Scalar product of two vectors
A porter carrying a load on his head and walking on a level road does no work on the load in the physics sense, because the force is upward and the motion is sideways. Link to earlier classes : work = force × distance of class 9 is now written with the angle between the force and the displacement.
The scalar (dot) product of two vectors is a number : the product of their magnitudes and the cosine of the angle between them.
Rule 1: Dot product — A · B = AB cos θ (θ = angle between A and B). It can be positive, zero or negative. It is commutative (A · B = B · A) and distributive. For unit vectors : î · î = ĵ · ĵ = k̂ · k̂ = 1 and î · ĵ = ĵ · k̂ = k̂ · î = 0. In components : A · B = AₓBₓ + A_yB_y + A_zB_z.
- Easy : A = 3, B = 4, θ = 60° ⇒ A · B = 12 × 0.5 = 6.
- Medium : A = 2î + 3ĵ and B = 4î − ĵ ⇒ A · B = 8 − 3 = 5.
- Exam : Find the angle between A = î + ĵ and B = î − ĵ. A · B = 1 − 1 = 0 ⇒ cos θ = 0 ⇒ θ = 90°.
Rule 2: Meaning and checks — A · B is the magnitude of A times the component of B along A. Two non-zero vectors are perpendicular if A · B = 0. A · A = A².
- Easy : A · B is a scalar; A × B (next chapter) is a vector.
- Medium : For θ = 0°, A · B = AB (largest); for θ = 180°, A · B = −AB.
- Exam : For A = 3î + 4ĵ, A · A = 9 + 16 = 25 ⇒ |A| = 5.
A · B = AB cos θ = AₓBₓ + A_yB_y + A_zB_z; result is a scalar; zero for perpendicular vectors.
Giving a direction to a dot product, and using sin instead of cos.
Memory clue : "Dot is Duplicate along (cos): same direction gives the full product."
Spot the difference
| Product | Result | Formula |
|---|---|---|
| Dot product | scalar | AB cos θ |
| Cross product | vector | AB sin θ (next chapter) |
Work
Pushing a stuck car moves nothing, so in physics no work is done even though you feel tired. Link to earlier classes : work done by a force of class 9 now includes direction and angle.
Work done by a force is the dot product of the force and the displacement. Unit : joule (J) = N m = kg m² s⁻².
Rule 1: Work by a constant force — W = F · d = F d cos θ (θ = angle between force and displacement). Work is positive when 0° ≤ θ < 90°, zero when θ = 90°, and negative when θ > 90°. Always say which force does the work.
- Easy : F = 10 N, d = 5 m, same direction ⇒ W = 50 J.
- Medium : A boy pulls a 20 N wagon handle at 60° over 10 m : W = 20 × 10 × cos 60° = 100 J.
- Exam : A 2 kg book is lifted 1.5 m at a steady speed. Work by the lifting force = mgh = 2 × 9.8 × 1.5 = 29.4 J; work by gravity = −29.4 J; net work = 0.
Rule 2: Variable force — For a force that changes with position, W = ∫ F dx, the area under the F–x graph.
- Easy : Force rises uniformly from 0 to 10 N over 4 m ⇒ area = ½ × 4 × 10 = 20 J.
- Medium : A spring is stretched from 0 to 0.1 m with k = 200 N m⁻¹ : W = ½ k x² = ½ × 200 × 0.01 = 1 J.
- Exam : F is 5 N for the first 2 m and 3 N for the next 4 m ⇒ W = 10 + 12 = 22 J.
In the figure : The shaded area under the F–x line is the work done by the force.
W = F d cos θ; work is a scalar; it can be negative; unit joule.
Calling "work" without naming the force, and forgetting that perpendicular forces do no work.
Memory clue : "W = F d cos θ — Work follows the angle: acute positive, right-angle zero, obtuse negative."
Spot the difference
| Angle θ | Work | Example |
|---|---|---|
| 0° | positive, largest | pushing a cart forward |
| 90° | zero | carrying a bag level |
| 180° | negative | friction on a sliding block |
Kinetic energy and the work–energy theorem
A fast cricket ball is much more dangerous than a slow one of the same mass. Link to earlier classes : kinetic energy ½mv² of class 9 is now derived from the work done.
Kinetic energy is the energy a body has because of its motion. The work–energy theorem says the work done by the net force equals the change in kinetic energy.
Rule 1: Kinetic energy — K = ½mv² (always positive); in terms of momentum K = p² ÷ 2m.
- Easy : m = 2 kg, v = 3 m s⁻¹ ⇒ K = ½ × 2 × 9 = 9 J.
- Medium : Doubling the speed makes K four times larger.
- Exam : Two bodies have equal momentum; the lighter one has more kinetic energy because K = p² ÷ 2m.
Rule 2: Work–energy theorem — K_f − K_i = W_net. It holds in every situation where the net work is known, even if the force is not constant.
- Easy : A 1 kg body speeds up from 2 to 4 m s⁻¹ ⇒ W = ½ × 1 × (16 − 4) = 6 J.
- Medium : A 1000 kg car moving at 20 m s⁻¹ is stopped by a constant braking force over 50 m. Work = −½ × 1000 × 400 = −200000 J ⇒ F = 4000 N.
- Exam : A 0.2 kg ball is thrown up at 10 m s⁻¹. At the top K_f = 0 ⇒ −mgh = −½mv² ⇒ h = 100 ÷ 19.6 = 5.1 m.
K = ½mv² = p² ÷ 2m; W_net = ΔK; kinetic energy is never negative.
Using W of a single force instead of the net work in the theorem, and forgetting that K depends on v² (not v).
Memory clue : "Net work in, kinetic energy out."
Spot the difference
| Quantity | Sign | Depends on |
|---|---|---|
| Kinetic energy | always positive | mass and speed |
| Work | positive, zero or negative | force, displacement, angle |
Potential energy and conservative forces
A stretched catapult or a raised hammer stores energy because of its position. Link to earlier classes : potential energy mgh of class 9 now comes from the idea of a conservative force.
A force is conservative if the work it does between two points does not depend on the path (equivalently, the work over a closed path is zero). Then a potential energy can be defined.
Rule 1: Conservative forces — Gravity and the spring force are conservative. Friction is not : the work it does depends on the path and over a closed path is not zero. For a conservative force, F = −dV ÷ dx, and V_i − V_f = ∫ F dx.
- Easy : Lifting a 2 kg ball 3 m raises its gravitational potential energy by mgh = 2 × 9.8 × 3 = 58.8 J.
- Medium : The work done by gravity when the ball falls through 3 m is +58.8 J, the same along any path.
- Exam : A 5 kg box is pushed up a rough incline and brought back; the work done by friction over the round trip is not zero, so friction has no potential energy.
Rule 2: Two common potential energies — Gravitational (near the earth) : V = mgh. Spring : V = ½kx² (x = extension or compression from the natural length). The zero of potential energy is chosen for convenience.
- Easy : A 4 kg book on a shelf 2 m high has V = 4 × 9.8 × 2 = 78.4 J relative to the floor.
- Medium : k = 400 N m⁻¹, compressed 5 cm ⇒ V = ½ × 400 × 0.0025 = 0.5 J.
- Exam : Doubling the compression of a spring makes the stored energy four times larger.
conservative force ⇒ path-independent work ⇒ potential energy exists; V = mgh and V = ½kx².
Assigning a potential energy to friction, and taking "zero level" as fixed (only differences matter).
Memory clue : "Path doesn't matter — conservative; path matters — friction."
Spot the difference
| Force | Conservative? | Potential energy |
|---|---|---|
| Gravity | yes | mgh |
| Spring | yes | ½kx² |
| Friction | no | none |
Conservation of mechanical energy
A swing keeps exchanging height for speed; a pendulum bob reaches the same height on both sides. Link to earlier classes : energy transformation of class 9 is now an exact law.
If only conservative forces do work, the total mechanical energy (kinetic + potential) stays constant.
Rule 1: The law — K + V = constant, i.e. ½mv₁² + mgh₁ = ½mv₂² + mgh₂. Non-conservative forces (friction) change the total energy by the work they do.
- Easy : A 1 kg ball is dropped from 5 m : v at the ground = √(2gh) = √(2 × 9.8 × 5) = 9.9 m s⁻¹.
- Medium : A 0.5 kg pendulum bob released from a height 0.2 m above the lowest point passes the lowest point with v = √(2 × 9.8 × 0.2) = 1.98 m s⁻¹.
- Exam : A 2 kg block slides from rest down a smooth curve of height 3 m and compresses a spring of k = 600 N m⁻¹. mgh = ½kx² ⇒ 58.8 = 300x² ⇒ x = 0.443 m.
Rule 2: With friction — The energy lost to friction = work done by friction.
- Easy : A 1 kg block slides down a rough ramp from 2 m height and reaches the bottom at 4 m s⁻¹. Initial energy = 19.6 J; final = ½ × 16 = 8 J ⇒ lost = 11.6 J.
- Medium : The lost 11.6 J has become heat and sound at the contact surfaces.
- Exam : A body of mass m thrown up with speed u reaches a maximum height h = u² ÷ 2g; at half this height its kinetic energy is half of the initial one.
K + V is constant when only gravity or spring forces act; friction removes mechanical energy.
Applying conservation when friction acts, and forgetting that energy lost still exists as heat.
Memory clue : "Height down, speed up — the sum stays the same."
Spot the difference
| Situation | Mechanical energy |
|---|---|
| Free fall, no air | constant |
| Slide with friction | decreases |
| Mass on a spring (no friction) | constant |
Power
Two students carry the same load up the same stairs; the one who finishes sooner delivers more power. Link to earlier classes : power = work ÷ time of class 9 now gets instantaneous form.
Power is the rate of doing work. Unit : watt (W) = J s⁻¹; 1 horse power ≈ 746 W; 1 kWh = 3.6 × 10⁶ J.
Rule 1: Average and instantaneous power — P = W ÷ t (average); P = F · v = Fv cos θ (instantaneous).
- Easy : A crane does 5000 J in 10 s ⇒ P = 500 W.
- Medium : A 60 kg student climbs 4 m in 8 s ⇒ P = 60 × 9.8 × 4 ÷ 8 = 294 W.
- Exam : A car on a level road moves at 20 m s⁻¹ against a resistance of 500 N : the engine's power = 500 × 20 = 10000 W = 10 kW.
Rule 2: Energy bills — Energy in kWh = power in kW × time in hours.
- Easy : A 100 W bulb lit for 10 h uses 1 kWh.
- Medium : A 2 kW heater used for 3 h consumes 6 kWh = 6 × 3.6 × 10⁶ = 2.16 × 10⁷ J.
- Exam : A pump lifts 600 kg of water per minute through 15 m. Power = 600 × 9.8 × 15 ÷ 60 = 1470 W.
P = W ÷ t = Fv; unit watt; 1 kWh = 3.6 × 10⁶ J.
Treating kWh as a unit of power (it is energy).
Memory clue : "Power is speed of work — P = Fv."
Spot the difference
| Quantity | Unit | Meaning |
|---|---|---|
| Energy | J, kWh | capacity to do work |
| Power | W, kW | rate of doing work |
Collisions
Two carom coins strike and fly apart; two cars crash and crumple together. Link to earlier classes : conservation of momentum of the last chapter now meets energy.
In every collision the total momentum is conserved. In an elastic collision kinetic energy is also conserved; in an inelastic one it is not; in a perfectly inelastic collision the bodies stick together.
Rule 1: Elastic collision in one dimension — For bodies m₁, m₂ with speeds u₁, u₂ : v₁ = [(m₁ − m₂)u₁ + 2m₂u₂] ÷ (m₁ + m₂); v₂ = [(m₂ − m₁)u₂ + 2m₁u₁] ÷ (m₁ + m₂). For equal masses the bodies exchange velocities.
- Easy : Equal masses, m₁ moving at 4 m s⁻¹ and m₂ at rest ⇒ after the collision m₁ stops and m₂ moves at 4 m s⁻¹.
- Medium : m₁ = 2 kg at 6 m s⁻¹ hits m₂ = 1 kg at rest : v₁ = (2 − 1) × 6 ÷ 3 = 2 m s⁻¹; v₂ = 2 × 2 × 6 ÷ 3 = 8 m s⁻¹.
- Exam : A heavy ball hitting a light stationary ball barely slows down; a light ball hitting a heavy stationary wall bounces back with nearly the same speed.
Rule 2: Perfectly inelastic collision — The bodies move together : v = (m₁u₁ + m₂u₂) ÷ (m₁ + m₂); kinetic energy is lost.
- Easy : 3 kg at 4 m s⁻¹ sticks to 1 kg at rest : v = 12 ÷ 4 = 3 m s⁻¹.
- Medium : Kinetic energy before = ½ × 3 × 16 = 24 J; after = ½ × 4 × 9 = 18 J; loss = 6 J.
- Exam : Coefficient of restitution e = (relative speed after) ÷ (relative speed before); e = 1 for elastic, e = 0 for perfectly inelastic.
momentum is conserved in every collision; kinetic energy only in elastic ones.
Assuming kinetic energy is conserved in every collision.
Memory clue : "Momentum always, energy sometimes." Equal masses elastic : "they swap speeds."
Spot the difference
| Type | Momentum | Kinetic energy |
|---|---|---|
| Elastic | conserved | conserved |
| Inelastic | conserved | not conserved |
| Perfectly inelastic | conserved | maximum loss; bodies stick |