WAEC SSCE - Physics - 2000 (Essay)
Tambaya 1 Rahoto
(a)(i) Define the term linear momentum.
(ii) State the law of conservation of linear momentum.
(b) A ball P of mass 0.25 kg, loses one-third of its velocity when it makes a head on collision with an identical ball Q at rest. After the collision, Q moves off with a speed of 2ms\(^{-1}\) in the original direction of P. Calculate the initial velocity of R
(c)(i) State Newton's second law of motion.
(ii) Show that F = ma where F is the magnitude of the force acting on a body of mass m to give it an acceleration of magnitude a.
(iii) The engine of a vehicle moves it forward with a force of 9600 N against a resistive force of 2200 N. If the mass of the vehicle is 3400 kg, calculate the acceleration produced.
(a)(i) Linear momentum is the product of the mass of a body and its velocity, \(p = mv\). It is a vector quantity with SI unit \(\text{kg m s}^{-1}\).
(a)(ii) Law of conservation of linear momentum: In a closed system on which no external resultant force acts, the total linear momentum before an interaction (e.g. a collision) equals the total linear momentum after it.
(b) Let the initial velocity of P be \(u\). Both balls have equal mass \(m = 0.25\ \text{kg}\). P loses one-third of its velocity, so after collision P moves with \(v_P = u - \tfrac{1}{3}u = \tfrac{2}{3}u\). Q starts at rest and moves off at \(v_Q = 2\ \text{m s}^{-1}\).
Conservation of momentum (mass cancels):
\[ u = \tfrac{2}{3}u + 2 \;\Rightarrow\; \tfrac{1}{3}u = 2 \;\Rightarrow\; u = 6\ \text{m s}^{-1}. \]Initial velocity of P \(= 6\ \text{m s}^{-1}\).
(c)(i) Newton's second law: The rate of change of momentum of a body is directly proportional to the resultant force acting on it and takes place in the direction of that force.
(c)(ii) For a body of mass \(m\) whose velocity changes from \(u\) to \(v\) in time \(t\),
\[ F \propto \frac{mv - mu}{t} = \frac{m(v-u)}{t} = ma, \]since \(a = \dfrac{v-u}{t}\). Writing the proportionality with a constant \(k\), \(F = kma\); the newton is defined so that \(k = 1\), giving \(F = ma\).
(c)(iii) Net (resultant) force \(= 9600 - 2200 = 7400\ \text{N}\). Acceleration
\[ a = \frac{F}{m} = \frac{7400}{3400} = 2.18\ \text{m s}^{-2}. \]Bayanin Amsa
(a)(i) Linear momentum is the product of the mass of a body and its velocity, \(p = mv\). It is a vector quantity with SI unit \(\text{kg m s}^{-1}\).
(a)(ii) Law of conservation of linear momentum: In a closed system on which no external resultant force acts, the total linear momentum before an interaction (e.g. a collision) equals the total linear momentum after it.
(b) Let the initial velocity of P be \(u\). Both balls have equal mass \(m = 0.25\ \text{kg}\). P loses one-third of its velocity, so after collision P moves with \(v_P = u - \tfrac{1}{3}u = \tfrac{2}{3}u\). Q starts at rest and moves off at \(v_Q = 2\ \text{m s}^{-1}\).
Conservation of momentum (mass cancels):
\[ u = \tfrac{2}{3}u + 2 \;\Rightarrow\; \tfrac{1}{3}u = 2 \;\Rightarrow\; u = 6\ \text{m s}^{-1}. \]Initial velocity of P \(= 6\ \text{m s}^{-1}\).
(c)(i) Newton's second law: The rate of change of momentum of a body is directly proportional to the resultant force acting on it and takes place in the direction of that force.
(c)(ii) For a body of mass \(m\) whose velocity changes from \(u\) to \(v\) in time \(t\),
\[ F \propto \frac{mv - mu}{t} = \frac{m(v-u)}{t} = ma, \]since \(a = \dfrac{v-u}{t}\). Writing the proportionality with a constant \(k\), \(F = kma\); the newton is defined so that \(k = 1\), giving \(F = ma\).
(c)(iii) Net (resultant) force \(= 9600 - 2200 = 7400\ \text{N}\). Acceleration
\[ a = \frac{F}{m} = \frac{7400}{3400} = 2.18\ \text{m s}^{-2}. \]Tambaya 2 Rahoto
a) Define the term surface tension
(b) Calculate the force required to lift a needle 4cm long off the surface of water if the surface tension of water is 7.3 x 10\(^{-4}\) NM\(^{-1}\)
(a) Surface tension
Surface tension is the force per unit length acting along (tangent to) the surface of a liquid, at right angles to a line drawn in the surface. It arises from the net inward attraction on the surface molecules and makes the surface behave like a stretched elastic skin. Its unit is \(\text{Nm}^{-1}\).
(b) Force to lift the needle
A floating needle is held by the surface film on both sides (two surfaces), each of length \(L\). The force due to surface tension is:
\[ F = \gamma \times 2L \]Data: \(\gamma = 7.3\times10^{-4}\,\text{Nm}^{-1}\), \(L = 4\,\text{cm} = 0.04\,\text{m}\).
\[ F = 7.3\times10^{-4} \times 2 \times 0.04 = 7.3\times10^{-4}\times0.08 = 5.84\times10^{-5}\,\text{N} \]The force required is about \(5.84\times10^{-5}\,\text{N}\).
Bayanin Amsa
(a) Surface tension
Surface tension is the force per unit length acting along (tangent to) the surface of a liquid, at right angles to a line drawn in the surface. It arises from the net inward attraction on the surface molecules and makes the surface behave like a stretched elastic skin. Its unit is \(\text{Nm}^{-1}\).
(b) Force to lift the needle
A floating needle is held by the surface film on both sides (two surfaces), each of length \(L\). The force due to surface tension is:
\[ F = \gamma \times 2L \]Data: \(\gamma = 7.3\times10^{-4}\,\text{Nm}^{-1}\), \(L = 4\,\text{cm} = 0.04\,\text{m}\).
\[ F = 7.3\times10^{-4} \times 2 \times 0.04 = 7.3\times10^{-4}\times0.08 = 5.84\times10^{-5}\,\text{N} \]The force required is about \(5.84\times10^{-5}\,\text{N}\).
Tambaya 3 Rahoto
What is a projectile? Give four examples of projectiles in everyday life
Projectile
A projectile is any object that is given an initial velocity and then moves freely under the action of gravity alone (after release, no driving force acts on it apart from its weight), following a curved (parabolic) path.
Four everyday examples of projectiles
- A ball kicked or thrown into the air (e.g. a football or cricket ball).
- A bullet fired from a gun.
- An arrow shot from a bow.
- A stone thrown horizontally or at an angle; water spurting from a hose; a long-jumper in flight.
Bayanin Amsa
Projectile
A projectile is any object that is given an initial velocity and then moves freely under the action of gravity alone (after release, no driving force acts on it apart from its weight), following a curved (parabolic) path.
Four everyday examples of projectiles
- A ball kicked or thrown into the air (e.g. a football or cricket ball).
- A bullet fired from a gun.
- An arrow shot from a bow.
- A stone thrown horizontally or at an angle; water spurting from a hose; a long-jumper in flight.
Tambaya 4 Rahoto
(a) Define the following terms:
(i) Electric field intensity
(ii) Electric potential
(b) The diagram below illustrates two collinear electric charges of magnitudes + Q and -Q. The charges are equidistant from a point P at which a rest charge is placed.
Copy the diagram and use arrows to indicate, from the point P, the direction of the;
(i) electric force F\(_1\) due to + Q.
(ii) electric force F\(_2\) due to -Q.
(iii) electric field intensity E.
(c) What is meant by dielectric substance?
(ii) List the factors which determine the capacitance of a parallel plate capacitor and state the effect each of them has on the capacitance.The diagram above represents a section of a circuit. Calculate the effective capacitance in the section.
(iii)
The diagram above represents a section of a circuit. Calculate the effective capacitance in the section.
(a)(i) Electric field intensity at a point is the electric force experienced per unit positive charge placed at that point, \( E = \dfrac{F}{Q} \), measured in \( \text{N C}^{-1} \) (or \( \text{V m}^{-1} \)). It is a vector quantity.
(a)(ii) Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point against the electric field, \( V = \dfrac{W}{Q} \), measured in volts (\( \text{J C}^{-1} \)). It is a scalar quantity.
(b) The charges \(+Q\) and \(-Q\) are collinear, with the rest charge at \(P\) lying on the same line midway between them. Taking the rest charge as positive:
- (i) Force \(F_1\) due to \(+Q\): like charges repel, so the arrow points away from \(+Q\), that is from \(P\) directed towards \(-Q\).
- (ii) Force \(F_2\) due to \(-Q\): unlike charges attract, so the arrow also points from \(P\) directed towards \(-Q\).
- (iii) Resultant field intensity \(E\): both forces act in the same direction along the line, so \(E\) is directed from the \(+Q\) side towards the \(-Q\) side, with magnitude equal to the sum of the two contributions.
(c)(i) A dielectric substance is an insulating (non-conducting) material placed between the plates of a capacitor; it becomes polarised in the field and increases the capacitance.
(c)(ii) Factors determining the capacitance of a parallel-plate capacitor \(\left(C = \dfrac{K\,\varepsilon_0 A}{d}\right)\):
| Factor | Effect on capacitance |
|---|---|
| Common (overlap) area of the plates, \(A\) | Capacitance increases as the area increases (directly proportional). |
| Distance between the plates, \(d\) | Capacitance decreases as the separation increases (inversely proportional). |
| Nature of the dielectric (dielectric constant \(K\)) | Capacitance increases as the dielectric constant increases. |
(c)(iii) Effective capacitance of the section. \(C_2\) and \(C_3\) are in parallel:
\[ C_{2,3} = C_2 + C_3 = 20 + 20 = 40\ \mu\text{F}. \]
This combination is in series with \(C_1 = 40\ \mu\text{F}\):
\[ \frac{1}{C_T} = \frac{1}{C_{2,3}} + \frac{1}{C_1} \quad\Rightarrow\quad C_T = \frac{C_{2,3}\times C_1}{C_{2,3}+C_1} = \frac{40\times 40}{40+40} = 20\ \mu\text{F}. \]
The effective capacitance of the section is \(20\ \mu\text{F}\).
Bayanin Amsa
(a)(i) Electric field intensity at a point is the electric force experienced per unit positive charge placed at that point, \( E = \dfrac{F}{Q} \), measured in \( \text{N C}^{-1} \) (or \( \text{V m}^{-1} \)). It is a vector quantity.
(a)(ii) Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point against the electric field, \( V = \dfrac{W}{Q} \), measured in volts (\( \text{J C}^{-1} \)). It is a scalar quantity.
(b) The charges \(+Q\) and \(-Q\) are collinear, with the rest charge at \(P\) lying on the same line midway between them. Taking the rest charge as positive:
- (i) Force \(F_1\) due to \(+Q\): like charges repel, so the arrow points away from \(+Q\), that is from \(P\) directed towards \(-Q\).
- (ii) Force \(F_2\) due to \(-Q\): unlike charges attract, so the arrow also points from \(P\) directed towards \(-Q\).
- (iii) Resultant field intensity \(E\): both forces act in the same direction along the line, so \(E\) is directed from the \(+Q\) side towards the \(-Q\) side, with magnitude equal to the sum of the two contributions.
(c)(i) A dielectric substance is an insulating (non-conducting) material placed between the plates of a capacitor; it becomes polarised in the field and increases the capacitance.
(c)(ii) Factors determining the capacitance of a parallel-plate capacitor \(\left(C = \dfrac{K\,\varepsilon_0 A}{d}\right)\):
| Factor | Effect on capacitance |
|---|---|
| Common (overlap) area of the plates, \(A\) | Capacitance increases as the area increases (directly proportional). |
| Distance between the plates, \(d\) | Capacitance decreases as the separation increases (inversely proportional). |
| Nature of the dielectric (dielectric constant \(K\)) | Capacitance increases as the dielectric constant increases. |
(c)(iii) Effective capacitance of the section. \(C_2\) and \(C_3\) are in parallel:
\[ C_{2,3} = C_2 + C_3 = 20 + 20 = 40\ \mu\text{F}. \]
This combination is in series with \(C_1 = 40\ \mu\text{F}\):
\[ \frac{1}{C_T} = \frac{1}{C_{2,3}} + \frac{1}{C_1} \quad\Rightarrow\quad C_T = \frac{C_{2,3}\times C_1}{C_{2,3}+C_1} = \frac{40\times 40}{40+40} = 20\ \mu\text{F}. \]
The effective capacitance of the section is \(20\ \mu\text{F}\).
Tambaya 5 Rahoto
An electron of mass 9. 1 x 10\(^{-31}\) kg moves with a velocity of 4.2 x 10 ms\(^{-1}\) between the cathode and anode of an x-ray tube. Calculate the wavelength. (Take Planck's constant h = 6.6 x 10\(^{-34}\) J S)
The wavelength associated with a moving electron is the de Broglie wavelength
\[ \lambda = \frac{h}{mv}, \]where \(h = 6.6\times10^{-34}\ \text{J s}\), \(m = 9.1\times10^{-31}\ \text{kg}\) and \(v = 4.2\times10^{7}\ \text{m s}^{-1}\) (the electron speed in an x-ray tube).
\[ mv = 9.1\times10^{-31} \times 4.2\times10^{7} = 3.82\times10^{-23}\ \text{kg m s}^{-1}. \] \[ \lambda = \frac{6.6\times10^{-34}}{3.82\times10^{-23}} = 1.73\times10^{-11}\ \text{m}. \]Wavelength \(\approx 1.7\times10^{-11}\ \text{m}\) (about \(0.017\ \text{nm}\)), which lies in the X-ray region.
Bayanin Amsa
The wavelength associated with a moving electron is the de Broglie wavelength
\[ \lambda = \frac{h}{mv}, \]where \(h = 6.6\times10^{-34}\ \text{J s}\), \(m = 9.1\times10^{-31}\ \text{kg}\) and \(v = 4.2\times10^{7}\ \text{m s}^{-1}\) (the electron speed in an x-ray tube).
\[ mv = 9.1\times10^{-31} \times 4.2\times10^{7} = 3.82\times10^{-23}\ \text{kg m s}^{-1}. \] \[ \lambda = \frac{6.6\times10^{-34}}{3.82\times10^{-23}} = 1.73\times10^{-11}\ \text{m}. \]Wavelength \(\approx 1.7\times10^{-11}\ \text{m}\) (about \(0.017\ \text{nm}\)), which lies in the X-ray region.
Tambaya 6 Rahoto
(a) Differentiate between interference and polarisation as applied to waves.
(b) Mention two uses of polaroids
(a) Interference vs polarisation
Interference is the superposition of two waves of the same frequency travelling in the same region, so that they reinforce one another (constructive interference) or cancel one another (destructive interference) depending on their path difference. It occurs in all types of waves (both transverse and longitudinal).
Polarisation is the restriction of the vibrations of a wave to a single plane containing the direction of travel. It occurs only in transverse waves, because it depends on the vibrations being at right angles to the direction of propagation. The fact that light can be polarised is direct evidence that light is a transverse wave.
(b) Two uses of polaroids
- In polaroid sunglasses to cut down glare (reflected, horizontally polarised light).
- In photographic filters and in liquid-crystal displays (LCDs) to control the intensity and plane of transmitted light; also in stress analysis of transparent materials.
Bayanin Amsa
(a) Interference vs polarisation
Interference is the superposition of two waves of the same frequency travelling in the same region, so that they reinforce one another (constructive interference) or cancel one another (destructive interference) depending on their path difference. It occurs in all types of waves (both transverse and longitudinal).
Polarisation is the restriction of the vibrations of a wave to a single plane containing the direction of travel. It occurs only in transverse waves, because it depends on the vibrations being at right angles to the direction of propagation. The fact that light can be polarised is direct evidence that light is a transverse wave.
(b) Two uses of polaroids
- In polaroid sunglasses to cut down glare (reflected, horizontally polarised light).
- In photographic filters and in liquid-crystal displays (LCDs) to control the intensity and plane of transmitted light; also in stress analysis of transparent materials.
Tambaya 7 Rahoto
(a)(i) Mention two modes of heat transfer other than convection.
(ii) Explain land and sea breezes.
(b) An iron rod of length 30cm is heated through 50 kelvin. Calculate its increase in length. (linear expansivity of iron = 1.2 x 10\(^{-5}\)K\(^{-1}\)
(c) An electric heater immersed in some water raises the temperature of the water from 40°C to 100°C in 6 minutes. After another 25 minutes, it is noticed that half the water has boiled away. Neglecting heat losses to the surrounding, calculate the specific latent heat of vaporisation of water.
(a)(i) Two modes of heat transfer other than convection: conduction and radiation.
(a)(ii) Land and sea breezes. During the day the land heats up faster than the sea (land has a lower specific heat capacity). The air over the land becomes hotter, rises, and cooler air from over the sea flows in to replace it, giving a sea breeze (from sea to land). At night the land cools faster than the sea, so the air over the sea is now warmer and rises, and cooler air from the land flows out to sea, giving a land breeze (from land to sea).
(b) Increase in length \( \Delta L = L\,\alpha\,\Delta\theta \):
\[ \Delta L = 0.30 \times 1.2\times10^{-5} \times 50 = 1.8\times10^{-4}\ \text{m} = 0.18\ \text{mm}. \](c) The heater delivers energy at a constant rate \(P\).
Heating stage (6 min = 360 s), taking \(c = 4200\ \text{J kg}^{-1}\text{K}^{-1}\), \(\Delta\theta = 100-40 = 60\ \text{K}\):
\[ P = \frac{mc\,\Delta\theta}{t_1} = \frac{m \times 4200 \times 60}{360} = 700m. \]Boiling stage (25 min = 1500 s), half the water (\(m/2\)) vaporised:
\[ P\,t_2 = \tfrac{m}{2}L \;\Rightarrow\; 700m \times 1500 = \tfrac{m}{2}L. \] \[ L = \frac{2 \times 700 \times 1500 \times m}{m} = 2.1\times10^{6}\ \text{J kg}^{-1}. \]Specific latent heat of vaporisation \(\approx 2.1\times10^{6}\ \text{J kg}^{-1}\).
Bayanin Amsa
(a)(i) Two modes of heat transfer other than convection: conduction and radiation.
(a)(ii) Land and sea breezes. During the day the land heats up faster than the sea (land has a lower specific heat capacity). The air over the land becomes hotter, rises, and cooler air from over the sea flows in to replace it, giving a sea breeze (from sea to land). At night the land cools faster than the sea, so the air over the sea is now warmer and rises, and cooler air from the land flows out to sea, giving a land breeze (from land to sea).
(b) Increase in length \( \Delta L = L\,\alpha\,\Delta\theta \):
\[ \Delta L = 0.30 \times 1.2\times10^{-5} \times 50 = 1.8\times10^{-4}\ \text{m} = 0.18\ \text{mm}. \](c) The heater delivers energy at a constant rate \(P\).
Heating stage (6 min = 360 s), taking \(c = 4200\ \text{J kg}^{-1}\text{K}^{-1}\), \(\Delta\theta = 100-40 = 60\ \text{K}\):
\[ P = \frac{mc\,\Delta\theta}{t_1} = \frac{m \times 4200 \times 60}{360} = 700m. \]Boiling stage (25 min = 1500 s), half the water (\(m/2\)) vaporised:
\[ P\,t_2 = \tfrac{m}{2}L \;\Rightarrow\; 700m \times 1500 = \tfrac{m}{2}L. \] \[ L = \frac{2 \times 700 \times 1500 \times m}{m} = 2.1\times10^{6}\ \text{J kg}^{-1}. \]Specific latent heat of vaporisation \(\approx 2.1\times10^{6}\ \text{J kg}^{-1}\).
Tambaya 8 Rahoto
(a)(i) State the laws of refraction of light.
(ii) Describe an experiment to determine the refractive index, n, of the material of an equilaleral triangular glass prism using the minimum deviation method.
(b) A rectangular glass prism of thickness 12cm is placed on a mark on a piece of paper resting on a horizontal bench:
(i) Draw a ray diagram to show the apparent position of the mark in the glass prism.
(ii) If the refractive-index of the material of the prism is 1.5, calculate the apparent displacement of the mark.
(a)(i) Laws of refraction of light
- The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
- The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media (Snell's law): \( \dfrac{\sin i}{\sin r} = n \).
(a)(ii) Determining the refractive index by the minimum-deviation method
Place the equilateral triangular glass prism flat on a sheet of paper and trace its outline. Remove the prism, and on one refracting face mark a point; draw the normal at this point and rule an incident ray meeting the face at a chosen angle of incidence \(i\). Replace the prism exactly on its outline. Stick two pins \(P_1\) and \(P_2\) vertically on the incident ray. Looking through the second refracting face, stick two more pins \(P_3\) and \(P_4\) so that they appear in a straight line with the images of \(P_1\) and \(P_2\). Remove the prism, ring the pin holes, and join \(P_3P_4\) to the face to give the emergent ray. Produce the incident and emergent rays forward until they meet; the angle between them is the angle of deviation \(d\). Measure \(i\) and \(d\) with a protractor.
Repeat the whole procedure for at least five values of \(i\), then plot a graph of the deviation \(d\) (vertical axis) against the angle of incidence \(i\) (horizontal axis). The graph is a smooth U-shaped curve; the deviation falls to a lowest value and then rises again. The value of \(d\) at the lowest point of the curve is the minimum deviation \(D\), read here as \(D = 37.2^\circ\).
With the refracting (apex) angle of the equilateral prism \(A = 60^\circ\), the refractive index is obtained from
\[ n = \frac{\sin\!\left(\dfrac{A + D}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)} = \frac{\sin\!\left(\dfrac{60^\circ + 37.2^\circ}{2}\right)}{\sin 30^\circ} = \frac{\sin 48.6^\circ}{\sin 30^\circ} = \frac{0.750}{0.500} = 1.50. \]Precautions: keep pin traces neat and thin; fix all pins truly vertical; and avoid parallax when reading the protractor.
(b)(i) Ray diagram showing the apparent position of the mark
The mark \(O\) is on the paper directly beneath the glass. Rays leaving \(O\) strike the top surface and are refracted away from the normal as they pass from glass to air, so they diverge more steeply. To the eye above, these emergent rays appear to come from a point \(I\) which lies vertically above \(O\) but nearer the top surface. Thus the mark is seen raised: \(I\) is its apparent (virtual) position.
(b)(ii) Apparent displacement of the mark
For viewing normally through a parallel-sided block,
\[ n = \frac{\text{real depth}}{\text{apparent depth}} \quad\Rightarrow\quad \text{apparent depth} = \frac{\text{real depth}}{n} = \frac{12}{1.5} = 8\ \text{cm}. \]Therefore the apparent displacement (the amount by which the mark appears raised) is
\[ \text{apparent displacement} = \text{real depth} - \text{apparent depth} = 12 - 8 = 4\ \text{cm}. \]Bayanin Amsa
(a)(i) Laws of refraction of light
- The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
- The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media (Snell's law): \( \dfrac{\sin i}{\sin r} = n \).
(a)(ii) Determining the refractive index by the minimum-deviation method
Place the equilateral triangular glass prism flat on a sheet of paper and trace its outline. Remove the prism, and on one refracting face mark a point; draw the normal at this point and rule an incident ray meeting the face at a chosen angle of incidence \(i\). Replace the prism exactly on its outline. Stick two pins \(P_1\) and \(P_2\) vertically on the incident ray. Looking through the second refracting face, stick two more pins \(P_3\) and \(P_4\) so that they appear in a straight line with the images of \(P_1\) and \(P_2\). Remove the prism, ring the pin holes, and join \(P_3P_4\) to the face to give the emergent ray. Produce the incident and emergent rays forward until they meet; the angle between them is the angle of deviation \(d\). Measure \(i\) and \(d\) with a protractor.
Repeat the whole procedure for at least five values of \(i\), then plot a graph of the deviation \(d\) (vertical axis) against the angle of incidence \(i\) (horizontal axis). The graph is a smooth U-shaped curve; the deviation falls to a lowest value and then rises again. The value of \(d\) at the lowest point of the curve is the minimum deviation \(D\), read here as \(D = 37.2^\circ\).
With the refracting (apex) angle of the equilateral prism \(A = 60^\circ\), the refractive index is obtained from
\[ n = \frac{\sin\!\left(\dfrac{A + D}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)} = \frac{\sin\!\left(\dfrac{60^\circ + 37.2^\circ}{2}\right)}{\sin 30^\circ} = \frac{\sin 48.6^\circ}{\sin 30^\circ} = \frac{0.750}{0.500} = 1.50. \]Precautions: keep pin traces neat and thin; fix all pins truly vertical; and avoid parallax when reading the protractor.
(b)(i) Ray diagram showing the apparent position of the mark
The mark \(O\) is on the paper directly beneath the glass. Rays leaving \(O\) strike the top surface and are refracted away from the normal as they pass from glass to air, so they diverge more steeply. To the eye above, these emergent rays appear to come from a point \(I\) which lies vertically above \(O\) but nearer the top surface. Thus the mark is seen raised: \(I\) is its apparent (virtual) position.
(b)(ii) Apparent displacement of the mark
For viewing normally through a parallel-sided block,
\[ n = \frac{\text{real depth}}{\text{apparent depth}} \quad\Rightarrow\quad \text{apparent depth} = \frac{\text{real depth}}{n} = \frac{12}{1.5} = 8\ \text{cm}. \]Therefore the apparent displacement (the amount by which the mark appears raised) is
\[ \text{apparent displacement} = \text{real depth} - \text{apparent depth} = 12 - 8 = 4\ \text{cm}. \]Tambaya 9 Rahoto
A body of mass 0.6kg is thrown vertically upward from the ground with a speed of 20ms\(^{-2}\). Calculate its;
(i) potential energy at the maximum height reached.
(ii) kinetic energy just before it hits the ground.
A body of mass \(m = 0.6\,\text{kg}\) is thrown vertically upward from the ground with initial speed \(u = 20\,\text{ms}^{-1}\).
(i) Potential energy at maximum height
By conservation of energy, at the highest point all the initial kinetic energy has become potential energy (the body is momentarily at rest):
\[ \text{P.E.}_{max} = \frac{1}{2}mu^{2} = \frac{1}{2}(0.6)(20)^{2} = \frac{1}{2}(0.6)(400) = 120\,\text{J} \](ii) Kinetic energy just before it hits the ground
Neglecting air resistance, the body returns to the ground with the same speed it left, so its kinetic energy just before impact equals its initial kinetic energy (equal also to the P.E. at the top):
\[ \text{K.E.} = \frac{1}{2}mu^{2} = 120\,\text{J} \]Bayanin Amsa
A body of mass \(m = 0.6\,\text{kg}\) is thrown vertically upward from the ground with initial speed \(u = 20\,\text{ms}^{-1}\).
(i) Potential energy at maximum height
By conservation of energy, at the highest point all the initial kinetic energy has become potential energy (the body is momentarily at rest):
\[ \text{P.E.}_{max} = \frac{1}{2}mu^{2} = \frac{1}{2}(0.6)(20)^{2} = \frac{1}{2}(0.6)(400) = 120\,\text{J} \](ii) Kinetic energy just before it hits the ground
Neglecting air resistance, the body returns to the ground with the same speed it left, so its kinetic energy just before impact equals its initial kinetic energy (equal also to the P.E. at the top):
\[ \text{K.E.} = \frac{1}{2}mu^{2} = 120\,\text{J} \]Tambaya 10 Rahoto
(a)(1) State the energy transformations which take place during the operation of a modern x-ray tube.
(ii) Distinguish between hard and soft x-rays.
(iii) State three uses of x-rays.
(iv) Mention one hazard of over-exposure to x-rays in a radiological laboratory, and indicate any two safety precautions.
(b) A possible fusion reaction is \(^2_1 H + ^2_1 H \to ^3_1H + ^1_1H + Q\)
where Q is the energy released as a result of the reaction. If Q = 4.03 MeV, calculate the atomic mass of \(^3_1H\) in atomic mass units. (\(^2_1 H = 2.01410 U; ^1_1H = 1.00783 U; 1U = 931 MeV\))
(a)(i) Energy transformations in an x-ray tube: electrical energy at the cathode heats the filament (electrical → heat → light), the emitted electrons gain kinetic energy in the accelerating field (electrical → kinetic), and on striking the target this kinetic energy is converted mainly to heat with a small fraction to X-rays (kinetic → heat + X-ray/electromagnetic energy).
(a)(ii) Hard vs soft X-rays: Hard X-rays have very short wavelength, high frequency and high penetrating power (produced by high tube voltage); soft X-rays have longer wavelength, lower frequency and low penetrating power (produced by lower tube voltage).
(a)(iii) Three uses: medical diagnosis (imaging bones/fractures); radiotherapy (treatment of cancer); detection of flaws in metals and inspection of luggage/crystals (crystallography).
(a)(iv) Hazard: over-exposure damages living tissue and can cause cancer or radiation burns. Precautions: shield operators with lead screens/aprons, and limit exposure time / keep a safe distance.
(b) For \(^2_1H + {}^2_1H \to {}^3_1H + {}^1_1H + Q\), the mass defect equals the energy released:
\[ \Delta m = \frac{Q}{931} = \frac{4.03}{931} = 4.33\times10^{-3}\ \text{u}. \]Mass balance: \((2\times2.01410) - \big[m(^3_1H) + 1.00783\big] = \Delta m\).
\[ 4.02820 - 1.00783 - m(^3_1H) = 4.33\times10^{-3}, \] \[ m(^3_1H) = 3.02037 - 0.00433 = 3.01604\ \text{u}. \]Atomic mass of \(^3_1H \approx 3.016\ \text{u}\).
Bayanin Amsa
(a)(i) Energy transformations in an x-ray tube: electrical energy at the cathode heats the filament (electrical → heat → light), the emitted electrons gain kinetic energy in the accelerating field (electrical → kinetic), and on striking the target this kinetic energy is converted mainly to heat with a small fraction to X-rays (kinetic → heat + X-ray/electromagnetic energy).
(a)(ii) Hard vs soft X-rays: Hard X-rays have very short wavelength, high frequency and high penetrating power (produced by high tube voltage); soft X-rays have longer wavelength, lower frequency and low penetrating power (produced by lower tube voltage).
(a)(iii) Three uses: medical diagnosis (imaging bones/fractures); radiotherapy (treatment of cancer); detection of flaws in metals and inspection of luggage/crystals (crystallography).
(a)(iv) Hazard: over-exposure damages living tissue and can cause cancer or radiation burns. Precautions: shield operators with lead screens/aprons, and limit exposure time / keep a safe distance.
(b) For \(^2_1H + {}^2_1H \to {}^3_1H + {}^1_1H + Q\), the mass defect equals the energy released:
\[ \Delta m = \frac{Q}{931} = \frac{4.03}{931} = 4.33\times10^{-3}\ \text{u}. \]Mass balance: \((2\times2.01410) - \big[m(^3_1H) + 1.00783\big] = \Delta m\).
\[ 4.02820 - 1.00783 - m(^3_1H) = 4.33\times10^{-3}, \] \[ m(^3_1H) = 3.02037 - 0.00433 = 3.01604\ \text{u}. \]Atomic mass of \(^3_1H \approx 3.016\ \text{u}\).
Tambaya 11 Rahoto
Define (i) Elasticity; (ii) Young's modulus; (iii) Force constant.
(i) Elasticity
Elasticity is the ability of a material to regain its original shape and size after the deforming force applied to it has been removed.
(ii) Young's modulus
Young's modulus is the ratio of tensile stress to tensile strain for a material stretched within its elastic (proportional) limit:
\[ Y = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/L} = \frac{FL}{Ae} \]Its unit is \(\text{Nm}^{-2}\) (pascal).
(iii) Force constant
The force constant (elastic/spring constant) is the force required to produce unit extension in a material obeying Hooke's law:
\[ k = \frac{F}{e} \]Its unit is \(\text{Nm}^{-1}\).
Bayanin Amsa
(i) Elasticity
Elasticity is the ability of a material to regain its original shape and size after the deforming force applied to it has been removed.
(ii) Young's modulus
Young's modulus is the ratio of tensile stress to tensile strain for a material stretched within its elastic (proportional) limit:
\[ Y = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/L} = \frac{FL}{Ae} \]Its unit is \(\text{Nm}^{-2}\) (pascal).
(iii) Force constant
The force constant (elastic/spring constant) is the force required to produce unit extension in a material obeying Hooke's law:
\[ k = \frac{F}{e} \]Its unit is \(\text{Nm}^{-1}\).
Tambaya 12 Rahoto
(a) Define angle of contact
(b) Draw sketches to show angles of contact for a capillary tube dipped vertically in (i) water; (ii) mercury.
(a) Angle of contact
The angle of contact for a liquid in contact with a solid is the angle, measured through the liquid, between the solid surface (the wall of the container) and the tangent to the liquid surface (the meniscus) at the point where the liquid, the solid and the air meet.
Its value shows whether the liquid wets the solid or not:
- If the angle of contact is acute (\(\theta < 90^\circ\)) the liquid wets the surface, the adhesive forces between liquid and solid being stronger than the cohesive forces within the liquid.
- If the angle of contact is obtuse (\(\theta > 90^\circ\)) the liquid does not wet the surface, the cohesive forces being stronger than the adhesive forces.
(b) Sketches of the angles of contact
(i) Water in a clean glass capillary tube. Water wets glass because adhesion (glass–water) exceeds cohesion (water–water). The liquid therefore climbs the walls, giving a concave meniscus that curves upward at the edges, and the water rises in the tube. The tangent to the meniscus at the wall lies almost along the glass, so the angle of contact is acute, close to \(0^\circ\) for pure water on clean glass.
(ii) Mercury in a glass capillary tube. Mercury does not wet glass because cohesion (mercury–mercury) exceeds adhesion (glass–mercury). The liquid is pulled away from the walls, giving a convex (dome-shaped) meniscus, and the mercury is depressed below the outside level. The tangent to the meniscus at the wall turns steeply away from the glass, so the angle of contact is obtuse, about \(140^\circ\).
Bayanin Amsa
(a) Angle of contact
The angle of contact for a liquid in contact with a solid is the angle, measured through the liquid, between the solid surface (the wall of the container) and the tangent to the liquid surface (the meniscus) at the point where the liquid, the solid and the air meet.
Its value shows whether the liquid wets the solid or not:
- If the angle of contact is acute (\(\theta < 90^\circ\)) the liquid wets the surface, the adhesive forces between liquid and solid being stronger than the cohesive forces within the liquid.
- If the angle of contact is obtuse (\(\theta > 90^\circ\)) the liquid does not wet the surface, the cohesive forces being stronger than the adhesive forces.
(b) Sketches of the angles of contact
(i) Water in a clean glass capillary tube. Water wets glass because adhesion (glass–water) exceeds cohesion (water–water). The liquid therefore climbs the walls, giving a concave meniscus that curves upward at the edges, and the water rises in the tube. The tangent to the meniscus at the wall lies almost along the glass, so the angle of contact is acute, close to \(0^\circ\) for pure water on clean glass.
(ii) Mercury in a glass capillary tube. Mercury does not wet glass because cohesion (mercury–mercury) exceeds adhesion (glass–mercury). The liquid is pulled away from the walls, giving a convex (dome-shaped) meniscus, and the mercury is depressed below the outside level. The tangent to the meniscus at the wall turns steeply away from the glass, so the angle of contact is obtuse, about \(140^\circ\).
Tambaya 13 Rahoto
State any three properties of matter which are common to all substances.
Properties common to all substances (matter)
- It has mass.
- It occupies space (has volume).
- It is made up of tiny particles (molecules/atoms) and has weight in a gravitational field.
Bayanin Amsa
Properties common to all substances (matter)
- It has mass.
- It occupies space (has volume).
- It is made up of tiny particles (molecules/atoms) and has weight in a gravitational field.
Tambaya 14 Rahoto
A force of 40 N is applied at the end of a wire 4m long and produces an extension of 0.24mm. If the diameter of the wire is 2.00mm, calculate the;
(i) stress on the wire; (ii) strain in the wire.
A force \(F = 40\,\text{N}\) stretches a wire of length \(L = 4\,\text{m}\), producing an extension \(e = 0.24\,\text{mm} = 0.24\times10^{-3}\,\text{m}\); the wire diameter is \(2.00\,\text{mm}\), so radius \(r = 1.00\times10^{-3}\,\text{m}\).
Cross-sectional area:
\[ A = \pi r^{2} = 3.142\times(1.00\times10^{-3})^{2} = 3.142\times10^{-6}\,\text{m}^{2} \](i) Stress on the wire:
\[ \text{stress} = \frac{F}{A} = \frac{40}{3.142\times10^{-6}} = 1.27\times10^{7}\,\text{Nm}^{-2} \](ii) Strain in the wire:
\[ \text{strain} = \frac{e}{L} = \frac{0.24\times10^{-3}}{4} = 6.0\times10^{-5} \](Strain is a ratio and has no unit.)
Bayanin Amsa
A force \(F = 40\,\text{N}\) stretches a wire of length \(L = 4\,\text{m}\), producing an extension \(e = 0.24\,\text{mm} = 0.24\times10^{-3}\,\text{m}\); the wire diameter is \(2.00\,\text{mm}\), so radius \(r = 1.00\times10^{-3}\,\text{m}\).
Cross-sectional area:
\[ A = \pi r^{2} = 3.142\times(1.00\times10^{-3})^{2} = 3.142\times10^{-6}\,\text{m}^{2} \](i) Stress on the wire:
\[ \text{stress} = \frac{F}{A} = \frac{40}{3.142\times10^{-6}} = 1.27\times10^{7}\,\text{Nm}^{-2} \](ii) Strain in the wire:
\[ \text{strain} = \frac{e}{L} = \frac{0.24\times10^{-3}}{4} = 6.0\times10^{-5} \](Strain is a ratio and has no unit.)
Tambaya 15 Rahoto
During the electrolysis of copper (II) tetraoxosulphate (VI) solution, a steady current of 4.0 x 10\(^2\) A flowing for one hour liberated 0.48 g of copper. Calculate the mass of copper liberated by one coulomb of charge.
Data: current \(I = 4.0\times10^{2}\text{ A} = 400\text{ A}\), time \(t = 1\text{ hour} = 3600\text{ s}\), mass liberated \(m = 0.48\text{ g}\).
The total charge that passed is
\[ Q = It = 400 \times 3600 = 1.44\times10^{6}\ \text{C}. \]By Faraday's first law the mass liberated is proportional to the charge, so the mass liberated by one coulomb (the electrochemical equivalent, \(z\)) is
\[ z = \frac{m}{Q} = \frac{0.48}{1.44\times10^{6}} = 3.33\times10^{-7}\ \text{g C}^{-1}. \]Mass of copper liberated by one coulomb \(= 3.33\times10^{-7}\ \text{g}\) (i.e. \(3.33\times10^{-7}\) grams per coulomb).
Bayanin Amsa
Data: current \(I = 4.0\times10^{2}\text{ A} = 400\text{ A}\), time \(t = 1\text{ hour} = 3600\text{ s}\), mass liberated \(m = 0.48\text{ g}\).
The total charge that passed is
\[ Q = It = 400 \times 3600 = 1.44\times10^{6}\ \text{C}. \]By Faraday's first law the mass liberated is proportional to the charge, so the mass liberated by one coulomb (the electrochemical equivalent, \(z\)) is
\[ z = \frac{m}{Q} = \frac{0.48}{1.44\times10^{6}} = 3.33\times10^{-7}\ \text{g C}^{-1}. \]Mass of copper liberated by one coulomb \(= 3.33\times10^{-7}\ \text{g}\) (i.e. \(3.33\times10^{-7}\) grams per coulomb).
