WAEC SSCE - Physics - 2013 (Essay)
Ajụjụ 1 Ripọtì
The horizontal component of the initial speed of a particle projected at 30° to the horizontal is 50 ms\(^{-1}\). If the acceleration cf free fall due to gravity is 10ms\(^{-2}\), determine its: (a) initial speed; (b) speed at maximum height reached.
The particle is projected at \(30^\circ\) to the horizontal, and its horizontal component of speed is \(u_x = 50\,\text{ms}^{-1}\).
(a) Initial speed \(u\)
The horizontal component is \(u_x = u\cos\theta\), so:
\[ u = \frac{u_x}{\cos\theta} = \frac{50}{\cos 30^\circ} = \frac{50}{0.866} = 57.7\,\text{ms}^{-1} \](b) Speed at maximum height
At the maximum height the vertical component of velocity is zero, so the speed there equals the (unchanged) horizontal component:
\[ v = u\cos\theta = 50\,\text{ms}^{-1} \]Akọwa Nkọwa
The particle is projected at \(30^\circ\) to the horizontal, and its horizontal component of speed is \(u_x = 50\,\text{ms}^{-1}\).
(a) Initial speed \(u\)
The horizontal component is \(u_x = u\cos\theta\), so:
\[ u = \frac{u_x}{\cos\theta} = \frac{50}{\cos 30^\circ} = \frac{50}{0.866} = 57.7\,\text{ms}^{-1} \](b) Speed at maximum height
At the maximum height the vertical component of velocity is zero, so the speed there equals the (unchanged) horizontal component:
\[ v = u\cos\theta = 50\,\text{ms}^{-1} \]Ajụjụ 2 Ripọtì
A mass of 11.0 kg is suspended from a rigid support by an aluminum wire of length 2.0 m, diameter 2.0 mm and Young's modulus 7.0 x 10\(^{11}\) Nm\(^{-2}\). Determine the extension produced. [g = 10 ms\(^{-2}\); \(\pi\) = 3.142]
The extension of a stretched wire is given by \(e = \dfrac{FL}{AY}\).
Data: mass \(= 11.0\,\text{kg}\), so \(F = mg = 11.0\times10 = 110\,\text{N}\); length \(L = 2.0\,\text{m}\); diameter \(= 2.0\,\text{mm}\), so radius \(r = 1.0\times10^{-3}\,\text{m}\); \(Y = 7.0\times10^{11}\,\text{Nm}^{-2}\).
Cross-sectional area:
\[ A = \pi r^{2} = 3.142\times(1.0\times10^{-3})^{2} = 3.142\times10^{-6}\,\text{m}^{2} \]Extension:
\[ e = \frac{FL}{AY} = \frac{110\times2.0}{(3.142\times10^{-6})(7.0\times10^{11})} \] \[ e = \frac{220}{2.199\times10^{6}} = 1.0\times10^{-4}\,\text{m} = 0.1\,\text{mm} \]Akọwa Nkọwa
The extension of a stretched wire is given by \(e = \dfrac{FL}{AY}\).
Data: mass \(= 11.0\,\text{kg}\), so \(F = mg = 11.0\times10 = 110\,\text{N}\); length \(L = 2.0\,\text{m}\); diameter \(= 2.0\,\text{mm}\), so radius \(r = 1.0\times10^{-3}\,\text{m}\); \(Y = 7.0\times10^{11}\,\text{Nm}^{-2}\).
Cross-sectional area:
\[ A = \pi r^{2} = 3.142\times(1.0\times10^{-3})^{2} = 3.142\times10^{-6}\,\text{m}^{2} \]Extension:
\[ e = \frac{FL}{AY} = \frac{110\times2.0}{(3.142\times10^{-6})(7.0\times10^{11})} \] \[ e = \frac{220}{2.199\times10^{6}} = 1.0\times10^{-4}\,\text{m} = 0.1\,\text{mm} \]Ajụjụ 3 Ripọtì
(a) Explain briefly the purpose of earthing electrical appliance.
(b) Why does the light frorr bulb connected to a simple cell dim and eventually goes off after a while?
(c) A coil of incidence 0.007 H, a resistor of resistance 8 \(\Omega\) and a capacitor capacitance 0.001 F are connected in series an a.c. source of frequency \(\frac{500}{\pi}\)Hz. If the r.m.s voltages across the coil, the resistor and capacitor are 30v, 20v and 70v respectively;
(i) draw a vector diagram to illustrate the voltage across the components in the circuit.
(ii) Calculate the: (\(\alpha\)) r.m.s voltage of the source
(\(\beta\)) r.m.s current in the circuit;
(\(\gamma\)) power dissipated in the circuit.
iii) write down the sinusoidal equation for the r.m.s voltage, V, in terms of the time, t.
(a) Purpose of earthing an electrical appliance
The earth wire connects the metal casing of an appliance to the ground through a low-resistance path. If a fault causes the live wire to touch the casing, a large current flows through the earth wire rather than through a person touching the appliance. This causes the fuse to melt or the circuit breaker to operate, reducing the risk of electric shock.
(b) Why a bulb connected to a simple cell dims and eventually goes out
A simple cell suffers from polarisation and local action.
- During polarisation, hydrogen bubbles collect on the copper electrode. This produces a back e.m.f. and increases the effective internal resistance of the cell, so the current in the bulb decreases.
- Local action causes zinc to be used up by unwanted chemical reactions within the cell.
As the current falls, the bulb becomes dimmer. Eventually the cell can no longer provide enough current, so the bulb goes out.
(c) Series LCR circuit
The frequency is \(f=\dfrac{500}{\pi}\,\text{Hz}\), so the angular frequency is:
\[ \omega=2\pi f=2\pi\left(\frac{500}{\pi}\right)=1000\,\text{rad s}^{-1}. \]Important check on the data: the stated component voltages are not consistent with the stated values of \(L\), \(C\), and \(R\). From the resistor voltage, \(I=20/8=2.5\,\text{A}\). This would predict \(V_L=I\omega L=17.5\,\text{V}\) and \(V_C=I/(\omega C)=2.5\,\text{V}\), rather than \(30\,\text{V}\) and \(70\,\text{V}\). Therefore the reference calculation giving \(50\,\text{V}\), \(5.0\,\text{A}\), and \(200\,\text{W}\) cannot follow from all the values in the question.
The calculations below use the directly stated r.m.s. voltages across the components, together with \(R=8\,\Omega\). This is the internally consistent approach for the voltage phasor diagram.
(i) Voltage vector diagram
Take the current, and therefore \(V_R\), as the horizontal reference direction. \(V_L\) is \(90^\circ\) ahead of the current, while \(V_C\) is \(90^\circ\) behind it. Since \(V_C\) is larger than \(V_L\), the circuit is overall capacitive.
(ii) Calculations
R.m.s. source voltage
The resistor voltage is perpendicular to the net reactive voltage:
\[ V_{\text{rms}}=\sqrt{V_R^2+(V_C-V_L)^2} \] \[ V_{\text{rms}}=\sqrt{20^2+(70-30)^2} =\sqrt{400+1600} =\sqrt{2000} =44.7\,\text{V}. \]R.m.s. current
\[ I_{\text{rms}}=\frac{V_R}{R} =\frac{20}{8} =2.5\,\text{A}. \]Power dissipated
For an ideal inductor and capacitor, only the resistor dissipates average power:
\[ P=I_{\text{rms}}^2R =(2.5)^2(8) =50\,\text{W}. \](iii) Sinusoidal equation for the source voltage
The peak voltage is:
\[ V_0=\sqrt{2}\,V_{\text{rms}} =\sqrt{2}(44.7) =63.2\,\text{V}. \]If the source voltage is chosen to have zero phase at \(t=0\), its instantaneous voltage is:
\[ v=63.2\sin(1000t)\,\text{V}. \]Examination reminder: In a series LCR circuit, do not add \(V_R\), \(V_L\), and \(V_C\) as ordinary numbers. \(V_L\) and \(V_C\) act in opposite vertical directions on the phasor diagram, so first find their difference, then combine it with \(V_R\) using Pythagoras.
Akọwa Nkọwa
(a) Purpose of earthing an electrical appliance
The earth wire connects the metal casing of an appliance to the ground through a low-resistance path. If a fault causes the live wire to touch the casing, a large current flows through the earth wire rather than through a person touching the appliance. This causes the fuse to melt or the circuit breaker to operate, reducing the risk of electric shock.
(b) Why a bulb connected to a simple cell dims and eventually goes out
A simple cell suffers from polarisation and local action.
- During polarisation, hydrogen bubbles collect on the copper electrode. This produces a back e.m.f. and increases the effective internal resistance of the cell, so the current in the bulb decreases.
- Local action causes zinc to be used up by unwanted chemical reactions within the cell.
As the current falls, the bulb becomes dimmer. Eventually the cell can no longer provide enough current, so the bulb goes out.
(c) Series LCR circuit
The frequency is \(f=\dfrac{500}{\pi}\,\text{Hz}\), so the angular frequency is:
\[ \omega=2\pi f=2\pi\left(\frac{500}{\pi}\right)=1000\,\text{rad s}^{-1}. \]Important check on the data: the stated component voltages are not consistent with the stated values of \(L\), \(C\), and \(R\). From the resistor voltage, \(I=20/8=2.5\,\text{A}\). This would predict \(V_L=I\omega L=17.5\,\text{V}\) and \(V_C=I/(\omega C)=2.5\,\text{V}\), rather than \(30\,\text{V}\) and \(70\,\text{V}\). Therefore the reference calculation giving \(50\,\text{V}\), \(5.0\,\text{A}\), and \(200\,\text{W}\) cannot follow from all the values in the question.
The calculations below use the directly stated r.m.s. voltages across the components, together with \(R=8\,\Omega\). This is the internally consistent approach for the voltage phasor diagram.
(i) Voltage vector diagram
Take the current, and therefore \(V_R\), as the horizontal reference direction. \(V_L\) is \(90^\circ\) ahead of the current, while \(V_C\) is \(90^\circ\) behind it. Since \(V_C\) is larger than \(V_L\), the circuit is overall capacitive.
(ii) Calculations
R.m.s. source voltage
The resistor voltage is perpendicular to the net reactive voltage:
\[ V_{\text{rms}}=\sqrt{V_R^2+(V_C-V_L)^2} \] \[ V_{\text{rms}}=\sqrt{20^2+(70-30)^2} =\sqrt{400+1600} =\sqrt{2000} =44.7\,\text{V}. \]R.m.s. current
\[ I_{\text{rms}}=\frac{V_R}{R} =\frac{20}{8} =2.5\,\text{A}. \]Power dissipated
For an ideal inductor and capacitor, only the resistor dissipates average power:
\[ P=I_{\text{rms}}^2R =(2.5)^2(8) =50\,\text{W}. \](iii) Sinusoidal equation for the source voltage
The peak voltage is:
\[ V_0=\sqrt{2}\,V_{\text{rms}} =\sqrt{2}(44.7) =63.2\,\text{V}. \]If the source voltage is chosen to have zero phase at \(t=0\), its instantaneous voltage is:
\[ v=63.2\sin(1000t)\,\text{V}. \]Examination reminder: In a series LCR circuit, do not add \(V_R\), \(V_L\), and \(V_C\) as ordinary numbers. \(V_L\) and \(V_C\) act in opposite vertical directions on the phasor diagram, so first find their difference, then combine it with \(V_R\) using Pythagoras.
Ajụjụ 4 Ripọtì
The accelerating potential in a cathode ray oscilloscope is 2.5 kV. Calculate the maximum speed of the accelerated electrons. [ e = 1.6 x 10\(^{-19}\) C; Me = 9.1 x 10\(^{-31}\) kg]
An electron accelerated through a potential difference \(V\) gains kinetic energy equal to the electrical work done on it:
\[ eV = \frac{1}{2}m v^{2} \Rightarrow v = \sqrt{\frac{2eV}{m}} \]Data: \(V = 2.5\,\text{kV} = 2.5\times10^{3}\,\text{V}\); \(e = 1.6\times10^{-19}\,\text{C}\); \(m = 9.1\times10^{-31}\,\text{kg}\).
\[ v = \sqrt{\frac{2(1.6\times10^{-19})(2.5\times10^{3})}{9.1\times10^{-31}}} = \sqrt{\frac{8.0\times10^{-16}}{9.1\times10^{-31}}} \] \[ v = \sqrt{8.79\times10^{14}} = 2.97\times10^{7}\,\text{ms}^{-1} \]The maximum speed of the accelerated electrons is about \(3.0\times10^{7}\,\text{ms}^{-1}\).
Akọwa Nkọwa
An electron accelerated through a potential difference \(V\) gains kinetic energy equal to the electrical work done on it:
\[ eV = \frac{1}{2}m v^{2} \Rightarrow v = \sqrt{\frac{2eV}{m}} \]Data: \(V = 2.5\,\text{kV} = 2.5\times10^{3}\,\text{V}\); \(e = 1.6\times10^{-19}\,\text{C}\); \(m = 9.1\times10^{-31}\,\text{kg}\).
\[ v = \sqrt{\frac{2(1.6\times10^{-19})(2.5\times10^{3})}{9.1\times10^{-31}}} = \sqrt{\frac{8.0\times10^{-16}}{9.1\times10^{-31}}} \] \[ v = \sqrt{8.79\times10^{14}} = 2.97\times10^{7}\,\text{ms}^{-1} \]The maximum speed of the accelerated electrons is about \(3.0\times10^{7}\,\text{ms}^{-1}\).
Ajụjụ 5 Ripọtì
(a) What is a polarizer?
(b) With the aid of a diagram, explain how a polarizer can be used to polarize a beam of unpolarized light.
(a) What is a polarizer?
A polarizer is an optical device (for example a Polaroid sheet or a tourmaline crystal) that has a single characteristic direction called its transmission axis. When light falls on it, it transmits only the component of the light vibrations that is parallel to this transmission axis and absorbs (or blocks) all the other components. It therefore converts an unpolarized beam, whose vibrations occur in all planes perpendicular to the direction of travel, into a plane-polarized beam whose vibrations lie in only one plane.
(b) How a polarizer polarizes a beam of unpolarized light
In an unpolarized beam the electric-field vibrations take place in every direction in the plane perpendicular to the direction of propagation. When this beam strikes the polarizer, only the vibration component that is parallel to the polarizer's transmission axis is allowed through; every component perpendicular to that axis is absorbed. The light emerging from the polarizer therefore vibrates in one plane only, that is, it is plane-polarized. Because only one component is transmitted, the emerging intensity is about half that of the incident unpolarized light \(\left(I = \tfrac{1}{2}I_0\right)\).
The action is shown in the diagram below. The unpolarized light (vibrating in all directions) passes through the first polarizer and emerges polarized in the vertical plane. When a second polarizer (the analyser) is placed with its transmission axis crossed at \(90^\circ\) to the first, it blocks these vertical vibrations and no light passes, which confirms that the beam leaving the first polarizer is indeed plane-polarized.
Akọwa Nkọwa
(a) What is a polarizer?
A polarizer is an optical device (for example a Polaroid sheet or a tourmaline crystal) that has a single characteristic direction called its transmission axis. When light falls on it, it transmits only the component of the light vibrations that is parallel to this transmission axis and absorbs (or blocks) all the other components. It therefore converts an unpolarized beam, whose vibrations occur in all planes perpendicular to the direction of travel, into a plane-polarized beam whose vibrations lie in only one plane.
(b) How a polarizer polarizes a beam of unpolarized light
In an unpolarized beam the electric-field vibrations take place in every direction in the plane perpendicular to the direction of propagation. When this beam strikes the polarizer, only the vibration component that is parallel to the polarizer's transmission axis is allowed through; every component perpendicular to that axis is absorbed. The light emerging from the polarizer therefore vibrates in one plane only, that is, it is plane-polarized. Because only one component is transmitted, the emerging intensity is about half that of the incident unpolarized light \(\left(I = \tfrac{1}{2}I_0\right)\).
The action is shown in the diagram below. The unpolarized light (vibrating in all directions) passes through the first polarizer and emerges polarized in the vertical plane. When a second polarizer (the analyser) is placed with its transmission axis crossed at \(90^\circ\) to the first, it blocks these vertical vibrations and no light passes, which confirms that the beam leaving the first polarizer is indeed plane-polarized.
Ajụjụ 6 Ripọtì
(a) Define ionization potential.
(b)(i) State the three types of emission spectra.
(ii) Name one source each which produces each of the spectra stated in (b)(i).
(c) In an x-ray tube, electrons are accelerated the target by a potential difference of 80 A Calculate the:
(i) speed of the electron;
ii) threshold wavelength of the electron. [h=6.6 x 10\(^{-34}\) Js; e = 1.6 x 10\(^{-19}\) C; Me = 9.1 x 10\(^{-31}\)
d) An x-ray photon of frequency 4.5 x 10\(^{-18}\) strikes an. electron, assumed to be at rest. If t electron absorbs all the photon energy, calculate the speed acquired by the electron. [ h = 6.6 x 10\(^{-34}\) Js; Me = 9.1 x 10\(^{-31}\) kg ]
(a) Ionization potential
The ionization potential of an atom is the minimum potential difference (energy per unit charge) needed to remove the most loosely bound (outermost) electron completely from the atom in its ground state.
(b)(i) Three types of emission spectra:
- Continuous (band-free) spectrum.
- Line (atomic) spectrum.
- Band (molecular) spectrum.
(b)(ii) One source of each:
- Continuous spectrum: a glowing solid such as the filament of an electric lamp (incandescent solid).
- Line spectrum: a glowing (excited) gas of a single element, e.g. a sodium/mercury vapour lamp or hydrogen discharge tube.
- Band spectrum: glowing (excited) molecular gases or vapours, e.g. a discharge in a molecular gas.
(c) X-ray tube (electrons accelerated through \(V = 80\,\text{kV}\))
(i) Speed of the electron from \(eV = \tfrac{1}{2}mv^{2}\):
\[ v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2(1.6\times10^{-19})(8.0\times10^{4})}{9.1\times10^{-31}}} = \sqrt{2.81\times10^{16}} = 1.68\times10^{8}\,\text{ms}^{-1} \](ii) Threshold (minimum) wavelength from \(eV = \dfrac{hc}{\lambda}\):
\[ \lambda = \frac{hc}{eV} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{(1.6\times10^{-19})(8.0\times10^{4})} = \frac{1.98\times10^{-25}}{1.28\times10^{-14}} = 1.55\times10^{-11}\,\text{m} \](d) X-ray photon of frequency \(4.5\times10^{18}\,\text{Hz}\) absorbed by a stationary electron
Photon energy: \(E = hf = (6.6\times10^{-34})(4.5\times10^{18}) = 2.97\times10^{-15}\,\text{J}\).
All of it becomes electron kinetic energy, \(\tfrac{1}{2}mv^{2} = E\):
\[ v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2(2.97\times10^{-15})}{9.1\times10^{-31}}} = \sqrt{6.53\times10^{15}} = 8.08\times10^{7}\,\text{ms}^{-1} \]Akọwa Nkọwa
(a) Ionization potential
The ionization potential of an atom is the minimum potential difference (energy per unit charge) needed to remove the most loosely bound (outermost) electron completely from the atom in its ground state.
(b)(i) Three types of emission spectra:
- Continuous (band-free) spectrum.
- Line (atomic) spectrum.
- Band (molecular) spectrum.
(b)(ii) One source of each:
- Continuous spectrum: a glowing solid such as the filament of an electric lamp (incandescent solid).
- Line spectrum: a glowing (excited) gas of a single element, e.g. a sodium/mercury vapour lamp or hydrogen discharge tube.
- Band spectrum: glowing (excited) molecular gases or vapours, e.g. a discharge in a molecular gas.
(c) X-ray tube (electrons accelerated through \(V = 80\,\text{kV}\))
(i) Speed of the electron from \(eV = \tfrac{1}{2}mv^{2}\):
\[ v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2(1.6\times10^{-19})(8.0\times10^{4})}{9.1\times10^{-31}}} = \sqrt{2.81\times10^{16}} = 1.68\times10^{8}\,\text{ms}^{-1} \](ii) Threshold (minimum) wavelength from \(eV = \dfrac{hc}{\lambda}\):
\[ \lambda = \frac{hc}{eV} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{(1.6\times10^{-19})(8.0\times10^{4})} = \frac{1.98\times10^{-25}}{1.28\times10^{-14}} = 1.55\times10^{-11}\,\text{m} \](d) X-ray photon of frequency \(4.5\times10^{18}\,\text{Hz}\) absorbed by a stationary electron
Photon energy: \(E = hf = (6.6\times10^{-34})(4.5\times10^{18}) = 2.97\times10^{-15}\,\text{J}\).
All of it becomes electron kinetic energy, \(\tfrac{1}{2}mv^{2} = E\):
\[ v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2(2.97\times10^{-15})}{9.1\times10^{-31}}} = \sqrt{6.53\times10^{15}} = 8.08\times10^{7}\,\text{ms}^{-1} \]Ajụjụ 7 Ripọtì
Write down the name of:
(a) two particles used in explaining the wave nature of matter;
(b) one device whose invention is based on the wave nature of matter.
(a) Two particles used in explaining the wave nature of matter:
- The electron.
- The neutron (a proton is also acceptable).
(b) One device based on the wave nature of matter:
- The electron microscope.
Akọwa Nkọwa
(a) Two particles used in explaining the wave nature of matter:
- The electron.
- The neutron (a proton is also acceptable).
(b) One device based on the wave nature of matter:
- The electron microscope.
Ajụjụ 8 Ripọtì
In an electrolysis experiment, the ammeter records a steady current of 1 A. The mass of copper deposited in 30 minutes is 0.66 g. Calculate the error in the ammeter reading. [Electrochemical equivalent of copper = 0.00033 g C\(^{-1}\)]
By Faraday's law of electrolysis, the mass deposited is \(m = zIt\), where \(z\) is the electrochemical equivalent, \(I\) the true current and \(t\) the time.
Data: \(m = 0.66\,\text{g}\), \(z = 0.00033\,\text{g C}^{-1}\), \(t = 30\,\text{min} = 1800\,\text{s}\).
True charge that passed:
\[ Q = \frac{m}{z} = \frac{0.66}{0.00033} = 2000\,\text{C} \]True (actual) current:
\[ I_{true} = \frac{Q}{t} = \frac{2000}{1800} = 1.11\,\text{A} \]Error in the ammeter reading:
\[ \text{Error} = I_{true} - I_{reading} = 1.11 - 1.00 = 0.11\,\text{A} \]The ammeter reads about \(0.11\,\text{A}\) too low (a percentage error of about \(\dfrac{0.11}{1.11}\times100 \approx 10\%\)).
Akọwa Nkọwa
By Faraday's law of electrolysis, the mass deposited is \(m = zIt\), where \(z\) is the electrochemical equivalent, \(I\) the true current and \(t\) the time.
Data: \(m = 0.66\,\text{g}\), \(z = 0.00033\,\text{g C}^{-1}\), \(t = 30\,\text{min} = 1800\,\text{s}\).
True charge that passed:
\[ Q = \frac{m}{z} = \frac{0.66}{0.00033} = 2000\,\text{C} \]True (actual) current:
\[ I_{true} = \frac{Q}{t} = \frac{2000}{1800} = 1.11\,\text{A} \]Error in the ammeter reading:
\[ \text{Error} = I_{true} - I_{reading} = 1.11 - 1.00 = 0.11\,\text{A} \]The ammeter reads about \(0.11\,\text{A}\) too low (a percentage error of about \(\dfrac{0.11}{1.11}\times100 \approx 10\%\)).
Ajụjụ 9 Ripọtì
A projectile is released with a speed u at an angle \(\theta\) to the horizontal. With the aid of a diagram, show that the time of flight is equal to \(\frac{2uSin\theta}{g}\), where g is the acceleration of free fall.
Projectile motion and time of flight
A body is projected from ground level with speed \(u\) at an angle \(\theta\) to the horizontal. The diagram shows the parabolic path and the initial velocity resolved into its horizontal and vertical components.
Resolving the initial velocity:
Horizontal component: \(u_x = u\cos\theta\) (this stays constant because there is no horizontal force).
Vertical component: \(u_y = u\sin\theta\) (this is decelerated by gravity \(g\) acting downward).
Time to reach maximum height
Taking upward as positive, the vertical velocity at time \(t\) is
\[ v_y = u\sin\theta - g t \]At the maximum height the vertical velocity is momentarily zero, \(v_y = 0\). Let \(t\) be the time taken to reach this point:
\[ 0 = u\sin\theta - g t \]\[ t = \frac{u\sin\theta}{g} \]Time of flight
By the symmetry of the path, the time taken to fall from the maximum height back to the projection level equals the time taken to rise to it. Hence the total time of flight \(T\) is twice the time to reach maximum height:
\[ T = 2t = \frac{2u\sin\theta}{g} \]This is the required result: the time of flight \(T = \dfrac{2u\sin\theta}{g}\), where \(g\) is the acceleration of free fall.
Akọwa Nkọwa
Projectile motion and time of flight
A body is projected from ground level with speed \(u\) at an angle \(\theta\) to the horizontal. The diagram shows the parabolic path and the initial velocity resolved into its horizontal and vertical components.
Resolving the initial velocity:
Horizontal component: \(u_x = u\cos\theta\) (this stays constant because there is no horizontal force).
Vertical component: \(u_y = u\sin\theta\) (this is decelerated by gravity \(g\) acting downward).
Time to reach maximum height
Taking upward as positive, the vertical velocity at time \(t\) is
\[ v_y = u\sin\theta - g t \]At the maximum height the vertical velocity is momentarily zero, \(v_y = 0\). Let \(t\) be the time taken to reach this point:
\[ 0 = u\sin\theta - g t \]\[ t = \frac{u\sin\theta}{g} \]Time of flight
By the symmetry of the path, the time taken to fall from the maximum height back to the projection level equals the time taken to rise to it. Hence the total time of flight \(T\) is twice the time to reach maximum height:
\[ T = 2t = \frac{2u\sin\theta}{g} \]This is the required result: the time of flight \(T = \dfrac{2u\sin\theta}{g}\), where \(g\) is the acceleration of free fall.
Ajụjụ 10 Ripọtì
Name one use of 'LASER' in each of the following areas:
(a) communication;
(b) medicine;
(c) security
Uses of the LASER
- (a) Communication: transmission of signals (voice, data, TV) through optical fibre cables using laser light.
- (b) Medicine: laser surgery, for example bloodless cutting of tissue, welding a detached retina, or removing tumours.
- (c) Security: laser beams used in burglar-alarm/intruder-detection systems (and in barcode/scanner and printing security).
Akọwa Nkọwa
Uses of the LASER
- (a) Communication: transmission of signals (voice, data, TV) through optical fibre cables using laser light.
- (b) Medicine: laser surgery, for example bloodless cutting of tissue, welding a detached retina, or removing tumours.
- (c) Security: laser beams used in burglar-alarm/intruder-detection systems (and in barcode/scanner and printing security).
Ajụjụ 11 Ripọtì
(a) State the triangle law of vector addition.
(b) Name the four physical quantities that are associated with the equationq of linear motion.
(c) Using the same set of axes, sketch and label two graphs to illustrate the variation of potential energy and kinetic energy with time for a body in simple harmonic motion.
(d)
A light spiral spring of force constant K lies on a horizontal frictionless surface and has one end fixed to a vertical wall. A block P of mass 2.0 kg placed against the free end of the spring is pushed a distance 5 cm towards the wall with 10J of energy as illustrated in the diagram above. The block is released and after 0.25s, it collides inelastically with a stationary block Q of mass 4.0 kg. Calculate the:
(i) value of k;
(ii) force used to compress the spring;
(iii) acceleration of the block p after release;
(iv) common speed after collision of the blocks.
(a) Triangle law of vector addition
If two vectors are represented, in magnitude and direction, by two sides of a triangle taken in order (head-to-tail), their resultant is represented by the third side of the triangle drawn from the tail of the first vector to the head of the second vector.
This is closely related to the triangle-of-forces statement: three forces in equilibrium can be represented by the three sides of a triangle taken in order.
(b) Physical quantities in the equations of linear motion
The four quantities are:
- displacement, \(s\),
- initial and final velocity, \(u\) and \(v\),
- acceleration, \(a\),
- time, \(t\).
For example, \(v=u+at\) relates velocity, acceleration and time, while \(s=ut+\tfrac{1}{2}at^2\) also includes displacement.
(c) Potential energy and kinetic energy in SHM
For an ideal body in simple harmonic motion, total mechanical energy is constant. Kinetic energy is greatest at the equilibrium position, whereas potential energy is greatest at each extreme position. Thus, when one energy is zero, the other is at its maximum value \(E\).
The energy curves have a period of \(T/2\), because energy depends on the square of displacement or velocity. At every instant,
\[KE+PE=E.\]
(d) Spring, block and collision
Given:
- mass of block P: \(m_P=2.0\,\mathrm{kg}\),
- mass of block Q: \(m_Q=4.0\,\mathrm{kg}\),
- compression: \(x=5.0\,\mathrm{cm}=0.050\,\mathrm{m}\),
- elastic potential energy stored: \(E=10\,\mathrm{J}\).
(i) Force constant of the spring
The elastic potential energy in a compressed spring is
\[E=\frac{1}{2}kx^2.\]
Therefore,
\[10=\frac{1}{2}k(0.050)^2\]
\[k=\frac{2(10)}{(0.050)^2}=8.0\times10^3\,\mathrm{N\,m^{-1}}.\]
(ii) Force used to compress the spring
The spring force is greatest at the maximum compression of \(0.050\,\mathrm{m}\):
\[F=kx=(8.0\times10^3)(0.050)=400\,\mathrm{N}.\]
Therefore, the force needed to hold the spring at this compression is \(400\,\mathrm{N}\). The force is not constant while the spring is being compressed; it increases from zero to \(400\,\mathrm{N}\).
(iii) Acceleration of P immediately after release
Immediately after release, the spring force is \(400\,\mathrm{N}\), so Newton’s second law gives
\[a=\frac{F}{m_P}=\frac{400}{2.0}=200\,\mathrm{m\,s^{-2}}.\]
The acceleration is \(200\,\mathrm{m\,s^{-2}}\) immediately after release, directed away from the wall. It then decreases as the spring expands, because \(F=kx\) decreases.
(iv) Common speed after the inelastic collision
The supplied reference solution treats the acceleration as constant for \(0.25\,\mathrm{s}\), producing \(50\,\mathrm{m\,s^{-1}}\). This is not physically consistent: the spring contains only \(10\,\mathrm{J}\), whereas a \(2.0\,\mathrm{kg}\) block moving at \(50\,\mathrm{m\,s^{-1}}\) would have \(2500\,\mathrm{J}\) of kinetic energy.
Since the surface is frictionless, the \(10\,\mathrm{J}\) of elastic potential energy becomes kinetic energy of P when the spring returns to its natural length:
\[\frac{1}{2}m_Pu_P^2=10\]
\[\frac{1}{2}(2.0)u_P^2=10\]
\[u_P=\sqrt{10}=3.16\,\mathrm{m\,s^{-1}}.\]
After leaving the spring, P continues at this constant speed until it reaches Q. Momentum is conserved in the inelastic collision:
\[m_Pu_P+m_Qu_Q=(m_P+m_Q)V.\]
Since Q is stationary, \(u_Q=0\):
\[(2.0)(3.16)+(4.0)(0)=(2.0+4.0)V.\]
\[V=\frac{6.32}{6.0}=1.05\,\mathrm{m\,s^{-1}}.\]
The common speed of the blocks after collision is therefore \(1.05\,\mathrm{m\,s^{-1}}\).
Examination reminder: For a spring, do not use \(v=at\) over the whole motion unless the force, and therefore acceleration, is constant. Here the force falls as the spring expands, so conservation of energy gives the speed correctly.
Akọwa Nkọwa
(a) Triangle law of vector addition
If two vectors are represented, in magnitude and direction, by two sides of a triangle taken in order (head-to-tail), their resultant is represented by the third side of the triangle drawn from the tail of the first vector to the head of the second vector.
This is closely related to the triangle-of-forces statement: three forces in equilibrium can be represented by the three sides of a triangle taken in order.
(b) Physical quantities in the equations of linear motion
The four quantities are:
- displacement, \(s\),
- initial and final velocity, \(u\) and \(v\),
- acceleration, \(a\),
- time, \(t\).
For example, \(v=u+at\) relates velocity, acceleration and time, while \(s=ut+\tfrac{1}{2}at^2\) also includes displacement.
(c) Potential energy and kinetic energy in SHM
For an ideal body in simple harmonic motion, total mechanical energy is constant. Kinetic energy is greatest at the equilibrium position, whereas potential energy is greatest at each extreme position. Thus, when one energy is zero, the other is at its maximum value \(E\).
The energy curves have a period of \(T/2\), because energy depends on the square of displacement or velocity. At every instant,
\[KE+PE=E.\]
(d) Spring, block and collision
Given:
- mass of block P: \(m_P=2.0\,\mathrm{kg}\),
- mass of block Q: \(m_Q=4.0\,\mathrm{kg}\),
- compression: \(x=5.0\,\mathrm{cm}=0.050\,\mathrm{m}\),
- elastic potential energy stored: \(E=10\,\mathrm{J}\).
(i) Force constant of the spring
The elastic potential energy in a compressed spring is
\[E=\frac{1}{2}kx^2.\]
Therefore,
\[10=\frac{1}{2}k(0.050)^2\]
\[k=\frac{2(10)}{(0.050)^2}=8.0\times10^3\,\mathrm{N\,m^{-1}}.\]
(ii) Force used to compress the spring
The spring force is greatest at the maximum compression of \(0.050\,\mathrm{m}\):
\[F=kx=(8.0\times10^3)(0.050)=400\,\mathrm{N}.\]
Therefore, the force needed to hold the spring at this compression is \(400\,\mathrm{N}\). The force is not constant while the spring is being compressed; it increases from zero to \(400\,\mathrm{N}\).
(iii) Acceleration of P immediately after release
Immediately after release, the spring force is \(400\,\mathrm{N}\), so Newton’s second law gives
\[a=\frac{F}{m_P}=\frac{400}{2.0}=200\,\mathrm{m\,s^{-2}}.\]
The acceleration is \(200\,\mathrm{m\,s^{-2}}\) immediately after release, directed away from the wall. It then decreases as the spring expands, because \(F=kx\) decreases.
(iv) Common speed after the inelastic collision
The supplied reference solution treats the acceleration as constant for \(0.25\,\mathrm{s}\), producing \(50\,\mathrm{m\,s^{-1}}\). This is not physically consistent: the spring contains only \(10\,\mathrm{J}\), whereas a \(2.0\,\mathrm{kg}\) block moving at \(50\,\mathrm{m\,s^{-1}}\) would have \(2500\,\mathrm{J}\) of kinetic energy.
Since the surface is frictionless, the \(10\,\mathrm{J}\) of elastic potential energy becomes kinetic energy of P when the spring returns to its natural length:
\[\frac{1}{2}m_Pu_P^2=10\]
\[\frac{1}{2}(2.0)u_P^2=10\]
\[u_P=\sqrt{10}=3.16\,\mathrm{m\,s^{-1}}.\]
After leaving the spring, P continues at this constant speed until it reaches Q. Momentum is conserved in the inelastic collision:
\[m_Pu_P+m_Qu_Q=(m_P+m_Q)V.\]
Since Q is stationary, \(u_Q=0\):
\[(2.0)(3.16)+(4.0)(0)=(2.0+4.0)V.\]
\[V=\frac{6.32}{6.0}=1.05\,\mathrm{m\,s^{-1}}.\]
The common speed of the blocks after collision is therefore \(1.05\,\mathrm{m\,s^{-1}}\).
Examination reminder: For a spring, do not use \(v=at\) over the whole motion unless the force, and therefore acceleration, is constant. Here the force falls as the spring expands, so conservation of energy gives the speed correctly.
Ajụjụ 12 Ripọtì
Name the three basic components P, Q and R that make up a cathode ray tube, as illustrated in the diagram above
The diagram is a cathode-ray tube (CRT), and the three labelled sections are its three basic components:
- P - the electron gun: the assembly at the narrow end containing the heated filament/cathode and the anodes (focusing and accelerating electrodes). It produces, focuses and accelerates a narrow beam of electrons.
- Q - the deflecting system (deflecting plates): the pairs of parallel plates (the X- and Y-plates) through which the beam passes. They deflect the electron beam horizontally and vertically by means of the electric fields between the plates.
- R - the fluorescent screen: the flared front face of the tube, coated on the inside with a fluorescent (phosphor) material. It glows to produce a visible spot where the electron beam strikes it.
Akọwa Nkọwa
The diagram is a cathode-ray tube (CRT), and the three labelled sections are its three basic components:
- P - the electron gun: the assembly at the narrow end containing the heated filament/cathode and the anodes (focusing and accelerating electrodes). It produces, focuses and accelerates a narrow beam of electrons.
- Q - the deflecting system (deflecting plates): the pairs of parallel plates (the X- and Y-plates) through which the beam passes. They deflect the electron beam horizontally and vertically by means of the electric fields between the plates.
- R - the fluorescent screen: the flared front face of the tube, coated on the inside with a fluorescent (phosphor) material. It glows to produce a visible spot where the electron beam strikes it.
Ajụjụ 13 Ripọtì
(a) Define boiling point of a liquid.
(b) Describe how water in a round bottom flask could be made to boil without heating it. [diagram not necessary]
(c) State three applications of expansion of metals.
(d) A room with floor measurements 7m x 10 m contains air of mass 250 kg at a temperature of 34°C. The air is cooled until the temperature falls to 24°C. Calculate the: (i) height of the room;
(ii) quantity of energy extracted to cool the room;
(iii) which is higher: the calculated value or the actual energy needed to cool the room? Give a reason for your answer. [ Specific heat capacity of air = 1010 Jkg\(^{-1}\)K\(^{-1}\); density of air = 1.25 kg m\(^{-3}]
(a) Boiling point
The boiling point of a liquid is the constant temperature at which the liquid boils, i.e. at which its saturated vapour pressure equals the external atmospheric pressure.
(b) Boiling water without heating it
Put warm water in a round-bottomed flask, cork it, and connect it to a vacuum pump. As the pump lowers the pressure above the water, the boiling point falls; when the reduced pressure equals the water's vapour pressure at that temperature, the water boils without any further heating. (Alternatively, boil the water, cork the flask, invert it and pour cold water over the top: the vapour inside condenses, pressure drops, and the water boils again.)
(c) Three applications of expansion of metals
- The bimetallic strip in thermostats and fire alarms.
- Fitting (shrink-fitting) of metal tyres/rims onto wheels and rivets.
- Expansion gaps/rollers left in railway lines and bridges.
(d) Cooling the room air from 34°C to 24°C
Floor area \(= 7\times10 = 70\,\text{m}^{2}\); air mass \(= 250\,\text{kg}\); \(c = 1010\,\text{Jkg}^{-1}\text{K}^{-1}\); density \(= 1.25\,\text{kg m}^{-3}\).
(i) Height of the room: volume of air \(= \dfrac{m}{\rho} = \dfrac{250}{1.25} = 200\,\text{m}^{3}\).
\[ \text{height} = \frac{\text{volume}}{\text{floor area}} = \frac{200}{70} = 2.86\,\text{m} \](ii) Energy extracted:
\[ Q = mc\,\Delta\theta = 250\times1010\times(34-24) = 250\times1010\times10 = 2.525\times10^{6}\,\text{J} \](iii) Which is higher? The actual energy that must be extracted is higher than the calculated value. The calculation accounts only for cooling the air; in practice heat must also be removed from the walls, furniture and occupants, and heat continually leaks in from the warmer surroundings, so more energy is needed than the value found above.
Akọwa Nkọwa
(a) Boiling point
The boiling point of a liquid is the constant temperature at which the liquid boils, i.e. at which its saturated vapour pressure equals the external atmospheric pressure.
(b) Boiling water without heating it
Put warm water in a round-bottomed flask, cork it, and connect it to a vacuum pump. As the pump lowers the pressure above the water, the boiling point falls; when the reduced pressure equals the water's vapour pressure at that temperature, the water boils without any further heating. (Alternatively, boil the water, cork the flask, invert it and pour cold water over the top: the vapour inside condenses, pressure drops, and the water boils again.)
(c) Three applications of expansion of metals
- The bimetallic strip in thermostats and fire alarms.
- Fitting (shrink-fitting) of metal tyres/rims onto wheels and rivets.
- Expansion gaps/rollers left in railway lines and bridges.
(d) Cooling the room air from 34°C to 24°C
Floor area \(= 7\times10 = 70\,\text{m}^{2}\); air mass \(= 250\,\text{kg}\); \(c = 1010\,\text{Jkg}^{-1}\text{K}^{-1}\); density \(= 1.25\,\text{kg m}^{-3}\).
(i) Height of the room: volume of air \(= \dfrac{m}{\rho} = \dfrac{250}{1.25} = 200\,\text{m}^{3}\).
\[ \text{height} = \frac{\text{volume}}{\text{floor area}} = \frac{200}{70} = 2.86\,\text{m} \](ii) Energy extracted:
\[ Q = mc\,\Delta\theta = 250\times1010\times(34-24) = 250\times1010\times10 = 2.525\times10^{6}\,\text{J} \](iii) Which is higher? The actual energy that must be extracted is higher than the calculated value. The calculation accounts only for cooling the air; in practice heat must also be removed from the walls, furniture and occupants, and heat continually leaks in from the warmer surroundings, so more energy is needed than the value found above.
Ajụjụ 14 Ripọtì
(a) what is Brownian motion?
(b) State the two inferences that can be drawn from Brownian motion experiment.
(a) Brownian motion
Brownian motion is the continuous, random (zig-zag) movement of tiny particles suspended in a fluid (such as smoke particles in air or pollen grains in water), caused by their being bombarded unevenly by the fast-moving molecules of the surrounding fluid.
(b) Two inferences from the Brownian-motion experiment
- Matter (the fluid) is made up of tiny particles (molecules).
- These molecules are in constant, random motion and possess kinetic energy (the motion becomes more vigorous as temperature rises).
Akọwa Nkọwa
(a) Brownian motion
Brownian motion is the continuous, random (zig-zag) movement of tiny particles suspended in a fluid (such as smoke particles in air or pollen grains in water), caused by their being bombarded unevenly by the fast-moving molecules of the surrounding fluid.
(b) Two inferences from the Brownian-motion experiment
- Matter (the fluid) is made up of tiny particles (molecules).
- These molecules are in constant, random motion and possess kinetic energy (the motion becomes more vigorous as temperature rises).
Ajụjụ 15 Ripọtì
(a) On which day would sound wave travel faster: on a hot or cold day? Explain.
(b) Why are megaphones shaped like funnels?
(c) A ray of light is incident on a surface of a ectangular glass prism of refractive index 1.5 illustrated in the diagram below.
(i) Copy the diagram a label the angles of: (\(\alpha\)) Incidence (x); (\(\beta\)) Reflection (y); (\(\gamma\)) refraction (z); with t glass letters indicated.
(ii) Calculate the angle refraction to the nearest whole number.
(d) A sonomesr wire vibrates in simple harmoi motion with a maximum amplitude of 1.0 cm. Calculate the frequency of vibration of the wire, giv that the magnitade of the maximum acceleration of the wire is 980ms\(^{-2}\). [\(\pi \frac{22}{7}\)]
(a) Faster on a hot or cold day?
Sound travels faster on a hot day. The speed of sound in air increases with temperature because the air molecules move faster and transmit the disturbance more rapidly (the speed is proportional to \(\sqrt{T}\), where \(T\) is the absolute temperature).
(b) Why megaphones are shaped like funnels
The funnel (conical) shape channels the sound energy and directs it forward in a narrow beam instead of allowing it to spread out in all directions. This concentrates the sound waves, so the sound is louder and carries farther in the intended direction.
(c) Refraction at the air-glass surface (refractive index \(n = 1.5\), angle of incidence read from the diagram \(= 30^{\circ}\))
(i) In the copied diagram: the angle of incidence (x) is between the incident ray and the normal in the air; the angle of reflection (y) is between the reflected ray and the normal (in air, equal to x); the angle of refraction (z) is between the refracted ray and the normal inside the glass.
(ii) Applying Snell's law from air into glass:
\[ n = \frac{\sin i}{\sin r} \;\Rightarrow\; \sin r = \frac{\sin i}{n} = \frac{\sin 30^{\circ}}{1.5} = \frac{0.5}{1.5} = 0.3333 \]
\[ r = \sin^{-1}(0.3333) = 19.47^{\circ} \approx 19^{\circ} \]
(d) Frequency of the vibrating sonometer wire (SHM)
Amplitude \(A = 1.0\ \text{cm} = 0.01\ \text{m}\); maximum acceleration \(a_{max} = 980\ \text{m s}^{-2}\). For SHM, \(a_{max} = \omega^{2} A\):
\[ \omega^{2} = \frac{a_{max}}{A} = \frac{980}{0.01} = 98000 \]
\[ \omega = \sqrt{98000} = 313.05\ \text{rad s}^{-1} \]
Since \(\omega = 2\pi f\), with \(\pi = \dfrac{22}{7}\):
\[ f = \frac{\omega}{2\pi} = \frac{313.05}{2 \times \tfrac{22}{7}} = \frac{313.05}{6.286} \approx 49.8\ \text{Hz} \approx 50\ \text{Hz} \]
Akọwa Nkọwa
(a) Faster on a hot or cold day?
Sound travels faster on a hot day. The speed of sound in air increases with temperature because the air molecules move faster and transmit the disturbance more rapidly (the speed is proportional to \(\sqrt{T}\), where \(T\) is the absolute temperature).
(b) Why megaphones are shaped like funnels
The funnel (conical) shape channels the sound energy and directs it forward in a narrow beam instead of allowing it to spread out in all directions. This concentrates the sound waves, so the sound is louder and carries farther in the intended direction.
(c) Refraction at the air-glass surface (refractive index \(n = 1.5\), angle of incidence read from the diagram \(= 30^{\circ}\))
(i) In the copied diagram: the angle of incidence (x) is between the incident ray and the normal in the air; the angle of reflection (y) is between the reflected ray and the normal (in air, equal to x); the angle of refraction (z) is between the refracted ray and the normal inside the glass.
(ii) Applying Snell's law from air into glass:
\[ n = \frac{\sin i}{\sin r} \;\Rightarrow\; \sin r = \frac{\sin i}{n} = \frac{\sin 30^{\circ}}{1.5} = \frac{0.5}{1.5} = 0.3333 \]
\[ r = \sin^{-1}(0.3333) = 19.47^{\circ} \approx 19^{\circ} \]
(d) Frequency of the vibrating sonometer wire (SHM)
Amplitude \(A = 1.0\ \text{cm} = 0.01\ \text{m}\); maximum acceleration \(a_{max} = 980\ \text{m s}^{-2}\). For SHM, \(a_{max} = \omega^{2} A\):
\[ \omega^{2} = \frac{a_{max}}{A} = \frac{980}{0.01} = 98000 \]
\[ \omega = \sqrt{98000} = 313.05\ \text{rad s}^{-1} \]
Since \(\omega = 2\pi f\), with \(\pi = \dfrac{22}{7}\):
\[ f = \frac{\omega}{2\pi} = \frac{313.05}{2 \times \tfrac{22}{7}} = \frac{313.05}{6.286} \approx 49.8\ \text{Hz} \approx 50\ \text{Hz} \]
