RI 2025 summaries all chapters

H2 Physics 2025 Revision 1 Chapter 1: Measurement 1. S.I. Units • The seven base units are: metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol) and candela (cd). • Derived units are defined in terms of base units. They are expressed as products or quotients of base units. • An equation is said to be homogeneous or dimensionally consistent if every term on both sides of the equation has the same base units. • A homogeneous equation may not be true or correct. 2. Errors and Uncertainties Item Accuracy Precision Instrument Calibration of instrument Smallest division of instrument Measurements Closeness of average to true value Closeness of measurements to one another • Systematic errors result in all readings or measurements being always smaller or always larger than the true value by a fixed amount. • Random errors result in readings or measurements being scattered about the true value. • Accuracy is the degree of closeness of the average value of the measurements to the true value. It is affected by systematic error. • Precision is the degree of agreement between repeated measurements of the same quantity. It is affected by random error. • Calculations: If Q = a X ± b Y, then ∆Q = |a| ∆X + |b| ∆Y. If mnQ aX Y= × or m n XQa Y= , then Q XYmnQ XY ∆ ∆∆= + . Percentage uncertainty of Q = 100%Q Q ∆ × .  A bsolute uncertainty Q∆ should always be expressed to 1 s.f.  Q uantity Q should be expressed up to the same decimal place as Q∆ . Eg: ( ) 19.81 0.02 m sgg −±∆ = ±  Fractional and percentage uncertainty is expressed to 2 s.f. Eg: 0.02 0.0020 (2 s.f.) 0.20 % (2 s.f.)9.81 g g ∆ = = = 3. Scalars & Vectors • A scalar quantity has a magnitude only. • A vector quantity has both a magnitude and a direction.

H2 Physics 2025 Revision 2 Chapter 2: Kinematics 1. Interpretation of Graphs s – t graph v – t graph a – t graph Gradient instantaneous velocity ds vdt= = = Gradient instantaneous acceleration dv adt= = = Area under the graph 2 1 change in displacement t t v dt s= = ∆ = ∫ Area under the graph 2 1 change in velocity t t a dt v= = ∆ = ∫ 2. Rectilinear Motion (1-D) ( )gradient vua t −= = v u at∴= + (1) ( )1area 2s u vt= = + (2) substituting (1) into (2), 21 2s ut at= + (3) and 22 2v u as= + (4) Equations (1) to (4) apply only to motion in a straight line at constant acceleration. 3. Solving kinematics problems • Draw diagram(s) and transfer data from the question onto the diagram(s). • Indicate positive displacements along the vertical and horizontal directions. • Ensure that sign conventions are applied consistently to all quantities. • Apply the above equations independently along the vertical and horizontal directions. t / s s / m t1 0 t / s v / m s −1 t1 0 t2 a / m s −2 t / s t1 0 t2 velocity v t time u 0

H2 Physics 2025 Revision 3 4. Projectile Motion (2-D) In the absence of air resistance, the path/trajectory of a projectile is a parabola. The magnitude of acceleration due to free fall (acting vertically downwards) is g. Vertically Horizontal Initial velocity u uy = u sin θ ux = u cos θ Acceleration a (downwards)yag = ax = 0 Final velocity v 22 2 yyy y y yy v u at v u as = + = + xxx xx v u at vu = + ∴= Displacement s ( ) 21 2 1 2 yy y y yy s ut at s u vt = + = + Maximum height (max sy) is reached when vy = 0 sx = ux t Time t t t θ u ux uy θ1 v1 v1x = ux v1y θ4 v4 v4x = ux v4y θ3 v3 v3x = ux v3y v2 = ux y x 0

H2 Physics 2025 Revision 4 Chapter 3: Dynamics 1. Important Laws and Definitions i. Newton’s First Law states that every object continues in its state of rest or uniform motion in a straight line unless it is acted upon by a resultant external force. ii. Newton’s Second Law states that the rate of change of momentum of a body is proportional to the resultant force acting on it and the change occurs in the direction of the force. That is, net dpF dt= iii. Newton’s Third Law states that if a body A exerts a force on body B, then body B exerts an equal but opposite force on body A. iv. Since the internal forces of a system always add up to zero vectorally, Fnet = 0 as long as there is no resultant external force acting on the system. When this happens, 0 system net dp Fdt = = which means that the total momentum of the system is constant. The Principle of Conservation of Momentum states that when a system of bodies interact, the total momentum of the system remains constant, provided no net external force acts on it. v. Impulse is defined as the product of a force F acting on an object and the time ∆t for which the force acts. Impulse = F ∆t 2. Applications of Newton’s Second Law i. A System of Objects One example is shown in the figure on the right. 1. Identify forces acting on each object. 2. Write down the “F = ma” equation for each object. 3. Solve the simultaneous equations to get the answer. ii. Force due to a Fluid Jet Consider a continuous stream of fluid of density ρ moving with velocity v. After striking a surface, its velocity becomes v’. In most questions, v’ is either 0 or −v. So 2F Avρ= or 22F Av ρ= m T Mg mg T N M v A ρ ( ) ( ) ( )( ) Force experienced by the fluid mass of fluid flow per unit time its change in velocity 'Av v v = × = − ρ

H2 Physics 2025 Revision 5 3. Momentum and its Conservation i. Momentum is defined as the product of the mass of an object and its velocity. ii. Momentum is a vector quantity and its unit is kg m s−1 or N s. iii. The impulse of a force = dFt∫ = area under the F−t graph. The impulse exerted on a body is equal to the change in momentum of the body. iv. In collision problems in which external forces are absent or negligible, the total momentum is always conserved. On the other hand, the kinetic energy of the system is conserved only of the collision is perfectly elastic. We may classify the types of collisions as follows: Type of collision Momentum conserved? K.E. conserved? Elastic   Inelastic  × Completely inelastic (bodies coalesced)  × v. A head-on collision between two objects is shown below: Since momentum is always conserved in such a collision, we can always write down: 1 12 2 1 12 2mu m u mv mv+= + It is important to remember that the equation is written down with the figure above in mind, i.e., all the velocities are rightward. If, for example, the question specifies that m2 is moving at 5 m s−1 to the left, then you should substitute u2 by −5. If the collision is also perfectly elastic, then one can write down an additional equation 2 222 1 12 21 12 2 11 11 22 22mu m u mv mv+= + This new equation is slightly difficult to solve due to the squared terms. However, we can combine the two equations to obtain a third, linear equation: 12 21uu v v−=− relative speed of approach = relative speed of separation m1 Before collision u1 m1 m2 After collision v1 v2 m2 u2

H2 Physics 2025 Revision 6 Chapter 4: Forces 1. Important Formulae and Definitions i. Hooke’s Law states that the extension x of a spring (or wire) is proportional to the applied force F, if the limit of proportionality is not exceeded. F = kx where k is the force constant. ii. The weight of a body may be taken as acting at a single point known as its centre of gravity. iii. Pressure is the normal force acting per unit area, where the force is acting at right angles to the area. Fp A= A scalar quantity. S.I. units: N m−2 or Pascal (Pa) At a given depth h, pressure due to the fluid column above is given by p ghρ= , where ρ is the density of fluid. Note that the total pressure at depth h also includes the atmospheric pressure. iv. Upthrust is t