Edexcel IGCSE Physics: Forces and Motion — Revision Notes

Condensed for the final weeks. For the full explanation, use the Forces and Motion study guide.

Motion

speed = distance / time          acceleration = (v - u) / t
v^2 = u^2 + 2 a s

Distance–time graph: gradient = speed. Horizontal line = stationary. Curve = changing speed. Velocity–time graph: gradient = acceleration; area under = distance travelled.

Confusing gradient with area on a velocity–time graph is the commonest error in this topic.

Scalars have magnitude only (distance, speed, mass); vectors have magnitude and direction (displacement, velocity, force, momentum). Distance and displacement are easily confused: distance is the total path length travelled, while displacement is the straight-line distance from start to finish, in a stated direction.

Forces

Newton’s first law — an object stays at rest or moves at constant velocity unless a resultant force acts.

Newton’s second law — F = ma.

Newton’s third law — for every force there is an equal and opposite force, acting on a different body.

Terminal velocity is the standard extended question, and the sequence must be given in order:

1. Weight acts downwards; the object accelerates.
2. As speed rises, air resistance increases.
3. When air resistance equals weight, the resultant force is zero.
4. Acceleration is zero, so the object falls at constant velocity.

Note that at terminal velocity the object is still moving — it just stops speeding up. Saying “the forces cancel so it stops” is wrong.

Opening a parachute: air resistance increases suddenly, becoming greater than weight, so there is a resultant upward force and the skydiver decelerates. As she slows, air resistance falls until it again equals weight, giving a new, lower terminal velocity — not zero.

Momentum

p = m v          F = (mv - mu) / t

Momentum is conserved in collisions and explosions in a closed system. It is a vector, so assign a positive direction and treat opposite motion as negative.

Safety features — crumple zones, airbags, seatbelts — all work the same way: they increase the time over which momentum changes, and since force is the rate of change of momentum, the force on the passenger is reduced.

Worked example. A car of mass 1200 kg decelerates from 25 m s⁻¹ to 15 m s⁻¹ in 5.0 s. Change in momentum = mΔv = 1200 × 10 = 12,000 kg m s⁻¹. Force = Δp/Δt = 12,000 ÷ 5.0 = 2400 N.

Stopping distance

stopping distance = thinking distance + braking distance

• Thinking distance increases with speed, tiredness, alcohol, drugs, distraction.
• Braking distance increases with speed, poor brakes, worn tyres, wet or icy roads.

Braking distance increases with the square of speed, because kinetic energy is proportional to v². Doubling the speed quadruples the braking distance. That is the point most answers miss.

Moments

moment = force x perpendicular distance from pivot

Principle of moments: in equilibrium, clockwise moments = anticlockwise moments.

An object topples when the line of action of its weight falls outside its base.

Worked example. A 2.5 N weight is placed 40 cm from the pivot of a balanced rule. Moment = 2.5 × 40 = 100 N cm. To balance at 20 cm on the other side: weight × 20 = 100, so weight = 5.0 N.

Hooke’s law

F = k x

Valid up to the limit of proportionality. Beyond the elastic limit, deformation is permanent — the spring will not return to its original length even once the force is removed.

Exam traps

• Reading area as gradient on a velocity–time graph.
• Saying an object stops at terminal velocity.
• Treating braking distance as proportional to speed rather than speed squared.
• Ignoring the vector nature of momentum.
• Omitting “perpendicular” from the definition of a moment.
• Saying an airbag “absorbs force” instead of extending the time.
• Forgetting to convert momentum change into a force using the actual time interval, not just stating Δp as the final answer.

Self-test

1. What do gradient and area represent on a velocity–time graph?
2. Explain terminal velocity in four steps.
3. Why does braking distance increase so sharply with speed?
4. How does an airbag reduce injury?
5. State the principle of moments.
6. What happens to a skydiver’s velocity when she opens her parachute, and why?
7. A 1500 kg car decelerates from 22 m s⁻¹ to 10 m s⁻¹ in 4.0 s. Calculate the average braking force.

Answers: 1. Gradient is acceleration; area under the graph is distance travelled. 2. Weight causes acceleration; air resistance increases with speed; when air resistance equals weight the resultant force is zero; acceleration becomes zero and the object falls at constant velocity. 3. Braking distance depends on kinetic energy, which is proportional to v², so doubling speed quadruples the distance needed to stop. 4. It increases the time over which the passenger’s momentum changes, and since force equals rate of change of momentum, the force is reduced. 5. For a body in equilibrium, the sum of clockwise moments about a pivot equals the sum of anticlockwise moments. 6. Air resistance suddenly exceeds weight, so there is a resultant upward force and she decelerates, until air resistance again equals weight at a new, lower terminal velocity. 7. Δp = mΔv = 1500 × 12 = 18,000 kg m s⁻¹; F = Δp/Δt = 18,000 ÷ 4.0 = 4500 N.