OxfordAQA International AS and A-level Mathematics: Forces, Newton's laws and momentum (9660) – Revision Notes

Quick recall for sections M1.3 and M1.4 of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2 (International AS exams from May/June 2018, A-level from May/June 2019). The specification places this content in International AS, Unit PSM1 (Mechanics, M1). For explanations and full worked examples, use the study guide. Then test yourself with the practice questions. See also the course hub, the printable checklist, exam preparation and the free diagnostics.

Formulae to learn

The specification lists these as formulae you must learn (they are not in the formulae booklet).

Result	Meaning	Notes
W = mg	Weight in newtons	g = 9.8 m/s²; mass in kg
F = ma	Resultant force = mass × acceleration	Also: force = rate of change of momentum
F = μR	Dynamic friction	Sliding only; R is the normal reaction
p = mv	Momentum	Signed in a straight line
Impulse = mv − mu	Change in momentum	Unit N s; equals Ft for a constant force

Forces in one line each

• Weight: down, mg.
• Normal reaction R: perpendicular to the surface. Find it from vertical forces, not by assuming R = mg.
• Tension: pulls, along a string or rod. Thrust: pushes, along a rod only.
• Resistance / friction: opposes motion (friction opposes sliding).
• Light: massless, same tension at both ends. Inextensible: same speed and acceleration magnitude. Smooth: frictionless, so tension is the same either side of a peg or pulley.
• Resolution of forces is not required: all forces here are horizontal or vertical.

Newton’s laws

Law	Statement	Use
First	Zero resultant force means rest or constant velocity	Constant speed gives a = 0, so forces balance
Second	Net force = ma	The working equation
Third	Mutual forces are equal in size, opposite in direction	Pairs act on different bodies and never cancel

Method in steps

1. Sketch each particle with every force and the direction of acceleration.
2. Choose a positive direction for each particle (the direction it accelerates).
3. Write “resultant force = ma” for each particle.
4. Add the equations to eliminate T, or solve by substitution.
5. Check the answer in a particle’s equation you did not use to solve.

Worked reminders

Lift. A person of mass 70 kg in a lift accelerating upward at 0.6 m/s²: R − 70g = 70 × 0.6, so R = 70 × 10.4 = 728 N.

Friction. A 6 kg block is pushed along a rough floor by a horizontal force of 40 N, μ = 0.3.

R = 6 × 9.8 = 58.8 N,  F = 0.3 × 58.8 = 17.64 N
40 − 17.64 = 6a,  a = 3.7267 m/s²  (3.73 to 3 s.f.)

Pulley (both hanging). Masses m₁ > m₂: a = (m₁ − m₂)g / (m₁ + m₂) and T = m₂(g + a). For 8 kg and 3 kg: a = 5 × 9.8 / 11 = 4.4545… m/s², so keep the fraction until the end.

Table and hanging particle (smooth). Particle of mass M on the table, m hanging: a = mg / (M + m), T = Ma.

Car and trailer. Whole system: driving force − all resistances = (total mass)a. Then use the trailer alone for the tow bar: T − (trailer resistance) = (trailer mass)a. With a rigid rod, a negative T means a thrust of that size.

String goes slack. When the hanging particle lands, T becomes zero. The other particle now moves under friction or resistance alone, with deceleration μg for a rough horizontal surface.

Momentum and impulse

Conservation (two particles, straight line):

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

• Choose a positive direction and give every velocity a sign.
• “Join together” or “coalesce”: the same final velocity v for both, so (m₁ + m₂)v on the right.
• “Move apart from rest”: total momentum is zero.
• Restitution is not required.

Impulse: I = mv − mu (final minus initial, with signs). For a constant force, I = Ft, so F = I/t.

Fixed smooth surface. The particle moves at right angles to the surface. Speed u towards the wall, speed w back: impulse on the particle = m(w + u) in magnitude, directed away from the wall. Momentum of the particle alone is not conserved, because the wall supplies the impulse.

Pair of impulses. In a collision of two particles, the impulses on them are equal in size and opposite in direction. Use this to check your final velocities.

Collision and impulse. A 5 kg particle at 4 m/s meets a 3 kg particle moving the other way at 2 m/s, and they join.

Right positive: 5 × 4 + 3 × (−2) = 8v
20 − 6 = 8v, so v = 1.75 m/s to the right
Impulse on the 3 kg particle: 3 × (1.75 − (−2)) = 11.25 N s
Impulse on the 5 kg particle: 5 × (1.75 − 4) = −11.25 N s  (equal and opposite)

Wall. A 0.5 kg ball hits a wall at 8 m/s and rebounds at 6 m/s. Impulse on the ball = 0.5 × (6 + 8) = 7 N s away from the wall. If contact lasts 0.01 s, the average force is 700 N.

Vertical motion with resistance. Falling: mg − D = ma. Rising (resistance now acts downwards): mg + D = m × (deceleration). Resistance always opposes the motion, so its direction changes when the motion reverses.

Checks to run before you move on

1. Does every unknown force appear in at least two equations, or have you used the whole-system equation to remove it?
2. Is the sign of each acceleration consistent with the direction you chose for that particle?
3. Do the tension values from the two particles agree?
4. Does the answer make sense? A friction force larger than the pulling force means the particle would not move, so the acceleration cannot be positive.
5. Are the units right: N for force, N s or kg m/s for impulse and momentum?

Must-know distinctions

• Mass and weight: kg and N. Weight = mg.
• Tension and thrust: strings cannot push; rods can do either.
• Third-law pair and balanced forces: a pair involves two different bodies; balanced forces are all on the same body.
• Momentum and impulse: momentum belongs to a state of motion; impulse is a change of it.
• Speed and velocity in collisions: use signed velocities in the conservation equation, then quote speeds if asked.

Quick self-test

1. Find the weight of a 15 kg object.
2. A 4 kg particle has a resultant force of 18 N. Find its acceleration.
3. A 10 kg box slides on a rough floor with μ = 0.25. Find the friction force.
4. Particles of 5 kg and 3 kg hang over a smooth pulley. Find the acceleration.
5. In question 4, find the tension.
6. A 5 kg trolley at 3 m/s joins a stationary 1 kg trolley. Find the common speed.
7. A 0.2 kg ball hits a wall at 15 m/s and rebounds at 10 m/s. Find the impulse on the ball.
8. A constant force acts for 0.05 s and gives an impulse of 6 N s. Find the force.
9. Particle X (2 kg) at 6 m/s collides with Y (4 kg) at rest. After the collision X moves at 1 m/s in the same direction. Find the speed of Y.
10. A 3 kg particle is pulled up by a vertical string at constant speed. State the tension.

Answers

1. 15 × 9.8 = 147 N
2. 18 / 4 = 4.5 m/s²
3. R = 98 N, F = 0.25 × 98 = 24.5 N
4. a = 2 × 9.8 / 8 = 2.45 m/s²
5. T = 3 × (9.8 + 2.45) = 36.75 N
6. 15 = 6v, so v = 2.5 m/s
7. 0.2 × (10 + 15) = 5 N s, directed away from the wall
8. 6 / 0.05 = 120 N
9. 12 = 2 × 1 + 4v, so v = 2.5 m/s
10. Constant speed means a = 0, so T = 3 × 9.8 = 29.4 N

Where marks are usually lost

• Writing the equation for the whole system when the question asks for a tension, then never splitting into particles.
• Using the same sign for both particles’ accelerations when one goes up and the other goes down.
• Taking R = mg when a vertical force other than weight acts.
• Using g = 10 or 9.81 instead of the 9.8 the specification sets.
• Rounding g × mass early and carrying a rounded number into later lines.
• Giving a negative impulse magnitude, or omitting the direction when it was asked for.
• Applying conservation of momentum to a ball hitting a fixed wall.
• Forgetting that after the string goes slack the equation of motion of the remaining particle changes.
• Leaving out units: N, N s, m/s².

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA: Unit PSM1, sections M1.3 and M1.4.