RVHS H2 Physics P3 Soln

River Valley High School Page 1 of 12 J2 H2 Physics 9749 Preliminary Examinations RVHS JC2 H2 Physics Preliminary Examinations Paper 3 Mark Scheme Questions Answers Marks 1 (a) Repeated readings with error of different magnitude and sign. OR Repeated readings with scatter of readings about a mean value. By plotting a graph and drawing a line of best fit for the points/can be reduced by averaging B1 B1 (b) Units of 𝑘𝑘𝑟𝑟3(𝑝𝑝1−𝑝𝑝2) 𝑙𝑙 � 𝑀𝑀 𝑅𝑅𝑅𝑅 = 𝑚𝑚3𝑃𝑃𝑃𝑃 𝑚𝑚 � 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚 𝑙𝑙−1 𝐽𝐽 𝐾𝐾−1 𝑚𝑚𝑚𝑚 𝑙𝑙−1𝐾𝐾 = 𝑘𝑘𝑘𝑘𝑘𝑘𝑠𝑠−2� 𝑘𝑘𝑘𝑘 𝑠𝑠2 𝑘𝑘𝑘𝑘𝑚𝑚2 = 𝑘𝑘𝑘𝑘𝑠𝑠−1 (c) (i) Density = m/V m = density x V = density x π(D 2 / 4 ) L = ( 8 . 9 6 ) x π ( 6 . 02 /4) L = 7093.46 g = 7.09346 kg ∆m/m = 2∆D/D + ∆L/L = (2(0.1)/(6.0) + (0.1/28.0) = 0.0369 ∆m = m(2∆D/D + ∆L/L) = (7093.46)(0.0369) = 0.262 kg 7.1 +/- 0.3 kg (Correct s.f.) M1 M1 A1 A1 (ii) Same measurements, but smaller absolute uncertainties, so percentage uncertainty is smaller B1 Questions Answers Marks 2 (a) apply equation of motion 𝑣𝑣 = 𝑢𝑢 + 𝑎𝑎𝑎𝑎 vertically, we get −5 = +5 − 9.81𝑎𝑎 for substitution M1 𝑎𝑎 = 1.019 s for intermediate value A1 (b) 20 1.019 = 19.62 m s-1 B1 Alternative, Using 𝑣𝑣𝑦𝑦 = 𝑢𝑢0 + 𝑎𝑎𝑎𝑎 and substituting 𝑎𝑎 = 𝑥𝑥 𝑣𝑣𝑥𝑥 , we get 𝒗𝒗𝒚𝒚 = 𝑢𝑢0 − 𝑘𝑘 1 𝑣𝑣𝑥𝑥 𝒙𝒙. So gradient = − 1 2 = − 9.81 𝑣𝑣𝑥𝑥

River Valley High School Page 2 of 12 J2 H2 Physics 9749 Preliminary Examinations (c) 𝑎𝑎𝑎𝑎𝑡𝑡 𝜃𝜃 = 5.0 19.62 M1 𝜃𝜃 = 14.3𝑚𝑚 A1 (d) x-intercept at (20,0) (because at maximum height 𝑣𝑣𝑦𝑦 = 0) y-intercept above 5 (because with θ < 45o to increase range , θ needs to increase, and v sinθ needs to increase.) Example: Extra note As θ increase, and 𝑣𝑣𝑥𝑥 =v cosθ decreases. So gradient increases. B1 3 (a) The moment of a force is equal to the product of the force and the perpendicular distance of the line of action of the force from the pivot. It is the turning effect of a force. B1 (b) (i) -15 -10 -5 0 5 10 15 0 2 4 6 8 10 12 14 16 18 20

River Valley High School Page 3 of 12 J2 H2 Physics 9749 Preliminary Examinations Take moments about the hinge, (400)( sin60) 2000( sin60) sin30( )2 3810 N L L TL T += = B1 C1 A1

River Valley High School Page 4 of 12 J2 H2 Physics 9749 Preliminary Examinations (ii) (c) The direction of all forces remain unchange since the load remains submerged in water at the same location. The upthrust acting on the load will reduce the tension in the rope. Since the horizontal component of the tension reduces, the force by wall on rod also reduces. This will result in a lower value of force by rod on wall. B1 (directi on) B1 (magnit ude) Questions Answers Marks total weight of rod and load ON rod tension of rope ON rod force by wall ON rod force by rod ON wall total weight of rod and load ON rod tension of rope ON rod force by wall ON rod force by rod ON wall water

River Valley High School Page 5 of 12 J2 H2 Physics 9749 Preliminary Examinations 4 (a) kinetic energy was converted to gravitational potential energy and work done against frictional force. B1 (b) 1 2 𝑘𝑘𝑣𝑣2 = 𝑘𝑘𝑘𝑘Δℎ + 𝑤𝑤𝑤𝑤𝑤𝑤𝑘𝑘 𝑑𝑑𝑤𝑤 𝑡𝑡𝑑𝑑 𝑎𝑎 𝑘𝑘 𝑎𝑎𝑎𝑎𝑡𝑡 𝑠𝑠𝑎𝑎 𝑓𝑓𝑤𝑤 𝑎𝑎𝑓𝑓𝑎𝑎 𝑎𝑎𝑤𝑤𝑡𝑡 M1 Δℎ = 150 sin 0.57𝑚𝑚 = 1.492 m C1 𝑤𝑤𝑤𝑤𝑤𝑤𝑘𝑘 𝑑𝑑𝑤𝑤 𝑡𝑡 𝑑𝑑 𝑎𝑎 𝑘𝑘 𝑎𝑎𝑎𝑎𝑡𝑡 𝑠𝑠𝑎𝑎 𝑓𝑓𝑤𝑤 𝑎𝑎𝑓𝑓𝑎𝑎 𝑎𝑎𝑤𝑤𝑡𝑡 = 102 2 𝑘𝑘 − 9 . 81 × 1.492𝑘𝑘 = 3 5.36𝑘𝑘 M1 proportion of energy lost due to friction = 35.36 102/2 = 0.71 A1

River Valley High School Page 6 of 12 J2 H2 Physics 9749 Preliminary Examinations 5 (a) Since the mass is not moving in a straight line, there must be a net force acting on the mass according to Newton’s first law. Although the speed remains the same, its direction changes. Since, the change in velocity acts towards the centre of circular motion, the acceleration as well as the net force also acts towards the centre of circular motion according to Newton’s second law. B1 B1 (b) cos sin 0 cos 40 61sin40 8.0(9.81) 0 51.26 N 51N N fW N N N θθ+ −= +− = = = M1 A1 (c) 𝑓𝑓 𝑓𝑓𝑤𝑤𝑠𝑠 𝜃𝜃 − 𝑁𝑁 𝑠𝑠𝑎𝑎𝑡𝑡 𝜃𝜃 = 𝑘𝑘𝑤𝑤𝜔𝜔2 61 𝑓𝑓𝑤𝑤𝑠𝑠 4 0 − 51 𝑠𝑠𝑎𝑎𝑡𝑡 4 0 = (8.0)(12) �2𝜋𝜋 𝑇𝑇 � 2 13.9465 = (8.0)(12) �2𝜋𝜋 𝑇𝑇 � 2 𝑇𝑇 = 16.5 𝑠𝑠 C1 A1 6 (a) A field of force (around a mass) is a region of space that can be mapped with lines of gravitational force (or with lines of gravitational potential) (Answer must be contextualised to “gravitational force” for the first point.) Gravitational field strength at a point is defined as the gravitational force per unit mass acting on a small mass placed at that point. B1 B1 (b) Gravitational force provides the centripetal force for circular motion to take place. By Newton’s 2 nd Law, F = ma GMm/r2 = mrω2 GM/r3 = (2π/T)2 T2 = 4π2 r3/ GM B1 B1 B1 (c) TG = 7.16, lg TG = 0.85 From graph, lg (rG) = 9.05, rG = 1.12 x 109 m~ 1 x 109 m (estimate) M1 A1 (d) TT = (16.2 / 24) days, lg TT = -0.17 B1 θ θ θ N f W

River Valley High School Page 7 of 12 J2 H2 Physics 9749 Preliminary Examinations From graph, lg (rG) = 8.35, rG = 2.24 x 108 m rG calculated is the distance of the Thebe to the centre of Jupiter. We can only decide on the accuracy of the statement if the radius of Jupiter is known. B1 (e) (i) W = m.∆V = m.( Φ final – Φ initial) = 53.2(− 42.5 − (− 47.6)) =271.32 MJ = 271.32×106 J ~ 2.7(1)×108 J M1 A1 (ii) Φ = −GM/r Φ.r = constant 42.5(9.38) = 398.65 ~ 400 47.6(8.38) = 398.89 ~ 400 54.1(7.38) = 399.26 ~ 400 Need to show constant value, to 1 s.f., for all 3 orbits (shown) M1 7 (a) When there is no current through the 12 V battery, the potential difference across the 12 V battery and the light bulbs must be equal to 12 V. If the p.d. across each light bulb is 12 V, the current through each bulb is 12/3.0 = 4.0 A Current through resistor R = 4.0 + 4.0 = 8.0 A p.d. across R = 14.0 – 12.0 = 2.0 V resistance of R = 2.0 / 8.0 = 0.25 Ω B1 C1 A0 (b) Since the potential difference across R1 remains at 12 V, same amount of current will flow through R1 and no current will flow through R2. Hence, the current through the 14 V generator and resistor R will reduce to 4.0 A Since the p.d. across R remains the same at 2.0 V, the value of R should be adjusted to a lower value (new R = 2.0 / 4.0 = 0.50 Ω) B1 B1 B1 (c) There will be emf across the lamps even if either the generator or the battery is defective. The power from the generator can be used to charge the car’s battery when the light bulbs are removed. The light bulbs will light up longer since energy is delivered from both generator and battery. B1 B1 8 (a) (i) P = 236, Q = 92 R = 143 B1 B1 (ii) E = (∆m) c2 = [(235.1 + 1.009) – (148.0 + 84.9 + 3 x1.009)] u c2 = (0.182) (1.66 x 10–27) (3.00 x 108)2 = 2.719 x 10–11 J for only one uranium nucleus M1 A1 (b) (i) It is not triggered or affected by external conditions (temperature, pressure, et cetera) B1

River Valley High School Page 8 of 12 J2 H2 Physics 9749 Preliminary Examinations It is impossible to predict which nucleus will decay at a particular instant or when a particular nucleus will decay B1 (ii) 𝑆𝑆𝑤𝑤38 90 → 𝑌𝑌39 90 + 𝛽𝛽−1 0 Both correct yittrium and beta numbers B1 (iii) 𝑁𝑁 = 𝑁𝑁0𝑑𝑑−𝜆𝜆𝜆𝜆 where λ= 𝑙𝑙𝑙𝑙2 𝜆𝜆1/2 (equations provided in data sheet) = (2.36 × 1013)𝑑𝑑−�1𝑙𝑙2 2 8 �(112) = (2.36 × 1013)(0.0625) = 1.48 × 1012 Alternative Half life is 28 years so 112 years is 4 half lives Number present after this time is 1/16 of original Number present = 2.36x1013/16 = 1.47(5) x 1012 M0 M1 A1 Questions An