MSHS 2025 PHY PRELIM P1+P2 MS

1 2025 MSHS 6091Physics Prelim Exam Solutions PAPER 1 1 2 3 4 5 6 7 8 9 10 C B D A C A A D A B 11 12 13 14 15 16 17 18 19 20 C C A A B B D A C A 21 22 23 24 25 26 27 28 29 30 A A D B D A A A A D 31 32 33 34 35 36 37 38 39 40 D C D B D D B A D B

2 Deductions U (units) – 1M for each occurrence capped at 1 M per question SF (significant figures) & P (presentation) – 1M for each occurrence capped at 1M per paper PAPER 2 SECTION A 1(ai) Distance = Area under v-t graph 5 = 1 2 (∆𝑣 10) (8.4 + 8.4 + ∆𝑣) ∆𝑣2 + 16.8∆𝑣 − 100 = 0 v = 4.66 m/s v = 8.4 + 4.66 = 13 m/s C1 A1 1(aii) Acceleration = gradient of v-t graph 10 = 13 − 8.4 𝑡 t = 0.46 s C1 A1 1(b) reasonable shape suitable scale correctly plotted 1st and last points at (0,8.4) and (0.90,0) with non-vertical line at 0.46 s M1 A1 A1 0.20 0.40 0.60 0.80 10 -10

3 2(a) all correct any 3 correct A2 A1 2(bi) Taking moments about point A, sum of anti-clockwise moments = sum of clockwise moments TB x 4.0 = 4.0 x 10 x 1.0 +25.0 x 10 x 2.0 + 60.0 x 10 x 3.6 TB = 675N C1 A1 2(bii) Sum of forces acting on scaffold = 0 N Upward forces = downward forces TA + TB = 4.0g + 25.0g + 60.0g TA = 89 x 10 - 675 = 215 N C1 A1 2(c) As the painter walks towards point A, the tension in rope A will increase and the tension in rope B will decrease. By considering the moment of the forces about point A, as x decreases, the sum of clockwise moment decreases. (Since the beam is in equilibrium), tension in rope B must also decrease. Hence the tension in rope A increases, as total downward force remains the constant. B1 B1 B1 3(a) height difference = 32 - 15 = 17 cm Pgas = 76 + 17 = 93 cmHg Pgas = gh = 13600 x 0.93 x 10 = 126000 Pa M1 A1

4 3(b) The number of gas molecules per unit volume is equal both at the bung and at the walls. Hence, the frequency of collision of the gas molecules per unit area against the walls and the bung are equal, exerting equal force per unit area, which is the same gas pressure. B1 B1 B1 3(c) Use a liquid of smaller density. For any gas pressure increase, the height increase will be more because the density is lesser, according to the formula P = gh. B1 B1 4(a) During the change in state, energy is being transferred to the internal potential store of the molecules. Since there is no increase in energy in the internal kinetic store, temperature remains constant. B1 4(bi) Energy transferred from flame = Energy transferred to soup (Pt) x 30% = mlv (300) (3.0) (3600) (0.30) = m (2.26 x 106) m = 0.430 kg C1 A1 4(bii) When high flame is used, large amount of soup will be vaporized in 3 hours. Hence a large amount of