Apply Automotive Engineering Science Principles — Questions and Answers

CDACC Level 5 • March/April 2026 written assessment.

Work through each question before reading the answer. These worked revision answers are prepared for Jemshah learners; they are not an official marking scheme.

Question 1

2 m/s2. Calculate the mass of the vehicle. (7 Marks) d) Determine the gravitational force acting on an engine block of mass 200kg assuming g=9.81 m/s². (6 Marks)

Answer

Mass is the amount of matter/inertia of a body; its SI unit is the kilogram (kg). Weight is the gravitational force on that mass, W = mg; its SI unit is the newton (N). Mass does not change when local gravity changes, but weight does.

Question 2

State the of Principle of Moments as widely used in mechanical and marine engineering. (3 Marks)

Answer

For a body in rotational equilibrium, the sum of clockwise moments about a point equals the sum of anticlockwise moments about that point. Equivalently, the algebraic sum of moments is zero. Moment = force × perpendicular distance from the pivot.

Question 3

Engineering materials are manufactured with specific properties for the kind of application State the difference between elasticity and plasticity. (4 Marks)

Answer

Elasticity is the ability to return to the original shape when the load is removed, within the elastic limit. Plasticity is the ability to undergo permanent deformation without immediate fracture; the body does not fully regain its original shape after unloading.

Question 4

Design engineers need to select a material for a component that will operate at high temperature. As an engineer state FIVE factors to consider when selecting good engineering materials for this operation. (5 Marks)

Answer

  1. Strength retained at the operating temperature.
  2. Resistance to creep during sustained loading.
  3. Oxidation/corrosion resistance at that temperature.
  4. Melting/softening range safely above the service temperature.
  5. Thermal expansion and resistance to thermal shock appropriate to the component and its joints.

Question 5

Too much friction may increase the force needed to move an object. State FOUR factors that affect friction when vehicle is driven. (4 Marks)

Answer

  1. Nature and roughness of the contacting surfaces, including tyre and road material.
  2. Normal load pressing the contacting surfaces together.
  3. Surface condition, such as water, oil, ice or loose grit.
  4. Temperature and the presence/condition of lubrication where applicable.

Tyre-road friction is affected by these factors and tyre condition; it is not described completely by the simplest dry solid-friction model.

Question 6

The velocity of a body increases from 72 km/h to 144 km/h in 10 seconds. Calculate its acceleration. (4 Marks)

Answer

u = 72/3.6 = 20 m/s; v = 144/3.6 = 40 m/s; t = 10 s.
a = (v − u)/t = (40 − 20)/10 = 2 m/s².

Question 7

A force of 7.5 N stretches a certain spring by 5 cm. calculates work done in stretching this spring by

Answer

Assume Hooke’s law applies throughout the extension.
k = F/x = 7.5/0.05 = 150 N/m.
For an extension of 0.08 m, W = ½kx² = ½ × 150 × 0.08² = 0.48 J.
The force increases from zero as the spring is stretched, so using final force × distance without the factor ½ would overestimate the work.

Question 8

State Archimedes’ principle which is widely applied in marine technology. (3 Marks)

Answer

A body wholly or partly immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the body.

Question 9

Density and relative density are two important concepts in physics and chemistry that are often used to describe the properties of substances. Differentiate between density and Relative density. (4 Marks)

Answer

Density is mass per unit volume, ρ = m/V, measured in kg/m³. Relative density is the ratio of a material’s density to the density of a reference substance (normally water for solids/liquids at a stated temperature). It is dimensionless.

Question 10

A gas in a fixed volume container has a pressure of 1.6 × 105 Pa at a temperature of 27 0C. Calculate the pressure of the gas if the container is heated to a temperature of 2770C? (4 Marks)

Answer

Use absolute temperature: T₁ = 27 + 273.15 = 300.15 K; T₂ = 277 + 273.15 = 550.15 K.
At constant volume, P₁/T₁ = P₂/T₂.
P₂ = 1.6 × 10⁵ × 550.15/300.15 = 2.93 × 10⁵ Pa, approximately 293 kPa. Using the usual exam approximation 273 K gives 293.3 kPa.

Question 11

a) State the difference between the following terms as used in simple machine and state their unit. (4 Marks) i. Energy ii. Power b) A car travelling at a speed of 72 km/h is uniformly retarded by an application of brakes and comes to rest after 8 seconds. If the car with its occupants has a mass of 1,250 kg. Calculate; i. The breaking force ii. The work done in bringing it to rest. (12 Marks) c) State FOUR forms of energy. (4 Marks)

Answer

(a) Energy is the capacity to do work, measured in joules (J). Power is the rate of doing work or transferring energy, P = E/t, measured in watts (W = J/s).

(b) u = 72/3.6 = 20 m/s, v = 0 and t = 8 s.
a = (0 − 20)/8 = −2.5 m/s².
F = ma = 1250 × (−2.5) = −3125 N. The braking force has magnitude 3125 N and acts opposite the motion.
Initial kinetic energy = ½mu² = ½ × 1250 × 20² = 250,000 J = 250 kJ.
The brakes remove 250 kJ of mechanical energy; work done by the braking force on the car is −250 kJ.

(c) Four forms: kinetic, gravitational potential, chemical and electrical energy.

Question 12

a) State Boyle’s law. (3 Marks) b) Some gas occupies a volume of 1.5 m3 in a cylinder at a pressure of 250 kPa. A piston, sliding ²in the cylinder, compresses the gas isothermally until the volume is 0.5 m3. If the area of the piston is 300 cm2, calculate the force on the piston when the gas is compressed. (9 Marks) c) Explain FOUR factors on which transfer of heat through conduction depend on. (8 Marks)

Answer

(a) For a fixed mass of gas at constant temperature, pressure is inversely proportional to volume: PV = constant.

(b) P₂ = P₁V₁/V₂ = 250 × 1.5/0.5 = 750 kPa = 750,000 Pa.
A = 300 cm² = 0.03 m².
F = P₂A = 750,000 × 0.03 = 22,500 N = 22.5 kN.
This is the gas-pressure force on the piston. A net piston force would additionally require the opposing pressure, which is not given.

(c) For steady conduction, heat-transfer rate = kAΔT/L. It increases with higher thermal conductivity k, larger area A and larger temperature difference ΔT, and decreases with greater thickness/heat-flow length L.

Question 13

a) Air and liquids exert pressure in different ways depending on how they are used. State four Application of pressure in gases and liquids. (4 Marks) b) A body weighs 2.760 N in air and 1.925 N when completely immersed in water of density 1000 kg/m3. Calculate; i. The volume of the body. (7 Marks) ii. The density of the body. (6 Marks) iii. The relative density of the body. (3 Marks) Take the gravitational acceleration as 9.81 m/s²

Answer

(a) Hydraulic brakes transmit pressure through brake fluid; hydraulic jacks multiply force through pistons of different areas; pneumatic tools use compressed-air pressure; tyres use air pressure to support the vehicle load.

(b) Upthrust = 2.760 − 1.925 = 0.835 N.
i. V = upthrust/(ρwater g) = 0.835/(1000 × 9.81) = 8.51 × 10⁻⁵ m³, approximately 85.1 cm³.
ii. Mass = Wair/g = 2.760/9.81 = 0.28135 kg.
Density = m/V = 3305 kg/m³ (approximately).
iii. Relative density = ρbody/ρwater = 3305/1000 = 3.305. Alternatively, RD = Wair/(Wair − Wwater) = 2.760/0.835.

Question 14

a) Natural laws apply to all matter. State Newton’s second law of motion. (2 Marks) b) A motor vehicle of mass 2.1 Mg is travelling with a uniform velocity of 72 km/h down a local road; calculate the momentum of the vehicle. (5 Marks) c) A vehicle has a force of 1.5 KN acting on it due to the engine, and the acceleration produced is

Answer

(a) The resultant force on a body equals the rate of change of its momentum and acts in that direction. For constant mass, F = ma.

(b) m = 2.1 Mg = 2100 kg; v = 72/3.6 = 20 m/s.
Momentum p = mv = 2100 × 20 = 42,000 kg·m/s.

(c) F = 1.5 kN = 1500 N.
m = F/a = 1500/1.2 = 1250 kg.

(d) W = mg = 200 × 9.81 = 1962 N, acting downward.