circ\right)\]
\[a=10\left(0.5-0.3\times0.866\right)=2.40\ \text{m s}^{-2}.\]
Using \(s=ut+\tfrac12at^2\),
\[8.0=0+\frac12(2.40)t^2\]
\[t^2=\frac{16.0}{2.40}=6.67\]
\[t=2.58\ \text{s}\approx2.6\ \text{s}.\]
Hence, the body takes \(2.6\ \text{s}\) to reach the bottom of the plane.
