IB Physics flashcards · SL and HL · first assessment 2025

A.2 Forces and momentum flashcards: IB Physics SL and HL

Revision flashcards for A.2 Forces and momentum, written for the IB Diploma Physics course first assessed in 2025. Cards common to both levels are marked SL & HL; extension material is marked HL only. Use study mode to test yourself one card at a time, or read the full list below with every answer.

Newton's laws, friction, drag and buoyancy, momentum and impulse, collisions and circular motion.

83 cards
Definition: 20
Equation: 14
Concept/explain: 19
SL and HL

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All 83 A.2 Forces and momentum cards

DefinitionSL & HL
State Newton's first law of motion.
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An object remains at rest or in uniform motion in a straight line unless acted on by a resultant (net) external force. Equivalent: if ΣF = 0, velocity is constant (translational equilibrium).
DefinitionSL & HL
State Newton's second law of motion (general form).
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The resultant force on an object equals the rate of change of its momentum: F = Δp/Δt, in the direction of the force. For constant mass this reduces to F = ma.
DefinitionSL & HL
State Newton's third law and the properties of a force pair.
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When body A exerts a force on body B, B exerts a force on A that is equal in magnitude and opposite in direction. Pair forces: same type (e.g. both gravitational), act on DIFFERENT bodies, act simultaneously, never cancel each other.
DefinitionSL & HL
Define linear momentum.
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Product of mass and velocity: p = mv. Vector in direction of velocity. SI unit: kg m s⁻¹ (equivalently N s).
DefinitionSL & HL
Define impulse.
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Change in momentum of an object: J = Δp = FΔt (for constant force) = area under a force–time graph. Vector; unit N s or kg m s⁻¹.
DefinitionSL & HL
State the law of conservation of linear momentum, including its condition.
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The total momentum of a system is constant provided no resultant EXTERNAL force acts on the system. Applies to each vector component separately, and to all collisions and explosions.
DefinitionSL & HL
Distinguish elastic, inelastic and totally inelastic collisions.
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Elastic: total kinetic energy conserved (and momentum). Inelastic: momentum conserved, KE decreases (to thermal/sound/deformation). Totally (perfectly) inelastic: objects stick together; maximum KE loss consistent with momentum conservation.
DefinitionSL & HL
Distinguish mass and weight.
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Mass: measure of inertia/amount of matter, scalar, kg, constant everywhere. Weight: gravitational force on a mass, W = mg, vector (down), newton, depends on local g.
DefinitionSL & HL
Define static friction and dynamic (kinetic) friction.
Show answer
Static: force preventing relative sliding between surfaces at rest; adjusts up to a maximum μ_s F_N. Dynamic: force opposing relative motion of sliding surfaces, constant F_f = μ_d F_N. Generally μ_d < μ_s. Direction: opposes (tendency of) relative motion.
DefinitionSL & HL
Define translational equilibrium.
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State in which the resultant force on a body is zero (ΣF = 0 in every direction), so the body is at rest or moving with constant velocity.
DefinitionSL & HL
Define centripetal force.
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The resultant force directed towards the centre of a circle required to keep a body in circular motion. Not a new type of force—it is provided by tension, friction, gravity, normal force, etc. Magnitude mv²/r.
EquationSL & HLData booklet: Yes
Write Newton's second law for constant mass and identify each symbol.
Show answer
F = ma. F = resultant (net) force (N); m = mass (kg); a = acceleration (m s⁻²). Vector equation—apply to each component. Misuse: using a single force instead of the RESULTANT.
EquationSL & HLData booklet: Yes
Write the equations for static and dynamic friction.
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F_f ≤ μ_s F_N (static; equality only at the point of slipping); F_f = μ_d F_N (dynamic). μ = coefficient of friction (no unit); F_N = normal reaction force (N). Misuse: using F_N = mg on an incline (should be mg cosθ) or assuming static friction is always at its maximum.
EquationSL & HLData booklet: Yes
State Hooke's law as an equation.
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F_H = −kx. F_H = restoring force of spring (N); k = spring constant (N m⁻¹); x = extension/compression from natural length (m). Minus sign: force opposes displacement. Valid up to the limit of proportionality. Misuse: using total length instead of extension.
EquationSL & HLData booklet: Yes
Write Stokes' law for viscous drag on a sphere.
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F_d = 6πηrv. η = viscosity of fluid (Pa s); r = radius of sphere (m); v = speed (m s⁻¹). Valid for small, smooth spheres at low speed (laminar flow). Misuse: applying to fast/turbulent motion.
EquationSL & HLData booklet: Yes
Write the drag equation for turbulent (high-speed) flow.
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F_d = ½C_d ρ A v². C_d = drag coefficient (no unit); ρ = fluid density (kg m⁻³); A = cross-sectional area (m²); v = speed. Drag ∝ v² (doubling speed → 4× drag). Misuse: confusing with Stokes' law (∝ v).
EquationSL & HLData booklet: Yes
Write the equation for buoyancy (upthrust).
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F_b = ρVg. ρ = density of the FLUID (kg m⁻³); V = volume of fluid displaced (= submerged volume of object) (m³). Misuse: using density of the object, or the whole object's volume when only partly submerged.
EquationSL & HLData booklet: Yes
Write the momentum and impulse equations.
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p = mv; F = Δp/Δt; J = FΔt. p momentum (kg m s⁻¹); F average resultant force (N); Δt contact time (s). Misuse: forgetting that a reversal of direction means Δp = m(v + u) in magnitude, not m(v − u).
EquationSL & HLData booklet: Yes
Write the equations for uniform circular motion.
Show answer
v = ωr; a = v²/r = ω²r = 4π²r/T². ω = angular speed (rad s⁻¹) = 2π/T = 2πf; r = radius (m); T = period (s). Centripetal force F = mv²/r = mω²r (derived from F = ma). Misuse: using degrees instead of radians.
Concept/explainSL & HL
Explain why an object in uniform circular motion is accelerating even though its speed is constant.
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- Velocity is a vector with direction
- Direction changes continuously
- Change in velocity means acceleration
- Acceleration directed towards centre (centripetal), perpendicular to velocity
- So speed unchanged but velocity changes.
Concept/explainSL & HL
Explain how a car can go round a horizontal (unbanked) bend, and why it may skid.
Show answer
- Friction between tyres and road provides centripetal force towards centre
- Required force = mv²/r
- Maximum available = μ_s F_N = μ_s mg
- Skids when mv²/r > μ_s mg, i.e. v > √(μ_s g r)
- Independent of car's mass.
Concept/explainSL & HL
Explain why the tension in a string is greatest at the bottom and least at the top for a mass in a vertical circle.
Show answer
- Centripetal force must point to centre
- Bottom: T − mg = mv²/r, so T = mv²/r + mg
- Top: T + mg = mv²/r, so T = mv²/r − mg
- Also speed is greater at bottom (energy)
- Minimum speed at top when T = 0: v = √(gr).
Concept/explainSL & HL
Explain, using impulse, how crumple zones/airbags/bending knees reduce injury.
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- Change in momentum Δp is fixed by initial and final velocities
- Δp = FΔt
- Increasing contact time Δt
- Reduces the average force F = Δp/Δt for the same Δp
- Smaller force → less injury.
Concept/explainSL & HL
Explain why momentum is conserved in an inelastic collision but kinetic energy is not.
Show answer
- Forces between colliding objects are an N3 pair: equal and opposite
- Act for same time → equal and opposite impulses
- Momentum changes cancel: total momentum constant (no external force)
- KE converted to thermal energy, sound, permanent deformation
- KE is scalar, not required to be conserved.
Concept/explainSL & HL
Explain why a sphere falling through a viscous liquid reaches terminal speed, stating the force balance.
Show answer
- Forces: weight down; buoyancy and viscous drag up
- Drag increases with speed (6πηrv)
- Resultant force decreases as speed increases
- At terminal speed: W = F_b + F_d
- Then mg = ρ_fluid V g + 6πηrv, acceleration zero.
Concept/explainSL & HL
Explain why passengers feel 'thrown outward' when a car turns, using Newton's laws.
Show answer
- No outward force exists (in inertial frame)
- Passenger tends to continue in straight line (N1)
- Car turns because friction provides centripetal force
- Seat/door pushes passenger inward to make them turn
- Sensation is inertia, not a real outward force.
Worked problemSL & HLData booklet: Yes
A 2.0 kg trolley at 3.0 m s⁻¹ collides with a stationary 1.0 kg trolley and they stick together. Find the final velocity and the KE lost.
Show answer
Momentum: 2.0(3.0) = (3.0)v → v = 2.0 m s⁻¹ (same direction). KE before = ½(2)(9) = 9.0 J; after = ½(3)(4) = 6.0 J. KE lost = 3.0 J (to thermal/sound).
Worked problemSL & HLData booklet: Yes
A 0.20 kg ball hits a wall at 10 m s⁻¹ and rebounds at 8.0 m s⁻¹. Contact time 0.050 s. Find the average force.
Show answer
Take rebound direction positive: Δp = 0.20(8.0 − (−10)) = 0.20(18) = 3.6 kg m s⁻¹. F = Δp/Δt = 3.6/0.050 = 72 N, directed away from the wall.
Worked problemSL & HLData booklet: Yes
A block slides down a 30° incline with μ_d = 0.20. Find its acceleration.
Show answer
Along slope: mg sin30° − μ_d mg cos30° = ma → a = 9.8(0.50 − 0.20 × 0.866) = 9.8(0.327) = 3.2 m s⁻² down the slope. Mass cancels.
Worked problemSL & HLData booklet: Yes
A 0.50 kg mass swings in a horizontal circle of radius 1.2 m at 3.0 m s⁻¹. Find the centripetal force.
Show answer
F = mv²/r = 0.50 × 9.0/1.2 = 3.75 N ≈ 3.8 N towards the centre.
Worked problemSL & HLData booklet: Yes
A 60 kg person stands on scales in a lift accelerating upward at 2.0 m s⁻². What do the scales read?
Show answer
N − mg = ma → N = m(g + a) = 60(9.8 + 2.0) = 708 N ≈ 710 N (i.e. reads about 72 kg). Downward acceleration: N = m(g − a).
Worked problemSL & HLData booklet: Yes
Two forces, 6.0 N east and 8.0 N north, act on a 4.0 kg mass. Find the acceleration (magnitude and direction).
Show answer
Resultant = √(6² + 8²) = 10 N. Direction: tanθ = 8/6 → 53° north of east. a = F/m = 10/4.0 = 2.5 m s⁻² at 53° N of E.
Worked problemSL & HLData booklet: Yes
A 1.5 kg block is pulled along a rough horizontal floor by 12 N at 25° above horizontal, μ_d = 0.30. Find the acceleration.
Show answer
F_N = mg − 12 sin25° = 14.7 − 5.07 = 9.63 N. Friction = 0.30 × 9.63 = 2.89 N. Horizontal: 12 cos25° − 2.89 = 10.88 − 2.89 = 7.99 N. a = 7.99/1.5 = 5.3 m s⁻².
Graph/diagramSL & HL
What does the area under a force–time graph represent, and how is the average force found?
Show answer
Area = impulse = change in momentum Δp. Average force = area ÷ total contact time (the rectangle of equal area). Peak force is usually much larger than average.
Graph/diagramSL & HL
State the rules for drawing a free-body diagram.
Show answer
- Show ONLY forces acting ON the chosen body (not forces it exerts)
- Arrows start from the body/point of action, labelled by type (weight, normal, tension, friction, drag)
- Lengths roughly proportional to magnitude
- Do NOT draw 'centripetal force' or 'resultant' as separate arrows.
Graph/diagramSL & HL
Describe a force–extension graph for a spring and what its gradient and area represent.
Show answer
Straight line through origin up to the limit of proportionality (Hooke's law). Gradient = spring constant k (N m⁻¹). Area under graph = elastic potential energy stored = ½kx². Beyond the limit, curve bends (non-linear); may not return to original length (plastic).
Exam technique/trapSL & HL
Trap: Are the weight of a book and the normal force from the table a Newton's third law pair? Explain.
Show answer
No. Both act on the SAME body (the book) and are different types (gravitational vs contact). N3 pairs: Earth pulls book / book pulls Earth; table pushes book / book pushes table. They only balance because the book is in equilibrium (N1), not because of N3.
Exam technique/trapSL & HL
Trap: how should centripetal force appear in a circular-motion problem?
Show answer
Never add it as an extra force. Identify the real forces (tension, friction, gravity, normal), resolve towards the centre, and set their resultant equal to mv²/r. Writing 'T = mv²/r + mg' at the bottom of a vertical circle is correct; drawing a separate 'F_c' arrow loses marks.
Exam technique/trapSL & HL
Trap: how do you apply momentum conservation to a 2D (oblique) collision?
Show answer
Conserve momentum separately in two perpendicular directions (x and y). Choose axes along the initial motion. Write Σp_x before = after and Σp_y before = after. Use trigonometry to resolve final velocities. Never conserve 'total speed'.
Exam technique/trapSL & HL
Practical: describe how to measure the coefficient of static friction with an inclined plane.
Show answer
Place block on plane; slowly increase angle until block just starts to slide; record angle θ. At limiting equilibrium mg sinθ = μ_s mg cosθ → μ_s = tanθ. Repeat and average. Limitations: judging the exact moment of slipping; surfaces not uniform; angle measurement uncertainty. Mass of block does not matter.
Exam technique/trapSL & HL
Practical/NOS: how would you verify conservation of momentum in the lab and what limits accuracy?
Show answer
Air track or low-friction trolleys; light gates and interrupt cards measure v before/after collision; compare Σmv before and after. Limitations: residual friction/air-track tilt (external force), card length uncertainty, gate timing resolution, velcro/magnet collisions not instantaneous. Improvement: level the track, use video analysis.
DefinitionSL & HL
Define the newton and express it in SI fundamental units.
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One newton is the resultant force that gives a mass of 1 kg an acceleration of 1 m s⁻² in the direction of the force. 1 N = 1 kg m s⁻². Force is a vector (magnitude and direction). Exam tip: the mark is lost for saying "force needed to move 1 kg" — it must be accelerate, not move — and for omitting "resultant". Writing kg m/s² is accepted but kg m s⁻² is the expected IB style. Always quote a direction (or a sign convention) when a numerical force is a final answer.
DefinitionSL & HL
Define angular velocity and state its SI unit.
Show answer
Angular velocity ω is the rate of change of angular displacement: ω = Δθ/Δt, measured in radians per second (rad s⁻¹). It is directed along the axis of rotation, so in IB problems it is treated as a scalar magnitude with a stated sense (clockwise/anticlockwise). Exam tip: students lose the mark by writing "angle turned per second" in degrees — θ must be in radians for v = ωr and a = ω²r to work. ω is not the same as the linear speed v; ω is the same for every point on a rigid rotating body, v is not.
DefinitionSL & HLData booklet: Yes
Define the period and the frequency of circular motion and state the link between them.
Show answer
Period T is the time for one complete revolution (unit s); frequency f is the number of revolutions per unit time (unit hertz, Hz = s⁻¹). They are reciprocals: f = 1/T, and ω = 2πf = 2π/T. Both are scalars. Exam tip: a very common error is using f in revolutions per minute directly — convert rpm to Hz by dividing by 60 before substituting. Another is treating T as the time for half a circle in vertical-circle questions; T is always one full revolution.
DefinitionSL & HL
Define a conical pendulum and state the forces acting on the bob.
Show answer
A conical pendulum is a mass on a string that moves in a horizontal circle at constant speed, so the string sweeps out the surface of a cone at a constant angle θ to the vertical. Only two forces act on the bob: its weight mg vertically down and the tension T along the string toward the support. Their resultant is horizontal, points to the centre of the circle and is the centripetal force. Exam tip: adding a third "centripetal force" arrow to the free-body diagram loses the mark — the centripetal force is the resultant of the two real forces, not an extra one.
DefinitionSL & HLData booklet: Yes
Define the spring constant and state its unit and physical meaning.
Show answer
The spring constant (force constant) k is the force per unit extension of a spring obeying Hooke's law: k = F/x, unit N m⁻¹. It is a scalar and it measures stiffness — a large k means a large force is needed for a small extension. Exam tip: k is the gradient of a force–extension graph, not of an extension–force graph, and it is defined by extension x (the change in length), never by the total length of the spring. State the unit N m⁻¹; "N/cm" from raw data must be converted.
DefinitionSL & HL
Define terminal speed and state the condition that determines it.
Show answer
Terminal speed is the constant maximum speed reached by an object falling through a fluid, attained when the resultant force on it is zero — the drag force plus any upthrust becomes equal in magnitude to the weight. At that point the acceleration is zero, so the velocity stops changing. Exam tip: "the forces are equal" is not enough for the mark; say the resultant (net) force is zero, and state which forces balance. Do not say the object has "no force acting" or that drag is zero — drag is at its maximum. Terminal speed is a speed, so it has no direction requirement.
DefinitionSL & HLData booklet: Yes
State Archimedes' principle and define upthrust.
Show answer
Upthrust (buoyancy) is the upward force exerted on a body by a fluid, arising because pressure increases with depth so the pressure on the lower surface exceeds that on the upper surface. Archimedes' principle: the upthrust equals the weight of the fluid displaced by the body, F_b = ρ_f V_f g. Unit newton; it is a vector, always vertically upward. Exam tip: V_f is the volume of fluid displaced (the submerged volume), not the whole volume of the object, and ρ_f is the density of the fluid, not of the object — swapping these is the single most common error.
DefinitionSL & HL
Define the normal contact force and state what determines its magnitude.
Show answer
The normal force N is the contact force exerted by a surface on a body, acting perpendicular to the surface of contact. Its magnitude is whatever value is required by Newton's second law perpendicular to the surface — it is not defined as mg. On a horizontal surface with no vertical acceleration N = mg; on an incline of angle θ, N = mg cos θ; in a lift accelerating upward N = m(g + a). Unit newton, vector. Exam tip: writing N = mg on an incline or in an accelerating lift is the classic lost mark; always resolve perpendicular to the surface first.
DefinitionSL & HLData booklet: No – derive
Define the banking angle of a curved track and state what it depends on.
Show answer
The banking angle θ is the angle to the horizontal at which a track is tilted so that the horizontal component of the normal force supplies the centripetal force, allowing a vehicle to turn without relying on friction. For the frictionless design speed, tan θ = v²/(rg), so θ depends only on the design speed v and the radius r (and g) — not on the mass of the vehicle. Exam tip: students often claim heavier vehicles need a larger angle; mass cancels. Above the design speed friction must act down the slope, below it friction acts up the slope.
EquationSL & HLData booklet: Yes
Write Newton's second law in its most general form and state when this form is required.
Show answer
F_net = Δp/Δt, where F_net is the resultant force in newtons (N), Δp is the change in momentum in kg m s⁻¹ and Δt is the time interval in seconds (s). This is the general statement and is valid even when mass changes; F = ma follows only for constant mass. It is required for rockets losing fuel, conveyor belts being loaded, and fluid jets, where F = (Δm/Δt)v. Data booklet: yes (as F = Δp/Δt). Common misuse: using F = ma with the instantaneous mass of a rocket while ignoring the momentum carried away by the exhaust. Sanity check: 20 kg s⁻¹ of water stopped at 15 m s⁻¹ gives 300 N.
EquationSL & HLData booklet: No – derive
Write the equation for the banking angle of a frictionless banked curve and define each symbol.
Show answer
tan θ = v²/(rg), where θ is the banking angle to the horizontal (degrees or radians), v is the design speed in m s⁻¹, r is the radius of the circular path in metres (m) and g = 9.8 m s⁻² is the acceleration of free fall. It comes from N sin θ = mv²/r and N cos θ = mg, so mass cancels. Data booklet: no — derive it from the two component equations. Common misuse: writing sin θ or using the length of the slope instead of the horizontal radius r. Sanity check: v = 25 m s⁻¹ on r = 150 m gives tan θ = 0.425, θ ≈ 23°, a realistic motorway bank.
EquationSL & HLData booklet: No – derive
Write the two component equations for a conical pendulum and the resulting expression for its speed.
Show answer
Vertically: T cos θ = mg. Horizontally: T sin θ = mv²/r, with r = L sin θ. Dividing gives tan θ = v²/(rg), so v = √(rg tan θ). Symbols: T tension in N, m mass in kg, θ angle of string to the vertical, r radius of the horizontal circle in m, L string length in m, v speed in m s⁻¹, g = 9.8 m s⁻². Data booklet: no — derive. Common misuse: using L instead of r = L sin θ in the centripetal term. Sanity check: L = 1.5 m, θ = 20° gives r = 0.51 m and v ≈ 1.4 m s⁻¹.
EquationSL & HLData booklet: No – memorise
State the relative-velocity condition for a perfectly elastic collision and explain how it is used.
Show answer
For a head-on elastic collision the relative speed of approach equals the relative speed of separation: u₁ − u₂ = v₂ − v₁, where u are velocities before and v after, all in m s⁻¹ with a consistent sign convention. Combined with momentum conservation m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ it gives two linear equations, avoiding the quadratic from ½mv². Data booklet: no — memorise (it follows from conserving p and E_k). Common misuse: applying it to inelastic collisions, where separation speed is less than approach speed. Sanity check: an equal-mass elastic collision with one body at rest gives a simple exchange of velocities.
EquationSL & HLData booklet: No – derive
Write kinetic energy in terms of momentum and state a use for this form.
Show answer
E_k = p²/(2m), where E_k is kinetic energy in joules (J), p is momentum in kg m s⁻¹ and m is mass in kg. It follows from p = mv and E_k = ½mv². Valid for non-relativistic speeds only. Data booklet: no — derive in one line. Use: in an explosion the two fragments have equal magnitudes of momentum, so E_k ∝ 1/m — the lighter fragment carries most of the kinetic energy. Common misuse: assuming equal momenta means equal kinetic energies. Sanity check: p = 10 kg m s⁻¹ with m = 5 kg gives E_k = 100/10 = 10 J, matching ½ × 5 × 2² = 10 J.
EquationSL & HLData booklet: No – derive
Write the condition for the minimum speed at the top of a vertical circle and define each symbol.
Show answer
At the minimum speed the string tension or track normal force falls to zero, so gravity alone provides the centripetal force: mg = mv²/r, giving v_min = √(gr). Here v_min is the speed at the highest point in m s⁻¹, r is the radius of the circle in m, g = 9.8 m s⁻² and the mass m cancels. Data booklet: no — derive. Valid only at the top, and only for a string or an inside track (a rigid rod can push, so v can be zero). Common misuse: setting v = 0 at the top for a string. Sanity check: r = 4.0 m gives v_min ≈ 6.3 m s⁻¹.
Concept/explainSL & HL
Explain why the normal force on a body is often not equal to its weight.
Show answer
- The normal force is a contact force set by Newton's second law perpendicular to the surface, not by a definition N = mg
- On a horizontal surface with no vertical acceleration and no other vertical forces, N = mg only as a special case
- On an incline of angle θ, resolving perpendicular to the surface gives N = mg cos θ, since only that component of weight presses on the surface
- In a lift accelerating upward at a, N − mg = ma so N = m(g + a); accelerating downward gives N = m(g − a), and in free fall N = 0 (apparent weightlessness)
- An extra push at an angle changes N, which changes friction F = μN and so changes the acceleration. Exam tip: incomplete answers state N = mg by habit and then use the wrong friction force for the whole question.
Concept/explainSL & HL
Explain how a rocket accelerates in empty space, where there is nothing to push against.
Show answer
- The rocket and its fuel form an isolated system, so total momentum is conserved
- Burning fuel ejects exhaust gas backwards with momentum −Δp, so the rocket gains an equal and opposite momentum +Δp forwards
- Equivalently, by Newton's third law the rocket pushes the exhaust backwards and the exhaust pushes the rocket forwards with an equal-magnitude, opposite force on a different body
- The thrust is the rate of change of momentum of the exhaust, F = (Δm/Δt)v_ex, so it does not require any external medium
- As fuel is used the mass falls, so for constant thrust the acceleration increases (a = F/m)
- Nature of science: this application of Newton's third law was doubted publicly in the 1920s, showing how a well-tested law resolves an apparent paradox. Exam tip: saying the gas "pushes against the air" scores zero.
Concept/explainSL & HL
Explain why static friction can take a range of values whereas dynamic friction is essentially fixed.
Show answer
- Static friction is a self-adjusting response force: it takes whatever value is needed to prevent relative sliding, up to a maximum
- Hence F_s ≤ μ_s N, an inequality, not an equality
- If a 10 N push acts on a stationary block whose maximum static friction is 25 N, the friction force is 10 N, not 25 N, and the block stays at rest
- Once sliding begins the contact points no longer interlock or bond for as long, so F_d = μ_d N with μ_d < μ_s, and the force is approximately constant and independent of speed
- This is why a pushed object suddenly jerks forward as it breaks away, and why ABS brakes keep the wheels rolling to stay in the larger static regime. Exam tip: the common incomplete answer uses F = μ_s N for a body that is clearly not on the point of slipping.
Concept/explainSL & HL
Explain why a car travelling at constant velocity has zero resultant force even though the engine is working.
Show answer
- Newton's first law: constant velocity means zero acceleration, which requires zero resultant force
- The forward driving force (friction of the road on the driven wheels) is exactly balanced by air drag plus rolling resistance
- Zero resultant force does not mean zero force — several forces act, and their vector sum is zero
- Vertically the normal force balances the weight, so there is equilibrium in both directions
- The engine still does work and transfers energy, because work is done against the resistive forces at a rate P = Fv, dissipating energy as internal energy — energy transfer is not the same as net force
- If the driving force is increased the balance is broken and the car accelerates until drag, which grows as v², restores the balance at a higher speed. Exam tip: "a force is needed to keep it moving" is the Aristotelian misconception the mark scheme looks for.
Concept/explainSL & HL
Explain how banking a track allows a vehicle to turn a corner with no friction.
Show answer
- On a banked surface the normal force is perpendicular to the track, so it is tilted from the vertical by the banking angle θ
- Resolve N into a vertical component N cos θ and a horizontal component N sin θ
- The vertical component balances the weight: N cos θ = mg, so there is no vertical acceleration
- The horizontal component is unbalanced and points toward the centre of the circle, providing the centripetal force: N sin θ = mv²/r
- Dividing gives tan θ = v²/(rg), so mass cancels and one banking angle works for cars and lorries alike
- Only at this design speed is friction unnecessary; faster and friction must act down the slope, slower and it acts up the slope. Exam tip: many answers resolve the weight along the slope as if it were a static incline problem, forgetting the acceleration is horizontal, not along the surface.
Concept/explainSL & HL
Explain how momentum can be conserved in an explosion even though the total kinetic energy increases.
Show answer
- Momentum conservation requires only that no resultant external force acts on the system; internal forces cancel in pairs by Newton's third law
- Energy conservation is a separate law: the extra kinetic energy comes from stored chemical or elastic potential energy released inside the system, not from nowhere
- Starting from rest, total momentum is zero, so the fragments must fly apart with equal-magnitude, opposite momenta
- Since E_k = p²/(2m), the lighter fragment carries far more kinetic energy — this is why a bullet is fast and lethal while the rifle recoil is slow
- Momentum is a vector so opposite momenta cancel; kinetic energy is a scalar and can only add. Exam tip: incomplete answers claim momentum is "created" or that kinetic energy must also be conserved because the system is isolated.
Concept/explainSL & HL
Explain how Newton's third law leads to conservation of momentum for two interacting bodies.
Show answer
- During the interaction body A exerts a force F on B, and B exerts −F on A: equal magnitude, opposite direction, same type, acting on different bodies
- The two forces act for exactly the same contact time Δt
- So the impulses are FΔt and −FΔt, equal and opposite
- By the impulse–momentum theorem the momentum change of A is exactly minus that of B: Δp_A = −Δp_B
- Therefore Δp_A + Δp_B = 0 and the total momentum of the system is unchanged
- The result holds only if no resultant external force acts on the system during the interaction, which is why we call it an isolated system
- Nature of science: a law about forces yields a conserved quantity, and the conservation statement survives into relativity and quantum physics where F = ma does not. Exam tip: omitting "same time of contact" leaves the derivation incomplete.
Concept/explainSL & HL
Explain how a real spring departs from Hooke's law as the load is increased.
Show answer
- Hooke's law F = kx holds only up to the limit of proportionality, where the force–extension graph stops being a straight line through the origin
- Just beyond it lies the elastic limit: the spring still returns to its original length when unloaded, but extension is no longer proportional to force
- Beyond the elastic limit the spring deforms plastically and keeps a permanent extension when the load is removed
- The loading and unloading curves then no longer coincide, and the area between them represents energy dissipated as internal energy
- Eventually the spring reaches its breaking point
- So k is only constant over the linear region, and E_p = ½kx² is only valid there. Exam tip: students routinely swap the limit of proportionality and the elastic limit, and describe the plastic region as "the spring breaks".
Concept/explainSL & HL
Explain why two masses connected over a light frictionless pulley share the same acceleration magnitude and the same tension.
Show answer
- The string is inextensible, so if one mass moves a distance d the other moves the same distance d in the same time — speeds and hence acceleration magnitudes are equal, though the directions differ
- The string is light (negligible mass), so the resultant force on any element of it must be zero; therefore tension is transmitted unchanged along its whole length
- The pulley is frictionless and light, so it changes only the direction of the tension and not its magnitude
- Applying Newton's second law to each mass separately gives m₁g − T = m₁a and T − m₂g = m₂a, which solve to a = (m₁ − m₂)g/(m₁ + m₂)
- Treating the system as a whole gives the same a directly, with tension as an internal force that cancels. Exam tip: the classic error is writing T = m₁g for the descending mass; that would mean zero acceleration.
Concept/explainSL & HL
Explain why drag is proportional to v for a small slow sphere but to v² for a fast large object, and the effect on terminal speed.
Show answer
- At low speeds the flow is laminar: layers of fluid slide smoothly and the resistance comes from viscosity, giving Stokes' law F = 6πηrv, so F ∝ v
- At higher speeds the flow becomes turbulent: the object must accelerate a mass of fluid out of its path each second, and both the mass rate and the speed given to it scale with v, so F = ½CρAv² and F ∝ v²
- The change happens above a critical Reynolds number, set by speed, size, density and viscosity
- Terminal speed follows from setting drag equal to weight minus upthrust: v_t = (mg − F_b)/6πηr in the laminar case, but v_t = √(2(mg − F_b)/CρA) in the turbulent case
- So doubling the weight doubles v_t for laminar flow but only multiplies it by √2 for turbulent flow. Exam tip: incomplete answers use ½CρAv² for a ball bearing in glycerol.
Concept/explainSL & HL
Explain how friction propels a walking person forward, and why this does not contradict the idea that friction opposes motion.
Show answer
- The foot pushes backwards on the ground; by Newton's third law the ground pushes forwards on the foot with an equal-magnitude friction force
- This forward friction is the external force that accelerates the person, so an internal muscular force alone could never do it
- Friction opposes relative sliding at the contact point, not the motion of the body as a whole — the foot tends to slip backwards, so friction acts forwards
- Because the foot does not actually slide, this is static friction, and its maximum value μ_s N sets how hard one can accelerate
- On ice μ_s is small, the required friction exceeds the maximum, the foot slips and the person cannot accelerate
- The same reasoning explains a car's driven wheels and a rolling ball. Exam tip: the standard incomplete answer says friction must act backwards because "friction always opposes motion".
Concept/explainSL & HL
Explain why Newton's laws only apply in an inertial frame of reference.
Show answer
- An inertial frame is one in which a body with no resultant force acting moves with constant velocity — a frame that is not accelerating
- In an accelerating frame, such as a braking bus, objects appear to accelerate with no identifiable force acting on them, so Newton's first and second laws appear to fail
- To keep using F = ma in such a frame one must invent fictitious forces (centrifugal, Coriolis) which have no third-law partner and no physical agent
- The proper analysis works in the ground frame, where the only real horizontal force on a passenger is friction from the seat
- Nature of science: choosing the frame is part of building a model, and the fact that the laws take their simplest form in inertial frames motivated Einstein's later work on relativity. Exam tip: answers that treat centrifugal force as a real force acting on a cornering car lose the explanation marks.
Worked problemSL & HLData booklet: No – derive
Masses of 3.0 kg and 2.0 kg hang from a light inextensible string over a frictionless pulley. Find the acceleration and the string tension.
Show answer
Apply Newton's second law to each mass, taking the direction of motion as positive. For the 3.0 kg falling mass: m₁g − T = m₁a. For the 2.0 kg rising mass: T − m₂g = m₂a. Adding eliminates T: (m₁ − m₂)g = (m₁ + m₂)a, so a = (3.0 − 2.0)(9.8)/(3.0 + 2.0) = 9.8/5.0 = 1.96 ≈ 2.0 m s⁻². Substitute back into the second equation: T = m₂(g + a) = 2.0(9.8 + 1.96) = 23.5 ≈ 24 N. Check/Trap: verify with the other mass, T = m₁(g − a) = 3.0(7.84) = 23.5 N — consistent. Note T lies between 19.6 N and 29.4 N, as it must; a tension equal to either weight would mean zero acceleration.
Worked problemSL & HLData booklet: No – derive
A 0.30 kg bob on a 1.5 m string moves as a conical pendulum with the string at 20° to the vertical. Find the tension and the speed.
Show answer
Vertically there is no acceleration: T cos θ = mg, so T = (0.30 × 9.8)/cos 20° = 2.94/0.940 = 3.13 ≈ 3.1 N. The radius of the horizontal circle is r = L sin θ = 1.5 × sin 20° = 1.5 × 0.342 = 0.513 m. Horizontally the resultant is centripetal: T sin θ = mv²/r, so 3.13 × 0.342 = 1.07 N = (0.30)v²/0.513, giving v² = 1.07 × 0.513/0.30 = 1.83 and v = 1.35 ≈ 1.4 m s⁻¹. Check/Trap: v = √(rg tan θ) = √(0.513 × 9.8 × 0.364) = 1.35 m s⁻¹ ✓. The trap is using L = 1.5 m instead of r = 0.51 m in the centripetal equation, and forgetting that T > mg = 2.9 N always.
Worked problemSL & HLData booklet: No – derive
A 60 kg pilot flies a vertical loop of radius 4.0 m. Find the minimum speed at the top and the seat force at the bottom.
Show answer
At the minimum speed at the top the seat force is zero and gravity alone is centripetal: mg = mv_t²/r, so v_t = √(gr) = √(9.8 × 4.0) = √39.2 = 6.26 ≈ 6.3 m s⁻¹ (independent of mass). Using conservation of energy from top to bottom, with a drop of 2r = 8.0 m: v_b² = v_t² + 4gr = 39.2 + 4(9.8)(4.0) = 39.2 + 156.8 = 196, so v_b = 14.0 m s⁻¹. At the bottom the resultant is upward: N − mg = mv_b²/r, so N = 60(9.8) + 60(196)/4.0 = 588 + 2940 = 3528 ≈ 3.5 × 10³ N. Check/Trap: N ≈ 6.0mg, so the pilot experiences about 6g — realistic for aerobatics. The trap is setting v = 0 at the top, or forgetting to add mg at the bottom.
Worked problemSL & HLData booklet: No – derive
A curve of radius 150 m is to be banked so that a car travelling at 25 m s⁻¹ needs no friction. Find the banking angle.
Show answer
For frictionless banking the only forces are the weight and the normal force. Vertically: N cos θ = mg. Horizontally, toward the centre: N sin θ = mv²/r. Dividing gives tan θ = v²/(rg) = 25²/(150 × 9.8) = 625/1470 = 0.4252. So θ = tan⁻¹(0.4252) = 23.0 ≈ 23°. Check/Trap: the mass cancelled, so the same angle serves every vehicle — a very common examined point. The traps are (a) using sin θ = v²/(rg), (b) using the slope length rather than the horizontal radius, and (c) failing to note that above 25 m s⁻¹ friction must act down the slope to supply the extra centripetal force, and below it, up the slope.
Worked problemSL & HLData booklet: No – derive
A 2.0 kg trolley moving at 4.0 m s⁻¹ collides elastically head-on with a stationary 6.0 kg trolley. Find both final velocities.
Show answer
Take the initial direction as positive. Conservation of momentum: (2.0)(4.0) + 0 = 2.0v₁ + 6.0v₂, so 8.0 = 2.0v₁ + 6.0v₂. For an elastic collision the relative speed of approach equals the relative speed of separation: 4.0 − 0 = v₂ − v₁. Substituting v₂ = v₁ + 4.0 gives 8.0 = 2.0v₁ + 6.0v₁ + 24, so 8.0v₁ = −16, v₁ = −2.0 m s⁻¹ and v₂ = +2.0 m s⁻¹. The lighter trolley rebounds at 2.0 m s⁻¹; the heavier moves forward at 2.0 m s⁻¹. Check/Trap: E_k before = ½(2.0)(16) = 16 J; after = ½(2.0)(4.0) + ½(6.0)(4.0) = 4.0 + 12 = 16 J ✓. The trap is dropping the minus sign — a lighter body always rebounds off a heavier one in an elastic collision.
Worked problemSL & HLData booklet: Yes
A hose delivers water at 20 kg s⁻¹ with speed 15 m s⁻¹ onto a wall, where it runs down without rebounding. Find the force on the wall.
Show answer
Use Newton's second law in its general form, F = Δp/Δt. In one second, a mass Δm = 20 kg arrives with velocity 15 m s⁻¹ and leaves with zero velocity in the original direction. Change of momentum per second = (Δm/Δt)v = 20 × 15 = 300 kg m s⁻² = 300 N. This is the force of the wall on the water; by Newton's third law the water pushes the wall with 300 N, directed into the wall. Check/Trap: if the water rebounded elastically at 15 m s⁻¹ the momentum change would double to 600 N — always ask whether the fluid stops or rebounds. Do not use F = ma with a single "mass of water"; the mass is flowing, so a rate of change of momentum is required.
Worked problemSL & HLData booklet: Yes
A 1200 kg cannon fires a 5.0 kg shell horizontally at 300 m s⁻¹. Find the recoil speed and comment on the energy shares.
Show answer
The system is isolated horizontally, so momentum is conserved and the total before firing is zero: 0 = m_s v_s + m_c v_c. Therefore v_c = −(5.0 × 300)/1200 = −1500/1200 = −1.25 ≈ −1.3 m s⁻¹, i.e. 1.3 m s⁻¹ backwards. Kinetic energies: shell ½(5.0)(300²) = 2.25 × 10⁵ J; cannon ½(1200)(1.25²) = 9.4 × 10² J. Total 2.26 × 10⁵ J, supplied by the chemical energy of the propellant. Check/Trap: the momenta are equal in magnitude (1500 kg m s⁻¹ each) but the energies are not, because E_k = p²/(2m) so E_k ∝ 1/m — the cannon takes only about 0.4% of the energy. Kinetic energy is not conserved in an explosion; momentum is.
Graph/diagramSL & HL
Describe the velocity–time graph for an object falling from rest through air until it reaches terminal speed.
Show answer
Axes: velocity (m s⁻¹) on the y-axis, time (s) on the x-axis. Shape: starts at the origin with the steepest gradient, equal to g = 9.8 m s⁻², then curves over with a decreasing gradient and approaches a horizontal asymptote at the terminal speed. Gradient at any point is the instantaneous acceleration; the area under the curve is the distance fallen. Extract terminal speed by reading the asymptote value; extract initial acceleration from the tangent at t = 0. The curve is concave down because drag grows with speed, reducing the resultant force. If the mass is increased (same shape and size) the asymptote rises and it takes longer to reach; opening a parachute causes a sharp drop in velocity — a steep negative gradient — settling to a new, lower asymptote.
Graph/diagramSL & HL
Describe a momentum–time graph and state the meaning of its gradient.
Show answer
Axes: momentum p (kg m s⁻¹) on the y-axis, time t (s) on the x-axis. Since F = Δp/Δt, the gradient of the graph equals the resultant force acting on the body in newtons. A straight line of constant gradient means a constant resultant force; a horizontal line means zero resultant force, so the body is in translational equilibrium; a curve means a changing force, and the tangent gives the instantaneous value. A negative gradient means the resultant force is opposite to the chosen positive direction. To find the impulse over an interval, read the change in the y-value, Δp, not the area — the area under a p–t graph has units kg m, which is not a useful quantity. Steepening the graph corresponds to a larger applied force.
Graph/diagramSL & HL
How would you linearise data from a circular-motion experiment to find the mass of a whirled bung?
Show answer
The relationship is F = mv²/r. For a fixed radius r, plot the centripetal force F (y-axis, N) against v² (x-axis, m² s⁻²), not against v, so that the theory predicts a straight line through the origin with gradient m/r. Measure m from gradient × r. Plotting F against v gives a curve from which no quantity can be read directly. Alternatively, for a fixed speed plot F against 1/r to get a straight line of gradient mv². Error bars come from the absolute uncertainties in F and v²; note that the fractional uncertainty in v² is twice that in v. A straight line that does not pass through the origin indicates a systematic error, such as friction at the tube rim or an unaccounted vertical component of the string tension.
Graph/diagramSL & HL
Describe the free-body diagram for a block on a rough incline and the components used to analyse it.
Show answer
Draw the block as a dot or box with exactly three forces as labelled arrows from it: weight mg vertically downward, normal force N perpendicular to and away from the slope surface, and friction F parallel to the slope, opposing the direction of relative sliding (up the slope for a block sliding down). Choose axes along and perpendicular to the slope, not horizontal and vertical. Resolve only the weight: component down the slope mg sin θ, component into the slope mg cos θ. Perpendicular: N = mg cos θ (no acceleration in that direction). Along: mg sin θ − μ_d mg cos θ = ma, so a = g(sin θ − μ_d cos θ) and mass cancels. Increasing θ increases mg sin θ and decreases N, so the block slips once tan θ > μ_s. Do not draw the resolved components as extra forces alongside the weight.
Exam technique/trapSL & HLData booklet: Yes
Trap: when must you use F = Δp/Δt rather than F = ma?
Show answer
The trap: students automatically write F = ma, but that form assumes constant mass, since Δp/Δt = mΔv/Δt only if m is fixed. Any situation with flowing or changing mass — a rocket burning fuel, sand falling onto a moving conveyor, a water jet hitting a surface, a hose recoiling — needs F = (Δm/Δt)v instead. Students fall for it because F = ma is the memorised form and the data booklet lists both without flagging the distinction. Correct approach: identify what changes momentum per second. Command-term guidance: "determine" expects a numerical answer with a unit; "show that" requires every substitution to be visible and the given value quoted to one more significant figure than the data. Give final answers to the same number of significant figures as the least precise datum, usually 2 or 3.
Exam technique/trapSL & HL
Trap: how should signs and directions be handled in one-dimensional momentum problems?
Show answer
The trap: momentum is a vector, but students substitute speeds as positive numbers and lose the direction information, typically getting a rebound velocity wrong or claiming momentum is not conserved. It happens because the arithmetic "looks right" and the magnitude of the answer seems plausible. Correct approach: state the positive direction explicitly in words at the start ("taking motion to the right as positive"), give every velocity a sign, and interpret a negative answer as motion in the opposite direction — never quietly drop the minus sign. A ball rebounding from 10 m s⁻¹ to 8 m s⁻¹ has Δv = −18 m s⁻¹, not −2 m s⁻¹. Quote the final answer as a magnitude plus a stated direction. For a two-dimensional collision, conserve momentum separately in two perpendicular directions.
Exam technique/trapSL & HL
Practical: describe an experiment to test F = mv²/r using a whirling rubber bung, and evaluate it.
Show answer
Apparatus: rubber bung on nylon thread through a smooth glass tube, hanging slotted masses providing the tension (F = Mg), paper marker to fix the radius, ruler, stopwatch. Method: whirl the bung horizontally so the marker stays just below the tube, time 20 revolutions to find T, then v = 2πr/T; repeat for a range of hanging masses. Controls: keep r and the bung mass m constant; keep the circle horizontal. Analysis: plot F against v² and find m from gradient × r. Limitations: the string is never truly horizontal, so Mg overestimates the horizontal tension component; friction at the tube rim adds an unmeasured force; the radius drifts during whirling; human reaction time dominates the timing uncertainty. Improvements: time many revolutions to reduce fractional timing uncertainty, use a video or photogate, smooth the tube rim, and repeat readings.
Exam technique/trapSL & HL
Trap: does friction always act opposite to the direction of motion?
Show answer
The trap: students draw friction backwards on every diagram, so they get the wrong sign for walking, for a car's driven wheels, for a box on an accelerating truck bed, and for a ball rolling without slipping. Why they fall for it: the textbook phrase "friction opposes motion" is a shorthand for "friction opposes relative sliding at the contact surface", and the two differ whenever the contact point is not slipping. Correct approach: ask which way the contact surfaces would slide if friction vanished, then draw friction opposing that. A driven wheel would spin backwards against the road, so road friction pushes the car forwards. On a truck bed that accelerates forwards, the box would slide backwards relative to the bed, so static friction on the box acts forwards and is the only force accelerating it, with F ≤ μ_s N setting the maximum safe acceleration.

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All 83 A.2 Forces and momentum cards

State Newton's first law of motion.

State Newton's second law of motion (general form).

State Newton's third law and the properties of a force pair.

Define linear momentum.

Define impulse.

State the law of conservation of linear momentum, including its condition.

Distinguish elastic, inelastic and totally inelastic collisions.

Distinguish mass and weight.

Define static friction and dynamic (kinetic) friction.

Define translational equilibrium.

Define centripetal force.

Write Newton's second law for constant mass and identify each symbol.

Write the equations for static and dynamic friction.

State Hooke's law as an equation.

Write Stokes' law for viscous drag on a sphere.

Write the drag equation for turbulent (high-speed) flow.

Write the equation for buoyancy (upthrust).

Write the momentum and impulse equations.

Write the equations for uniform circular motion.

Explain why an object in uniform circular motion is accelerating even though its speed is constant.

Explain how a car can go round a horizontal (unbanked) bend, and why it may skid.

Explain why the tension in a string is greatest at the bottom and least at the top for a mass in a vertical circle.

Explain, using impulse, how crumple zones/airbags/bending knees reduce injury.

Explain why momentum is conserved in an inelastic collision but kinetic energy is not.

Explain why a sphere falling through a viscous liquid reaches terminal speed, stating the force balance.

Explain why passengers feel 'thrown outward' when a car turns, using Newton's laws.

A 2.0 kg trolley at 3.0 m s⁻¹ collides with a stationary 1.0 kg trolley and they stick together. Find the final velocity and the KE lost.

A 0.20 kg ball hits a wall at 10 m s⁻¹ and rebounds at 8.0 m s⁻¹. Contact time 0.050 s. Find the average force.

A block slides down a 30° incline with μ_d = 0.20. Find its acceleration.

A 0.50 kg mass swings in a horizontal circle of radius 1.2 m at 3.0 m s⁻¹. Find the centripetal force.

A 60 kg person stands on scales in a lift accelerating upward at 2.0 m s⁻². What do the scales read?

Two forces, 6.0 N east and 8.0 N north, act on a 4.0 kg mass. Find the acceleration (magnitude and direction).

A 1.5 kg block is pulled along a rough horizontal floor by 12 N at 25° above horizontal, μ_d = 0.30. Find the acceleration.

What does the area under a force–time graph represent, and how is the average force found?

State the rules for drawing a free-body diagram.

Describe a force–extension graph for a spring and what its gradient and area represent.

Trap: Are the weight of a book and the normal force from the table a Newton's third law pair? Explain.

Trap: how should centripetal force appear in a circular-motion problem?

Trap: how do you apply momentum conservation to a 2D (oblique) collision?

Practical: describe how to measure the coefficient of static friction with an inclined plane.

Practical/NOS: how would you verify conservation of momentum in the lab and what limits accuracy?

Define the newton and express it in SI fundamental units.

Define angular velocity and state its SI unit.

Define the period and the frequency of circular motion and state the link between them.

Define a conical pendulum and state the forces acting on the bob.

Define the spring constant and state its unit and physical meaning.

Define terminal speed and state the condition that determines it.

State Archimedes' principle and define upthrust.

Define the normal contact force and state what determines its magnitude.

Define the banking angle of a curved track and state what it depends on.

Write Newton's second law in its most general form and state when this form is required.

Write the equation for the banking angle of a frictionless banked curve and define each symbol.

Write the two component equations for a conical pendulum and the resulting expression for its speed.

State the relative-velocity condition for a perfectly elastic collision and explain how it is used.

Write kinetic energy in terms of momentum and state a use for this form.

Write the condition for the minimum speed at the top of a vertical circle and define each symbol.

Explain why the normal force on a body is often not equal to its weight.

Explain how a rocket accelerates in empty space, where there is nothing to push against.

Explain why static friction can take a range of values whereas dynamic friction is essentially fixed.

Explain why a car travelling at constant velocity has zero resultant force even though the engine is working.

Explain how banking a track allows a vehicle to turn a corner with no friction.

Explain how momentum can be conserved in an explosion even though the total kinetic energy increases.

Explain how Newton's third law leads to conservation of momentum for two interacting bodies.

Explain how a real spring departs from Hooke's law as the load is increased.

Explain why two masses connected over a light frictionless pulley share the same acceleration magnitude and the same tension.

Explain why drag is proportional to v for a small slow sphere but to v² for a fast large object, and the effect on terminal speed.

Explain how friction propels a walking person forward, and why this does not contradict the idea that friction opposes motion.

Explain why Newton's laws only apply in an inertial frame of reference.

Masses of 3.0 kg and 2.0 kg hang from a light inextensible string over a frictionless pulley. Find the acceleration and the string tension.

A 0.30 kg bob on a 1.5 m string moves as a conical pendulum with the string at 20° to the vertical. Find the tension and the speed.

A 60 kg pilot flies a vertical loop of radius 4.0 m. Find the minimum speed at the top and the seat force at the bottom.

A curve of radius 150 m is to be banked so that a car travelling at 25 m s⁻¹ needs no friction. Find the banking angle.

A 2.0 kg trolley moving at 4.0 m s⁻¹ collides elastically head-on with a stationary 6.0 kg trolley. Find both final velocities.

A hose delivers water at 20 kg s⁻¹ with speed 15 m s⁻¹ onto a wall, where it runs down without rebounding. Find the force on the wall.

A 1200 kg cannon fires a 5.0 kg shell horizontally at 300 m s⁻¹. Find the recoil speed and comment on the energy shares.

Describe the velocity–time graph for an object falling from rest through air until it reaches terminal speed.

Describe a momentum–time graph and state the meaning of its gradient.

How would you linearise data from a circular-motion experiment to find the mass of a whirled bung?

Describe the free-body diagram for a block on a rough incline and the components used to analyse it.