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CIE iGCSE Co-Ordinated Science P1.2 Motion Exam Style Questions Paper 4

Question

(a) (i) Circle all the vector quantities.
energy       gravitational field strength       temperature       time       weight
(a) (ii) Define the term velocity.
(b) Fig. 9.1 shows the speed–time graph for a cyclist travelling along a straight horizontal road.
Calculate the acceleration of the cyclist during the first 12 seconds.
(c) (i) In a crash test, a car experiences a deceleration of \(35\text{ m/s}^2\). Calculate the ratio:
\(\frac{\text{deceleration of car}}{\text{acceleration due to gravity}}\)
(c) (ii) Before the crash, the car has a velocity of \(28\text{ m/s}\).
       The kinetic energy of the car is \(470\text{ kJ}\).
       Calculate the mass of the car.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P1.1 — Physical quantities and measurement techniques (Part (a)(i))
• Topic P1.2 — Motion (Parts (a)(ii) & (b))
• Topic P1.3 — Mass and weight (Part (c)(i))
• Topic P1.6.2 — Work / Kinetic energy (Part (c)(ii))

▶️ Answer/Explanation

(a)(i) Vector quantities: gravitational field strength, weight.
Vector quantities have both magnitude and direction. Gravitational field strength and weight are vectors; energy, temperature, and time are scalars (magnitude only).

(a)(ii) Velocity is speed in a given direction.
Velocity is a vector quantity that describes the rate of change of displacement (distance travelled per unit time in a specific direction).

(b) Acceleration = 0.45 m/s².
Acceleration is the gradient of a speed-time graph. During the first 12 seconds, the speed increases from 0 to 5.4 m/s.
\(a = \frac{\Delta v}{\Delta t} = \frac{5.4 – 0}{12 – 0} = \frac{5.4}{12} = 0.45\text{ m/s}^2\)

(c)(i) Ratio = 3.6 (or –3.6).
\(\text{ratio} = \frac{35}{9.8} = 3.57 \approx 3.6\)
(The negative sign indicates deceleration/opposite direction.)

(c)(ii) Mass = 1200 kg.
\(E_k = \frac{1}{2}mv^2\)
\(470000 = \frac{1}{2} \times m \times 28^2\)
\(470000 = \frac{1}{2} \times m \times 784\)
\(470000 = 392m\)
\(m = \frac{470000}{392} = 1198.98 \approx 1200\text{ kg}\)

Question

(a) A torch (flashlight) consists of a battery, a switch and a lamp connected in series.
(i) State the energy store which decreases when the battery powers the lamp.
(ii) State the energy transfer from the battery to the lamp.
(iii) State the energy transfer from the lamp to the surroundings.
(b) A diver with mass 70 kg stands 5.0 m above a swimming pool as shown in Fig. 9.2.
(i) The diver falls 5.0 m. Calculate the change in gravitational potential energy of the diver.
(ii) As the diver falls toward the water, there are no frictional forces acting on the diver. State the kinetic energy of the diver just before entering the water.
(iii) Calculate the speed of the diver just before entering the water.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P1.6.1 — Energy (Part (a))
• Topic P1.6.2 — Work (Part (b)(i))
• Topic P1.2 — Motion (Part (b)(ii), (b)(iii))

▶️ Answer/Explanation

(a)(i) Chemical energy store
The battery stores chemical energy, which is converted to electrical energy when the circuit is complete.

(a)(ii) Chemical → Electrical
Energy is transferred from the battery to the lamp as electrical energy.

(a)(iii) Electrical → Light and Thermal
The lamp transfers electrical energy to the surroundings as light and thermal (heat) energy.

(b)(i) Change in gravitational potential energy:
ΔEₚ = mgΔh = 70 × 9.8 × 5.0 = 3430 J (≈ 3400 J)

(b)(ii) Kinetic energy just before entering the water = 3430 J
By conservation of energy, the gravitational potential energy lost is converted to kinetic energy (assuming no energy loss to friction).

(b)(iii) Speed:
Eₖ = ½mv²
3430 = 0.5 × 70 × v²
v² = 3430 ÷ 35 = 98
v = √98 = 9.9 m/s

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