Aristotle's fallacy, inertia and Newton's first law
When a bus suddenly starts, you fall backwards; when it brakes, you lurch forward. Your body "wants" to keep its old state. Link to earlier classes : the first law of motion of class 9 is now used to see why a force is needed to change motion, not to keep it going.
Inertia is the tendency of a body to stay at rest or in uniform motion in a straight line. Newton's first law : a body stays at rest or moves with constant velocity unless an external force acts on it. In short : if the net external force is zero, the acceleration is zero.
Rule 1: Aristotle's mistake — Aristotle thought a force is always needed to keep a body moving. In real life a push is needed only to cancel friction. Galileo rolled balls down one incline and up another and reasoned that on a perfectly smooth surface the ball would never stop. This is the law of inertia.
- Easy : A hockey puck slides far on smooth ice because friction is small.
- Medium : A coin placed on a card over a glass drops into the glass when the card is flicked away; the coin stays at rest.
- Exam : In a car moving at constant velocity on a straight road, the engine force exactly balances the resistive forces; the net force is zero.
Rule 2: Force changes velocity — A force is needed to start, stop, speed up, slow down or turn a body. Constant velocity needs no net force. Mass measures inertia : a larger mass resists change of motion more.
- Easy : An empty cart is easier to push than a loaded cart.
- Medium : A passenger standing in a bus is thrown forward when the bus brakes, because the feet stop but the upper body keeps moving.
- Exam : A stone is dropped from a car moving at constant speed. Neglecting air, it keeps the horizontal velocity of the car, so it stays directly below the car while falling (seen from the ground it follows a parabola).
Net force zero ⇒ acceleration zero ⇒ rest or uniform motion; inertia is measured by mass.
Thinking that a moving body always needs a force; only a change of velocity needs a net force.
Memory clue : "Lazy body" — every body is lazy; it keeps doing what it is doing until a force makes it change.
Spot the difference
| Situation | Net force | Motion |
|---|---|---|
| At rest | zero | stays at rest |
| Constant velocity | zero | stays in uniform motion |
| Changing velocity | not zero | accelerates |
Momentum and Newton's second law
A cricket ball hit by a bat and a tennis ball moving at the same speed are not equally hard to stop; the heavier, faster one is. Link to earlier classes : force = mass × acceleration from class 9 is now seen as the rate of change of momentum.
Momentum is mass times velocity. Newton's second law : the rate of change of momentum of a body equals the net external force on it, and takes place in the direction of the force.
Rule 1: Momentum and force — p = mv. Force = rate of change of momentum : F = dp ÷ dt. For constant mass F = ma. Unit of force : newton, 1 N = 1 kg m s⁻². It is a vector equation, and F is the net external force on the body or the system.
- Easy : m = 2 kg, v = 5 m s⁻¹ ⇒ p = 10 kg m s⁻¹.
- Medium : A 1500 kg car speeds up at 2 m s⁻² ⇒ net force = 1500 × 2 = 3000 N.
- Exam : A force of 20 N acts on a 4 kg block for 3 s starting from rest. a = 20 ÷ 4 = 5 m s⁻²; v = 5 × 3 = 15 m s⁻¹; displacement = ½ × 5 × 9 = 22.5 m.
Rule 2: Force and motion are local — The force at a given instant decides the acceleration at that instant; a body does not carry force from its past. If v = 0 for a moment, the force need not be zero (a ball at the top of its path still has weight mg and a = g).
- Easy : At the top of its flight a ball has v = 0 but a = 9.8 m s⁻² downwards.
- Medium : Force may act along, against, or at any angle to the velocity, but it is always along the acceleration.
- Exam : A 0.5 kg ball moving at 10 m s⁻¹ is hit back at 10 m s⁻¹ in 0.02 s. Change of momentum = 0.5 × (10 − (−10)) = 10 kg m s⁻¹; average force = 10 ÷ 0.02 = 500 N.
F = dp ÷ dt = ma; the net force is the vector sum of all external forces; ma is not an extra force.
Drawing "ma" as one more force on a free-body diagram, and thinking zero velocity means zero force.
Memory clue : "F = ma — Force makes Acceleration; Mass stands in the way." Same force, double mass ⇒ half acceleration.
Spot the difference
| Quantity | Definition | Unit |
|---|---|---|
| Momentum | mass × velocity | kg m s⁻¹ |
| Force | rate of change of momentum | N = kg m s⁻² |
| Impulse | force × time = change in momentum | N s = kg m s⁻¹ |
Impulse
A cricketer pulls his hands back while catching a fast ball; a high jumper lands on a soft mat. Link to earlier classes : the idea of momentum now explains why a longer stopping time means a smaller force.
Impulse is force multiplied by the time it acts. When a large force acts for a very short time, the impulse equals the change in momentum.
Rule 1: Impulse–momentum relation — J = F × Δt = Δp = mv − mu (J = impulse, u = initial velocity, v = final velocity). For a varying force, impulse is the area under the F–t graph. Average force = Δp ÷ Δt.
- Easy : F = 50 N for 0.2 s ⇒ J = 10 N s.
- Medium : A 0.15 kg ball arrives at 20 m s⁻¹ and is caught in 0.05 s. Δp = 0.15 × 20 = 3 kg m s⁻¹; F = 3 ÷ 0.05 = 60 N. If the catch takes 0.15 s, F = 20 N.
- Exam : A 60 kg person lands at 6 m s⁻¹ and stops in 0.3 s on soft sand : F = 60 × 6 ÷ 0.3 = 1200 N; on hard ground in 0.03 s : F = 12000 N, ten times larger.
Rule 2: Why cushioning works — The same change in momentum over a longer time gives a smaller force. This is why cricket fielders draw their hands back, vehicles have airbags, and packing uses foam.
- Easy : Landing on bent knees takes more time than on stiff legs, so the force is less.
- Medium : A hammer blow on a nail is an impulsive force : large force, very short time.
- Exam : Doubling the stopping time halves the average force for the same Δp.
impulse = F Δt = change in momentum; unit N s = kg m s⁻¹.
Forgetting the sign of Δp when the ball reverses direction.
Memory clue : "Long time, small force." Increase Δt to reduce F.
Spot the difference
| Case | Time of stopping | Force |
|---|---|---|
| Stiff-legged landing | very short | very large |
| Bent-knee landing | longer | smaller |
Newton's third law and conservation of momentum
Swimmers push the water backwards and move forwards; a gun recoils when a bullet is fired. Link to earlier classes : "every action has an equal and opposite reaction" of class 9 is now made exact with momentum.
Third law : forces always come in pairs; the force on A by B is equal and opposite to the force on B by A. They act at the same time on different bodies. Conservation of momentum : the total momentum of an isolated system stays constant.
Rule 1: Third law pair — F₁₂ = −F₂₁. Action and reaction act on different bodies, so they never cancel each other. Either force can be called the action; neither causes the other.
- Easy : You push a wall with 30 N; the wall pushes you with 30 N.
- Medium : A book on a table : the weight (earth pulls the book) and the book's pull on the earth form a pair; the normal force and the book's push on the table form another pair.
- Exam : A horse pulls a cart. The cart pulls the horse back equally, yet the pair moves forward because the ground exerts a forward friction on the horse that is larger than the backward forces on it.
Rule 2: Conservation of momentum — If the external force on a system is zero, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. It follows from the second and third laws.
- Easy : A 0.02 kg bullet leaves a 4 kg rifle at 300 m s⁻¹. Recoil speed = 0.02 × 300 ÷ 4 = 1.5 m s⁻¹, opposite to the bullet.
- Medium : A 2 kg trolley at 3 m s⁻¹ hits a stationary 1 kg trolley and they stick together : v = (2 × 3) ÷ 3 = 2 m s⁻¹.
- Exam : A 60 kg skater at rest throws a 3 kg ball at 4 m s⁻¹. Her recoil speed = 3 × 4 ÷ 60 = 0.2 m s⁻¹.
forces in a pair act on different bodies at the same time; momentum is conserved only when the external force is zero.
Saying that action and reaction cancel (they act on different bodies), and mixing up "reaction" with "later".
Memory clue : "Total momentum before = total momentum after." Write the equation first, then insert signs.
Spot the difference
| Point | Action–reaction pair | Balanced forces |
|---|---|---|
| Act on | different bodies | the same body |
| Cancel? | no | yes (net force zero) |
| Equal and opposite | yes | yes |
Equilibrium of a particle and common forces
A lamp hanging from two ropes does not move; a book on a table stays still. Link to earlier classes : balanced forces of class 9 now become the condition ΣF = 0.
A particle is in equilibrium when the vector sum of all forces on it is zero. Common forces : weight, normal force, tension, spring force, friction.
In the figure : A block on an incline has weight mg, normal force N and friction f; the weight is split along and across the slope.
Rule 1: Equilibrium — ΣF = 0, which means the sum of the x-components and the sum of the y-components are both zero.
- Easy : A 5 kg block hangs at rest from a string ⇒ tension T = 5 × 9.8 = 49 N.
- Medium : A 10 kg lamp hangs from two equal ropes, each at 30° to the horizontal. 2T sin 30° = 98 ⇒ T = 98 N.
- Exam : For the same lamp with ropes at 60° to the horizontal : 2T sin 60° = 98 ⇒ T = 56.6 N. Steeper ropes bear smaller tension.
Rule 2: Common forces — Weight W = mg (pull of the earth, downwards). Normal force N acts perpendicular to the surface; it adjusts itself and is not always equal to mg. Tension T acts along a string and is the same throughout a light string over a smooth pulley. Spring force F = −kx (k = spring constant, x = extension), opposite to the extension.
- Easy : A 2 N/cm spring stretched by 3 cm pulls with 6 N.
- Medium : On an incline of 30°, the weight component along the slope is mg sin 30° = 0.5 mg and across the slope mg cos 30° = 0.866 mg.
- Exam : A 70 kg man in a lift accelerating upwards at 2 m s⁻² : N − mg = ma ⇒ N = 70 × (9.8 + 2) = 826 N; in a lift accelerating downwards at 2 m s⁻² : N = 70 × (9.8 − 2) = 546 N.
ΣF = 0 in equilibrium; N = mg only when there is no vertical acceleration.
Writing N = mg on an incline or in a lift; N must come from the force balance.
Memory clue : "Draw the free-body diagram first — every arrow is 'force on the body BY something'."
Spot the difference
| Force | Direction | Value |
|---|---|---|
| Weight | towards earth | mg |
| Normal force | perpendicular to the surface | adjusts to the situation |
| Tension | along the string, pulling | same along a light string |
| Spring force | opposite to extension | kx |
Friction
A heavy almirah is hard to start pushing but easier to keep moving; shoes grip the road because of friction. Link to earlier classes : friction of class 8 now has numbers : coefficients of static and kinetic friction.
Friction is the force along the surface of contact that opposes relative motion (or its attempt). Static friction acts while there is no sliding; kinetic friction acts when the surfaces slide.
In the figure : Static friction rises with the applied force up to a limit; once the body slides, kinetic friction is smaller and steady.
Rule 1: Laws of friction — Static : fₛ ≤ μₛN (a self-adjusting force up to its maximum). Kinetic : fₖ = μₖN. μ values depend only on the two surfaces; friction does not depend on the area of contact; μₖ < μₛ.
- Easy : A 20 kg box on the floor, μₛ = 0.5 : maximum static friction = 0.5 × 20 × 9.8 = 98 N.
- Medium : A 10 N push on this box does not move it, so friction is 10 N, not 98 N.
- Exam : A 20 kg box slides with μₖ = 0.4 under a 100 N push. Friction = 0.4 × 196 = 78.4 N; net force = 21.6 N; a = 21.6 ÷ 20 = 1.08 m s⁻².
Rule 2: Angle of repose — A block on an incline just begins to slide when tan θ = μₛ.
- Easy : μₛ = 0.58 ⇒ θ = tan⁻¹(0.58) ≈ 30°.
- Medium : At 30° incline a block slides with acceleration a = g (sin 30° − μₖ cos 30°). With μₖ = 0.2 : a = 9.8 × (0.5 − 0.173) = 3.2 m s⁻².
- Exam : A car's wheels grip because of static friction on the road; the car's maximum braking deceleration is μₛg, e.g. 0.8 × 9.8 = 7.84 m s⁻².
fₛ ≤ μₛN; fₖ = μₖN; μₖ < μₛ; friction is independent of area (for the ordinary surfaces).
Putting fₛ = μₛN without checking that the body is just about to slide.
Memory clue : "Static is a shy helper — it gives only as much as needed, up to a limit."
Spot the difference
| Property | Static friction | Kinetic friction |
|---|---|---|
| When | no sliding yet | during sliding |
| Value | adjusts, up to μₛN | fixed at μₖN |
| Size | larger limit | smaller than the limit |
Circular motion : level road and banked road
A scooter taking a sharp turn leans inwards; highway bends are tilted. Link to earlier classes : centripetal acceleration of the previous chapter now gets its cause : a real force.
The centripetal force is simply the name for whichever real force (tension, friction, gravity, normal force) provides the acceleration v² ÷ R towards the centre.
Rule 1: Centripetal force — F = mv² ÷ R = mω²R. It is not a new force; do not add it to a free-body diagram as an extra arrow.
- Easy : A 0.5 kg stone on a 1 m string whirled at 4 m s⁻¹ : T = 0.5 × 16 ÷ 1 = 8 N.
- Medium : A car of 1000 kg takes a bend of radius 50 m at 10 m s⁻¹ : friction must provide 1000 × 100 ÷ 50 = 2000 N.
- Exam : A 2 kg mass whirled in a vertical circle of radius 1 m has tension at the bottom with speed 5 m s⁻¹ : T − mg = mv² ÷ r ⇒ T = 2 × 9.8 + 2 × 25 = 69.6 N.
Rule 2: Level road — Friction provides the force : v² ≤ μₛRg, so v_max = √(μₛRg), independent of the mass.
- Easy : μₛ = 0.4, R = 50 m ⇒ v_max = √(0.4 × 50 × 9.8) = 14 m s⁻¹ (50 km h⁻¹).
- Medium : On a wet road μₛ = 0.1 : v_max = √(0.1 × 50 × 9.8) = 7 m s⁻¹.
- Exam : Doubling the radius increases v_max by a factor √2.
Rule 3: Banked road — Without friction the horizontal component of N gives the force : tan θ = v² ÷ (Rg). Banking lets a car turn safely at the design speed without depending on friction.
- Easy : R = 100 m, v = 20 m s⁻¹ ⇒ tan θ = 400 ÷ 980 = 0.408 ⇒ θ = 22.2°.
- Medium : Banking angle 15°, R = 300 m ⇒ design speed v = √(Rg tan 15°) = √(300 × 9.8 × 0.268) = 28.1 m s⁻¹.
- Exam : A bend of R = 150 m, θ = 10°, μₛ = 0.2 (maximum speed with friction) : v_max = √[Rg (tan θ + μₛ) ÷ (1 − μₛ tan θ)] = √[150 × 9.8 × (0.1763 + 0.2) ÷ (1 − 0.0353)] = 23.9 m s⁻¹.
centripetal force is provided by a real force; v_max = √(μₛRg) on a level road; tan θ = v² ÷ Rg on a banked road.
Drawing a "centrifugal" force in the ground frame; there is only the inward net force.
Memory clue : "Level road — friction; banked road — normal force; both give v²/R."
Spot the difference
| Road | Force providing v²/R | Maximum or design speed |
|---|---|---|
| Level | friction | √(μₛRg) |
| Banked (no friction) | horizontal part of N | √(Rg tan θ) |
Solving problems with free-body diagrams
Two buckets joined by a rope over a pulley, a lift starting to go up — each needs the same careful method. Link to earlier classes : all earlier laws now combine into one repeatable method.
A free-body diagram shows one body alone with every external force acting on it. Writing F = ma along each axis for each body gives equations to solve.
Rule 1: The method — (1) Choose the body. (2) Draw all forces on it. (3) Pick axes and resolve forces. (4) Write ΣF = ma for each axis. (5) Use constraints (a connecting string means equal acceleration). (6) Solve and check the units.
- Easy : A 3 kg block is pulled by 12 N on a smooth table : a = 12 ÷ 3 = 4 m s⁻².
- Medium : Two blocks 2 kg and 3 kg touch each other on a smooth surface and a 10 N force pushes the 2 kg block. a = 10 ÷ 5 = 2 m s⁻²; contact force on the 3 kg block = 3 × 2 = 6 N.
- Exam : Atwood machine, 2 kg and 3 kg over a light pulley : a = (3 − 2) × 9.8 ÷ (3 + 2) = 1.96 m s⁻²; T = 2 × (9.8 + 1.96) = 23.5 N.
Rule 2: Block pulled over a table by a hanging mass — On a smooth table, M on the table and m hanging : a = mg ÷ (M + m); T = Mmg ÷ (M + m).
- Easy : M = 4 kg, m = 1 kg ⇒ a = 9.8 ÷ 5 = 1.96 m s⁻²; T = 4 × 1 × 9.8 ÷ 5 = 7.84 N.
- Medium : With friction μₖ = 0.2 on the table : a = (mg − μₖMg) ÷ (M + m) = (9.8 − 7.84) ÷ 5 = 0.39 m s⁻².
- Exam : Tension is always less than mg while the hanging mass accelerates downwards.
one body, one free-body diagram, one equation for each direction.
Using the weight of the hanging mass as the tension, and mixing forces on different bodies in one equation.
Memory clue : "Isolate, Draw, Resolve, Equate, Solve."
Spot the difference
| Case | Acceleration | String tension |
|---|---|---|
| Atwood machine | (m₂ − m₁)g ÷ (m₁ + m₂) | 2m₁m₂g ÷ (m₁ + m₂) |
| Block + hanging mass | mg ÷ (M + m) | Mmg ÷ (M + m) |