2023 RI H2 Physics Prelims P2 Answers

1 © Raffles Institution [Turn over 2023 RI Preliminary Examinations H2 Physics Paper 2 Solutions 1 (a) To hit the coconut horizontally, vertical component of the velocity = 0. Initial vertical velocity = 20 sin . Using v2 = u2 + 2as, and setting v to 0 m s−1 and s = 18.0 − 2.2 = 15.8 m: 0 = (20sin )2 – 2(9.81)(15.8) sin = 0.880  = 61.7 (b) By v = u + at, 0 = 20 (0.880) – 9.81t t = 1.79 s (c) Horizontal displacement = 20  cos(61.7)  1.79 = 17.0 m (d) The downward acceleration will be larger and hence the same initial vertical velocity will be reduced to zero over a shorter vertical displacement. Hence,  has to be larger so that the initial vertical velocity is larger. Since initial horizontal velocity is smaller and it will decrease due to the air resistance in the horizontal direction, the horizontal displacement will be lower. 2 (a) product of mass and (linear) velocity (b) (i) Since collision is elastic, total kinetic energy is the same before and after collision. 1 2( ) 2 12 2mu + 2 1 2mu = ( ) 2 A 12 2mv + ( ) 2 B 2 2 2 AB3 2 shown mv u v v=+ (ii) 1. Let the scattering angle of B be  respectively. By principle of conservation of linear momentum, Considering horizontal motion, taking right as positive, ( ) ( ) ( ) A A 2 2 cos 90 2 cos ... 1 m u mu mv uv   − = =  = Considering vertical motion, taking up as positive, ( ) AB BA 0 2 sin sin 2 sin ... 2 mv mv vv   =− = (1)2 + (2)2, ( ) ( ) ( ) ( ) ( ) 2 2 2 2 B A A 2 2 2 BA 2 cos 2 sin 4 ... 3 u v v v u v v + = + +=

2 © Raffles Institution [Turn over From equation in (b)(i), ( ) 2 2 2 AB 22 2 B A 32 3 ... 42 u v v uvv =+ −= Substituting (4) into (3), ( ) 2 2 2 BA 22 B 22 B 22 B B 5 5 5 1 4 34 2 62 35 5 3 5 3.5 10 4.5185 10 4.5 10 m s3 u v v uv uv vu vu − +=  −=  =− = = =  =  =  Alternatively, Let the horizontal and vertical components of the velocity of particle A after collision be vA,x and vA,y respectively, and the vertical (downward) velocity of particle B after collision be vB. By principle of conservation of linear momentum, Considering horizontal motion, taking right as positive, ( ) ( ) ( ) A, B, A, A, 2 2 0 2 ... *2 xx x x m u mu mv v uv uv − = = = = Considering vertical motion, taking up as positive, ( ) A, B B A, B A, 02 2 ... ^2 y y y mv mv vv vv =− = = By Pythagorean theorem, ( ) ( )( ) 2 2 2 A A, A, 22 2 B A 2 2 2 BA from * and ^22 4 xyv v v vuv u v v =+ =+   += which is identical to equation (3). (b) (ii) 2. where u, vB and v are the magnitudes of the vectors. vB u  v vB u v Alternatively, 

3 © Raffles Institution [Turn over By Pythagorean theorem, ( ) 22 B 2 2 5 5 5 1 5 3 88 3.5 10 5.7155 10 5.7 10 m s33 v u v uu u −  = + =+  = =  =  =  B 11 B tan 5tan tan 52 3 v u v u   −− = = = =  52 below horizontal ( )( ) 27 5 22 1 1.7 10 5.7155 10 9.7 10 kg m s p m v − −−  =  =   = change in momentum is 9.7  10−22 kg m s−1 (b) (iii) 22 6 16 9.7 10 1.2 10 8.1 10 N pF t − − − = =  = by Newton’s third law, average force exerted by particle B on particle A is 8.1  10−16 N at 52 above horizontal (in the opposite direction of change in momentum of particle B). 3 (a) 1. The resultant force acting on the body is zero. 2. The resultant moment/torque on the body about any axis/point is zero. (b) (i) ( )( )tension in spring 21 0.015 0.315 N = = ( )( )( ) ( )( )( ) ( )( )( ) sum of clockwise moments sum of anticlock wise moments 0.30 9.81 0.50 21 0.015 0.50 0.250 9.81 0.363 kg M M = += = A B u path of particle B path of particle A  u 52

4 © Raffles Institution [Turn over (ii) ( )( ) 150 3 3 0.36 8.9 10 0.36 9.81 2 8.9 10 g m V V U   = =  =  ( ) ( ) ( )( ) ( )( ) ( )( ) ( )( )3 33 0.50 0.50 0.360.250 9.81 21 0.015 0.36 9.81 9.81 2 8.9 10 7.03 10 kg m mg T Mg U   − − = − − = −  = (iii) The metre rule is tilted anti-clockwise at an angle from its original position with less than half of block M submerged in the liquid. OR The metre rule is tilted at an angle from its original position with less than half of block M submerged in the liquid. Additional information: ( )( ) ( )( ) ( )( )( ) 3 3 0.360.250 9.81 0.36 9.81 7.0266 10 9.81 8.9 10 0.387 mg Mg U b b =− = −   = When the rule is balanced, 39% of block M is submerged in liquid. 4 (a) A longitudinal wave is one in which its particles oscillate in a direction parallel to the direction of energy transfer. (b) (i) 1. = = 1 wavelength 0.080 m2 wavelength 0.160 m 2. = = = speed 343frequency 2144 2140 Hzwavelength 0.160 (ii) (iii) 13 10 14 11 12 9 1 7 2 8 5 4 3 6 B A C R y t 0 y0 −y0

5 © Raffles Institution [Turn over (iv) ( ) I II II    =  = =  =  = − =  =  = = = 2 22 2 2 1 2 12 2 1 22 2 1 2 1 4 14 2 4 2 2 2 0.25 0.35355 m distance moved by detector 0.35355 0.25 0. 10355 0.10 m P r P r P r r r r r r r 5 (a) The resistance of a resistor is the ratio of potential difference across the resistor to the current in it. (b) (i) XV E r=− I From graph of V against I, gradient = – r 5.40 4.20 gradient 0.40 1.20 1.5 r −=− =−  − = OR Using substitution of point 4.80 = 6.0 – 0.80 r r = 1.5  (ii) When I = 0.40 A, p.d. across fixed resistor = VX – VY = 5.40 – 4.00 = 1.40 V . . 1.40 3.50.40I= = = S pdR (shown) (c) There is a maximum value of the resistance in the circuit, and from the given electromotive force of 6.0 V, there will be a minimum current from I = e.m.f. R . (d) When VX = 5.25 V, I = 0.50 A. VY = 3.50 V P = IV = (0.50)(3.50) = 1.75 W

6 © Raffles Institution [Turn over (e) (i) (ii) 0.70 A (intersection of graph Z and voltmeter Y) 6 (a) The charge is stationary. The charge is moving parallel to the direction of the magnetic field. (b) (i) 5 5 14.7 10 sin20 1.61 10 m sv − ⊥ =   =  (ii) Magnetic force provides centripetal force ( ) 2 5 19 27 1.6075 100.12 3.2 10 1 0.0562 6.69 10 kg mvBqv r m m ⊥ ⊥ − − =   =  = (iii) 6 5 2 0.056 1.0944 10 s 1.6075 10 rT v  − ⊥ = = =   0.0 1.0 2.0 3.0 4.0 5.0 6.0 0.2 0.4 0.6 0.8 1.0 1.2 V / V I / A voltmeter X voltmeter Y Z

7 © Raffles Institution [Turn over 56 cos20 4.7 10 cos20 1.0944 10 0.483 m x v T − =  =     = (iv) Electrons will travel in a helical path with smaller radius. Electrons will travel in a helical path with smaller period. Electrons will travel in a helical path with smaller x. Direction of rotation of the helical path of the electrons will be opposite to that of the particles. Additional Information: ( ) 2 mvBqv r mvr Bq ⊥ ⊥ ⊥ = = Mass of electron is 9.11  10−31 kg and charge is −1.60  10−19 C. Hence, re will be much smaller. Time 2 rT v  ⊥  =  taken for the electrons to complete one revolution in the helix will also be much smaller and x will be smaller. 7 (a) (i) 2 (25.0 1.0) 6.25 4.0V == 2.50 VrmsV = (ii) 2 22.50 0.781 W8.0 rmsVP R= = = (b) (i) Since average power is the same, Vrms of sinusoidal function = Vrms of original source = 2.50 V V0 = 2  2.50 = 5.00