Question 1 Part I: Candidates were required to answer any five questions from this part. 


ANSWER
V =d/t
OR t= d/V
= 30/40
= 0.75s
h = ½gt^{2}
= ½ x gt^{2}
=½ x 10x(O.75)^{2}
=2.81m
Question 2 

List three methods for the polarization of light 
ANSWER
Methods of polarization of light
– Double refraction
– Selective absorption
– Reflection
– Scattering
List three situations in which polarized glasses are used.
ANSWER
Uses of polarized glasses
– Welding
– Driving
– Sailing
– Skiing
– Viewing (solar) eclipse
Question 4 

(a) State two differences between solid friction and viscosity. 
ANSWER


Question 5 

Explain how the addition of detergents of water reduces surface tension

ANSWER
Surface tension is due to the intermolecular forces among the surface molecules of a liquid. Addition of detergents to water reduces/weakens theses forces; hence the reduction in its surface tension
Question 6 

(a) Define tensile strain. 
ANSWER
(a) Tensile strain is the ratio of extension to original length. (mathematical definition is acceptable jf symbols are defmed).
(b)Spring balance
(c) Elastic constant/stiffness
Question 7 

Use a labelled diagram to illustrate how cathode rays are deflected by a magnet. 
ANSWER
Question 8 

In electroplating an iron spoon with gold, state which material is made the 
ANSWER
a) Gold rod/plate;
(b) Iron spoon;
(c) Solution/molten gold salt.
Question 9 

List the three components of the electron gun’ in a cathode ray tube

ANSWER
Components of an electron gun in cathode ray tube
– Heated filamentlheater/cathode
– Grid
– Anode
Question 10 

List three phenomena that illustrate the particle nature of light. 
ANSWER
Phenomena that illustrates particle nature of light
– Photoelectric effect
– Radiation of energy from heated bodies
– (general) emission/absorption of light
– Compton effect.


ANSWER
(a)(i) Work is the product ofa force (applied) and the distance moved in the direction of the force
OR
The product of displacement and the component of the force in the direction of the displacement.
(ii) Power is the time rate at which energy is expended.
OR
Power is the time rate of doing work.
NOTE: Mathematical definitions are acceptable provided symbols are correctly defined.
(b) Any correct difference
e.g.
Static Friction  Dynamic Friction 
It exists between surfaces of stationary bodies  It exists between surfaces in relative motion 
– For the same system the static frictional force is greater than dynamic frictional force
(c) The tyres have treads in them to provide better grip of roads due to increased friction thus preventing vehicles from skidding
(ii) Retardation d = slope (oflast segment of graph)
= v – 0
T3 – t2
= v
t3 – t2
F = md
= Mv/t3 – t2
Question 12 Part II : Candidates are expected to answer any three form this part 

(a) Explain: (c) A ball of mass 4.0 kg. hits a smooth platform vertically with a speed of 3 ms^{1} and rebounds with a speed of 2 ms^{1}. Calculate the impulse experienced by the ball. 
ANSWER
(a)(i) Linear momentum of a body moving in a straight line is the product of its mass and its velocity.
Accept product of mass and linear velocity.
Any valid additional information
e.g.
– Its unit is kgms” – Do not accept Ns
– it is a vector quantity.
(ii) Impulse of a force is the product of the (large) force acting on a body and the (short)
time during which it acts. Any valid additional information e.g.
Any valid additional information
e.g.
– Its unit is Ns – Do not accept kgms”
– it is a vector quantity.
(b) Instances where Newton’s third law of motion is applicable
– recoil of a gun
– jet propulsion/rocket propulsion
– action of sprinkler/lawn spray
– Accept any other valid instance(s)
(c) Impulse = Ft = m(v u)
= 4[2(3)]
= 20Ns
(d) Weight, W=mg
= (5000 + 4000) x 10
= 90,000N.
Downward action = upward reaction
90,000 = 50(v0)
v = 90,000
50
= 1.8 X 10^{3} ms^{1}.
Question 13 Part II : Candidates are expected to answer any three form this part 

(a) With the aid of a labelled diagram, explain. (d) A wave is represented by the equation y = 0.20 sin 0.407t (x – 60 t) where aU distances are measured in centimeters and time in seconds.

ANSWER
(a) (i)
(Reflected wave)
N’
Incident wave).
Correct diagram
N=Node
A = Antinode
(Do not penalize for barriers not shown)
A node is a point on a stationary wave where there is no movement ofthe medium/points
of zero displacement.
OR
A Node is a point of destructive interference
An antinode is a point on a stationary wave where there is maximum displacement of
the medium.
OR
An antinode is a point of constructive interference.
(b)(i) Standing wave occurs when a progressive wave is reflected from a hard surface and
superimposed on the incident wave.
OR
Standing wave occurs when waves of the same frequency and amplitude traveling in opposite direction are superimposed.
(ii) Persistence of vision is a property of the eye whereby the sensation of vision lasts for a short but definite time after the object has been removed.
(c)(i) In dim light, the iris contracts increasing the size of the pupil to let in sufficient light.
(ii) In excessive brightness the iris quickly expands in size thus reducing the size of the pupil which reduces the amount of light entering the eye .
(d)(i) Either y = A sin2Π/λ (x – vt) ………………………………… (α)
OR y = A sin[2Πx/λ – 2Πvt/λ ]……………………………………… (β)
= 0.20 sin (0.40Πx – 0.40Π x 60t) …………… (γ)
Comparing equations (β) and (γ)
2Πx/λ= 0.40Πx
λ = 5 cm
(ii) 2Πvt/λ = 0.40Πx x 60t
v = fλ
2Πvt = 0.40Π x 60t
f = 12Hz


ANSWER
(a)(i)A capacitor is a device for storing charges/electrical energy.
OR
A capacitor is a device consisting of two parallel metal plates, carrying equal but opposite charges separated by a dielectric/an insulator.
(ii) A short circuit
An electrical connection of relatively very low resistance made between two points (at different potentials) in a circuit
OR
When two wires touch and the resistor in the circuit is bypassed/there is not enough resistance in the circuit.
(b) More current passes through the 200 watt bulb than the 60 watt bulb and the more current that passes through the bulb the brighter and more powerful the bulb.
(c)
Ca = 5 µF
1/Cb = 1/C3 + 1/C4 =1/10 + 1/10 + 1/5
Cb = 5 µF
Cc = Ca + Cb = 5+ 5 =10 µF
1/Cxy = 1/10 + 1/10 = 2/10
Cxy = 5 µF
(d)(i)
Question 15 Part II : Candidates are expected to answer any three form this part 

(a) Explain: 
ANSWER
(a) (i) Stopping potential is the (negative) minimum potential difference of the anode which is required to stop the most energetic electron from reaching the anode. Any valid additional information e.g Unit is volt.
(ii) Work function is the minimum amount of energy required to liberate/eject an electron from the surface of a metal when irradiated by electromagnetic radiation. Any valid additional information e.g. joule.
(b) Factors affecting kinetic energy of ejected electrons
– work function of surface/metal irradiated
– wavelength/frequency of incident radiation
(c) Uses of photocell
– burglar alarms
– switches for street lights
– automatic doors
– television cameras
– solar calculators
– counters
Accept any other valid example
(d) Work function = Wo = E – eVs or = hf/λ – eVs
= hc/ – eVs
= hc/5.893 x 10 ^{7} – 0.36e …………..(α)
= hc/5.969 x 10^{7} 1.38e………….(β)
Equation (β) – Equation (α) gives
1.38e – 0.36e = hc/5.969 x 10^{7} – hc/5.893 x 10 ^{7}
1.02e = hc (0.2520 – 0.1697) x 10^{7}
Taking 1.6 x 10^{19} as e and 3.0 x 10^{8} as c,
1.02 x 1.6 x 10^{19} = h x 3 X 10^{8} x 0.0823 X 10^{7}
h = 6.61 X 10^{34} Js
From equation (α)
Wo = 6.61 X 10^{34}x 3.0 X 108
5.893 X 107 – 0.36 x 1.6 x 10^{19}
= 3.365 X 10^{19} – 0.576 x 10^{19}
= 2.79 X 10^{19} J
= 2.79 X 10^{19}
6.61 X 10^{34}
= 4.22 x 10^{14} Hz