2023 RI Prelim P2 AS

1 © Raffles Institution [Turn over 2023 RI Preliminary Examinations H1 Physics Paper 2 Solutions 1 (a) 2 2 2 2 2 2 0.8504 4 9.858 9.86 m s 36.9 20 LT g Lg T                       (b) 2 2 0.2 0.00129.858 36.9 0.850 0.118 0.1 m s g T L g T L g g                     g = (9.9  0.1) m s2 2 (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. 3 (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 2m u  2 1 2mu    2 A 12 2m v    2 B 2 2 2 A B3 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,    A2 2 cos 90m u mu mv      

2 © Raffles Institution  A2 cos ... 1u v  OR A,x A cos (1') 2 uv v    Considering vertical motion, taking up as positive, A B0 2 sin sinmv mv     B A 2 sin ... 2v v  OR B A,y A sin (2') 2 vv v    (1)2  (2)2,           2 2 2 2 B A A 2 2 2 B A 2 cos 2 sin 4 ... 3 u v v v u v v       OR By Pythagorean theorem        2 2 2 A A, A, 22 2 B A 2 2 2 B A from 1' and 2'2 2 4 ... 3 x yv v v vuv u v v               From equation in (b)(i),   2 2 2 2 2 2 B A B A 33 2 ... 4 2 u vu v v v     Substituting (4) into (3),   2 2 2 2 B B 2 2 B 5 5 5 1 B 34 2 6 2 5 5 3.5 10 4.5185 10 4.5 10 m s3 3 u vu v u v v u                   (b) (ii) 2. where u, vB and v are the magnitudes of the vectors. By Pythagorean theorem,   2 2 B 2 2 5 5 5 1 5 3 8 8 3.5 10 5.7155 10 5.7 10 m s3 3 v u v u u u                   B 1 1 B tan 5tan tan 52 3 v u v u          52 below horizontal vB u  v vB u v Alternatively, 

3 © Raffles Institution [Turn over    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 52 above horizontal (in the opposite direction of change in momentum of particle B). 4 (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     (ii) F + 0.315 = (0.250 + 0.363) g F = 5.70 N 5 (a) The charge is stationary. The charge is moving in the direction of the magnetic field. (b) (i) Out of plane of paper (ii) Magnetic force provides centripetal force 2 5 19 27 4.7 100.12 3.2 10 0.080 6.54 10 kg mvBqv r m m          A B u path of particle B path of particle A  u 52

4 © Raffles Institution (iii) 1. 2. Bqv = qE 0.12  4.7  105 = E E = 5.64  104 N C1 3. Since the velocity of the other particles are the same , the path is a straight line. 6 (a) (i) While the speed is constant, the direction of the velocity is changing continuously. Since velocity is a vector, a change in direction means a change in the quantity. Therefore, the train is accelerating. (ii) Horizontal distance between rails = 2 2w E   2 2 vertical distancetan horizontal distance E w E     (iii)         2 2 2 2 2 1 2 2 2 Vertically, sin 1 Horizontally, the horizontal component of normal contact force provides the centripetal force, cos 2 1 2 tan N mg mvN r v rg E v rgw E Ergv w E                 (b) (i) Passenger trains are lighter/carry a smaller weight. (ii) CDmax is 110 mm for 1435 rail gauge. Ev = 110 + 95 = 205 mm region of uniform magnetic field, flux density 0.12 T path of particles, speed 4.7  105 m s1 0.16 m

5 © Raffles Institution [Turn over     0.5 0 2 2 0.53 2 2 3 1 205 10 1500 9.81 1435 205 10 46.086 46.1 m s Ergv w E                    (ii) 1. 2.         2 2 15000019.6 60 60 1500 23 N mv r 3.      95sin 1435 3.8 4.                 Vertically, cos sin 1 Horizontally, sin cos 23 sin 23 cos 2 2 ,1 23 costan 19.6 9.81 sin 19.6 9.81 tan sin tan 23 cos 10.2 N 9.98 N if 22.69 N is used in the calculation                  c N mg f N f N f f f f f f f F             Since f < fmax, luggage will not slide OR Faster method the component of weight parallel to the slope and friction should provide the component of centripetal force that is parallel to the slope.    sin cos 19.6 9.81 sin3.8 23cos3.8 10.2 N       cmg f F f f   luggage bag carriage floor frictional force weight normal contact force 95 mm 1435 mm 

6 © Raffles Institution (c) (i) As the circumference of the outer wheel is larger, it will be able to travel a greater distance even though it rotates as the same rate as the inner wheel, allowing both wheels to stay on the track. (ii) For a wheel of diameter D and angular velocity of W moving at speed v along a curved rail of radius R, linear speed at rim of wheel = speed of wheel w rain 2 t D R        Since both the inner and outer wheels are rotating with the same angular velocity of W, and both wheels are moving along the curved rails with the same angular velocitytrail, outer inner 200 1.524 1.150 200 1.1588 1.16 m o i o o D R D R D D     (d) (i)  Easier maintenance (of rails)  Cheaper or easier or faster to build (as there is no need to build tunnels or viaducts) (ii)  Allows land surface to be used for other purposes  Less noise to residents / less noise pollution  Train operations is not affected by weather conditions (rain or snow) 7 (a) (i) The resistance of a resistor is the ratio of potential difference across the resistor to the current in it. (i) 1. 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  2. 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 p dR (shown) w v D train

7 © Raffles Institution [Turn over (iii) There is a