2025 DHS H2 Prelim P3 Ans w MC

2025 DHS Prelim Paper 3 Suggested Solutions 1 (a) Gravitational field strength at a point is defined as the gravitational force exerted per unit mass placed at that point. B1 (b) Since gravitational field strength g is the gravitational force F exerted per unit mass, Fg m= and the gravitational force exerted by the point mass on another body with mass m is 2 GMmF r= 2 2 GMm F GM rg mm r  = = = Note: Do not accept answers with just working but no explanations. M1 M1 A0 Marker’s Comment: Credit requires explanation of the physics steps. Candidates must state Newton's law, reference the field strength definition from part (a), and explain the substitution process. Simply writing equations and algebra without explaining the reasoning shows m athematical manipulation but not physics understanding. (c) (at surface,) lines (of force) are radial Earth has large radius / height above surface is small, so lines are (approximately) parallel and equally spaced B1 B1 B1 Marker’s Comment: Candidates must answer the question as asked. The requirement is to explain using lines of gravitational force (field lines), not mathematical expressions. Credit requires describing how field lines near the surface appear parallel and uniformly spaced, sh owing constant field strength. Using g = GM/r² arguments ignores the specified approach. 2 (a) The rate of change of momentum of a body is directly proportional to the resultant force acting on the body and occurs in the direction of the resultant force. B1 Marker’s Comment:

Many did not write the correct Newton’s 2nd law of motion. A few stated the force is ma. (b) (i) p = ∆(mv) = −65 x 10-3 (5.2 + 3.7) = −0.58 N s C1 A1 (ii) F = (0.58) / (7.5 x 10-3) = 77 N A1 (c) (i) 1. Force on the wall from the ball is equal to the force on ball from the wall B1 but in the opposite direction B1 (general statement of Newton’s third law only score 1 mark) 2. momentum change of ball is equal B1 and opposite to momentum change of the wall OR since there is no external resultant force acting on the system (ball and wall), the total momentum of the system is conserved OR total change in momentum of ball and wall is zero Marker’s Comment: Common mistakes such as the word ‘resultant’ was missed out in stating Principle of Conservation of Linear Momentum. Change in momentum was wrongly calculated without considering momentum as a vector. (ii) kinetic energy (of ball and wall) is reduced / not conserved, inelastic. (allow relative speed of approach does not equal to relative speed of separation) B1 3 (a) The moment of a force about a point is the product of the force and the perpendicular distance from that point to the line of action of the force. B1 (b) (i) Taking moments about point A and since the sum of the clockwise moments must equal the sum of the anticlockwise moments, +=007.58.0(9.81)( )cos37 12.0(9.81)(7.5cos37 ) (3.8 )2 T Hence T = 2.5 × 102 N (to 2 s.f.) C1 A1 Marker’s Comment: Kinetic energy was ‘different/not equal’ will not be credited.

(ii) Since net horizontal force must be zero, FH = 2.5 ×102 N Since net vertical force must be zero, FV = 8.0× 9.81 + 12.0× 9.81 = 2.0 × 102 N F = 3.2×102 N tan  = Fv / FH = 39°, anticlockwise above horizontal. A1 C1 A1 Marker’s Comment: Most were unable to give the right definition of moment. Quite a number were also unable to take moments of forces correctly and solve for the value of the tension. 4 (a) (i) The internal energy of a substance is the sum of the random distribution of kinetic and potential energies associated with the molecules of a system. B1 Marker’s Comment: Many responses were not specific enough to associate internal energy with the molecules of a system. A handful stated the first law of thermodynamics rather than the definition of internal energy (ii) At the same temperature, water vapour and water have the same total kinetic energy of molecules but water vapour has higher total potential energy of molecules (due to weaker intermolecular forces of attraction) B1 Hence the internal energy per unit mass of water vapour is higher than that of water B1 Marker’s Comment: Not very well done, many misconceptions were surfaced. Most common misconception was reasoning that having stronger intermolecular forces of attraction would result in higher potential energy. Another common misconception was to compare the particle motion of liquid vs solid (molecules sliding past each other vs moving around at high speeds) . Students would often reason that because of this difference, water vapour molecules would have higher kinetic energy instead of linking it to the concept of temperature.

(b) (i) 1 1 2 2 12 2 6 2 63 constant 890101000(15000) 288 228 1.3476 10 1.35 10 m pV nRT pV p V TT V V == = = = = C1 A1 Marker’s Comment: Most common error was not realising that T is not a constant, and using p1V1 = p2V2 for the calculation instead. (ii) 23 21 3 2 3 (1.38 10 )(228)2 4.72 10 J avgKE kT − − = = = B1 (iii) 6 8 8 33 22 3 890(1.3476 10 ) (101000)(15000)2 4.7345 10 4.73 10 J fi f f i i U U U p V pV  = − =− =  − =−  =−  C1 A1 Marker’s Comment: (b)(ii) was well done in general. However, (b)(iii) was not , with a fairly wide range of errors. Common errors include: • Using (initial − final) to compute change. • Using wrong formula (e.g. ∆U = p1V1 – p2V2) • Finding difference between average KE rather than U 5 (a) (i) Constant phase difference B1 Marker’s Comment: Many did not know the definition with many different variants of wrong answers. These include comparing the frequency, wavelength, type of wave, speed or stated that the sources must always be in / have the same phase for coherence.

(ii) The vapour produces photons when electrons de-excite from a higher energy state to a lower energy state. B1 The de-excitation of electrons is a random process. B1 The photons emitted are produced at random time intervals and therefore the phase difference between the emitted photons is not constant. Note: Mechanism must be correct to be awarded the last two B1 marks. B1 Marker’s Comment: Poorly done despite being very lenient. Most tried to smoke their way through by using the explanation for absorption/emission spectra wholesale to explain. (e.g. vapour absorbs white light to undergo excitation and re-emits photon in all directions during de-excitation) Most did not understand that monochromatic meant only 1 wavelength is being seen/produced, which led to subsequent explanations contradicting monochromatism (e.g. different wavelengths / frequencies produced) More serious misconceptions include: • Associating the low pressure of the vapour to the production of light. • Associating the motion of the vapour molecules to the production of light. (e.g. photons are emitted when vapour molecules move around at high speeds) • Diffraction occurred which caused a difference in phase. • Using X -ray production or photoelectric effect to explain the production of photons. (b) (i) Any 2 of the following: • Waves have equal or approximately equal amplitude • The slit separation must be of the same order of wavelength as the wave • The distance between the two slits must be small compared to the distance between the source and screen. • The waves from the two sources must be unpo