YIJC 2024 JC2 PRELIM H2 Phy P3 Soln

1 ©YIJC 9749/02/YIJC/23 2024 JC2 Prelim Exam H2 Physics Paper 3 Solution 1 (a) (i) Because the speed at B is the same at C, the gain in KE is due to the loss of GPE from A to B only. Gain in KE in the end = Loss in GPE from A to B 21 2 =mv mgh == 2 2(9.81)(2.5)v gh = 7.0 m s-1 C1 A1 (ii) From A to B, the gravitational potential energy drop is converted to kinetic energy of the child. From B to C, the further loss in gravitational potential energy is not converted into kinetic energy but dissipated as work done against contact friction (or thermal energy and sound). Thus, the child’s speed is unchanged. B1 B1 (iii) ( ) ( )( ) ( ) 2 Work done against friction loss of KE from C to D 17.6 54 7.0 02 168 N 3sf CD Fs Fs KE KE F F = = =− =− = OR ( ) ( )( ) 22 2 2 2 0 7.0 2 7.6 average accel., 3.22 m s Average frictional force 52 3.22 168 N ave ave ave ave v u a s a a ma − =+ =+ =− == = M1 A1 (M1) (A1) (b) (i) When the carousel rotates, the chair and its occupant will travel in a straight-line if not for the cable attached to it. A horizontal (centripetal) force is required in order to pull the man towards the centre. This is provided by the horizontal components of the tension on the cable holding the chair. Thus, the cable has to tilt to provide for the needed centripetal force. M1 A1 (ii) All forces must be indicated and labelled in words (not in letters or symbols). Tension on cable and gravitational force (or weight) on person and seat. Deduct mark if FBD includes centripetal or centrifugal force, normal force or any irrelevant forces. B1 weight tension

2 ©YIJC 9749/02/YIJC/23 (iii) Along the y-direction: cosT mg = (balanced) Along the x-direction: 2sinT mr= (unbalanced due to centripetal rotation) ( ) 22 2 1 4.7 8.4sin35 9.518 m tan 9.518tan35 9.81 0.85 rad s r mr r mg g    − = + = == = = C1 C1 A1 2 (a) The first law of thermodynamics states that the increase in internal energy of a system is the sum of the heat (thermal energy) supplied to the system and the work done on the system. B1 (b) (i) 1. ( ) ( ) ( ) 562.79 10 1125 950 10 48.825 48.8 J 3sf onW P V − =−  =−  −   =− =− Max 1m for positive answers. C1 A1 2. ( )( ) ( )( ) ( ) 5 6 5 6 0, since no volume change for Q R 3 2 3 2.10 10 1125 10 2.79 10 1125 102 116.44 116 J 3sf in on on in loss in U Q W W Q U pV QQ −−  = + =→ = =  =   −   =− =− = Max 2m for negative answers, but do not penalize if same type of mistake in 1. C1 C1 A1 (ii) Since there is no heat loss/gained, and there is (positive) work done on the gas through the compression (decrease in volume), the internal energy will increase. Hence, the temperature will increase. M1 A1

3 ©YIJC 9749/02/YIJC/23 (iii) At point P: ( )( ) ( )( ) ( ) 562.79 10 950 10 8.31 350 0.09113 0.091 mol 2sf pV nRT n n − =   = == OR At point Q: ( )( ) ( )( ) ( ) 562.79 10 1125 10 8.31 414 0.09124 0.091 mol 2sf pV nRT n n − =   = == ( ) ( )( ) ( ) 0.091 20.8 414 350 121 J 3sf Q nC T= =− = OR ( )( )( ) ( ) 3 2 3 0.091 8.31 414 350 48.82 121 J 3sf on in in in U nR T W Q Q Q  =  = + − =− + = B1 (B1) C1 A1 (C1) (A1) 3 (a) The graph shows a linear line passing through the origin. This shows that acceleration is proportional to displacement. The graph shows a negative gradient. This means that acceleration and displacement is always in opposite direction. B1 B1 (b) 2ax =− Hence, using one of the data point (−0.04,0.32), ( )( ) 2 2 2 2 0.32 ( 0.04) 8.0 8.0 1.24 9.92 m s k kLL k    − =− − = =  = = = C1 C1 A1 (c) Increasing the length decreases the angular frequency. Since total energy of oscillation = 22 0 1 2 mx , and total energy does not change, the amplitude is inversely proportional to angular frequency2. increasing the length will increase the amplitude of oscillation. M1 A1 4 (a) Coulomb’s law states that the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of their separation. B1

4 ©YIJC 9749/02/YIJC/23 (b) (i) Electric force, FE = 9 12 2 96 10 4 8.85 10 0.080 YQ  − −      ( ) 1 22 0.080angle to the vertical, sin 3.823 1.2 tan (0.013)(9.81)tan(3.823 ) 0.080 OR (0.013)(9.81) 1.2 0.080 E o F mg   − = =   = = =  − 9 12 2 96 10 (0.013)(9.81)tan(3.823 )4 8.85 10 0.080 63 nC (shown) oY Y Q Q  − −  =    = M1 M1 A0 (ii) 99 12 12 4 96 10 63 10 4 8.85 10 0.040 4 8.85 10 0.040 21580 14162 3.57 10 V net  −− −− =+         =+ = C1 A1 (c) (i) ( ) ( )( ) ( ) ( ) 41 19 4 31 15 2 250 1.389 10 V m0.018 1.6 10 1.389 10 9.11 10 2.4 10 m s 2sf VE d F qEa mm − − − − = = =   = = =  = M1 M1 A0 (ii) time taken to travel through the parallel plates = length/ speed = ( ) 8 7 20 cm 1.0 10 s2.0 10 −= ( )( ) ( ) ( ) ( ) 15 8 71 2277 71 final vertical speed 0 2.4 10 1.0 10 2.4 10 m s speed 2.0 10 2.4 10 3.1 10 m s yy Fu a t t m − − − = + = + =   = =  +  = C1 C1 A1 (iii) The polarity of the charge is different, so the force will be in the opposite direction to that of the electron. Hence, the proton will be deflected down instead of up. The mass of the proton is heavier than that of the electron, so the acceleration is lower. Hence, the deflection will be less than that of the electron. B1 B1 5 (a) (i) Resistance is infinite / very high B1 (ii) Resistance decreases as V increases (Ratio of V/I increases) B1 (b) (i) R = L/A  = (18  (0.15 10−3)2/0.94 = 1.4  10−6  m B1 Weight Electric Force

5 ©YIJC 9749/02/YIJC/23 (ii) p.d. across wire = 18 (6.2)18 2.0+ = 5.58 V Since S is at the mid-point, pd across half the length of the wire = ½ (5.58) Voltmeter reading = 2.8 V C1 A1 (iii) Current in the battery: increase Voltmeter reading: decrease B1 B1 (c) (i) I = Anvq q = 0.93/[((0.15 10−3)2)  (9.0 1028)  (1.3 10−3)] = 1.1  10−19 C A1 (ii) The charge carriers in a metal wire are the electrons which has a charge of 1.6  10−19 C. Charge q is less than 1.6  10−19 C the elementary charge. So the value must be wrong B1 6 (a) (i) Since the velocity is always right angle to the magnetic field, by Fleming’s left hand rule, the magnetic force always be right angle to its velocity. The force only changes the direction of the motion and not its speed. This results in circular motion. B1 B1 (ii) Out of page B1 (iii) The magnetic force provides the centripetal force 2 27 5 19 (20 1.66 10 )(5.6 10 ) (1.6 10 )(0.051) 2.3 T mvBqv r mvB qr B − − =   ==  = B1 C1 A1 (b) (i) Upwards B1 (ii) 5 61 (2.3)(5.6 10 ) 1.3 10 V m qE Bqv E Bv − = = =  = C1 A1 7 (a) (i) ( ) ( ) ( ) ( )50 4.5 2.8 50 1.8 1.4 75 g x x Y Y x x Y Ym u m u m v m v MM M + = + + − = − + = M1 A1 (ii) Total kinetic energy of the system before and after collision is the same. B1

6 ©YIJC 9749/02/YIJC/23 (iii) Using KE: ( ) ( ) ( ) ( ) 2222 2222 1 1 1 0.050 4.5 0.075 2.8 0.80 J2 2 2 1 1 1 0.050 1.8 0.075 1.4 0.15 J2 2 2 x x x x x x x x m u m u m v m v + = + − =  + = − + =  Since final KE < initial KE, inelastic collision OR Using relative speed ( ) ( ) -1 -