VJC 2026 H3 Physics Prelim (Ans)

VJC 2026 9814/01/J2Prelim/26 [Turn over VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 3 CANDIDATE NAME CLASS TUTOR NAME PHYSICS 9814/01 Paper 1 22 September 2026 3 hours Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, class, Centre number and index number in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Section A Answer all questions. You are advised to spend about 1 hour and 50 minutes on Section A. Section B Answer two questions only. You are advised to spend about 35 minutes on each question in Section B. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 / 6 2 / 13 3 / 14 4 / 10 5 / 17 6 / 20 7 / 20 8 / 20 Total / 100 This document consists of 26 printed pages H

2 VJC 2026 9814/01/J2Prelim/26 Data speed of light in free space c = 3.00 × 108 m s−1 permeability of free space  = 4 × 10−7 H m−1 permittivity of free space  = 8.85 × 10−12 F m−1 . 91 0 1 8 99 10 m F4 − = elementary charge e = 1.60 × 10−19 C 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 Avogadro constant NA = 6.02 × 1023 mol−1 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 Formulae

3 VJC 2026 9814/01/J2Prelim/26 [Turn over

4 VJC 2026 9814/01/J2Prelim/26 Section A Answer all questions in this section. You are advised to spend about 1 hour 50 minutes on this section. 1 A thorium nucleus decays by the emission of an -particle to form a radium nucleus. This is represented by the nuclear equation: 232 228 90 88Th Ra + α→ The thorium nucleus is initially stationary. (a) (i) Show that =speed of radium nucleus 1speed of α-particle 57 . [1] (ii) Show that =kinetic energy of radium nucleus 1kinetic energy of α-particle 57 . [1] Conservation of momentum: Final momentum of system = initial momentum mRavRa - mv = 0 Rav v = Ra m m  = 4 228 = 1 57 Rake ke = 21 2 21 2 Ra Ramv mv = 2 Ra Ramv mv          = 2 228 1 4 57   = 1 57

5 VJC 2026 9814/01/J2Prelim/26 [Turn over (b) The total energy released in the decay is 4.08 MeV. Assume that all this energy goes into the kinetic energies of the -particle and the radium nucleus. The specific charge (charge per unit mass) of an -particle is 4.81  107 C kg-1. Calculate the speed of the -particle. speed = m s-1 [4] [Total: 6] ½ mv2 = E (where E = 57x4.08/58 MeV, since  particle has 57/58 of total energy) v = 2E m = 6757 4.08 10 4.81 102 58 2 e e    = 1.39 x 107 m s-1

6 VJC 2026 9814/01/J2Prelim/26 2 A point mass m is held at the centre of a circular plate of radius r = 20 cm attached to a fixed board. The board is tilted at an angle of 30o to the horizontal, as shown in Fig. 2.1. The point mass is given an initial velocity u = 0.50 m s-1 in the direction shown in Fig. 2.1. (a) Determine the time taken for the point mass to fall off the edge of the circular plate. Ignore the efforts of friction. time = s [5] Consider the motion of the mass on the plane of the slope. sx : horizontal displacement from starting point sy : displacement down-slope from starting point Let T be the time taken for the mass to reach the edge of the plate. Then xs uT= and 2 2211( sin 30 ) (0) ( sin 30 )2 2 4yy gTs u T g T T g T= + = + =  At time T, 2 2 2 xys s r+= 22 22() 4gTuT r += 2 4 2 2 2 016g T u T r+ − = Solve the quadratic equation for T2: ( ) 2222 2 2 2 4 2 2 22 9.814 ( ) (0.50) 0.50 4 (0.20 )16 16 9.8122 16 16 gu u r T g    −  − − −  +      ==            Since T2 can only be positive, 2 0.25 1.012354 0.06337412.0295T −+==  T = 0.252  0.25 s

7 VJC 2026 9814/01/J2Prelim/26 [Turn over (b) The circular plate is removed. A solid ball of mass m is released at the top of the sloping face of the board The ball rolls down the sloping board without slipping, as shown in Fig. 2.2. A constant frictional force F exerts a torque  on the ball. By considering the forces on the ball, show that the linear acceleration a of the ball down the slope is 3.5 m s-2. [6] There are 3 forces acting on the ball: weight (W), frictional force (F) and normal force (N) Applying Newton’s 2nd law for translational motion ( netF ma= ): Perpendicular to the slope: cos30WN =  (this equation isn’t useful here) Parallel to the slope (downslope as +ve): sin30W F ma −=  2 mg F ma−= -----(1) Applying Newton’s 2nd law for rotational motion ( net I = ): 22 5FR I mR ==  , where  is the angular acceleration of the ball and R is its radius For pure rolling motion, linear acceleration a = R  222 55 aFR mR mRa R  ==     2 5F ma= -----(2)

8 VJC 2026 9814/01/J2Prelim/26 (c) A solid cylinder of mass 80 g is released from rest at the top of the board. It rolls down the slope without slipping, with its central axis perpendicular to the slope, as shown in Fig. 2.3. When the cylinder reaches point A, the linear speed v of the cylinder is 1.84 m s-1. Calculate the rotational kinetic energy of the rolling cylinder as it passes point A. rotational kinetic energy = J [2] [Total: 13] For pure rolling motion, the linear velocity v of the cylinder is related to its angular velocity  by v = R , where R is the radius of the cylinder. Moment of inertia of the solid cylinder: 21 2I MR= Rotational KE at A = 2 2 2 2 21 1 1 1 1 (0.080)(1.84)2 2 2 4 4 vI MR Mv R   = = =     = 6.77  10−2 J Substitute (2) into (1) : 2 25 mg ma ma−=  5 5(9.81) 14 14 ga= = = 3.5036 m s−2  a  3.5 m s−2

9 VJC 2026 9814/01/J2Prelim/26 [Turn over 3 A parallel plate capacitor has circular parallel plates. The capacitance C of a parallel plate capacitor is given by the equation: = 0 AC d where 0 is the permittivity of free space, A is the surface area of each capacitor plate and d is the separation of the capacitor plates. (a) (i) Express the volt in SI base units. nits = [2] (ii) Express the units for the permittivity of free space in SI base units. units = [2] (b) The diameter of the circular capacitor plates is measured to be 6.50 cm and their separation is measured to be 5.42 mm. State and explain which instruments are used for making these measurements. instrument for measuring diameter instrument for measuring separation explanation [2] SI base units of volt, base units of Energy base units of Charge= 22 2 3 -1kgm s kg