2024 JPJC Prelim H2 Physics P3 Solutions Printed

2024/JPJC/Prelim/9749/03 Answers to 2024 JC2 H2 Preliminary Examinations Paper 3 Suggested Solutions: No. Solution Remarks 1(a) v = u + at = −2.5 + (9.81)(2.0) = 17.1 Speed is 17 m s−1 [1] substitution [1] answer 1(b) v / m s−1 0 1.0 2.0 t / s [1] helicopter: horizontal straight line (-ve value) [1] parcel: diagonal straight line (+ve gradient) (reverse sign accepted) [-1] if not labelled 1(c) Distance between helicopter and parcel = area of right-angled triangle = 1 2 (2.0)(17.1 2.5) 19.6 m+= OR From point of release, For helicopter, upwards s = ut = (2.5)(2.0) = 5.0 m For parcel, downward s = ut + 21 2 at = (−2.5)(2.0) + 21 2 (9.81)(2.0) = 14.6 m Hence their separation = 5.0 + 14.6 = 19.6 m [1] working [1] answer 2(a) The change in height is negligible compared with radius of the planet. Thus, the field strength 2 GMg R= remains relatively constant for any small change in R. The gravitational field lines are effectively parallel. [1] 2(b)(i) 2 GMY R = where G is the gravitational constant [1] P H –2.5 17

2 2024/JPJC/Prelim/9749/03 2(b)(ii) [1] for correct shape of curve from 3R− to R− ; ending at ( , 1.0 )RY− [1] for correct shape of curve from R to 3R ; starting at ( , 1.0 )RY− [1] for curves passing through ( 3 , 0.11 )RY− ; ( 2 , 0.25 )RY− ; (2 , 0.25 )RY− ; (3 , 0.11 )RY− (don’t need to mark for ±3R) 2(b)(iii) Initial total energy = Final total energy 221 1 1 2 2 2 GMm GMmmv mv Rx − = −  ( ) ( )( ) 2 23 11 23 6 6 1 1 1 4 1 1 14.7 10 6.67 10 6.4 104 3.4 10 6.070 10 m v GM Rx x x − =−   =   −   = Therefore distance travelled 66 6 6 6.070 10 3.4 10 2.670 10 m 2.7 10 m xR=− =  −  = = [1] for correct equation for conservation of energy [1] for correct substitution [1] for correct answer 3(a) The electric field strength in both sphere A (between x = 0 and x = 1.4 cm) and sphere B is zero (between x = 11.4 and x = 12.0 cm). [1] 3(b)(i) The resultant field strength is zero at a point between the spheres and this shows that electric fields are in opposite directions in the region between the two spheres. This shows that the polarity of the two charges are the same. Hence, since sphere A is positively charged, sphere B must also be positively charged. [1] expl [1] state

3 2024/JPJC/Prelim/9749/03 [Turn over 3(b)(ii) At x = 0.08 m, the electric field strength due to sphere A cancels out the electric field strength due to sphere B. EA = EB ( ) ( ) 2 o B 2 o A 04.0 408.0 4  QQ = 404.0 08.0 2 B A =    =Q Q [1] sub [1] ans 3(c)(i) The electric field strength is negative of the electric potential gradient, i.e. dx dVE −= [1] don’t accept dx dVE −= 3(c)(ii) From dx dVE −=  x 8cm 2 x cm V E dx = =  =− Hence, change in potential, ∆V = negative of area under E−x graph (from x = 2.0 cm to x = 8.0 cm) By approximation, area of triangle  area under E−x graph (from x = 2.0 cm to x = 8.0 cm) [1] ∆V = area under E−x graph [1] value of ∆V -200 -150 -100 -50 0 50 100 150 0 2 4 6 8 10 12

4 2024/JPJC/Prelim/9749/03 ( )( ) 2 6 51 2.6 10 60 10 7.8 10 V2V − =−   =−  Work done ( )( ) 567.8 10 2.0 10 1.6 J−= −  −  = [1] ans accept 1.4 J to 1.8 J 4(a) The magnetic flux linkage is the product of the number of turns of the coil of wire and the magnetic flux through (each turn of) the coil of wire. The magnetic flux through an area is defined as the product of that area and the component of the magnetic flux density normal to the plane of that area. [1] [1] 4(b)(i) ( ) max max max 2 73 5 5 1800(1000 4 ) (1. 0.0305 )0.12 4 2.014 610 10 10 10 Wb 2.0 Wb BA n A    −− − − = = =   = =      I [1] substitution for Bmax only [1] unrounded value for max flux 4(b)(ii) By Faraday’s law of EMI, induced 0 0 3 3 2 7 induced [1.0 ( 1.0)] 10 (2 () (1000 ) 1000 (1000)(24 4 56) 10 (0.036) 1800(4 104 0.00 1)) 2 2.88 V Q Q Q dE dt d N BA dt dnNA dt dN A n dt d dt dE dt     − − −  − −  −    =− =− =− =− = =−  = I I I I [1] for correct substitution of nA [1] correct gradient from Fig. 4.2 [1] for correct substitution of QN [1] for final answer

5 2024/JPJC/Prelim/9749/03 [Turn over 4(b)(iii) [1] for ϕ – t graph [1] for V – t graph [1] for correct scale on vertical axes of both graphs 5(a) 2 . . . 2 Mean power 2.7 (50) 2 182 W 180 W r m sPR= = == I [1] sub [1] ans 5(b) . . .r.m.s voltage across secondary coil 2.7 (50) 2 95.5 V S r m sVR= = = I [1] value -3 -2 -1 0 1 2 3 0 20 40 60 80 100 120 -3 -2 -1 0 1 2 3 0 20 40 60 80 100 120 ϕ / × 10–5 Wb t / ms t / ms V / V

6 2024/JPJC/Prelim/9749/03 Turns ratio, 25 20 95.46 (25)(95.5) 20 119 120 PP SS S S NV NV N N = = = == [1] ans 5(c) For the current in the secondary coil, Period T = 3 250 10 1.67 103 − − = 2 1Frequency 1. 167 6z 0 0 H f −= =  Hence, the frequency of the alternating voltage supply is 60 Hz since it is the same as the frequency of the current through the secondary coil. [1] correct T [1] ans and statement 5(d) The values of the root -mean-square current and voltage of the alternating voltage supply are independent of its frequency. The mean power due to the alternating voltage supply is constant. Since the transformer is ideal, the mean power dissipated across R remains unchanged. [1] [1] 6(a) Threshold frequency refers to the minimum frequency of the illuminating electromagnetic radiation that will cause a photoelectron to be ejected for a particular metal. [1] 6(b)(i) Energy of a photon, ( )( ) 34 8 19 9 6.63 1 0 3.00 1 0 4.42 1 0 450 1 0 hcE  − − −  = = =   J [1] sub [1] ans 6(b)(ii) Power incident on metal, P = (2.7  103)(3.0  10−4) = 0.81 W 18 1 19 0.81 1.83 1 0 s 4.42 1 0 NPE t NP tE − − =  = = =   [1] value [1] ans 6(b)(iii) Max. K.E. = eVs = (1.6  10−19)(1.6) = 2.56  10−19 J Applying Einstein Photoelectric equation, Work function, 19 19 19max. K.E. 4.42 1 0 2.56 1 0 1.86 1 0 Jhf − − −= − =  −  =  [1] value [1] value

7 2024/JPJC/Prelim/9749/03 [Turn over Threshold wavelength, ( )( ) 34 8 6 19 6.63 1 0 3.00 1 0 1.07 1 0 m 1.86 1 0 hc  − − −  = = =   [1] ans 6(c)(i) From the graph, min is the same for both spectra. minmin  e hcVhceV == [1] 6(c)(ii) From the graph, m 1016 12 min −= 34 8 4 19 12 min 6.63 10 3.00 10 7.8 10 V1.60 10 16 10 hcV e − −−   = = =     [1] sub [1] ans 7(a)(i) The half -life of a radioactive nuclide is the time taken for the activity of a sample to reduce to half its initial value. [1] answer 7(a)(ii) The decay constant is the fraction of the total number of nuclei in a sample that decay per unit time. [1] answer 7(b)(i) E = m c2 = 0.7060 MeV = (0.7060)(1.6  10−13) = 1.1296  10−13 J → m = 2 E c = 13 82 1.1296 10 (3 10 ) −  = 1.2551  10−30 kg = 30 27 1.2551 10 1.66 10 − −   = 7.5609  10−4 u MN + Mn − (MC + MX) = 7.5609  10−4 u → MX = 1.00858 u − 7.5609  10−4 u = 1.007825 u = 1.01u [1] m in kg [1] convert kg to u [1] answer 7(b)(ii)1. Number produced = 7500 14 (6.02  1023) = 3.2  1026 [1] answer 7(b)(ii)2. The probability of decay of the nucleus in a time of 1.0 year is the decay constant of the nucleus in that period of time. Decay constant,  = 1/2 ln 2 t = 3 ln2 5.7 10 = 1.22  10−4 year−1 [1]