ASRJC 2022 Prelim P3 Guide

1 9749/03/ASRJC/2022PRELIM [Turn Over Anderson Serangoon Junior College 2022 JC2 H2 Physics Prelim Mark Scheme Paper 3 1a The rate of change of momentum of a body is proportional to the resultant force acting on the body and takes place in the direction of the resultant force. B1 B1 1bi The acceleration of object X is dependent on the resultant of the two forces, weight and tension, acting on object X. Since the weight and tension are in opposite directions, the resultant force must be less than its weight and hence, has an acceleration less than g. B1 1bii Let T be the tension in the rope. Considering the FBD of object X, by N2L, ↓+: mXg – T = mXa T = mXg – mXa ………. (1) Considering the FBD of object Y, by N2L, ↑+: T – mYg = mYa T = mYg + mYa ………. (2) Equating (1) and (2), we have mXg – mXa = mYg + mYa g (mX – mY) = a (mX + mY) Hence, XY XY mmag mm −= + M1 M1 A1 1biii Since the motion is linear with uniform acceleration, using s = ut + ½ at2 ↓+: 21 6.0 3.05.0 0 [ (9.81)]2 6.0 3.0 t−=+ + t = 1.749 s = 1.7 s C1 A1 2a The object is in equilibrium and hence resultant force is zero. Before lowering into water, sF mg= After lowering into water ' ' ' ( ' ) s ss ss F Up mg F Up F Up F F k x x += += = − = − Hence, upthrust = decrease in spring force = k∆x [Note: cannot just be substituting numbers, but must indicate it is representing ∆x] = 42 (0.0045) = 0.189 N = 0.19 N (to 2 s.f.) M1 A1 2b Upthrust = Vρg V(1000)(9.81) = 0.19 V = 1.94 x 10—5 m3 M1 A1

2 9749/03/ASRJC/2022PRELIM 2c By Newton’s 3rd Law, force of water on mass = - force of mass on water. Since force of water on mass is the upthrust, magnitude of force of mass on water is therefore same magnitude as upthrust. Hence magnitude of force of mass on water = 0.19 N M1 A1 2d When fully inside water, 1 () sF V g mg ke V g mg e mg V gk    += += =− So, first part of graph while inside water is a horizontal flat line since e is independent of s. While it is moving out of water, 1 () sF V g mg ke V g mg e mg LA gk    += += =− L is the submerged height of the mass and thus varies linearly with s. So, the second part of the graph is a linear line with increasing e. When the mass is fully out of water, the spring force is thus equal to weight and there is no upthrust, so the graph is a horizontal flat line independent of s. B1 B1 B1 3ai The gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point. B1 B1 3aii Gravitational potential is always negative because the gravitational potential is taken to be zero at infinity B1 e s 0 s1 s2

3 9749/03/ASRJC/2022PRELIM [Turn Over and gravitational forces are attractive, work done by the external agent on the point mass moving it from infinity is negative. OR (test) mass getting closer (from infinity) loses potential energy. B1 B1 3b one curve with negative field strength near planet with gradient of decreasing magnitude at 2R and finishing with positive field strength near the moon with gradient of increasing magnitude at D – R field strength shown as zero (only) near the point of maximum potential (within the 4 th big square from left, acceptable range of 4th to 8th small square). B1 B1 3ci At Earth’s surface, g = GM RE 2 ➔ gRE2 = GM where RE is radius and M is mass of Earth gravitational force on Moon, F = GMm r2 = gRE 2m r2 = 9.81(6.37×106) 2 (7.35×1022) (3.84×108) 2 = 1.9841 × 1020 = 1.98 × 1020 N M1 M1 A0 3cii Since the increase in orbit is significantly smaller than radius of orbit, the gravitational force remains constant. Hence, change in potential energy = F × 0.040 = 1.98 × 1020 × 0.040 = 7.9 × 1018 J do not accept alternative method of finding change in p.e. if not using answer in (c)(i) B1 C1 A1 4a thermal energy per unit mass required to produce unit rise of temperature of the substance. B1

4 9749/03/ASRJC/2022PRELIM 4bi p =  2 3 1 cV Nm Since pV = NkT, 21 3NkT Nm c=   231 22 kT mc=   M1 A1 4bii Internal energy is the sum of the kinetic energy (due to random motion) of the molecules and the potential energy (due to intermolecular forces) between the molecules. However, for an ideal gas, there is no potential energy due to absence of intermolecular forces between the molecules. B1 B1 4c ( ) 2 23 .. 27 31 13 22 3 1.38 10 25 273.15 3.34 10 1.9 10 r m s kTmc c m s − − −   =    +=  = C1 A1 4di At constant volume, no work is done by the gas. Thermal energy required to increase the temperature of unit mass of gas by one unit is solely to increase its internal energy. At constant pressure, work is done by the gas as it expands. Thermal energy is required for the increase of its internal energy and the work done by the gas. Hence the specific heat capacity of ideal gas at constant pressure is greater than the specific heat capacity at constant volume. B1 B1 4dii Small volume change/decrease and hence work done on ice is negligible/positive, Thermal energy is absorbed to break lattice structure / increase molecular potential energy. Hence, by first law of thermodynamics, internal energy increases. B1 B1 5a As current passes through the filament, mobile charge carriers (electrons) transfer kinetic energy to the lattice ions of the filament causing the filament to heat up gradually Filament produces light when it is at a sufficiently high temperature Heating process takes time and hence, full brightness is not immediately achieved. B1 B1 5b By potential divider principle, 16 16 1216 14 6.4 V V  = + = C1 A1 5ci 6.0one lamp VV = Potential difference across XY = 6.4 – 6.0 = 0.4 V A1

5 9749/03/ASRJC/2022PRELIM [Turn Over 5cii Point Y is at a higher potential. A1 5d Current through resistors = 12 14 16 0.40 A + = Current through lamps = 0.50 – 0.40 = 0.10 A M1 A0 5e total power dissipated by the lamps total power produced by battery lamps battery VI VI= 12 0.10 12 0.50 0.20 =  = C1 A1 5f No change to potential difference across lamps and hence (by 2V R ), power of lamps unchanged (Since effective resistance of circuit increases,) total current decreases and so power produced by battery decreases Ratio increases. M1 A1 6a 7 7 5.0 10 2.0 3.3 10 3.03 3.0 lR A  − − = =  = = M1 A1 6b Current through wire = 1.5 3.0 0.75 0.40 A + = I = nAqv v = 28 7 19 5 0.40 8.5 10 3.3 10 1.6 10 8.91 10 -1m s −− −      = C1 A1 6ci 3.0 1.53.0 0.75 1.2 STV V = + = E = VSP 1.4 1.22.0 0.84 V = = M1 A1

6 9749/03/ASRJC/2022PRELIM 6cii Second wire has higher resistance Potential difference (per unit length) across second wire increases (For the same E), balance length decreases. M1 M1 A1 7a Marking points: 1. Direction of field lines indicated correctly on every line 2. Overall pattern, resembling that of a solenoid, extending over the whole area of the board. 3. In region A, field is approximately uniform, with li