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IGCSE Physics · guided topic map

Mechanics for Cambridge IGCSE Physics

Mechanics for IGCSE Physics, organized into 3 syllabus topics and 3 mapped concept guides.

Syllabus topics
3
Mapped concept guides
3
Educational level
Cambridge IGCSE Core and Extended

Syllabus to lesson

Choose the exact concept

Work in order or jump to the concept named in your specification, course outline, or assignment.

1.2

Motion

Motion, forces and energy

1 guide
  1. 01MotionMapped lesson
1.3

Mass and weight

Motion, forces and energy

1 guide
  1. 01Mass and weightMapped lesson
1.5.1

Effects of forces

Motion, forces and energy

1 guide
  1. 01Effects of forcesMapped lesson

Diagrams

Mechanics as IGCSE Physics draws it

The figures from the IGCSE Physics practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Fig. 7.1Motion · Effects of forcesIGCSE
Car on a level road with the forward force and the resistive force markedmass = 1200 kgforward force 4500 Ntotal resistive forcedirection of travellevel road

Figure comment

Fig. 7.1A car on a straight, level road, labelled mass = 1200 kg, with a faint arrow above it showing the direction of travel to the right. A long horizontal arrow points forwards from the front of the car and is labelled forward force 4500 N. A shorter horizontal arrow points backwards from the rear of the car and is labelled total resistive force, with no value given.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

1.2 Motion1.5.1 Effects of forcesOpen IGCSE question · Motion, forces and energy Q7
Guided questions 5 parts

Reading cue. Only horizontal forces are drawn: the weight and the road's upward push are missing, and the shorter arrow carries no number, so its size cannot be read off the drawing at all.

  1. aIdentify Identify what the difference in length between the two arrows in Fig. 7.1 shows about the motion of the car at the moment drawn.

    recall2 marks

    Check answer 2 marks
    1. there is a resultant force on the car, acting forwards (1)
    2. so the car is accelerating, gaining speed in its direction of travel (1)

  2. bDetermine The car starts from rest and covers 100 m of road in the first 8.0 s. Determine the average power developed by the forward force shown in Fig. 7.1 over this time.

    routine3 marks

    Check answer 3 marks
    1. work done by forward force = force x distance = 4500 × 100 (1)
    2. = 4.5 × 10⁵ J (1)
    3. power = work done / time taken = 4.5 × 10⁵ / 8.0 = 5.6 × 10⁴ W, that is about 56 kW (1)

  3. cDetermine The car has a speed of 24 m/s at the end of the 8.0 s. Determine the energy transferred to the surroundings by the resistive force drawn in Fig. 7.1 during that time.

    demanding3 marks

    Check answer 3 marks
    1. kinetic energy gained = (1/2) x 1200 × 24² = 3.456 × 10⁵ J (1)
    2. energy to surroundings = work done by forward force - kinetic energy gained = 4.5 × 10⁵ - 3.456 × 10⁵ (1)
    3. = 1.044 × 10⁵ J, that is 1.0 × 10⁵ J to two significant figures (1)

  4. dExplain No value is printed beside the resistive arrow in Fig. 7.1. Explain why an acceleration worked out from a starting speed of 0, a final speed of 24 m/s and a time of 8.0 s is only an average value, and describe how the resistive force behaves over that time.

    top of the paper4 marks

    Check answer 4 marks
    1. 24/8.0 uses only the total change in velocity divided by the total time, so it is a mean value for the whole 8.0 s (1)
    2. the resistive force grows as the car goes faster, because air resistance increases with speed, while the forward force stays at 4500 N (1)
    3. the resultant force therefore falls during the 8.0 s, and the mass is fixed, so the acceleration falls too: greater than the mean at the start, smaller at the end (1)
    4. a constant acceleration would give an average speed of (0 + 24)/2 = 12 m/s and so 96 m in 8.0 s; the car in fact covers 100 m, which is what a larger acceleration early on gives (1)

Transfer challenge

A lift and its passengers have a total mass of 800 kg. The cable pulls the lift upwards with a tension of 9000 N. Determine the acceleration of the lift, and explain why the weight must appear in this calculation although no vertical force is drawn in Fig. 7.1. Take the gravitational field strength as 9.8 N/kg.

Check answer 4 marks
  1. weight = 800 × 9.8 = 7840 N (1)
  2. resultant force = 9000 - 7840 = 1160 N upwards (1)
  3. a = F/m = 1160/800 = 1.45 m/s², that is 1.5 m/s² upwards to two significant figures (1)
  4. in Fig. 7.1 the road pushes up on the car with a force equal to its weight, so the vertical forces cancel and can be left out; here the motion is vertical and the weight is one of the forces along that line (1)