DHS 2023 JC1 Promos Physics (Solutions)

DUNMAN HIGH SCHOOL 2023 PHYSICS H2 (YEAR 5) 1 Answers to Promotional Exam Section A 1 2 3 4 5 6 7 8 9 10 A B C B C C D D A D 11 12 13 14 15 A D B C B 1 A Energy produced = Pt = 3.0 x 109 x 2.0 x 10-12 = 6.0 x 10-3 J = 6.0 x 10-15 TJ 2 B Recall: v = vf - vi The horizontal component of vi = vf since there’s no horizontal acceleration, so v is vertical. v = 12 sin 25o = 5.1 m s-1 3 C Using: 21s 2ut at=+ On Moon: ( ) 2 moon 131 2 at= −−− On Earth: ( ) ( ) 2 moon Earth 12 6 22 at= −−− 25o vf −vi = 12 m s−1 v Solving (1) and (2), 2 moon Earth 2 moon Earth 62 3 3 at a t tt =  =

DUNMAN HIGH SCHOOL 2023 PHYSICS H2 (YEAR 5) 2 4 B Before time instant B, the cars move towards each other. After time instant B, the cars move away from each other. 5 C The acceleration of the system can be found by mtotala = 12. ∴ a = 2.0 m s−1 Considering the forces acting on the 1.0 kg mass, 2 on 1 1 2 on 1 12 12 1.0(2.0) 10 N N m a N −= =− = Considering the forces acting on the 3.0 kg mass, 2 on 3 3 3.0(2.0) 6.0 N N m a= = = 6 C At equilibrium, the 3 forces on the object formed a closed triangle. By Pythagoras theorem, √F1 2 +F2 2 = 100. That would rule out A and B. Since the net force on the object is zero, then horizontally, F1 cos a = F2 cos b Since b > a, cos b < cos a, so F2 > F1 1.0 kg 2.0 kg 3.0 kg 12 N F1 F2 W a b N2 on 1 N2 on 3

DUNMAN HIGH SCHOOL 2023 PHYSICS H2 (YEAR 5) 3 7 D Fully submerged: Ufull = Vdisp waterwaterg = Vicewaterg Floating (at equilibrium) Ufloat = Wice = miceg = Viceiceg Hence upthrust when fully submerged upthrust when floating = water ice = 1.1 8 D D = kv2 where k is a constant of proportionality At 12 m s-1, the driving force provided by the 2 engines is equal and opposite to the total drag. 3 3 36000 2 (12) 720000 (12) P Dv k k = = = If only 1 engine is on, the new maximum speed v’ can be found by: 3 3 33 3 1 720000360000 ( ') (12) 360000' (12 )720000 12 2 9.5 m s v v − = = = = 9 A Since they are rotating together with the horizontal turntable without slipping, the angular velocity of the two coins is the same as that of the horizontal turntable, 12 1 2 1    = = 11 22 1 2 vr vr r r  = =

DUNMAN HIGH SCHOOL 2023 PHYSICS H2 (YEAR 5) 4 10 D Consider forces acting on man of mass m when in orbit, W – N = mv2 r (1) Since the space station and man (total mass M) is in orbit around the Earth, GMm r 2 = Mv2 r (2) or Gm r 2 = v2 r i.e. g = ac (3) Sub (3) into (1) gives N = 0 i.e. weight provides the centripetal force 11 A gravitational force F = GMm r 2 = (mr2 = mv2 r ) (1) gravitational potential energy U = −GMm r As radius of orbit r decreases, F increases, U decreases (as it is more negative). That would rule out C and D. Since F increases, from equation (1),  increases. That would rule out B. Also from equation (1), linear speed v increases. 12 D Use x = −xo cos  t = −10 cos 2 (1.25) 2 = 7.1 cm Hence the particle is displaced 7.1 cm to the right from the equilibrium position (60 cm). weight W normal N Earth Earth satellite initial orbit final orbit r 50 cm 70 cm 60 cm t = 0 t = 1.0 s

DUNMAN HIGH SCHOOL 2023 PHYSICS H2 (YEAR 5) 5 13 B EK = 1 2mv2 = 1 2mvo2(sin2t) From the graph, period T of motion = 2.0 s and maximum EK = 88 J So 1 2(3.0)vo2 = 88 or vo = xo = 7.66 m s−1 Since  = 2 T = , then xo = 2.44 m Maximum acceleration = 2xo = ( 2 2 )2 (2.44) = 24 m s−2 14 C Option D is incorrect since the current is the same in both sections. Since l ll =  =  RRR A A A 1 (ρ is constant for same material), resistance per unit length of the narrow section is twice that of wide section since the constant current flows through the narrower section. Since VR for constant l, the resistor with the larger resistance has a larger p.d. across it. Hence, the narrower section (with larger resistance per unit length), has larger p.d. per unit length across it. Hence options A and B are wrong. 15 B The decrease in temperature will cause an increase in the resistance of the thermistor while the increase in light intensity will cause a drop in the resistance of the LDR. Since V1 is independent of the change in resistance of the thermistor, most of the current must have passed through a real diode for a short time duration instead of the LDR or a fixed resistor.

DUNMAN HIGH SCHOOL 2023 PHYSICS H2 (YEAR 5) 6 Section B 1 (a) Speed is a physical quantity and can only be defined in terms of other physical quantities. B1 Distance is a physical quantity, but second is a unit for time which cannot be used to define speed. B1 (b) Time to reach maximum height, sinut g = M1 Time of flight, 2 sin2 uTt g == A1 2 ( cos )( ) 2 sin ( cos ) 2 sin cos R u T uu g u g    = =  = M1 (c) Maximum R occurs when sin2 1 = . Hence 45o = A1 (d) 2 0 uR g= = -145.36 km 45.36 (1000) m 12.6 m sh (60)(60) s== C1 2 0 2 2 (12.6) 16.3 9.7399 m s ug R − = = = 0 0 2 Rgu g u R  =+ 2 2(3%) 4% 9.7399 100% 0.97 1 m s g g − + = = = Therefore, ( )10 1g = m s−2 A1 A0 A1 A1

DUNMAN HIGH SCHOOL 2023 PHYSICS H2 (YEAR 5) 7 2 (a) (i) 80 km h−1 A1 (ii) 1.25 hours (between 09.00 and 10.15) A1 (b) (i) Between 10.15 and 10.30, the lorry accelerates steeply from rest but appears to hit a speed limit of 80 km h−1 beyond which it is not able to exceed. A1 (ii) Between 12.45 and 13.00, the lorry keeps speeding up and slowing down very frequently. The average speed is extremely low, around 20 km h−1. A1 (c) (d) Average speed between 12.00 and 12.45 is 40 km h −1, hence the distance travelled within this time of 45 minutes is (40 km h−1)(0.75 h) = 30 km. Average speed between 12.45 and 13.00 is 20 km h −1, hence the distance travelled within this time of 15 minutes is (20 km h−1)(0.25 h) = 5 km. M1 The total distance travelled between 12.00 and 13.00 is 30 km + 5 km = 35 km. A1 (e) (i) 14.15 is the time the tachograph reading shoots to the maximum of 100 km h−1. A1 (ii) This is to distinguish the switched-off period from periods when the lorry is not moving (e.g. at red traffic lights). B1 There is no way the lorry could have reached a maximum speed of 100 km h −1 since it is limited to 80 km h−1. Hence a recorded maximum speed of 100 km h−1 can only mean that the tachograph has been switched off. A1 - for three correct traces A2 - for five correct traces

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