WAEC SSCE - Physics - 2011 (Essay)

For a projectile launched horizontally, the horizontal and vertical motions are independent.

Step 1 - find the time of flight from the horizontal motion. Horizontally the speed is constant at \(u = 4\,\text{m s}^{-1}\):

\[ R = u\,t \;\Rightarrow\; 25 = 4t \;\Rightarrow\; t = 6.25\,\text{s}. \]

Step 2 - find the height from the vertical motion. The stone starts with zero vertical velocity and falls freely:

\[ h = \tfrac{1}{2} g t^{2} = \tfrac{1}{2}\times10\times(6.25)^{2}. \]

\[ h = 5\times39.06 \approx 195.3\,\text{m}. \]

The height of the tower is approximately 195.3 m.

The shape of a meniscus depends on the relative strengths of two forces: cohesion (the attraction between molecules of the same liquid) and adhesion (the attraction between the liquid molecules and the molecules of the glass).

Water (concave meniscus): For water in a glass tube, the adhesive force between water and glass is greater than the cohesive force between the water molecules. The water is therefore pulled up the walls of the tube, wetting the glass, and its surface curves upward at the edges to give a concave meniscus.

Mercury (convex meniscus): For mercury, the cohesive force between the mercury molecules is greater than the adhesive force between mercury and glass. The mercury therefore does not wet the glass; it is pulled together and depressed at the walls, so its surface bulges upward in the middle to give a convex meniscus.

Three properties of cathode rays which suggest the particle (corpuscular) nature of matter are:

1. They travel in straight lines and cast sharp shadows of objects placed in their path, as would a stream of moving particles.
2. They are deflected by electric and magnetic fields, showing that they consist of charged particles (electrons carrying negative charge) that possess mass and charge.
3. They possess momentum and kinetic energy - they can exert a mechanical force, for example turning a small paddle wheel placed in their path, which shows the moving stream carries the properties of particles.

(They also produce heat on striking a target and can be deflected/behave as though made of discrete massive particles.)

(a)(i) Transmutation: Transmutation is the change of one element into another as a result of a nuclear reaction, in which the number of protons in the nucleus (the atomic number) changes, for example during radioactive decay or artificial nuclear bombardment.

(a)(ii) Stopping potential: The stopping potential is the minimum negative (retarding) potential that must be applied to the collecting electrode of a photocell to just stop the most energetic photoelectrons from reaching it, so that the photocurrent becomes zero. \(eV_s = KE_{max}\).

(b) Nuclear equations:

\[ {}^{23}_{11}A + {}^{2}_{1}B \rightarrow {}^{p}_{q}C + {}^{1}_{1}\text{(proton)}. \]

Conserving nucleon number: \(23 + 2 = p + 1 \Rightarrow p = 24\). Conserving charge: \(11 + 1 = q + 1 \Rightarrow q = 11\).

\[ {}^{24}_{11}C \rightarrow {}^{r}_{s}E + {}^{0}_{-1}\beta. \]

Conserving nucleon number: \(24 = r + 0 \Rightarrow r = 24\). Conserving charge: \(11 = s + (-1) \Rightarrow s = 12\).

Therefore \(p = 24,\; q = 11,\; r = 24,\; s = 12\).

(c) Energy of the incident photon:

\[ E = \frac{hc}{\lambda} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{3.6\times10^{-7}} = 5.5\times10^{-19}\,\text{J}. \]

Work function \(W_0 = 1.6\,\text{eV} = 1.6\times(1.6\times10^{-19}) = 2.56\times10^{-19}\,\text{J}\).

(i) Maximum kinetic energy:

\[ KE_{max} = E - W_0 = 5.5\times10^{-19} - 2.56\times10^{-19} = 2.94\times10^{-19}\,\text{J}. \]

(ii) Speed of the emitted electron: from \(KE = \tfrac{1}{2}mv^{2}\),

\[ v = \sqrt{\frac{2\,KE}{m}} = \sqrt{\frac{2\times2.94\times10^{-19}}{9.1\times10^{-31}}} = \sqrt{6.46\times10^{11}} \approx 8.04\times10^{5}\,\text{m s}^{-1}. \]

(d) Moving the source of ultra-violet light farther from the metal only reduces the intensity (number of photons per second arriving), so fewer electrons are emitted per second. It does not change the frequency/energy of each photon, so the maximum kinetic energy and hence the maximum speed of the ejected electron remain unchanged.

(a)(i) Eclipse: An eclipse is the total or partial disappearance of a heavenly body (the sun or the moon) from view, caused when the shadow of one body falls on another as the sun, earth and moon come to lie nearly in a straight line.

(a)(ii) Three types of eclipse:

• Eclipse of the sun (solar eclipse)
• Eclipse of the moon (lunar eclipse)
• Annular eclipse

(b)(i) Eye defect: The student sees distant objects (the board) clearly but cannot focus on near objects (the book), so the defect is long sight (hypermetropia).

(b)(ii) Correcting lens: A convex (converging) lens.

(b)(iii) Power of the lens: \(f = 10\,\text{cm} = 0.10\,\text{m}\).

\[ P = \frac{1}{f} = \frac{1}{0.10} = +10\,\text{D (dioptres)}. \]

(c) Distance between the cars: The ultrasonic beep travels from B to A and is reflected back to B, so it covers a total path of \(2d\) in \(t = 5.6\times10^{-3}\,\text{s}\).

\[ 2d = vt = 340\times5.6\times10^{-3} = 1.904\,\text{m}, \]

\[ d = \frac{1.904}{2} = 0.952\,\text{m}. \]

The two cars are 0.952 m apart at that instant.

(d) Focal length of the convex mirror: The image is virtual, formed \(9\,\text{cm}\) behind the mirror, so \(v = -9\,\text{cm}\). The magnification \(m = \dfrac{-v}{u} = 0.6\) gives \(u = \dfrac{9}{0.6} = 15\,\text{cm}\).

\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-9} + \frac{1}{15} = -0.1111 + 0.0667 = -0.0444, \]

\[ f = -22.5\,\text{cm}. \]

The focal length of the convex mirror is 22.5 cm, the negative sign confirming that it is a diverging (convex) mirror.

The energy gained by a charge \(Q\) accelerated through a potential difference \(V\) is the work done on it:

\[ E = QV. \]

Substituting \(Q = 1.60\times10^{-19}\,\text{C}\) and \(V = 1.0\times10^{5}\,\text{V}\):

\[ E = (1.60\times10^{-19})\times(1.0\times10^{5}) = 1.60\times10^{-14}\,\text{J}. \]

The energy of the electron is \(1.60\times10^{-14}\,\text{J}\).

(a)(i) Two advantages of alcohol over mercury as a thermometric liquid: (1) Alcohol has a much lower freezing point (about -115 °C), so it can measure very low temperatures where mercury would freeze; (2) Alcohol expands more (larger expansivity) for a given temperature rise, making the thermometer more sensitive. (It is also cheaper and can be coloured for easy reading.)

(a)(ii) Mercury level first falls then rises: When the bulb is placed in hot water, the glass bulb is heated first and expands before the heat reaches the mercury. The expansion of the bulb increases the space available, so the mercury level first falls. Shortly afterwards the heat reaches the mercury, which expands much more than the glass, and the level then rises steadily.

(b) Two factors (other than temperature) affecting the rate of evaporation: (1) Surface area exposed; (2) Draught/movement of air over the surface. (Humidity of the surrounding air and nature of the liquid are also acceptable.)

(c) Heating the lead:

(i) Heat to raise the lead to its melting point:

\[ Q = mc\,\Delta\theta = 100\times120\times(320-40) = 100\times120\times280 = 3.36\times10^{6}\,\text{J}. \]

(ii) Additional heat to melt the lead:

\[ Q = mL = 100\times2.5\times10^{4} = 2.5\times10^{6}\,\text{J}. \]

(iii) Time to supply this additional energy (furnace power \(P = 10\,\text{kW} = 10000\,\text{W}\)):

\[ t = \frac{Q}{P} = \frac{2.5\times10^{6}}{10000} = 250\,\text{s}. \]

(d) Two precautions in determining specific latent heat of vaporization of water: (1) Lag (insulate) the apparatus to minimise heat loss to the surroundings; (2) Ensure steady conditions and collect the condensed water only after the water has been boiling steadily (take readings at constant rate). (Correcting for heat losses by repeating at two heating rates is also acceptable.)

By Heisenberg's uncertainty principle for energy and time,

\[ \Delta E \,\Delta t \geq \frac{h}{2\pi} = \hbar. \]

Taking the limiting (minimum) case, the uncertainty in energy is

\[ \Delta E = \frac{\hbar}{\Delta t} = \frac{1.054\times10^{-34}}{2.0\times10^{-9}}. \]

\[ \Delta E = 5.27\times10^{-26}\,\text{J}. \]

The uncertainty in determining the energy of that level is about \(5.27\times10^{-26}\,\text{J}\).

Three methods of polarizing an unpolarized beam of light are:

1. Polarization by reflection - when ordinary light is reflected from a transparent surface (for example water or glass) at the polarizing (Brewster) angle, the reflected light becomes plane polarized.
2. Polarization by selective absorption (using a polaroid/dichroic material) - a polaroid sheet transmits vibrations in only one plane and absorbs those in all other planes, producing plane polarized light.
3. Polarization by double refraction (using a doubly-refracting crystal such as calcite/Nicol prism) - the crystal splits the light into two plane polarized rays vibrating in mutually perpendicular planes.

(Polarization by scattering, as occurs with sunlight in the atmosphere, is also acceptable.)

(a) The force causing the spring to shrink in fig.(b)

When the load is lowered into the water it experiences an upthrust (buoyant force) from the water, acting vertically upwards. This upthrust partly supports the weight of the load, so the tension pulling on the spring is reduced and the spring contracts (shrinks). The force responsible for the shrink is therefore the upthrust of the water on the load.

(b) Work done by the force in causing the shrink

Reading the contraction from the metre rule. The pointer attached to the load marks the position of the lower end of the spring against the fixed rule:

• In air (fig. a) the pointer reads \(20\ \text{cm}\) (the spring is fully stretched by the whole weight \(W_a\)).
• In water (fig. b) the upthrust lifts the load, so the pointer rises to \(18\ \text{cm}\).

The contraction (shrink) of the spring is the difference of the two readings:

\[x = 20\ \text{cm}-18\ \text{cm}=2\ \text{cm}=2\times10^{-2}\ \text{m}.\]

Work done. For a spring obeying Hooke's law, the work done in changing its length by \(x\) against a force constant \(k\) is

\[W=\tfrac{1}{2}kx^{2}.\]

With \(k=2.0\times10^{11}\ \text{N m}^{-1}\) and \(x=2\times10^{-2}\ \text{m}\):

\[W=\tfrac{1}{2}\times(2.0\times10^{11})\times(2\times10^{-2})^{2}\]

\[W=\tfrac{1}{2}\times(2.0\times10^{11})\times(4\times10^{-4})\]

\[W=\mathbf{4.0\times10^{7}\ \text{J}}.\]

The key step is to read \(x\) as the difference of the two rule readings (\(20\ \text{cm}\) in air and \(18\ \text{cm}\) in water), giving \(x = 2\ \text{cm}\), and only then substitute into \(W=\tfrac12 kx^{2}\).

(a) Difference between plane polarized light and ordinary light: In ordinary (unpolarized) light the vibrations of the electric field take place in all directions (all planes) perpendicular to the direction of propagation. In plane polarized light the vibrations are restricted to a single plane containing the direction of propagation.

(b) Two uses of polaroids:

1. In polaroid sunglasses, to reduce glare from light reflected off surfaces such as water and roads.
2. In the control of light intensity and in photography (polarizing filters), and in liquid-crystal displays (LCDs) and stress analysis of transparent materials.

(a) Two applications of electrolysis in industry:

1. Electroplating - coating objects with a thin layer of metal (for example chromium, silver or gold plating) to protect them from corrosion or to improve appearance.
2. Extraction and purification (refining) of metals - such as the extraction of aluminium from molten alumina and the purification of copper by electrolytic refining. (Electrolytic production of chemicals such as caustic soda and chlorine is also acceptable.)

(b) One application of electrolysis in a school laboratory: Electrolysis of acidified water to prepare and collect hydrogen and oxygen gases (for example in a Hofmann voltameter). (Electroplating of a small object or determining the chemical equivalent of a metal are also acceptable.)

(a) Diffusion: Diffusion is the net movement of the molecules of a substance from a region where they are more concentrated to a region where they are less concentrated, arising from the continuous random motion of the molecules, until they become uniformly distributed.

(b) Why diffusion is faster in gases than in liquids (at the same temperature): In a gas the molecules are much farther apart and the intermolecular forces are very weak, so the molecules move freely at high speeds over large distances between collisions. In a liquid the molecules are close together with strong intermolecular forces, so their movement is slower and more restricted. Hence, at the same temperature, molecules of a gas diffuse much faster than those of a liquid.