NSSC Physics Advanced Subsidiary Level Study Notes

Essential Physical Constants and Formulae Reference

In the study of Physics at the Advanced Subsidiary level, specific fundamental constants are utilized across various calculations. The speed of light in free space, denoted as cc, is defined as 3.00×108 m s−13.00 \times 10^8\,\text{m}\,\text{s}^{-1}. The elementary charge, represented by ee, is 1.60×10−19 C1.60 \times 10^{-19}\,\text{C}. Mass is quantified using the unified atomic mass unit, where 1 u=1.66×10−27 kg1\,u = 1.66 \times 10^{-27}\,\text{kg}. Particle masses are specifically noted, with the rest mass of an electron mem_e being 9.11×10−31 kg9.11 \times 10^{-31}\,\text{kg} and the rest mass of a proton mpm_p being 1.67×10−27 kg1.67 \times 10^{-27}\,\text{kg}. The standard acceleration of free fall, gg, is taken as 9.81 m s−29.81\,\text{m}\,\text{s}^{-2}.

Mathematical relationships for uniformly accelerated motion include the equations s=ut+12at2s = ut + \frac{1}{2}at^2 and v2=u2+2asv^2 = u^2 + 2as. For thermodynamics and fluids, work done on or by a gas is expressed as W=pΔVW = p\Delta V, and hydrostatic pressure change is determined by Δp=ρgΔh\Delta p = \rho g\Delta h. Wave phenomena involve the Doppler effect formula fo=fsvv±vsf^o = f_s \frac{v}{v \pm v_s}. Electrical principles are guided by the equation for electric current I=AnvqI = Anvq, and the combination of resistors. For series circuits, total resistance is R=R1+R2+…R = R_1 + R_2 + \dots, while for parallel circuits, it is calculated using 1R=1R1+1R2+…\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \dots.

Physical Quantities, Units, and Measurements

Physical quantities are categorized as either base or derived, and must be expressed in SI base units. For example, mass has the SI base unit of the kilogram, symbolized as kg\text{kg}, while velocity is a derived quantity with the base units m s−1\text{m}\,\text{s}^{-1}. It is crucial to distinguish between scalar quantities, which possess only magnitude, and vector quantities, which possess both magnitude and direction. Mass and distance are defined as scalars, whereas moment of force is categorized as a vector. When selecting measuring instruments, precision is a primary factor; a micrometer is the appropriate tool for measuring the diameter of a thin copper wire when a precision of 0.01 mm0.01\,\text{mm} is required.

Measurement analysis involves dealing with absolute and percentage uncertainties. For a rectangular tank with dimensions L=(2.14±0.04) mL = (2.14 \pm 0.04)\,\text{m}, w=(1.98±0.05) mw = (1.98 \pm 0.05)\,\text{m}, and h=(1.50±0.04) mh = (1.50 \pm 0.04)\,\text{m}, the volume is found by multiplying the three dimensions, resulting in 6.36 m36.36\,\text{m}^3. The uncertainty in a product is determined by summing the fractional uncertainties: ΔVV=0.042.14+0.051.98+0.041.50\frac{\Delta V}{V} = \frac{0.04}{2.14} + \frac{0.05}{1.98} + \frac{0.04}{1.50}. Converting this to a percentage yields an uncertainty of approximately 7%7\%. Therefore, the volume of water needed to fill the tank is summarized as 6.36 m3±7%6.36\,\text{m}^3 \pm 7\%.

Kinematics and Dynamics of Motion

Motion can be analyzed through velocity-time graphs, where the area under the curve represents the total distance covered. For a cyclist accelerating from rest to a velocity of 12 m s−112\,\text{m}\,\text{s}^{-1} in a certain period and continuing for a total of 25 s25\,\text{s}, the displacement or distance is the integral of the velocity over time. In a specific scenario where a cyclist's velocity increases and then sustains, the distance covered in 25 s25\,\text{s} may be calculated as 150 m150\,\text{m}. Vertical motion under gravity requires consideration of time and acceleration. If a ball is thrown vertically upwards and takes 4.10 s4.10\,\text{s} to return to its starting point with negligible air resistance, the total time of flight is divided by two to find the time to reach maximum height (2.05 s2.05\,\text{s}). Using v=u+atv = u + at where v=0v = 0 at the peak and a=−9.81 m s−2a = -9.81\,\text{m}\,\text{s}^{-2}, the initial velocity uu is calculated as approximately 20.1 m s−120.1\,\text{m}\,\text{s}^{-1}.

Force application on Inclined planes involves decomposing weight into components. For a 50 kg50\,\text{kg} trolley on an 8.0∘8.0^{\circ} frictionless slope pulled by a rope with a tension T=150 NT = 150\,\text{N}, the resultant force is the difference between the tension and the component of weight acting down the slope (mgsin⁡(8.0∘)mg\sin(8.0^{\circ})). The acceleration aa is determined by the equation Fnet=maF_{net} = ma, leading to a value of 1.6 m s−21.6\,\text{m}\,\text{s}^{-2}. Momentum is conserved in collisions. When two identical gliders, QQ and SS, travel toward each other at 70 cm s−170\,\text{cm}\,\text{s}^{-1} and 40 cm s−140\,\text{cm}\,\text{s}^{-1} respectively and stick together upon collision, the final velocity is found using m1u1+m2u2=(m1+m2)vm_1u_1 + m_2u_2 = (m_1 + m_2)v. Calculating with direction (70−4070 - 40) for identical masses gives a combined speed of 15 cm s−115\,\text{cm}\,\text{s}^{-1}.

Equilibrium, Moments, and Fluid Mechanics

Statics and the principle of moments are used to maintain system balance. A metal rod of total length 1.5 m1.5\,m pivoted at 0.75 m0.75\,m from one end (the center) requires the sum of clockwise moments to equal the sum of anticlockwise moments. If a force FF is applied at 0.25 m0.25\,m from the pivot and a 5.0 N5.0\,N force is applied at 0.75 m0.75\,m on the opposite side, the balance is achieved when (F×0.25)=(5.0×0.75)(F \times 0.25) = (5.0 \times 0.75), meaning F=15.0 NF = 15.0\,N. For a couple, which consists of two equal and opposite forces not acting along the same line, the torque is the product of one of the forces and the perpendicular distance between them. For a 150.0 cm150.0\,cm (1.5 m1.5\,m) bar with forces of 45.0 N45.0\,N at each end, the torque is 45.0×1.5=67.5 N m45.0 \times 1.5 = 67.5\,N\,m. An object in equilibrium must satisfy two conditions: there is no resultant force acting on it, and there is no resultant torque.

Fluid properties like density are additive when mixing substances without chemical reactions. If oils MM, NN, and OO are mixed, with mass and density values of (670 kg670\,kg, 650 kg m−3650\,kg\,m^{-3}), (1200 kg1200\,kg, 840 kg m−3840\,kg\,m^{-3}), and (2000 kg2000\,kg, 950 kg m−3950\,kg\,m^{-3}) respectively, the mixture density is calculated as ρmix=∑m∑mρ\rho_{mix} = \frac{\sum m}{\sum \frac{m}{\rho}}. Total mass is 3870 kg3870\,kg and total volume is the sum of individual volumes (1.03+1.43+2.11≈4.57 m31.03 + 1.43 + 2.11 \approx 4.57\,m^3), resulting in a mixture density of 850 kg m−3850\,kg\,m^{-3}. In thermodynamics, if a gas at atmospheric pressure (101 kPa101\,kPa) in a syringe expands from 3.0 cm33.0\,cm^3 to 9.0 cm39.0\,cm^3 upon heating, the work done by the gas is W=pΔV=101×103×(9.0−3.0)×10−6W = p\Delta V = 101 \times 10^3 \times (9.0 - 3.0) \times 10^{-6}, which equals 606 mJ606\,mJ.

Work, Energy, Power, and Material Properties

Energy conversion and power in systems like hydroelectricity involve gravity and efficiency. For water falling 300 m300\,m at a rate of 2500 kg2500\,kg per minute (41.67 kg/s41.67\,kg/s), the input power is Pin=mtgh=41.67×9.81×300=122610.3 WP_{in} = \frac{m}{t}gh = 41.67 \times 9.81 \times 300 = 122610.3\,W. With an efficiency of 75%75\%, the average output power is 0.75×122.6 kW≈92 kW0.75 \times 122.6\,kW \approx 92\,kW. Kinetic energy change occurs when velocity varies; for a 900 kg900\,kg car where speed doubles from 6.00 m s−16.00\,\text{m}\,\text{s}^{-1} to 12.0 m s−112.0\,\text{m}\,\text{s}^{-1}, the change in kinetic energy is ΔEk=12m(v2−u2)=12(900)(144−36)=48600 J\Delta E_k = \frac{1}{2}m(v^2 - u^2) = \frac{1}{2}(900)(144 - 36) = 48600\,J. Speed can also be determined from power and force: if an engine does 36000 J36000\,J of work in 1.3 s1.3\,s to move a car against a driving force of 960 N960\,N, the power is P=360001.3P = \frac{36000}{1.3}. Since P=FvP = Fv, the speed vv is calculated as approximately 29 m s−129\,\text{m}\,\text{s}^{-1}.

Material deformation is described by the Young modulus EE. For a wire of length ll and diameter dd extending by ee under a mass mm, the formula for the Young modulus is E=stressstrain=mgAelE = \frac{\text{stress}}{\text{strain}} = \frac{\frac{mg}{A}}{\frac{e}{l}}. Since the cross-sectional area A=πd24A = \frac{\pi d^2}{4}, substituting gives E=4mglπd2eE = \frac{4mgl}{\pi d^2 e}, which rearranges to solve for extension: e=4mglπd2Ee = \frac{4mgl}{\pi d^2 E}. The work done on a spring is represented by the area under its force-extension graph. If a spring reaches an extension of 39.0 cm39.0\,cm with a force of 4.0 N4.0\,N, the work done is 12Fx=0.5×4.0×0.39=0.78 J\frac{1}{2}Fx = 0.5 \times 4.0 \times 0.39 = 0.78\,J.

Waves, Sound, and the Electromagnetic Spectrum

Wave mechanics are visualized through displacement-time or displacement-distance graphs. On a displacement-time graph, the maximum displacement from the equilibrium position represents the amplitude, while the time for one complete cycle represents the period. A fundamental relationship exists between a wave's power and its amplitude, where power is proportional to the square of the amplitude (P∝A2P \propto A^2), thus A∝PA \propto \sqrt{P}. Sound wave frequency can be determined using a cathode-ray oscilloscope (c.r.o). If a trace shows one full cycle across 2 cm2\,cm and the time-base is set to 0.20 ms cm−10.20\,ms\,cm^{-1}, the period TT is 0.40 ms0.40\,ms. The frequency is then f=1T=10.0004=2500 Hzf = \frac{1}{T} = \frac{1}{0.0004} = 2500\,Hz.

The Doppler effect occurs when a source and observer are in relative motion. For a train sounding a 500 Hz500\,Hz horn while moving at 15.0 m s−115.0\,\text{m}\,\text{s}^{-1} away from a stationary observer (speed of sound 336 m s−1336\,\text{m}\,\text{s}^{-1}), the observed frequency is fo=500×336336+15.0=479 Hzf^o = 500 \times \frac{336}{336 + 15.0} = 479\,Hz. Stationary waves in a tube closed at one end are longitudinal and stationary in nature. The electromagnetic spectrum is ordered by decreasing frequency (increasing wavelength) as follows: gamma rays, X-rays, ultraviolet, visible light, infrared, microwaves, and radio waves. Light behavior includes diffraction, which is the spreading of a wave as it passes through a narrow gap. For double-slit interference, fringe spacing is determined by the wavelength, distance to screen DD (e.g.50 cme.g. 50\,cm), and slit separation aa (e.g.0.50 mme.g. 0.50\,mm). Diffraction gratings produce patterns where the maximum order nn is limited by n≤dλn \le \frac{d}{\lambda}. For 750 lines mm−1750\,lines\,mm^{-1}, d=1×10−3750 md = \frac{1 \times 10^{-3}}{750}\,m. For a wavelength of 590 nm590\,nm, the maximum order is 22.

Electricity, Resistance, and Circuits

Electric charge and fields exert forces on particles. A charge of 3.0 μC3.0\,μ C between plates separated by 30 cm30\,cm with a potential difference of 6.0 V6.0\,V experiences a force F=qE=qVd=3.0×10−6×6.00.30=6.0×10−5 NF = qE = q\frac{V}{d} = 3.0 \times 10^{-6} \times \frac{6.0}{0.30} = 6.0 \times 10^{-5}\,N. In a metal wire, current II is related to drift velocity vv by I=AnvqI = Anvq. For a current of 2.0 A2.0\,A, drift velocity of 0.4 mm s−10.4\,mm\,s^{-1}, and 2.0×10282.0 \times 10^{28} electrons per unit volume, the cross-sectional area is A=2.0(2.0×1028×0.0004×1.6×10−19)=1.6×10−6 m2A = \frac{2.0}{(2.0 \times 10^{28} \times 0.0004 \times 1.6 \times 10^{-19})} = 1.6 \times 10^{-6}\,m^2. Electrical power dissipation in parallel circuits involves four bulbs of 960 Ω960\,\Omega each connected to 240 V240\,V. The power for one bulb is P=V2R=2402960=60 W=0.060 kWP = \frac{V^2}{R} = \frac{240^2}{960} = 60\,W = 0.060\,kW. The property of resistivity has SI units of Ω m\Omega\,m.

In circuit theory, electromotive force (e.m.f.) and internal resistance influence potential difference (p.d.p.d.). When a switch is closed in a circuit with internal resistance, the potential difference across the battery terminals decreases because work must be done against the internal resistance. Kirchhoff's first law, based on the conservation of charge, dictates that the sum of currents entering a junction equals the sum of currents leaving it (e.g.I1=I2+I3e.g. I_1 = I_2 + I_3, so I3=I1−I2I_3 = I_1 - I_2). Potential dividers allow for specific voltage outputs; in a circuit with a 9.0 V9.0\,V battery and resistors of 3.0 kΩ3.0\,k\Omega and 2.0 kΩ2.0\,k\Omega in series, a voltmeter across the 2.0 kΩ2.0\,k\Omega resistor will read Vout=9.0×2.03.0+2.0=3.6 VV_{out} = 9.0 \times \frac{2.0}{3.0 + 2.0} = 3.6\,V.

Particle and Nuclear Physics

Radioactive decay transforms nuclei through types of radiation. Thorium-232 (90232Th^{232}_{90}Th) emitting an alpha particle (24He^4_{2}He) forms a daughter nucleus XX with a nucleon number of 228228 and proton number of 8888 (X=88228RaX = ^{228}_{88}Ra). If XX then undergoes beta-minus decay (−10e^0_{-1}e), the granddaughter nucleus YY maintains a nucleon number of 228228 but the proton number increases to 8989 (Y=89228AcY = ^{228}_{89}Ac). Alpha radiation is characterized by a short range (a few millimeters) in air and deflection by electric fields. Isotopes of an element are defined by having the same proton number but different neutron numbers. Fundamental particles are those not composed of smaller components, such as electrons, neutrinos, and quarks; neutrons and protons are not fundamental as they are composed of quarks.

Quark composition determines the charge of composite particles. Up quarks (uu) have a charge of +23e+\frac{2}{3}e and down quarks (dd) have a charge of −13e-\frac{1}{3}e. To form a particle with a 2+e2+e charge, a combination of four up quarks and one down quark is required calculation: 4(+23)+1(−13)=83−13=734(+\frac{2}{3}) + 1(-\frac{1}{3}) = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} (adjustment: 4u4u and 2d2d gives 4(2/3)+2(−1/3)=6/3=+2e4(2/3) + 2(-1/3) = 6/3 = +2e). Neutrinos are fundamental particles classified as leptons, possess no charge (neutral), and are emitted during processes such as beta-minus decay alongside an electron.