Mechanics

Mechanics is about a third of the Physics paper (our estimate: 8 of 27 questions).

The official syllabus lists seven sub-areas: kinematics and vectors, Newton's laws and friction, work-energy, circular motion with gravitation and SHM, rotation, elasticity, and fluids.

The Section B questions combine two ideas, typically energy with circular motion, momentum with energy loss, or continuity with Bernoulli.

1.1 Physical quantities, vectors and kinematics

Dimensions. [\text{force}]=\text{MLT}^{-2}, [\text{work, energy}]=\text{ML}^2\text{T}^{-2}, [\text{power}]=\text{ML}^2\text{T}^{-3}, [\text{pressure, stress, }Y]=\text{ML}^{-1}\text{T}^{-2}, [\text{surface tension}]=\text{MT}^{-2}, [\text{viscosity }\eta]=\text{ML}^{-1}\text{T}^{-1}, [G]=\text{M}^{-1}\text{L}^3\text{T}^{-2}, [h]=\text{ML}^2\text{T}^{-1}.

Only quantities of the same dimensions can be added; dimensional analysis checks a formula but cannot fix numerical constants.

Vectors. Resultant of \vec A,\vec B at angle \theta: R=\sqrt{A^2+B^2+2AB\cos\theta}, direction \tan\alpha=\frac{B\sin\theta}{A+B\cos\theta}. Two equal forces F at 120^\circ give F.

Polygon law: vectors drawn head to tail close the polygon when the sum is zero. Resolution: A_x=A\cos\theta, A_y=A\sin\theta. Dot product (work, projection), cross product (torque, magnetic force).

Equations of uniformly accelerated motion: v=u+at, s=ut+\frac{1}{2}at^2, v^2=u^2+2as, s_n=u+\frac{a}{2}(2n-1). Free fall uses a=g=9.8\ \text{m s}^{-2}. Area under the v–t graph is displacement; slope of the x–t graph is velocity.

Projectile (speed u, angle \theta): T=\dfrac{2u\sin\theta}{g}, H=\dfrac{u^2\sin^2\theta}{2g}, R=\dfrac{u^2\sin2\theta}{g} (maximum at 45^\circ, equal for \theta and 90^\circ-\theta), \tan\theta=\frac{4H}{R}.

Trajectory y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta} is a parabola. Horizontal launch from height h: t=\sqrt{2h/g}.

Relative motion: \vec v_{AB}=\vec v_A-\vec v_B. River crossing: shortest time when heading perpendicular to the bank (time d/v_b); shortest path when \sin\theta=v_r/v_b (needs v_b>v_r).

Rain appears to fall at \tan\alpha=v_{\text{man}}/v_{\text{rain}} from the vertical.

1.2 Newton's laws of motion and friction

Laws: (1) inertia; (2) \vec F=\frac{d\vec p}{dt}=m\vec a; (3) action–reaction on different bodies. Impulse =F\Delta t=\Delta p (area under the F–t graph).

Momentum conservation holds when no external force acts: m_1u_1+m_2u_2=m_1v_1+m_2v_2; a gun recoils with v_{\text{gun}}=\frac{m_{\text{bullet}}}{M}v.

Equilibrium: \sum\vec F=0 (e.g. Lami's theorem for three concurrent forces). Inclined plane (angle \theta, frictionless): a=g\sin\theta. Atwood machine: a=\dfrac{(m_1-m_2)g}{m_1+m_2}, T=\dfrac{2m_1m_2g}{m_1+m_2}.

Lift accelerating up at a: apparent weight m(g+a); down: m(g-a); free fall: zero.

Solid friction. Static f_s\le\mu_sN; kinetic f_k=\mu_kN (\mu_k<\mu_s). Laws (verified with an inclined plane): independent of the area of contact, proportional to normal reaction, independent of speed.

Angle of friction \lambda: \tan\lambda=\mu; angle of repose \alpha (a block just slips): \tan\alpha=\mu. On a rough incline a=g(\sin\theta-\mu\cos\theta). Horizontal push P on a block: a=\dfrac{P-\mu mg}{m}.

1.3 Work, energy and power

W=\vec F\cdot\vec s=Fs\cos\theta; area under the F–x graph; spring work \frac{1}{2}kx^2. Work–energy theorem: W_{\text{net}}=\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2. K=\frac{p}{2m}\cdot p=\frac{p^2}{2m}.

Gravitational PE =mgh; elastic PE =\frac{1}{2}kx^2. Mechanical energy is conserved under conservative forces (gravity, spring, electrostatic: work independent of path, zero over a closed loop); friction and drag are non-conservative.

Power P=\frac{W}{t}=\vec F\cdot\vec v; 1 hp = 746 W; kWh = 3.6\times10^6 J.

Collisions. Momentum is always conserved.

Elastic kinetic energy also conserved, e=1; for equal masses the velocities exchange; for m_1 on a stationary m_2: v_1=\frac{m_1-m_2}{m_1+m_2}u, v_2=\frac{2m_1}{m_1+m_2}u.

Perfectly inelastic bodies stick, v=\frac{m_1u_1+m_2u_2}{m_1+m_2} and the maximum kinetic energy is lost. Coefficient of restitution e=\frac{v_2-v_1}{u_1-u_2}; a ball dropped from h rebounds to e^2h.

1.4 Circular motion, gravitation and SHM

Circular motion. \omega=\frac{2\pi}{T}=2\pi f, v=\omega r, a_c=\frac{v^2}{r}=\omega^2r toward the centre, F_c=\frac{mv^2}{r} (supplied by tension, friction, gravity, normal reaction — not a separate force).

Conical pendulum T\cos\theta=mg, T\sin\theta=\frac{mv^2}{r}, period 2\pi\sqrt{\frac{L\cos\theta}{g}}.

Level turn v_{\max}=\sqrt{\mu rg}. Banking (no friction): \tan\theta=\frac{v^2}{rg}.

Vertical circle T_{\text{top}}=\frac{mv_t^2}{r}-mg; v_{\text{top,min}}=\sqrt{gr}, v_{\text{bottom,min}}=\sqrt{5gr}. Over a convex bridge the apparent weight is m(g-\frac{v^2}{r}).

Gravitation. F=\dfrac{Gm_1m_2}{r^2}, G=6.67\times10^{-11}. g=\dfrac{GM}{R^2}; at height h: g'=g\big(\frac{R}{R+h}\big)^2 (=\frac{g}{4} at h=R); at depth d: g'=g\big(1-\frac{d}{R}\big); with latitude \lambda: g'=g-\omega^2R\cos^2\lambda (zero effect at the poles).

Gravitational potential V=-\frac{GM}{r}, PE =-\frac{GMm}{r}. Escape speed v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}=11.2 km/s; v_{\text{orbit}}=\sqrt{\frac{GM}{r}} (near Earth \sqrt{gR}=7.9 km/s), so v_e=\sqrt{2}v_o.

Kepler: T^2\propto r^3; T=2\pi\sqrt{\frac{r^3}{GM}}. Orbital energy: K=\frac{GMm}{2r}, E=-\frac{GMm}{2r}.

A geostationary satellite has T=24 h, orbit radius \approx42{,}400 km (height \approx36{,}000 km), moves west to east above the equator. Weightlessness in orbit: free fall.

Rocket: thrust =v_{\text{rel}}\frac{dm}{dt}; multi-stage design sheds dead mass to reach the required speed.

SHM. a=-\omega^2x, x=A\sin(\omega t+\phi), v=\omega\sqrt{A^2-x^2}, v_{\max}=A\omega, a_{\max}=A\omega^2.

Spring–mass T=2\pi\sqrt{\frac{m}{k}} (series springs k=\frac{k_1k_2}{k_1+k_2}, parallel k_1+k_2); simple pendulum T=2\pi\sqrt{\frac{l}{g}} (a seconds pendulum has T=2 s, l\approx0.994 m).

Energy E=\frac{1}{2}kA^2=\frac{1}{2}m\omega^2A^2; K=\frac{1}{2}k(A^2-x^2), U=\frac{1}{2}kx^2; both oscillate at twice the frequency; K=U at x=\frac{A}{\sqrt{2}}.

Damped oscillations lose amplitude to friction; forced oscillations have the driver's frequency and the amplitude peaks at resonance (f_{\text{driver}}=f_0).

1.5 Rotational dynamics

I=\sum mr^2; K_{\text{rot}}=\frac{1}{2}I\omega^2; \tau=I\alpha=\vec r\times\vec F; L=I\omega; \tau=\frac{dL}{dt}; P=\tau\omega. Radius of gyration k=\sqrt{I/M}.

Body (axis)	I
ring / hollow cylinder (central axis)	MR^2
disc / solid cylinder (central axis)	\frac{1}{2}MR^2
solid sphere (diameter)	\frac{2}{5}MR^2
hollow sphere (diameter)	\frac{2}{3}MR^2
thin rod (centre, perpendicular) / (end)	\frac{ML^2}{12} / \frac{ML^2}{3}

Parallel-axis theorem I=I_{cm}+Md^2; perpendicular-axis (planar) I_z=I_x+I_y. Rolling without slipping: K=\frac{1}{2}Mv^2\big(1+\frac{k^2}{R^2}\big); down an incline a=\frac{g\sin\theta}{1+k^2/R^2}.

Conservation of angular momentum: I_1\omega_1=I_2\omega_2 (a skater pulling arms in spins faster). Centre of mass \frac{\sum m\vec r}{\sum m}; for uniform gravity the centre of gravity coincides with it.

A body is in equilibrium when \sum F=0 and \sum\tau=0.

1.6 Elasticity

Stress =\frac{F}{A}, strain =\frac{\Delta L}{L}.

Hooke's law: stress \propto strain within the elastic limit. Y=\dfrac{FL}{A\,\Delta L}; bulk modulus K=-\dfrac{\Delta P}{\Delta V/V} (compressibility 1/K); rigidity \eta=\dfrac{F/A}{\theta}.

Poisson's ratio \sigma=\frac{\text{lateral strain}}{\text{longitudinal strain}}, between -1 and 0.5. Energy stored =\frac{1}{2}F\Delta L per unit volume \frac{1}{2}\times\text{stress}\times\text{strain}=\frac{\text{stress}^2}{2Y}.

Extension \propto\frac{L}{r^2} for the same material and force. Relations: Y=3K(1-2\sigma)=2\eta(1+\sigma). Steel has a higher Y than rubber (it is more elastic in the physics sense).

Elastic after-effect, fatigue, yield point, breaking stress are read from the stress–strain graph. Pressure at depth h in the sea decreases the volume of a ball by \frac{\rho gh}{K} (fraction).

1.7 Fluid mechanics

Pressure P=P_0+\rho gh; Pascal's law (hydraulic press, force ratio = area ratio).

Archimedes: buoyant force =\rho_{\text{fluid}}Vg; a body floats when its weight equals the buoyancy; ice (900 kg m⁻³) floats with \frac{9}{10} submerged; apparent weight in a liquid =W-F_B.

Surface tension T=\frac{F}{l} (N m⁻¹) =\frac{\text{energy}}{\text{area}}. Excess pressure: bubble with one surface \frac{2T}{r}, soap bubble (two surfaces) \frac{4T}{r}. Work to blow a soap bubble of radius r: 8\pi r^2T.

Capillary rise h=\dfrac{2T\cos\theta}{r\rho g} (h\propto1/r; mercury, with \theta>90^\circ, is depressed). Surface tension falls with temperature and with detergent.

Viscosity. F=\eta A\frac{dv}{dx} (Newton). Stokes' law F=6\pi\eta rv; terminal velocity v_t=\dfrac{2r^2(\rho-\sigma)g}{9\eta} (\propto r^2). Poiseuille \dfrac{dV}{dt}=\dfrac{\pi Pr^4}{8\eta l} (\propto r^4).

Reynold number R_e=\dfrac{\rho vD}{\eta}; below about 2000 the flow is streamline (laminar), above about 3000 turbulent.

Continuity A_1v_1=A_2v_2 (incompressible). Bernoulli P+\frac{1}{2}\rho v^2+\rho gh=\text{constant} — conservation of energy along a streamline. Uses: Venturi meter, aerofoil lift, atomiser, speed of efflux v=\sqrt{2gh} (Torricelli).

(a) A hydraulic jack: one pressure, two piston areas

(b) Force gain 16x, distance loss 16x, same work

Chapter summary

• Check units by dimensions; use \sin2\theta for range, \sin^2\theta for height.
• Momentum is conserved in every collision; kinetic energy only in elastic ones.
• For circular and orbital motion identify what supplies the centripetal force.
• SHM periods depend only on the system constants (m,k or l,g), not on amplitude.
• Angular momentum conservation changes \omega when I changes.
• Elastic energy density is \frac{1}{2}\times stress \times strain.
• Fluids: continuity first for speeds, then Bernoulli for pressure.

Practice questions

Set A — Recall (14 questions)

1. The dimensional formula of work isa[MLT^{-2}]b[ML^2T^{-2}]c[ML^2T^{-3}]d[ML^{-1}T^{-2}]
2. A projectile is launched at 45^\circ with a speed of 20 m s⁻¹ (g=10 m s⁻²). Its range isa20 mb80 mc40 md10 m
3. A force of 10 N acts for 3 s on a 5 kg body initially at rest. Its final speed isa6 m s⁻¹b2 m s⁻¹c15 m s⁻¹d50 m s⁻¹
4. The kinetic energy of a 2 kg body moving at 3 m s⁻¹ isa6 Jb18 Jc12 Jd9 J
5. The power needed to lift a 100 kg load through 10 m in 10 s (g=10) isa100 Wb1000 Wc10000 Wd10 W
6. A body moves at 10 m s⁻¹ on a circle of radius 5 m. Its centripetal acceleration isa2 m s⁻²b50 m s⁻²c20 m s⁻²d10 m s⁻²
7. The escape speed from the Earth's surface is nearlya7.9 km s⁻¹b9.8 km s⁻¹c8 km s⁻¹d11.2 km s⁻¹
8. If the length of a simple pendulum is made four times, its time periodahalvesbdoublescquadruplesdis unchanged
9. The moment of inertia of a solid sphere about a diameter isa\frac{1}{2}MR^2b\frac{2}{3}MR^2c\frac{2}{5}MR^2dMR^2
10. The maximum theoretical value of Poisson's ratio isa0.5b1c0.25d0.33
11. Ice of density 900 kg m⁻³ floats in water with the fraction submerged equal toa1/10b1/2c3/4d9/10
12. Bernoulli's theorem expresses the conservation ofaenergybmomentumcmassdangular momentum
13. The SI unit of surface tension isaN mbN m⁻²cN m⁻¹dJ m⁻³
14. A block just starts to slide when a plane is tilted to 30^\circ. The coefficient of static friction isa0.866b0.5c1.73d0.577

Answers and explanations — Set A

1. (b) — force × distance. (a) is force.

2. (c) — R=\frac{u^2\sin90^\circ}{g}=\frac{400}{10}=40 m.

3. (a) — impulse: v=\frac{Ft}{m}=\frac{30}{5}=6.

4. (d) — \frac{1}{2}\cdot2\cdot9=9 J.

5. (b) — \frac{mgh}{t}=\frac{100\cdot10\cdot10}{10}=1000 W.

6. (c) — v^2/r=100/5=20.

7. (d) — \sqrt{2gR}=11.2 km s⁻¹; 7.9 km s⁻¹ is the orbital speed near the surface.

8. (b) — T\propto\sqrt{l}.

9. (c) — \frac{2}{5}MR^2; (b) is the hollow sphere.

10. (a) — theory limits: -1<\sigma<0.5.

11. (d) — \rho_{ice}/\rho_{water}=0.9.

12. (a) — it is the work–energy theorem for a flowing fluid.

13. (c) — force per unit length.

14. (d) — \mu=\tan30^\circ=0.577.

Set B — Understanding (10 questions)

1. A ball is thrown at 60^\circ to the horizontal with speed 20 m s⁻¹ (g=10). Its maximum height isa10 mb15 mc20 md30 m
2. Two forces of 10 N each act at 120^\circ to each other. Their resultant isa20 Nb10\sqrt{3} Nc10 Nd0
3. A 2 kg block on a rough floor (\mu=0.2, g=10) is pushed by a 10 N horizontal force. Its acceleration isa5 m s⁻²b3 m s⁻²c4 m s⁻²d2 m s⁻²
4. A 2 kg body moving at 4 m s⁻¹ collides and sticks to a stationary 2 kg body. The kinetic energy lost isa4 Jb16 Jc0d8 J
5. A frictionless banked road must let a car take a bend of radius 10 m at 10 m s⁻¹ (g=10). The banking angle isa45^\circb30^\circc60^\circd37^\circ
6. A particle moves as x=0.1\sin(10t) (SI). Its maximum speed isa0.1 m s⁻¹b10 m s⁻¹c1 m s⁻¹d100 m s⁻¹
7. A 2 kg mass hangs on a spring of constant 200 N m⁻¹. The period of its oscillation isa0.2\pi sb2\pi sc0.02\pi sd\pi s
8. The time period of a geostationary satellite isa12 hb90 minc6 hd24 h
9. A disc of mass 2 kg and radius 0.5 m is acted on by a torque of 0.5 N m about its axis. Its angular acceleration isa2 rad s⁻²b1 rad s⁻²c0.5 rad s⁻²d4 rad s⁻²
10. A wire has twice the length and twice the radius of another of the same material. Under the same force its extension isa2 timesb\frac{1}{2} timesc\frac{1}{4} timesdthe same

Answers and explanations — Set B

1. (b) — H=\frac{u^2\sin^2\theta}{2g}=\frac{400\cdot0.75}{20}=15.

2. (c) — equal vectors at 120^\circ give a resultant equal to each.

3. (b) — a=\frac{10-0.2\cdot2\cdot10}{2}=3.

4. (d) — v=2 m s⁻¹; K_i=16 J, K_f=\frac{1}{2}\cdot4\cdot4=8 J; loss 8 J.

5. (a) — \tan\theta=\frac{v^2}{rg}=1.

6. (c) — v_{\max}=A\omega=0.1\times10.

7. (a) — T=2\pi\sqrt{2/200}=0.2\pi s \approx0.63 s.

8. (d) — it must match the Earth's rotation.

9. (a) — I=\frac{1}{2}MR^2=0.25; \alpha=\frac{0.5}{0.25}=2.

10. (b) — \Delta L\propto\frac{L}{r^2}=\frac{2}{4}.

Set C — Application (8 questions)

1. A stone is thrown horizontally at 10 m s⁻¹ from a 20 m high tower (g=10). It lands at a horizontal distance ofa10 mb14 mc40 md20 m
2. A 1000 kg car moving at 20 m s⁻¹ is stopped by brakes in 40 m. The average braking force isa5000 Nb10000 Nc2500 Nd500 N
3. At a height equal to the Earth's radius, the acceleration due to gravity is (g=9.8 at the surface)a4.9 m s⁻²b3.27 m s⁻²c2.45 m s⁻²d1.1 m s⁻²
4. A stone on a 0.5 m string is whirled in a vertical circle. The minimum speed at the bottom to complete the circle is (g=10)a5 m s⁻¹b2.5 m s⁻¹c7.07 m s⁻¹d10 m s⁻¹
5. A skater spinning at 2 rad s⁻¹ with I=4 kg m² pulls in her arms so that I=1 kg m². Her new angular speed isa0.5 rad s⁻¹b4 rad s⁻¹c8 rad s⁻¹d16 rad s⁻¹
6. Water (\rho=1000) flows at 2 m s⁻¹ in a pipe of area 4 cm² that narrows to 1 cm². The pressure drop in the narrow part isa28 kPab30 kPac32 kPad60 kPa
7. Water rises 3 cm in a capillary tube. In a tube of half the radius it risesa6 cmb3 cmc12 cmd1.5 cm
8. Two identical spheres fall through the same liquid; one has twice the radius. Ratio of terminal speeds (larger : smaller) isa2b8c16d4

Answers and explanations — Set C

1. (d) — t=\sqrt{2h/g}=2 s; x=10\cdot2=20 m.

2. (a) — F=\frac{\frac{1}{2}mv^2}{d}=\frac{200000}{40}.

3. (c) — g'=g\big(\frac{R}{2R}\big)^2=\frac{g}{4}=2.45.

4. (a) — v=\sqrt{5gr}=\sqrt{25}.

5. (c) — L=I\omega: 4\times2=1\times\omega.

6. (b) — v_2=8 m s⁻¹ by continuity; \Delta P=\frac{1}{2}\cdot1000\,(64-4)=30000 Pa.

7. (a) — h\propto\frac{1}{r}: halving the radius doubles the rise.

8. (d) — v_t\propto r^2, ratio 4.

Group D — Model answers for long questions (10 marks each, for deeper revision)

1. Derive the time of flight, maximum height and range of a projectile and solve a numerical. [10]
2. Explain motion in a horizontal circle, a banked road and a vertical circle, with the condition for completing the loop. [10]
3. State and prove the equation of continuity and Bernoulli's principle; give four applications and a numerical. [10]

(a) Hydraulic jack

(b) Kite

(c) Beam balance

(d) Vernier caliper

Model answer D1: Projectile motion (10 marks)

Introduction. A projectile is a body thrown into the air that then moves only under gravity (air resistance neglected). The path is a parabola, because the horizontal and vertical motions are independent.

Derivation. Let the body be thrown with speed u at angle \theta above the horizontal.

1. Components: u_x = u\cos\theta (constant) and u_y = u\sin\theta (changing by -g each second).
2. Time of flight: the vertical displacement is zero at landing, so 0 = u\sin\theta\,T - \frac{1}{2}gT^2 and

3. Maximum height: at the top v_y = 0, so 0 = u^2\sin^2\theta - 2gH and

4. Range: R = u\cos\theta \times T = \frac{2u^2\sin\theta\cos\theta}{g}, so

5. Maximum range is at \theta=45^\circ, where R_{\max}=u^2/g. Angles \theta and (90^\circ-\theta) give the same range.
6. Equation of path: eliminating t gives y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}, which is a parabola.

Numerical. A ball is thrown at 40\ \text{m s}^{-1} at 30^\circ (g=10\ \text{m s}^{-2}).

Quantity	Working	Answer
Time of flight	2\times40\times0.5/10	4\ \text{s}
Maximum height	1600\times0.25/20	20\ \text{m}
Range	1600\times0.866/10	138.6\ \text{m}
Speed at the top	40\cos30^\circ	34.6\ \text{m s}^{-1}

Conclusion. The horizontal speed never changes, and the height and time depend only on the vertical component, so a 45^\circ throw gives the longest range.

Model answer D2: Circular motion (10 marks)

Introduction. A body in uniform circular motion has constant speed but changing direction, so it has a centripetal acceleration a = \frac{v^2}{r} = \omega^2 r towards the centre, and needs a force F = \frac{mv^2}{r}.

1. Horizontal circle. A stone whirled on a string has tension as the centripetal force. A car on a flat road uses friction: \mu mg \ge \frac{mv^2}{r}, so the safe speed is v \le \sqrt{\mu r g}.

2. Banked road. For a road banked at angle \theta the normal reaction has a horizontal part N\sin\theta = \frac{mv^2}{r} and a vertical part N\cos\theta = mg. Dividing:

For v = 20\ \text{m s}^{-1} and r = 100\ \text{m}: \tan\theta = 400/(100\times10) = 0.4, so \theta \approx 21.8^\circ.

3. Vertical circle (mass m on a string of length r).

1. At the top: T + mg = \frac{mv_{t}^2}{r}. The least speed is when T = 0, so v_{t} = \sqrt{gr}.
2. At the bottom: T - mg = \frac{mv_b^2}{r}.
3. By conservation of energy, \frac{1}{2}mv_b^2 = \frac{1}{2}mv_t^2 + mg(2r), so v_b^2 = v_t^2 + 4gr.
4. For the minimum case v_b = \sqrt{5gr}.
5. The tension difference between the bottom and top is 6mg.

Numerical. For r = 0.9\ \text{m} and g = 10: v_{t} = \sqrt{9} = 3\ \text{m s}^{-1} and v_b = \sqrt{45} = 6.7\ \text{m s}^{-1}.

Case	Centripetal force from	Condition
Stone on string	Tension	T = mv^2/r
Flat road	Friction	v \le \sqrt{\mu rg}
Banked road	Component of N	\tan\theta = v^2/rg
Vertical loop	Tension + weight	v_b \ge \sqrt{5gr}
Satellite	Gravity	v = \sqrt{GM/r}

Conclusion. Every circular motion needs a net inward force, and for the vertical loop the string must stay taut at the top.

Model answer D3: Equation of continuity and Bernoulli's principle (10 marks)

Introduction. For an ideal fluid (incompressible, non-viscous, steady flow) two laws describe the flow: conservation of mass and conservation of energy.

Equation of continuity. The mass entering a pipe per second equals the mass leaving, so A_1v_1 = A_2v_2. The fluid speeds up where the pipe narrows.

Bernoulli's principle. Along a streamline

Proof (outline). The work done by the pressure difference on a small volume V is (P_1 - P_2)V. This equals the gain in kinetic energy, \frac{1}{2}\rho V(v_2^2 - v_1^2), plus the gain in potential energy, \rho V g(h_2 - h_1). Dividing by V gives the result.

Applications.

1. Torricelli's theorem: the speed of efflux from a hole at depth h is v = \sqrt{2gh}.
2. Venturi meter: measures flow speed from the pressure drop in a narrow throat.
3. Aerofoil lift: faster air over the curved top gives lower pressure than below.
4. Atomiser and spray gun: fast air over a tube lowers the pressure and draws liquid up.

Numerical. In a pipe A_1 = 20\ \text{cm}^2, v_1 = 2\ \text{m s}^{-1} and A_2 = 5\ \text{cm}^2.

Then v_2 = 20\times2/5 = 8\ \text{m s}^{-1}, and (horizontal pipe, water) P_1 - P_2 = \frac{1}{2}\times1000\times(64 - 4) = 30{,}000\ \text{Pa}.

For a tank with a hole at 1.8\ \text{m} depth, v = \sqrt{2\times10\times1.8} = 6\ \text{m s}^{-1}.

Law	Conserves	Equation
Continuity	Mass	Av = \text{constant}
Bernoulli	Energy	P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}
Pascal	Pressure transmission	F_1/A_1 = F_2/A_2 (hydraulic jack)

Conclusion. Where a fluid moves faster its pressure is lower, which explains the flight of aeroplanes and the working of sprays and flow meters.