CJC 2024.H2.Phy.PRELIM.P2 - Solns

CANDIDATE NAME MARK SCHEME CLASS 2T PHYSICS 9749/02 Paper 2 Structured Questions 23 August 2024 2 hours Candidates answer on the Question Paper. READ THESE INSTRUCTIONS FIRST Write your name and class 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 an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Answer all questions. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 34 printed pages and zero blank page. [Turn over FOR EXAMINER’S USE Q1 / 6 Q2 / 12 Q3 / 5 Q4 / 5 Q5 / 7 Q6 / 11 Q7 / 8 Q8 / 6 Q9 / 20 PAPER 2 / 80 Catholic Junior College JC2 Preliminary Examinations Higher 2

2 DATA speed of light in free space c = 3.00 x 108 m s-1 permeability of free space 0 = 4 x 10-7 H m-1 permittivity of free space 0 = 8.85 x 10-12 F m-1 (1/(36)) x 10-9 F m-1 elementary charge e = 1.60 x 10-19 C the Planck constant h = 6.63 x 10-34 J s unified atomic mass constant u = 1.66 x 10-27 kg rest mass of electron me = 9.11 x 10-31 kg rest mass of proton mP = 1.67 x 10-27 kg molar gas constant R = 8.31 J K-1 mol-1 the Avogadro constant NA = 6.02 x 1023 mol-1 the Boltzmann constant k = 1.38 x 10-23 mol-1 gravitational constant G = 6.67 x 10-11 N m2 kg-2 acceleration of free fall g = 9.81 m s-2

3 [Turn over FORMULAE uniformly accelerated motion s = u t + ½ a t2 v2 = u2 + 2as work done on / by a gas W = p V hydrostatic pressure p = gh gravitational potential  = - Gm r temperature T / K = T / ˚C + 273.15 pressure of an ideal gas p = 1 3 Nm V 〈c2〉 mean translational kinetic energy of an ideal gas molecule E = 3 2 kT displacement of particle in s.h.m. x = x0 sin t velocity of particle in s.h.m. v = v0 cos t = 22 0 xx − electric current I = Anvq resistors in series R = R1 + R2 + ... resistors in parallel 1/R = 1/R1 + 1/R2 + ... electric potential V = Q 4πεor alternating current / voltage x = x0 sin t magnetic flux density due to a long straight wire B = μoI 2πd magnetic flux density due to a flat circular coil B = μoNI 2r magnetic flux density due to a long solenoid B = μonI radioactive decay x = x0 exp(-t) decay constant λ = 1 2 ln2 t

4 Answer all questions in the spaces provided. 1 A car of mass 1700 kg travels over a curved hump in the road as shown in Fig. 1.1. The radius of curvature of the hump is 45 m. Fig. 1.1 (a) The speed of the car at the top of the hump is 19 m s-1. Determine, for the car at the top of the hump, (i) the magnitude of the centripetal force acting on the car, centripetal force = ………………..……………… N [1] L1 Fc = mv2 r = (1700)(19)2 45 = 13638 = 14000 N A1 (ii) the magnitude of the normal contact force exerted by the road on the car. normal contact force = ………………………………. N [2] L2 Free body diagram: The resultant force of the downward weight of the car W and the upward normal contact force exerted by the road N provide for the centripetal force required. W – N = 13638 (1700)(9.81) – N = 13638 N = 3039 = 3000 N (magnitude) Direction of N: vertically upwards M1 A1 (b) Determine the maximum speed vmax that the car can travel at without losing contact with the top of the hump. Explain your working. vmax = ……………………………. m s-1 [3] L2 The resultant force of the downward weight of the car W and the upward normal contact force exerted by the road N provide for the centripetal force Fc required. Thus, W – N = Fc N = W – Fc B1 car hump N W

5 [Turn over For the car not to lose contact with the top of the hump, the normal contact force N with the road must be greater than zero, i.e. N > 0 W – Fc > 0 mg – mv2 r > 0 v < √rg v < √(45)(9.81) vmax = 21.011 = 21 m s-1 B1 A1 [Total: 6] 2 A long, straight wire W carrying a direct current of 3.0 A flows in the direction as shown in Fig. 2.1. Fig. 2.1 (a) Draw on Fig. 2.1, the pattern of the magnetic field produced by wire W in the regions indicated by the dotted boxes. Use the symbol x to represent magnetic field directed into the page and use the symbol • to represent magnetic field directed out of the page. [3] L1 Solution: Marks scheme: 1 mark for: • Correct direction of magnetic flux density on left side (into plane of paper) and right side (out of plane of paper) of wire W. 1 mark for: • Increasing spacing with increasing distance from wire W. 1 mark for (Pattern shown sufficiently) • At least 4 columns each of x and • shown to show the trend of increasing spacing between consecutive circular field lines. B1 B1 B1 wire W 3.0 A wire W 3.0 A

6 • At least 3 rows each of x and • shown to show the trend of equal spacing parallel to the wire. (b) A similar wire Y is placed parallel to wire W, separated by a distance of 40.0 cm as shown in Fig. 2.2. Initially, there is no current in wire Y. Fig. 2.2 (i) Show that the magnetic flux density at wire Y due to the current in wire W is 1.5  10-6 T. [1] L1 Magnetic flux density 0 2 IB d  = where I is the current in wire W and d the distance of Y from W. ( )( ) ( ) 7 2 4 10 3.0 2 40.0 10   − −  =  B = 1.5  10-6 T (Shown) M1 A0 (ii) A current of 1.0 A is now switched on in wire Y and flows in the opposite direction as the direction of current flow in wire W. Use your answer in (b)(i) to calculate the force per unit length acting on wire Y. force per unit length = …………………….…….. N m-1 [2] L2 Magnetic force on Y at Y due to W in Y of Y sin=F B I L where θ = 90° since B and I are perpendicular. F BIL = ( )( ) 61.5 10 1.0−=F L = 1.5  10-6 N m-1 M1 A1 (iii) Explain why the force that the two wires exert on each other is repulsive. …………………………………………………………………………………………………... …………………………………………………………………………………………………... …………………………………………………………………………………………………... wire W 3.0 A wire Y 40.0 cm

7 [Turn over …………………………………………………………………………………………………... …………………………………………………………………………………………………... …………………………………………………………………………………………………... …………………………………………………………………………………………… [3] L2 By Right Hand Grip Rule ( or, using Fig. 2.1), the magnetic field produced by the current in wire W acts perpendicular to wire Y and out of the plane of paper. By Fleming’s Left Hand Rule, the direction of the magnetic force on Y by W acts to the right / away from wire W. By Newton 3rd law of motion, the direction of the magnetic force on W by Y is opposite to that on wire Y / towards the left / away from wire Y. [OR, Apply Fleming’s Left Hand Rule a second time to determine the force on W.] Therefore there is a repulsive force acting between the two wires. M1 M1 M1 A0 (iv) Determine a possible position, other than at infinity, where the resultant magnetic flux density due to the magnetic fields of both wires is zero. position: ………………………………….…………………………. [3] L2 First, consider directions of the two fields in different regions: