Tiêu đề Physics P42 qp .pdf ✅

2 © UCLES 2025 9702/42/F/M/25 Data acceleration of free fall g = 9.81 m s–2 speed of light in free space c = 3.00 × 108 m s–1 elementary charge e = 1.60 × 10–19 C unified atomic mass unit 1 u = 1.66 × 10–27 kg rest mass of proton mp = 1.67 × 10–27 kg rest mass of electron me = 9.11 × 10–31 kg Avogadro constant NA = 6.02 × 1023 mol–1 molar gas constant R = 8.31 J K–1 mol–1 Boltzmann constant k = 1.38 × 10–23 J K–1 gravitational constant G = 6.67 × 10–11 N m2 kg–2 permittivity of free space ε 0 = 8.85 × 10–12 F m–1 ( 4 1 0 rf = 8.99 × 109 m F–1) Planck constant h = 6.63 × 10–34 J s Stefan–Boltzmann constant σ = 5.67 × 10–8 W m–2 K–4 Formulae uniformly accelerated motion s = ut + 2 1 at 2 v 2 = u 2 + 2as hydrostatic pressure ∆p = ρg∆h upthrust F = ρgV Doppler effect for sound waves f o = v v f v s s ! electric current I = Anvq resistors in series R = R1 + R2 + ... resistors in parallel R 1 = R 1 1 + R 1 2 + ... * 0000800000002 * , , ĬÕĊ3⁄4Ġ ́íÈõÏĪÅĊàû·þ× ĬÂüôØģîěÖöùĂÙûčċĦĂ ĥåÅÕμÕĥÕÕÅÅÅąõÅμĕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN
3 © UCLES 2025 9702/42/F/M/25 [Turn over 2 gravitational potential φ = – r GM gravitational potential energy EP = – r GMm pressure of an ideal gas p = V Nm 3 1 〈c 2 〉 simple harmonic motion a = – ω 2 x velocity of particle in s.h.m. v = v0 cos ωt v = ! ω ( ) x x 0 2 2 - electric potential V = r Q 4 0 rf electrical potential energy EP = r Qq 4 0 rf capacitors in series C 1 = C 1 1 + C 1 2 + ... capacitors in parallel C = C1 + C2 + ... discharge of a capacitor x = x e RC t 0 - Hall voltage VH = I ntq B alternating current/voltage x = x0 sin ωt radioactive decay x = x0 e –λt decay constant λ = . t 0 693 2 1 intensity reflection coefficient I IR 0 = ( ) ( ) Z Z Z Z 1 2 2 1 2 2 + - Stefan–Boltzmann law L = 4πσr 2T 4 Doppler redshift ∆λ λ á f Df á c v * 0000800000003 * , , Ĭ×Ċ3⁄4Ġ ́íÈõÏĪÅĊàù·þ× ĬÂûóÐĥòīãĄĈÇýăÉċĖĂ ĥåμĕõμąμÅμĕÅąĕåõąÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN
4 © UCLES 2025 9702/42/F/M/25 1 A steel ball is placed on the inside surface of a hollow circular cone. The ball moves in a horizontal circle at constant speed, as shown in Fig. 1.1. 52° 52° path of steel ball steel ball cone Fig. 1.1 The angle of the side of the cone to the horizontal is 52°. There is no friction between the ball and the cone. (a) Fig. 1.2 shows a cross‐section through the cone and the steel ball. Fig. 1.2 On Fig. 1.2, draw labelled arrows to show the two forces acting on the ball. [1] (b) Describe how the forces acting on the ball cause its acceleration to be centripetal. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] * 0000800000004 * , , ĬÕĊ3⁄4Ġ ́íÈõÏĪÅĊÞû·Ā× ĬÂûòÐğĄĞØþÿÀßßīěĞĂ ĥĕĥĕμμąĕåĕĥÅÅĕąõĕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN