(VJC) 2024 H2 Prelim P3 Soln

2024 Physics Prelim Exam H2 Paper 3 suggested solutions 1(a) 2 22 2 2 1 2 From Fig. 1.1, 1.75 m and = 0.600 s [ 1] 2 2(1.75) (0.600) 9.722 m s 9.72 m s [1] − − =+ =  = = = = yy y y s u t gt st sg t (b) % uncertainty = actual uncertainty/ data point. [1] Hence the larger the data point, the smaller the % uncertainty since the absolute uncertainty is fixed. Hence more reliable. [1] (c) 2 2 2 [1] 0.001 0.006[ 2( )](9.722) 1.75 0.600 =0.2 m s [1] 9.7 0.2 m s [1] − −  =+  = + = y y sgt g s t g g (d) Since sy = ½gt2, plot a graph of sy against t2, where, sy = vertical distance travelled by the sphere, t = time taken to travel sy. The gradient = ½g. [1] Random error is reduced when a best fit line is drawn using all the data points. [1] 2(a) By the principle of conservation of momentum, since there is no external force acting, the total change in momentum of the of ball and wall = 0. [1] Therefore the change in momentum (impulse) of the wall is equal and opposite to the change in momentum of the ball . [1] Therefore, Student A is wrong. (b) Taking values from Fig 2.2, Total momentum before collision = (1.2 x 4) + 0 = 4.8 kg m s-1 [1] Total momentum after collision = (1.6 x 3.6) + (-0.8 x 1.2) = 4.8 kg m s-1 [1] Since total momentum before collision is equal to the total momentum after collision, momentum is conserved in this collision. [1]

(c) Relative speed of approach = 4.0 – 0 = 4.0 m s-1 Relative speed of separation = 1.6 – (-0.8) = 2.4 m s-1 [1] Since relative speed of approach is not equal to relative speed of separation, the collision is inelastic. [1] 3 (a) (i) Sinusoidally shaped graph [1] Maximum intensity value at 0°, 180° and 360° and zero intensity at 90° and 270°. [1] (ii) Using Malus’s Law, intensity of transmitted beam, 2costoII = [1] Therefore angle, 11 4.2cos cos 42 7.6 t o I I −−  = = =    [1] (b) Diffraction refers to the spreading or bending of plane waves when they encounter an aperture or obstacle, [1] whose linear dimension is comparable to the wavelength of the waves. [1] (c)(i) For diffraction grating, sindn = [1] Therefore, line spacing, ( ) 7 63 4.3 10 1.4 10 msin sin 68 nd   − − = = =   [1] (c)(ii) For diffraction grating, sindn = Different visible wavelength, ( )( ) 6 71.4 10 sin 68sin 6.5 10 m2 d n  − − = = =  [1 for n = 2; 1 for answer] 4(a) Electric field strength at a point is electric force per unit positive charge at that point. [1] Intensity 0 90 180 270 360 Angle / o

5(a) Magnetic flux density is defined to be the magnetic force acting per unit current and per unit length on a conducting wire [1] placed at right angles to the direction of the magnetic field. [1] (b)(i) Direction of the magnetic flux density is into the plane of the page. [1] (b)(i) ( )( ) 19 4 14 (2 ) [1 ] 2 1.6 10 7.5 10 2.4 10 N [1 ] F e E − − = =    = (b)(ii) Time taken for alpha particles to travel 1 m in the horizontal direction, 7 1.0 1.50 10 x x st u==  86.67 10 [1 ]s−= (b)(iii) Acceleration in the vertical direction, 14 13 2 27 2.4 10 0.3614 10 [1 ]4 4 1.66 10 FFa msmu − − − = = = =   Displacement in vertical-direction during time t, ( )( ) −= =   =  22 13 811 0.3614 10 6.67 10 0.008039 0.0080 [1 ]22s at m The particles will not hit any of the plates as the vertical displacement of the electron is less than 0.0125 m when it is travellling between the two parallel plates. (b)(iv) Parabolic path curves downward inside the plates Straight path outside the plates + V – V Beam

(ii) Magnetic force on a charge particle, sinBF Bqv = [1] ( )( )( ) ( ) 73 19 14 4.8 10 1.6 10 1.7 10 sin 90 1.3 10 N −− − =     = [1] (iii) For circular motion, magnetic force provides for the circular motion, 2 B mvF r= [1] Therefore, the electron will move in a circular motion of radius, 2 B mvr F= ( )( ) 231 7 14 9.11 10 1.7 10 0.020 m1.3 10 − −  ==  [1] Required distance, ( )2 2 0.020 0.040mdr= = = [1] 6(a)(i) Vrms = 𝑉0 √2 = 170 √2 = 120 V [1] (ii) 𝜔 = 2𝜋/𝑇 = 314 𝑇 = 0.0200 s [1] (b)(i) 3500 170 2000 298 V SS PP S S VN VN V V = = = (ii) 𝑉𝑆 = 𝐼𝑆𝑅 298 = 𝐼𝑆(130) 𝐼𝑆 = 2.288 A 𝐼𝑃 𝐼𝑆 = 𝑁𝑆 𝑁𝑃 = 3500 2000 IP = 4.00 A [1] [1] [1] [1]

(c) 7(a) 90 38Sr → 90 0 39 1Ye −+ [1; no need neutrino] (b) Energy released = (mass defect) c2 = (mSr – mY - me) c2 [1] = [(89.907738 – 89.907151) x 1.66 x 10-27 – 9.11 x 10-31] x (3.00 x 108)2 = 5.7078 x 10-15 [1] = 5.71 x 10-15 J (c)(i) Total energy that needs to be released per second, E = Power to be supplied Efficiency = 155 0.070 = 2214.3 J [1] Activity = Energy released in one reaction E = 15 2214.3 5.71 10 − = 3.8779 x 1017 [1] = 3.88 x 1017 Bq (ii) A = N Number of strontium-90 needed N = A  [1] = 1/2 ln2 A  = 173.8779 10 29 365 24 3600 ln2      = 5.1165 x 1026 = 5.12 x 1026 [1] I/A t/s 0 0.020 0.040 2.29 2 = 4.58 -2.29 Shape of graph: 1 mark. Correct I values: 1 mark Correct T: 1 mark

(iii) Mass of strontium-90 needed = number of nuclei x mass of 1 nuceus [1] = 5.1165 x 1026 x 89.907738 x 1.66 x 10-27 = 76.4 kg [1] 8(a)(i) 51 24 hours [1] 22 (24 60 60) =7.3 10 rad s [1]  −− = ==   T T (ii) 2 2 11 24 3 32 5 2 7 Gravitational force provides centripetal force [1] (6.67 10 )(5.9 10 ) [1](7.3 10 ) 4.2 10 m [1]   − − = ==  = GMm mrr GMr (iii) Communication, weather forecasting or navigation (GPS) (iv) Application Advantage Disadvantage Communication No break in the signal transmissions as it is fixed position in sky. High altitude so there is a significant lag time in the signal transmissions. Weather Navigation (b)(i) Gravitational field strength is equal to the negative gravitational potential gradient i.e. g = - d / dr (ii) Potential gradient at surface of star S1 is steeper than that of S2. [1] Using relationship in (b)(i), gravitational field strength at the surface of star S1 is greater than that of star S2. [1] (iii) From the graph, when the particle travels from S 2 to S1, it loses gravitational pe (since it experiences a drop in gravitational potential). [1] As the total energy of the particle is constant, it gains ke. So its ke at the surface of S 1 is larger than Ek. [1] (iv) It is the point in which the resultant gravitational field strength is zero. [1]

(v) 12 2 10 2 [1]x (1.2 10 - x)=  GM GM From Fig. 8.2, x = 4.8 x 109 m [1] 9 21 10 9 2 4.8 10[ ](1.2 10 - 4.8 10 ) = 0.44 [1] =  M M (vi) Correct shape of curve with gravitational field strength at surface of star S1 greater than S2. [1] Field strength is zero at the point of maximum potential. [1] Surface of star S2 Surface of star S1 g / N kg -1

9(a)(i) Resistance = L A  [1] = 2 4 L d     Cross sectional area of wire, A = 2 4 d = ( ) 62 23 1.50 10 6.0 10 0.30 10 4  −− −       = 1.273 [1] = 1.3  (ii)1. e.m.f. is the amount of other forms of energy converted to electrical energy per unit charge delivered by a source of e.m.f. [1] p.d. is the amount of electrical energy converted to other forms of energy per