(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 pla