RVHS 2022 Prelim P3 Guide

River Valley High School Page 1 of 8 J2 H2 Physics 9749 Preliminary Examinations 2020 RVHS JC2 H2 Physics Preliminary Examinations Paper 3 Mark Scheme 1 (a) Distance travelled before hitting the ground = Area under graph from 0 to 0.6 s. = ½ (4 + 10) * 0.60 = 4.2 m M1 A1 (b) Distance that the ball bounce upwards = ½ (1.0 * 10.0) = 5.0 m Height above original = 0.8 m M1 A1 (c) Acceleration = 10/1.0 = 10 m s-2 A1 (d) Time taken to travel upwards = time taken to travel downwards Next time the ball will reach ground = 2.6 s A1 2 (a) (i) conservation of linear momentum gives 10𝑚 + 2(5𝑚) = 𝑚𝑣 + 4.5(5𝑚) C1 𝑣 = −2.5 m s-1 A1 (ii) kinetic energy before collisioin 1 2 𝑚102 + 1 2 (5𝑚)22 = 60𝑚 B1 kinetic energy after collision 1 2 𝑚(−2.5)2 + 1 2 (5𝑚)4.52 = 53.75𝑚 B1 percentage loss = 60−53.75 60 × 100% = 10% B1 (b) conservation of momentum gives 10𝑚 + 2(5𝑚) = 𝑚𝑣1 + (5𝑚)𝑣2 -------(1) B1 for elastic collsion, relative speed of approach = relative speed of separation. So 10 − 2 = 𝑣2 − 𝑣1-------- (2) Or conservation of kinetic energy gives 1 2 𝑚22 + 1 2 (5𝑚)2.02 = 1 2 𝑚𝑣1 2 + 1 2 (5𝑚)𝑣2 2 -------(2) B1 solving simultaneous (1) and (2) M1 𝑣1 = −3.3 m s-1 , 𝑣2 = +4.7 m s-1 A1 3 (a) loss in gravitational potential energy = gain in kinetic energy + gain in elastic potential energy B1 (b) (i) 4.0 × 9.81 × 0.090 = 1 2 × 200 × 0.092 + 𝐸𝑘 B1 𝐸𝑘 = 2.7 J B1 (ii) Since both blocks have same acceleration and hence same velocity, 𝐸𝑘 ∝ 𝑚 So taking ratio, 𝐸𝑘,𝑄 𝐸𝑘 = 4 6 or other appropriate method M1 𝐸𝑘 = 1.8 J A1 (c) When blocks come to rest 𝐸𝑘 = 0. Let distance required by 𝑥, then conservation of energy gives 4 × 9.81 × 𝑥 = 1 2 × 200 × 𝑥2 M1 𝑥 = 39 cm A1

River Valley High School Page 2 of 8 J2 H2 Physics 9749 Preliminary Examinations 2020 4 (a) 1 mark for correct direction of arrows 1 mark for similar length of arrows in vertical direction B1 B1 (b) Since frictional force provides for the centripetal force, 2 2 -1 (80)70 55 6.9 m s mvf R v v = = = C1 A1 (c) (i) For a surface which is banked, the horizontal component of the normal reaction is an additional source of the centripetal force. Since Centripetal Force = r mv 2 , an increase in the centripetal force would allow the rider to turn the corner at a higher speed without slipping, provide the mass of cyclist and bicycle and radius of turn remain constant. B1 B1 (ii) Considering forces acting on the rider/bicycle in the horizontal direction: 2 sin 20 cos 20 mv Nf r  +  = ---------------------- (1) For the equilibrium in the vertical direction: += 20sinfmg20cosN --------------------(2) Solving (1) & (2), ( ) 2 tan 20 sin 20 cos 20 mv mg f f r  +  = −  →max velocity during turning v = 15.7 m s-1 C1 for bot h C1 A1 reaction force weight normal contact force friction weight

River Valley High School Page 3 of 8 J2 H2 Physics 9749 Preliminary Examinations 2020 5 (a) (i) A = 3.00 ω = 2π / T = 2π / 0.785 = 8.00 x = Acos( B)t − 2.00 = 3.00cos(8.00(0) B) 0.841B − = B1 B1 B1 (ii) Maximum velocity = 0x = (8.00)(3.00) = 24.0 Maximum acceleration = 2 0x = (8.00)(3.00)2 = 192 A1 A1 (iii) ½ mvmax2 = ½ mω2x02 = ½ (5.0 × 10-6)(8.002)(3.002) =1.44 × 10−3 J 1 for correct maximum kinetic energy 1 for correct shape (2 periods with peak slightly to the left of origin) B1 B1 (b) An oscillator with damping can vibrate at resonance with amplitude that remains constant with time. This is because the energy supplied by driver is equal to energy lost due to dissipative forces. M1 A1 6 (a) When a p.d. of 240 V is applied across it, then the power dissipated by the lamp is 100 W. B1 (b) Resistance of lamp 2 2240 576100 VR P= = =  Current flowing through the filament , I = 280 0.486576 V R == A In one second, quantity of charge flowing through it, Q = 0.500 C 18 -19 0.486= = =3.04×101.6×10 nQ te s-1 C1 A1 (c) The temperature (or internal energy) of the filament rises with work done by the electrons on the filament. This causes more vigorous lattice vibration of the atoms resulting in greater collision frequency between electrons and atoms, causing the resistance to rise. To include explanation that just involve collisions of electrons wi th lattice ions. B1 B1 7 (a) 𝐴 = 𝐴0𝑒−𝜆𝑡 3.7 × 1011 = 𝐴0 𝑒−ln 2 5.2 (2.5) 𝐴0 = 5.16 × 1011 M1 A1

River Valley High School Page 4 of 8 J2 H2 Physics 9749 Preliminary Examinations 2020 (b) 𝐴0 = 𝜆𝑁0 𝑁0 = 5.16 × 1011 5.2 × 365 × 24 × 60 × 60 = 1.2216 × 1020 mass = 1.2216 × 1020 6.02 × 1023 × 60 = 0.012 𝑔 C1 C1 A1 (c) Rate of emission of energy = (3.7 × 1011 × (0.31 + 1.17 + 1.33)(1.60 × 10−19) = 0.166 W B1 B1 8 (a) (i) potential divider method gives 𝑉 = 9.0 ( 470 470+1650) C1 𝑉 = 2.00 V A1 (ii) downward sloping with decreasing gradient B1 cuts at (0,9) and graph below 2 V (or the value in (a)(i)) when curve reaches 𝑅1 = 2000 Ω B1 Example Justification 𝑉 = 9 ( 470 470+𝑅) (b) (i) 1 𝑅∥ = 1 470 + 1 100 000, 𝑅∥ = 467.8 Ω B1 𝑉 = 9 ( 467.8 467.9+1650) = 1.988 B1 voltmeter reading / V R1 /

River Valley High School Page 5 of 8 J2 H2 Physics 9749 Preliminary Examinations 2020 1.988−1.955 1.955 × 100% = −0.35% A1 (ii) No significant changes / very slight differences / voltmeter may not have enough precision to detect the difference. B1 comparison using numbers. E.g. internal resistance is 3 orders of magnitude smaller than resistance in circuit. Or, take internal resistance to be maximum at 10 Ω, fraction ≈ 10 1645+640 = 0.004. internal resistance is small compared to resistance in circuit (1 mark maximum) B1 9 (a) (i) The electric field strength at a point is defined as the electric force per unit positive charge placed at that point: B1 (ii) Electron gains speed as it moves from point A of 200V to 300V. Its speed remains approximately constant when it is between the region of 300V. Its speed decreases (to the same initial speed) as it moves from 300V to point B B2 for all 3 (iii) Loss in kinetic energy = gain in electric potential energy = q (VC – VA) = (–1.60 x 10-19) (0 – 200) = 3.20 x 10-17 J To deduct one mark if answer does not reflect negative value which represents the loss in KE. C1 A1 (iv) Distance between 300V and 100V beside B is 2.7×10-2 m Magnitude of field strength = |∆V/∆x| = (300 – 100)/(2.7×10-2) = 7400 V m-1 (7100 – 7700) C1 A1 (v) Vertically downwards B1 (vi) -19 -15 =(1.60×10 )(7400) =1.18×10 N F qE= Vertically upwards. A1 A1 (b) (i) The gravitational force acting on the object provides the centripetal force. GMm 2mR2R = B1

River Valley High School Page 6 of 8 J2 H2 Physics 9749 Preliminary Examinations 2020 3 GM R = (ii) A longer arrow acts towards the Sun. A shorter arrow acts towards the Earth B1 (iii) The centripetal force acting on the SOHO depends on both the gravitational forces due to the Sun and the Earth. Hence, the expression in (i) does not apply to the angular velocity for SOHO. Due to the gravitational force acting on SOHO by the Earth, the net centripetal force acting on SOHO is lower. This will result in a lower angular velocity for SOHO that can be comparable to that of the Earth at a further distance from the Sun. B1 B1 (iv) 2 2 Earth Earth Earth Earth GM mGMm 2mR( )22RR GMGM 2R( )22RR T T   −= −= where m is the mass of SOHO, MEarth is the mass of the Eart