All 22 questions from the West African Examinations Council (WAEC) Physics 2010 Theory paper, with the correct answer and a full explanation for each. Free, no signup needed.
PART I (FOR CANDIDATES IN NIGERIA, SIERRA LEONE AND THE GAMBIA)
1. A stone is projected vertically upward with a speed of 30 ms⁻¹ from the top of a tower of height 50 m. Neglecting air resistance, determine the maximum height it reached from the ground. [g = 10 ms⁻²]
Model answer
The stone is projected vertically upwards, so its initial speed is entirely vertical: uy=30m/s. Height of tower, H1=50m.
Maximum height reached from the point of release, H2 = uy²/(2g) = 30²/(2x10) = 900/20 = 45m.
Maximum height reached from the ground = H1+H2 = (50+45)m = 95m.
2. A force of 40 N is applied at one end of a wire fixed at one end to produce an extension of 0.24 mm. If the original length and diameter of the wire are 3 m and 2.0 mm respectively, calculate the (a) stress on the wire (b) strain in the wire.
Model answer
Given: Force (F)=40N, extension (e)=2.4x10⁻⁴m, original length=3m, diameter (d)=2.0mm=2x10⁻³m.
(a) Area = πd²/4 = π(2x10⁻³)²/4. Stress on the wire = Force/Area of wire = 40/(3.14x(2x10⁻³)²/4) = 40/(3.14x4x10⁻⁶/4) = 1.27x10⁷ N/m².
(b) Strain in the wire = Extension/original length = 2.4x10⁻⁴/3 = 8x10⁻⁵.
3. When a lead-acid accumulator is fully charged, evolution of gases occurs at the respective electrodes at which they are given off. State two factors which affect the mass of elements deposited during electrolysis.
Model answer
When a lead-acid accumulator is fully charged, the gases given off are: (i) Hydrogen (at the anode, which is lead peroxide) (ii) Oxygen (at the cathode, which is lead plate).
Factors that affect the mass of elements deposited during electrolysis: (i) quantity of electricity, (ii) nature/e.c.e of electrochemical equipment.
4(a)(i) State two factors which affect the angle of deviation of a ray of light through a triangular glass prism.
Model answer
Factors that affect the angle of deviation of a ray of light through a triangular glass prism are: (i) angle of incidence, (ii) refracting angle (A) of the prism, (iii) refractive index of the material of the glass prism (any two).
4(b) By means of a ripple tank, a student was able to generate series of transverse waves by varying the frequency of the dipper and all the waves so generated covered a distance of 0.80 m in 0.2 s. Copy and complete the table given above in your answer booklet. [Table with F/Hz, λ/m, λ⁻¹/m⁻¹ columns] (i) Plot a graph with f on the vertical axis and λ⁻¹ on the horizontal axis. (ii) What does the slope of the graph represent?
Model answer
Speed of waves generated = v = 0.8/0.2 = 4 m/s.
Using v=fλ, λ=v/f, so for each frequency f, wavelength λ=v/f and λ⁻¹=f/v.
F(Hz): 2.0, 4.0, 6.0, 8.0, 10.0. λ(m): 2.0, 1.0, 0.667, 0.5, 0.4. λ⁻¹(m⁻¹): 0.5, 1.0, 1.5, 2.0, 2.5.
The slope of the graph of f against λ⁻¹ represents the speed/velocity of the waves.
14a) When a positively charged conductor is placed near a candle flame, the flame spreads out as shown in the diagram above. Explain this observation.
Model answer
When a positively charged conductor is placed near a candle flame, the flame spreads out because: (i) the candle flame ionizes the air around it; (ii) the pathway charged conductor attracts the negative charges in the air and repels the positive charges. This makes the flame spread out.
14(b) A proton moving with a speed of 5.0x10⁵ ms⁻¹ enters a magnetic field of flux density 0.2 T at an angle of 30° to the field. Calculate the magnitude of the magnetic force exerted on the proton. [Proton charge = 1.6x10⁻¹⁹ C]
Model answer
Velocity of proton = 5.0x10⁵ m/s; magnetic flux density, B=0.2T; angle to field, θ=30°.
F=qvBsinθ = 1.6x10⁻¹⁹ x 5.0x10⁵ x 0.2 x sin30° = 8x10⁻¹⁵N.
15(a) State three conclusions that can be drawn from Rutherford's experiment on the scattering of alpha particles by a thin metal foil in relation to the structure of the atom.
Model answer
Conclusions from Rutherford's model:
(i) The electrons are revolving around the nucleus at high speed.
(ii) The volume of the nucleus is negligibly small compared to that of the volume of the atom.
(iii) Most of the space in an atom is empty.
(iv) The number of electrons in an atom is equal to the number of protons, hence it is electrically neutral.
(v) The atom of an element consists of a small, positively charged nucleus in the centre, which centres almost the entire mass of the atom (any three).
15(b) [Energy level diagram, n2=-2.0eV, n0=-12.0eV] If an excited electron moves from n2 to n0, calculate the: (i) frequency; (ii) wavelength of the emitted radiation. [h=6.6x10⁻³⁴Js; 1eV=1.6x10⁻¹⁹J; C=3.0x10⁸ms⁻¹]
Model answer
Energy level of n2=-2.0eV; Energy level of n0=-12.0eV. Energy change as it moves from n2 to n0 = E20 = E2-E0 = -2-(-12) = 10eV.
(i) E=hf. f=E/h = (10x1.602x10⁻¹⁹)/(6.63x10⁻³⁴) = 2.42x10¹⁵ Hz.
(ii) c=λf; λ=c/f = (3x10⁸)/(2.42x10¹⁵) = 1.24x10⁻⁷ m.
15(c) The following nuclear equations represent two types of radioactivity: ²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂α; ¹⁴₇N + ⁴₂α → ¹⁷₈O + ¹₁P. Identify each type and explain briefly the difference between them.
Model answer
²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂α represents a natural radioactivity decay (alpha decay occurring spontaneously).
¹⁴₇N + ⁴₂α → ¹⁷₈O + ¹₁P represents an artificial radioactivity decay (induced by bombarding nitrogen with an alpha particle).
Difference between Natural and Artificial Radioactivity:
(1) Natural radioactivity is a spontaneous disintegration of atomic nuclei of an element, while artificial radioactivity is achieved by bombarding a non-radioactive element with a particle e.g. neutron, proton, alpha particle etc.
(2) In natural radioactivity, the alpha particle is a by-product, while it is used as a bombarding particle in artificial radioactivity.
(3) Heavy nuclei are involved in natural radioactivity, but in artificial radioactivity, light atoms are also involved.
PART II (FOR ALL CANDIDATES)
11(a) Distinguish between perfectly elastic collision and perfectly inelastic collision.
Model answer
Perfectly Elastic Collision: (i) Directions of the bodies are reversed (in a head-on collision). (ii) Kinetic energy is conserved. (iii) Colliding bodies do not stick together after collision.
Perfectly Inelastic Collision: (i) Directions of bodies are not necessarily reversed. (ii) Kinetic energy is not conserved (it decreases). (iii) Colliding bodies stick together and move as a unit with the same velocity after collision.
11(b) Sketch a distance-time graph for a particle moving in a straight line with: (i) uniform speed; (ii) variable speed.
Model answer
(i) For uniform speed, the distance-time graph is a straight line with a constant (positive) slope.
(ii) For variable (non-uniform) speed, the distance-time graph is a curve, with the slope changing continuously over time.
11(c) A body starts from rest and travels distances of 120m, 300m and 180m in successive equal time intervals of 12 s. During each interval the body is uniformly accelerated. (i) Calculate the velocity of the body at the end of each successive time interval. (ii) Sketch a velocity-time graph for the motion.
Model answer
For the first motion: distance covered=120m, time interval=12s, initial velocity=0. Using s=ut+½at²: 120=½xax12²; a=(120x2)/144=1.67ms⁻². Using v=u+at: v=(1.67x12)=20m/s.
For the second motion, the final velocity of the first motion becomes the initial velocity of the second (20m/s); distance covered=300m, time interval=12s. Using s=ut+½at²: 300=20(12)+½xax12²; 300-240=½xax144; a=(60x2)/144=0.83ms⁻². v=u+at=20+(0.83x12)=20+10=30m/s.
Also, for the third motion, initial velocity=final velocity of the second motion=30m/s; distance covered=180cm... [as given, 180m], time interval=12s. s=ut+½at²: 180-30(12)=½a(144); 180-360=½a(144); a=(-180x2)/144=-2.5ms⁻² (deceleration). v=u+at=30+(-2.5x12)=30-30=0ms⁻¹, meaning the body decreases to rest at the end of the third interval.
12(a) Explain the terms: (i) inertia; (ii) inertia mass.
Model answer
(i) Inertia: the tendency of a body or object to continue in its state of rest or uniform motion unless acted upon by an external force.
(ii) Inertia mass: represents resistance to any type of force whatever. The more the mass of a body, the more the force that is required to give it acceleration, in the formula f=ma, where m is the inertia mass of the body.
12(c) Two ice cubes pressed together for some time were found to stick together when the pressure was removed. Explain this observation.
Model answer
When two ice cubes are pressed together for some time, they stick together after releasing them, showing that the melting point of ice is lowered by an increase in pressure. When the ice cubes are pressed together, the parts of ice with high pressure melt. When the pressure is released, the water refreezes and therefore joins the two ice cubes together (regelation).
12(d) Two vertical capillary tubes of the same diameter are lowered into beakers situated at the same level, containing liquids A and B of densities 9.2x10² kgm⁻³ and 1.30x10³ kgm⁻³ respectively. A suction pump is used to withdraw air from the top of the liquid columns in the tubes by means of a T-piece arrangement until the liquid in A rises to a height of 26.0 cm. Calculate the height of the liquid in tube B.
Model answer
Density of A, PA=9.2x10²kg/m³; density of B, PB=1.3x10³kg/m³; height of A, HA=26cm; height of B, hB=?
Since pressure=ρhg, and pressure at same level is equal, then: ρAhAg=ρBhBg; hB=(PA x hA)/PB = (9.2x10² x 26)/(1.3x10³) = 18.4 cm.
13(a) State three factors that affect the angle of deviation of a ray of light through a triangular glass prism.
Model answer
Factors that affect the angle of direction of a ray of light through a triangular glass prism are: (i) angle of incidence, (ii) refracting angle (A) of the prism, (iii) refractive index of the material of the glass prism.
13(c) By means of a ripple tank, a student was able to generate series of transverse waves by varying the frequency of the dipper and all the waves so generated covered a distance of 0.80 m in 0.2 s. Determine the speed, v, of the waves.
Model answer
Speed, v = distance/time = 0.8/0.2 = 4 m/s.
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