RVHS H2 Physics P2 Soln

River Valley High School Page 1 of 8 J2 H2 Physics 9749 Preliminary Examinations 2022 RVHS JC2 H2 Physics Preliminary Examinations Paper 2 Mark Scheme 1 (a) error that does not have a fixed magnitude or direction or error that scattered around the true value. or equivalent statement B1 (b) (i) 13 cos 35𝑜 = 10.6 or 11 m s-1 B1 (ii) 𝑣𝑚𝑎𝑥 = 13 × 1.1 cos 34𝑜 = 11.9 m s-1 or 𝑣𝑚𝑖𝑛 = 13 × 0.9 cos 36𝑜 = 9.47 m s-1 B1 uncertainty = 11.9 − 10.6 ≈ ± 1 m s-1 or 10.6 − 9.47 or 11.9−9.47 2 B1 (11 ± 1) m s-1 uncertainty given to 1 sf and value given to the same precision as uncertainty B1 2 (a) No net force No net torque/moment about any point B1 B1 (b) (i) Downwards weight from centre of block Force by ball at upper left hand corner of block, follows direction of velocity of ball. Upwards normal reaction force at A (equal length to weight) Frictional force acting at A to the left B1 for bot h B1 (ii) Taking pivot about A, clockwise moments by force from ball = anticlockwise moments by weight. 9.0 × F = 300/1000 × 9.81 × 5.0 F = 1.64 N C1 M1 A1 3 (a) (i) When the gas is heated, the molecules gains kinetic energy (or speed), leading to a larger change in momentum per collision with the container per unit time. This leads to a larger force on the walls of container and piston. This leads to an increase in pressure in the container. To keep the pressure constant, the piston moves up that leads to an increase in volume of gas until the rate of change of momentum of gas molecules with the walls of the container decreases to the original level. B1 B1

River Valley High School Page 2 of 8 J2 H2 Physics 9749 Preliminary Examinations 2022 (ii) 2 3 2 0.60 28 273.15 57 273.15 0.66 m V V =++ = 1 mark for correct graph 1 mark for correct showing correct values on both axes B2 (iii) Energy from heater = (24)(15)(60) = 21600 J Using U Q W = + , Q = − 7000 + 21600 = 14600 J W = 101000(0.66 – 0.60) = 6060 J U = 14600 − 6060 = 8540 J Accept slight variation in values if using raw values from (ii) Max of 1 mark if students using alternative methods B1 C1 A1 (b) (i) Using 21 3pV Nm c= and pV NkT= , 221 32N m c NkT = 213 22E m c kT== B1 (ii) Using 213 22m c kT= , A Rk N= and AM N m= 2 3 -1 3 3 3 3(8.31)(57 273.15) 1.43 10 m s4.0 10 rms kT RTcc mM − += = = = =   C1 A1 pressure / kPa volume /m3 0.60 0.66 101

River Valley High School Page 3 of 8 J2 H2 Physics 9749 Preliminary Examinations 2022 4 (a) A progressive wave is a wave in which energy is carried from one point to another by means of vibrations or oscillations within the wave. Particles within the wave are not transported along the wave. B1 (b) (i) λ = 0.40 m since distance between P and Q is 1.5 wavelength. v = f λ = 0.40 × 850 = 340 m s-1 C1 A1 (ii) Intensity of sound at man 2 -2 1500 4 (80.0) 0.01865 W m PI A == = Power intercepted by man = I × A = 0.01865 × 2.1 × 10-3 = 3.9 × 10-5 W C1 C1 A1 5 (a) Destructive interference is when two waves arrive at the same point anti-phase (phase difference of π / 3π etc rad) they superpose to produce a resultant wave with minimum (or zero) amplitude or a minimum is produced. B1 B1 (b) (i) f v= 330 0.1851780 v mf = = = A1 (ii) 22 1S D 12 4=+ = 12.649 m Path difference = 12.649 – 12 0.649 0.18539= 3.5= C1 A1 (iii) The path difference of 3.5λ would lead to a phase difference of π rad. Since sound from both speakers have a phase difference of π rad, this would mean that there is no net phase difference at D. Constructive interference of sound waves will take place and loud sound will be heard at D. B1 B1 (c) (i) d sinθ = nλ Since sin θ cannot be greater than 1 For max order, sin θ < 1 n λ / d <1 n < 2.7 Since n must be an integer, and cannot exceed 2.7, the max order =2 C1 A1

River Valley High School Page 4 of 8 J2 H2 Physics 9749 Preliminary Examinations 2022 (ii) 1 mark for correct shape grid of spots 5x5 1 mark for 2nd order dots are further apart than 1st order dots B1 B1 (iii) The new light source has shorter wavelength. Using d sinθ = nλ, the diffracted angle for each wave will be smaller. Hence, more bright spots can be seen on the screen and the closer spacing between spots. B1

River Valley High School Page 5 of 8 J2 H2 Physics 9749 Preliminary Examinations 2022 6 (a) (i) 100𝜋 = 2𝜋𝑓, so frequency 𝑓 = 50 Hz. So period T= 1/50 = 0.020 s B1 (ii) 1 mark for shape 1 mark for labelling of period B1 B1 (b) (i) mean power = 𝑣𝑟𝑚𝑠2 640 = 40 B1 For half-rectified voltage, mean square voltage = 𝑉02 4 C1 Hence 𝑉02 4 = 640 × 40, giving 𝑉0 = 320 V A1 (ii) 1 mark for shape B1 1 mark for labelling of peak power at 160 W B1 0.02 0.04 t / s

River Valley High School Page 6 of 8 J2 H2 Physics 9749 Preliminary Examinations 2022 7 (a) (i) photons and electrons undergo a one-to-one interaction. B1 (ii) photons could have interacted with electrons below surface; energy is used to take electron to the surface. or some electrons are more tightly bound to nuclei than others B1 (iii) 1. 𝜙 = −𝐸𝑚𝑎𝑥 when 1 𝜆 = 0 (use of the x-intercept ) Allow range (3.8 ± 0.2) × 10−19 J (allow use of substitution) B1 2. Use of gradient = 1 ℎ𝑐 C1 Allow range of gradient (4.8 − 5.2 ) × 1024 Allow range of h (6.4 − 6.9) × 10−34 J s B1 alternative (allow use of substitution of points) Allow range of h (6.4 − 6.9) × 10−34 J s C1 A1 (iv) increase intensity only increases the rate / number per unit time of the emitted electrons B1 Emax depends on frequency of photons and work function of metal only. Hence, no change. B1

River Valley High School Page 7 of 8 J2 H2 Physics 9749 Preliminary Examinations 2022 (b) (i) energy difference Δ𝐸 = ℎ𝑐 𝜆 = 6.63×10−34×3.0×108 663×10−19 = 3.0 × 10−19 M1 So energy levels associated with the transition is between n = 2 and n = 3 B1 Since dark line is observed, n = 2 to n = 3 A1 (ii) 21.8 × 1 = 21.8 5.4 × 22 = 21.6 2.4 × 32 = 21.6 1 mark for working to find constant 1 mark for comparing the any pair of values B2 8 (a) (i) Similarity: Both nuclear reactions produce energy or Both produces products that are more stable/higher binding energy per nucleon. B1 Difference: Nuclear fusion combines two lighter nuclei into a heavier nuclei while nuclear fission splits a heavier nuclei into two lighter nuclei of similar sizes. Nuclear fission produces radioactive waste while nuclear fusion does not. B1 (ii) The process is able to use the energy produced to keep itself going without any external help/guidance. B1 (iii) It does not produce any radioactive waste / produces very little short- lived radioactive wastes. It produces more energy per unit mass compared to nuclear fission. Nuclear fusion reactors will not melt down compared to nuclear fission reactors. The raw ingredients required for fusion is more readily available compared than that for fission. B1 (b) (i) High amount of kinetic energy required for the positively charged nuclei to overcome the Coulombic/electrostatic repulsion B1 B1 (ii) To increase the probability of collision between the nuclei B1 (c) (i) 4 mH – (mHe + 2 me) = 4(1.007825) – (4.002604 + 2(0.000549))u = 4.58(1) × 10-29 kg. M1 A1 (ii) E = Δmc2 = (4.581 × 10-29)(3.00 × 108)