NYJC 2026 Timed Practice (Answers)

1 NANYANG JUNIOR COLLEGE JC2 Timed Practice 2026 General Certificate of Education Advanced Level Higher 3 CANDIDATE NAME CIVICS GROUP - REGISTRATION NUMBER PHYSICS Paper 1 9814/01 09 July 2026 1 hour 30 min Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, civics group and registration number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. The use of an approved scientific calculator is expected where appropriate. Answer all questions. This document consists of 14 printed pages For Examiner’s Use 1 10 2 12 3 10 4 20 Total 52

2 Data Formulae uniformly accelerated motion s = ut + ½at2 work done on/by a gas W = p V v2 = u2 + 2as Pressure p = F A moment of inertia of rod through one end I = ML21 3 gravitational potential  = Gm r− moment of inertia of hollow cylinder through axis I = M r r22 12 1 ()2 + temperature T/K = T / oC + 273.15 moment of inertia of solid sphere through centre I = MR22 5 pressure of an ideal gas p = 21 3 Nm cV moment of inertia of hollow sphere through centre I = MR22 3 mean translational kinetic energy of an ideal gas particle E = 3 2 kT speed of light in free space, c = 3.00  108 m s–1 permeability of free space, μo = 4π  10–7 H m–1 permittivity of free space, εo = 8.85  10–12 F m–1 ( . −= 91 0 1 8 99 10 m F4 ) elementary charge, e = 1.60  10–19 C the Planck constant, h = 6.63  10–34 J s unified atomic mass constant, u = 1.66  10–27 kg rest mass of electron, me = 9.11  10–31 kg rest mass of proton, mp = 1.67  10–27 kg molar gas constant, R = 8.31 J K–1 mol–1 the Avogadro constant, NA = 6.02  1023 mol–1 the Boltzmann constant, k = 1.38  10–23 J K–1 gravitational constant, G = 6.67  10–11 N m2 kg–2 acceleration of free fall, g = 9.81 m s–2

3 displacement of particle in s.h.m. x = xo sin ωt alternating current/voltage, x = xo sin ωt velocity of particle in s.h.m. v = vo cos ωt magnetic flux density due to a long straight wire B = 2 o d   I = ( )22 xxo − magnetic flux density due to a flat circular coil B = 2 oN r  I electric current, I = Anvq magnetic flux density due to a long solenoid B = on I resistors in series, R = R1 + R2 + … energy in an inductor U = L 21 2 I resistors in parallel, R 1 = RR12 11 ...++ RL series circuits 𝜏 = L R capacitors in series C 1 = CC12 11 ...++ RLC series circuits (underdamped) ω = R LC L − 2 2 1 4 capacitors in parallel C = C1 + C2 + … energy states for quantum particle in a box nE = h nmL 2 2 28 energy in a capacitor U = QV1 2 radioactive decay x0 = txe0 − = Q C 21 2 radioactive decay constant  = t1/2 ln2 = CV 21 2 Lorentz factor  = v c 2 1 1 − charging a capacitor Q = t Qe0 1 −−  length contraction L = L0  discharging a capacitor Q = t Qe0 − time dilation T = T0 RC time constant  = RC Lorentz transformation equations (1 dimension) x ' = ( )x vt − electric potential, V = 4 o Q r t ' = vxt c2  − electric field strength due to a long straight wire E = or  2 velocity addition u' = uv uv c2 () 1 −  − electric field strength due to a large sheet E = o  2 mass-energy equivalence E 2 = pc mc2 2 2( ) ( ) +

4 Answer all questions. 1 Two perfectly elastic balls, A and B, are held above the ground at a height h, such that h is much greater than the diameter of each ball. The balls are almost touching, with ball A directly above ball B. The balls are released simultaneously. Ball A rebounds to a maximum height H above the ground. (a) Sketch a labelled diagram to show the velocities of the balls relative to the ground just before they collide with each other. Hence show that when the balls collide with each other, they do so with relative velocity relv gh= 8 [3] (b) The mass of the lower ball mB is greater than or equal to the mass of the upper ball mA so that: B A m nnm = and 1 Determine an expression for the velocity of the centre of mass frame vCoM relative to the ground immediately before the two balls collide. [2] By conservation of energy, mv mgh v gh = = 2 Gain in KE = Loss in GPE 1 2 2 [B1] Diagram showing balls moving in opposite directions with same speed [B1] Taking upwards as positive. relv gh gh gh gh= − − = = Since the balls have the same speed just before collision, AND the ball collides elastically with the ground and changes direction, 2 ( 2 ) 2 2 8 (shown) [B1] It’s a “show” question, clear explanation of the steps used is important to score. Don’t just show the maths! CM CM CM m m v m v m v m v m vv mm mm mm nnv v v gh nnmm mm + = − −= + − − + −   = = =    ++      A B A B AB AB AB AA AB AA ( ) ( ) + ( ) [C1] + () + 11 2 (upwards) [A1] 11+

5 (c) (i) Determine an expression for the ratio of the rebound height H to the drop height h. [3] (ii) Deduce the maximum value of H h for n 1 . maximum value of H h =………………………………. [2] [Total: 10] '' CM ' A CM A ' A ' A v v v v u u u ngh gh un ngh u n − = = −= −− − =  +  − − + =  + Using (where velocity of body rel ative to CM) 122 1 12 1 [B1]1 In CM frame, to cons     '' AA ' A ' A CM A vu nv gh n v v + v nn gh ghnn ngh n ngh ghnn hv H h =−  − =+  + = −  −    = + +    ++      − =+  + −   = = −    ++     = 2 erve momentum, [B1] 121 1 11 2 2 1 11 1 2 1 2 1 3 1 4 2 2 3 11 since    ( ) gh n ngh   − +  =− + 2 2 2 423 1 43 [A1]( 1)2 Rearrange answer in 1(c)(i) to give H hn =− + 2 43 1 [M1] As n tends to infinity, the expression tends to 9 [A1]

6 2 A consequence of special relativity is that observers in different inertial reference frames disagree on the lengths of objects and the times between events. (a) State what is meant by an inertial observer. ………………………………………………………………………………………………………. …………………………………………………………………………………………………… [1] (b) An inertial reference frame (x, y, z, t) has Earth as its origin. Ignore the rotational motion of Earth as negligible on the length scale of interest. In this reference frame, the star Proxima Centauri is a distance of 4.25 light years (ly) along the positive x-axis, as shown in Figure 2.1. Figure. 2.1 A spacecraft leaves Earth at t = 0 and travels directly to Proxima Centauri at a speed of c 2 . Assume that there is negligible relative motion between Earth and Proxima Centauri. The origin of an inertial reference frame (x ', y ', z ', t ' ) moving with the spacecraft coincides with the origin of the Earth’s reference frame at the moment the spacecraft leaves the Earth. an observer, moving at constant velocity / not accelerating [B1]

7 (i) Write down the x and t coordinates of the spacecraft in Earth’s reference frame at the moment of departure from Earth and at the moment of arrival at Proxima Centauri. Use units of light years (ly) for x and years (yr) for t. In these units, c = 1. x at departure = ……………………………. ly t at departure = ……………………………. yr x at arrival = ……………………………. ly t at arrival = ……………………………. yr [3] (ii) Use the Loren