WAEC SSCE - Physics - 2016 (Essay)

Antwoorddetails

Identifying the components of the cathode ray tube. Following the electron beam from left (the electron gun) to right (the screen end):

• X - the filament/cathode (heater of the electron gun). X is connected to the coil at the far left. When heated by the low-voltage supply it emits electrons by thermionic emission, providing the source of electrons for the beam.
• Y - the anode (accelerating and focusing anodes). Y points to the cylindrical electrodes held at a high positive potential. They accelerate the electrons and focus them into a fine beam directed along the axis of the tube.
• Z - the deflecting plates. Z points to the pair of parallel plates. A potential difference applied across them sets up an electric field that deflects the electron beam, steering the spot on the fluorescent screen.

In summary: X = cathode/filament (electron source), Y = accelerating anode, Z = deflecting plates.

Antwoorddetails

(a)(i) Mass defect. The mass defect is the difference between the sum of the separate masses of all the protons and neutrons (nucleons) in a nucleus and the actual (smaller) measured mass of the nucleus:

\[ \Delta m = \big(Z m_p + (A-Z) m_n\big) - M_{\text{nucleus}} \]

(a)(ii) Binding energy of a nucleus. It is the energy equivalent of the mass defect (\(E = \Delta m\, c^2\)); that is, the energy released when the free nucleons come together to form the nucleus, or equivalently the energy that must be supplied to separate the nucleus completely into its individual nucleons.

(b)(i) Effect of decreasing the speed of a photoelectron (de Broglie relation \(\lambda = h/mv\), kinetic energy \(E = \tfrac{1}{2}mv^2\)):

• (\(\alpha\)) Wavelength: since \(\lambda = \dfrac{h}{mv}\), decreasing \(v\) increases the wavelength.
• (\(\beta\)) Energy: since \(E = \tfrac{1}{2}mv^2\), decreasing \(v\) decreases the kinetic energy.

(b)(ii) de Broglie wavelength. Data: \(m = 4.4 \times 10^{-23}\ \text{kg}\), \(v = 10^{5}\ \text{m s}^{-1}\), \(h = 6.6 \times 10^{-34}\ \text{J s}\).

\[ \lambda = \frac{h}{mv} = \frac{6.6 \times 10^{-34}}{(4.4 \times 10^{-23})(10^{5})} \]\[ \lambda = \frac{6.6 \times 10^{-34}}{4.4 \times 10^{-18}} = 1.5 \times 10^{-16}\ \text{m} \]

(c) The decay-series graph. The graph plots nucleon number \(A\) (vertical, 208 to 232) against proton number \(Z\) (horizontal, Pb 82, Bi 83, Po 84, At 85, Rn 86, Fr 87, Ra 88, Ac 89, Th 90). Each down-left diagonal step (\(Z\) falls by 2, \(A\) falls by 4) is an alpha decay; each horizontal step to the right (\(Z\) rises by 1, \(A\) unchanged) is a beta decay. This is the thorium (4n) series running from Th-232 down to Pb-208.

(c)(i) A pair of isotopes. Isotopes have the same proton number but different nucleon numbers. From the graph, examples are the two lead points \(^{208}_{82}\text{Pb}\) and \(^{212}_{82}\text{Pb}\) (both at Z = 82); equally acceptable: \(^{216}_{84}\text{Po}\) and \(^{212}_{84}\text{Po}\) (Z = 84), or \(^{228}_{88}\text{Ra}\) and \(^{224}_{88}\text{Ra}\) (Z = 88), or \(^{232}_{90}\text{Th}\) and \(^{228}_{90}\text{Th}\) (Z = 90).

(c)(ii) Isotope with which the series starts. The top-right point, at Z = 90 and A = 232, is thorium-232, \(^{232}_{90}\text{Th}\).

(c)(iii) Nuclear equations (two examples of each).

(\(\alpha\)) Alpha decay (emits \(^{4}_{2}\text{He}\); Z falls by 2, A by 4):

\[ ^{232}_{90}\text{Th} \rightarrow\ ^{228}_{88}\text{Ra} + ^{4}_{2}\text{He} \]\[ ^{224}_{88}\text{Ra} \rightarrow\ ^{220}_{86}\text{Rn} + ^{4}_{2}\text{He} \]

(\(\beta\)) Beta decay (emits \(^{0}_{-1}e\); Z rises by 1, A unchanged):

\[ ^{228}_{88}\text{Ra} \rightarrow\ ^{228}_{89}\text{Ac} + ^{0}_{-1}e \]\[ ^{212}_{82}\text{Pb} \rightarrow\ ^{212}_{83}\text{Bi} + ^{0}_{-1}e \]

Antwoorddetails

An electron moving perpendicular to a uniform magnetic field experiences a force

\[F=Bev\sin\theta,\qquad \theta=90^\circ,\ \sin 90^\circ=1.\]

Making the speed the subject:

\[v=\frac{F}{Be}.\]

Substituting \(F=9.6\times10^{-12}\text{ N}\), \(B=0.12\text{ T}\) and \(e=1.6\times10^{-19}\text{ C}\):

\[v=\frac{9.6\times10^{-12}}{0.12\times1.6\times10^{-19}}=\frac{9.6\times10^{-12}}{1.92\times10^{-20}}=5.0\times10^{8}\text{ m s}^{-1}.\]

The speed of the electron as it enters the field is \(5.0\times10^{8}\text{ m s}^{-1}\).

Antwoorddetails

The initial velocity components are \(u_x = u\cos\theta\) (horizontal) and \(u_y = u\sin\theta\) (vertical), with \(u = 20\ \text{m s}^{-1}\), \(\theta = 40^\circ\).

At maximum height the vertical component of velocity is momentarily zero (the projectile stops rising). Only the horizontal component remains, and it is unchanged throughout the flight:

\[ v_x = u\cos\theta = 20 \times \cos 40^\circ = 20 \times 0.766 = 15.3\ \text{m s}^{-1}. \] \[ v_y = 0. \]

At maximum height: horizontal component \(\approx 15.3\ \text{m s}^{-1}\), vertical component \(= 0\).

Antwoorddetails

(a) Net force is the single resultant force obtained when all the forces acting on a body are combined (added as vectors). It is the force that determines the body's acceleration, \(F_{net}=ma\).

(b) Principle of conservation of linear momentum: in a closed system on which no external force acts, the total linear momentum before an interaction (collision or explosion) equals the total linear momentum after it. Example: the recoil of a gun when a bullet is fired - the forward momentum of the bullet equals the backward momentum of the gun.

(c) Impulse received by the floor.
Mass \(m = 200\,\text{g}=0.2\,\text{kg}\).
Speed on hitting the floor: \(v_1=\sqrt{2gh_1}=\sqrt{2\times10\times2.0}=\sqrt{40}=6.32\,\text{ms}^{-1}\) (downward).
Speed on rebound: \(v_2=\sqrt{2gh_2}=\sqrt{2\times10\times1.8}=\sqrt{36}=6.0\,\text{ms}^{-1}\) (upward).
Taking upward as positive, impulse on the ball

\[ J = m(v_2-(-v_1)) = 0.2\,(6.0+6.32)=0.2\times12.32 = 2.46\,\text{Ns} \]

By Newton's third law the impulse received by the floor is equal and opposite: 2.46 Ns directed downward.

(d) Simple harmonic motion.
\(m=20\,\text{g}=0.02\,\text{kg}\), \(f=5\,\text{Hz}\Rightarrow \omega=2\pi f=2\times3.14\times5=31.4\,\text{rads}^{-1}\).
At \(x=10\,\text{cm}=0.1\,\text{m}\), \(v=200\,\text{cms}^{-1}=2.0\,\text{ms}^{-1}\).

(i) Maximum displacement (amplitude) A. Using \(v=\omega\sqrt{A^2-x^2}\):

\[ v^2=\omega^2(A^2-x^2)\Rightarrow A^2 = x^2 + \frac{v^2}{\omega^2}=0.01+\frac{4}{(31.4)^2}=0.01+0.00406=0.01406 \]\[ A = 0.119\,\text{m}\;(\approx 11.9\,\text{cm}) \]

(ii) Maximum velocity.

\[ v_{max}=\omega A = 31.4\times0.119 = 3.73\,\text{ms}^{-1} \]

(iii) Maximum potential energy (equals the total energy of the oscillation):

\[ P.E._{max}=\tfrac{1}{2}m\omega^2A^2=\tfrac{1}{2}(0.02)(31.4)^2(0.01406)=0.139\,\text{J} \]

So \(P.E._{max}\approx 0.14\,\text{J}\).

Antwoorddetails

(a) Critical angle: the angle of incidence, in the denser medium, for which the angle of refraction in the less dense medium is \(90^{\circ}\). For angles greater than this, total internal reflection occurs.

(b) Formation of antinodes: antinodes are points of maximum displacement in a stationary wave. They are formed where the incident wave and the reflected wave meet in phase, so their displacements always add up (constructive superposition), giving maximum vibration.

(c) Refractive index of the prism. For a prism of refracting angle \(A=60^{\circ}\) (equilateral) at minimum deviation \(D=46.2^{\circ}\):

\[ n = \frac{\sin\!\left(\frac{A+D}{2}\right)}{\sin\!\left(\frac{A}{2}\right)} = \frac{\sin\!\left(\frac{60+46.2}{2}\right)}{\sin 30^{\circ}} = \frac{\sin 53.1^{\circ}}{0.5} = \frac{0.7997}{0.5} = 1.60 \]

(d) Concave mirror gives no image on the screen - two possible reasons:

• The object is placed at (or inside) the focal point, so the reflected rays are parallel or diverge and form a virtual image that cannot be caught on a screen.
• The screen is not placed at the correct image distance (it is nearer or farther than the position where the rays actually converge).

(e) Converging lens.
(i) \(u=75\,\text{cm}\), \(f=30\,\text{cm}\). Using \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\):

\[ \frac{1}{v}=\frac{1}{30}-\frac{1}{75}=\frac{5-2}{150}=\frac{3}{150}=\frac{1}{50}\Rightarrow v=50\,\text{cm} \]

(ii) The object is moved \(25\,\text{cm}\) farther, so \(u_2 = 75+25 = 100\,\text{cm}\), and the sharp image is formed on the same screen, i.e. \(v_2 = 50\,\text{cm}\).

\[ \frac{1}{f_2}=\frac{1}{u_2}+\frac{1}{v_2}=\frac{1}{100}+\frac{1}{50}=\frac{1+2}{100}=\frac{3}{100} \]\[ f_2 = \frac{100}{3} = 33.3\,\text{cm} \]

Antwoorddetails

(a) Impulse. Impulse \(=\) force \(\times\) time \(=\) change in momentum. Its dimension is that of momentum:

\[ [\text{impulse}] = MLT^{-1}. \]

(ii) Acceleration. Acceleration \(=\) velocity/time \(= \dfrac{LT^{-1}}{T}\):

\[ [\text{acceleration}] = LT^{-2}. \]

(iii) Work. Work \(=\) force \(\times\) distance \(= (MLT^{-2}) \times L\):

\[ [\text{work}] = ML^{2}T^{-2}. \]

Antwoorddetails

Materials that can demonstrate Brownian motion:

• Smoke particles suspended in air (in a smoke cell), viewed under a microscope.
• Pollen grains suspended in water.
• Fine dust or graphite particles suspended in water or air.

Note: a suspension of gamboge or sulphur particles in water is also acceptable. In each case the tiny visible particles are seen to move in continuous, random, zig-zag paths because they are bombarded unequally by the much smaller, fast-moving molecules of the surrounding fluid.

Antwoorddetails

(a) Dielectric strength: the maximum electric field (or maximum p.d. per unit thickness) that an insulator (dielectric) can withstand without breaking down and allowing charge to conduct through it.

(b) The wave is infrared radiation.

• (i) Infrared (heat) radiation.
• (ii) Detector: a thermopile / bolometer / blackened thermometer (a special infrared-sensitive photodiode is also acceptable).
• (iii) Source: any hot body, e.g. the Sun, an electric heating element or a warm object.

(c) Weight of the oil drop. Since the drop is at rest, the electric force balances the weight:

\[ W = qE = (1.0\times10^{-19})(1200) = 1.2\times10^{-16}\,\text{N} \]

(d) RLC series circuit. \(R=100\,\Omega\), \(L=0.05\,\text{H}\), \(C=25\,\mu\text{F}=25\times10^{-6}\,\text{F}\), \(f=50\,\text{Hz}\).

\[ X_L = 2\pi f L = 2(3.14)(50)(0.05)=15.7\,\Omega \]\[ X_C = \frac{1}{2\pi f C}=\frac{1}{2(3.14)(50)(25\times10^{-6})}=\frac{1}{7.85\times10^{-3}}=127.4\,\Omega \]

(i) Impedance:

\[ Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{100^2+(127.4-15.7)^2}=\sqrt{10000+12477}=\sqrt{22477}=150\,\Omega \]

(ii) Current:

\[ I=\frac{V}{Z}=\frac{220}{150}=1.47\,\text{A} \]

Antwoorddetails

(a)(i) Thermal equilibrium: the state reached when two (or more) bodies in thermal contact are at the same temperature so that there is no net flow of heat between them.

(a)(ii) Fundamental interval: the temperature range between the lower fixed point (ice point) and the upper fixed point (steam point) of a thermometer scale.

(b) Two uses of the hydraulic press:

• Lifting heavy loads (e.g. hydraulic jack / car lift).
• Compressing or baling materials such as cotton, scrap metal and refuse (also forging and stamping metals).

(c) Material used to reset the steel index in the Six's maximum and minimum thermometer: a magnet (the steel indices are drawn back with a small magnet).

(d)(i) Specific heat capacity of the mixture.
Heat lost by the hot milk mixture = heat gained by the water + heat gained by the aluminium bowl.
Mixture: \(m=300\,\text{g}=0.3\,\text{kg}\), cools \(85^{\circ}\text{C}\to40^{\circ}\text{C}\), \(\Delta\theta=45\,\text{K}\).
Water: \(0.5\,\text{kg}\), warms \(26^{\circ}\text{C}\to40^{\circ}\text{C}\), \(\Delta\theta=14\,\text{K}\).
Bowl heat capacity \(=90\,\text{JK}^{-1}\), warms by \(14\,\text{K}\).

\[ 0.3\,c\,(45) = (0.5)(4200)(14) + (90)(14) \]\[ 13.5\,c = 29400 + 1260 = 30660 \]\[ c = \frac{30660}{13.5} = 2271\,\text{Jkg}^{-1}\text{K}^{-1} \]

So \(c \approx 2.27\times10^{3}\,\text{Jkg}^{-1}\text{K}^{-1}\).

(d)(ii)(\(\alpha\)) Two ways the bottle loses heat: conduction and convection (radiation and evaporation are also acceptable).

(d)(ii)(\(\beta\)) Two industrial processes using a heat exchanger: electricity generation in power stations (steam condensers/boilers) and refrigeration / air-conditioning (also petroleum refining and food pasteurization).

Antwoorddetails

(a) Doping is the deliberate addition of a small, controlled amount of a suitable impurity (a pentavalent or trivalent element) to a pure (intrinsic) semiconductor in order to increase the number of free charge carriers and hence its electrical conductivity.

(b) How doping improves conductivity:

A pure semiconductor such as silicon has very few free charge carriers, so it conducts poorly.

• Adding a pentavalent impurity (e.g. phosphorus, arsenic) donates extra free electrons to the crystal, producing an n-type semiconductor in which electrons are the majority carriers.
• Adding a trivalent impurity (e.g. boron, aluminium) creates vacancies called holes that act as positive carriers, producing a p-type semiconductor in which holes are the majority carriers.

In both cases the impurity greatly increases the concentration of mobile charge carriers, so more charge can flow for a given voltage and the conductivity rises sharply.

Antwoorddetails

Uses of rockets:

• Launching artificial satellites and spacecraft into orbit (space exploration).
• As military weapons, i.e. carrying missiles/warheads.
• Scientific research, e.g. carrying instruments to study the upper atmosphere and outer space.

Other acceptable uses: signalling and distress flares, and fireworks/display.