2024 HCI H2 Physics Paper 3 Solutions

2024 HCI Preliminary Examination Paper 3 Suggested Solutions Q1 (a) Random errors are deviations of the measured value from the mean value, with varying signs and magnitudes. Systematic errors are deviations of the mean value from the true value, with same sign and similar magnitude. B1 B1 (b) The units of P are (kg m s−2)(m) s = kg m2 s-3 The units of pqk Av are 2 1 3 2 3 s kg( )(m ) (m ) kg () m1 m p q p q q s− − + + −= For the equation to be homogeneous, the units of P must be equal to the units of pqk Av . Comparing power: seconds: -q = -3  q = 3 metres: -3 + 2p + q = 2  p =1 M1 M1 A1

Q2 (a) The acceleration of an object is its rate of change of velocity with respect to time. A1 (b)(i) Taking upward to be positive and using 22 2y y y yv u a s=+ , where vy is the vertical component of the final velocity, uy is the vertical component of the initial velocity, ay is the vertical component of the acceleration and sy is the vertical component of the displacement. At the highest point, the vertical component of the velocity is zero. Furthermore, ignoring air resistance, the vertical component of the acceleration is ay = -g = -9.81 m s-2. Hence, 0 = (210 sin (30⁰))2 + 2(-9.81) h h = 561.9 m = 560 m (1, 2 or 3 s.f.) M1 A1 (b)(ii) Taking upwards as positive and using y y yv u a t=+ 0 = (210 sin (30⁰)) + (-9.81) t t = 10.703 s = 11 s (1, 2 or 3 s.f.) M1 A1 (b)(iii) From (b)(i), we know that the highest point for P is 561.9 m. Using 21 2 y y ys u t a t=+ , 561.9 = (210 sin (60⁰))tQ + (0.5)(-9.81) tQ2 t = 3.4 s or t = 33.7 s Since t must be less than 10.7 [from part (b)(ii)], t = 33.7 s should be rejected. t = 3.4 s (shown) B1 A0 (b)(ii) B1 – parallel to and above original line. (P starts at -1210sin30 105 m s= while Q starts at -1210sin60 182 m s= . Values not needed in sketch) B1 – starts after mid-point of original line and ends at the same time (3.4 s not needed in sketch) vertical velocity / m s -1 time / s 3.4 s 105 182

Q3 (a)(i) Fluid pressure increases with depth. The upward forces due to the fluid pressure acting on the lower surface of the material are larger than the downward forces due to the fluid pressure acting on the upper surface of the material, resulting in a net upward force called the upthrust. B1 B1 (a)(ii) Upthrust on cup = weight of water displaced = V ρwater g = (6.8 × 10-5) × (1000) × (9.81) = 0.66708 N Weight of cup = V ρcup g = (6.8 × 10-5) × (2200) × (9.81) = 1.4676 N Since the weight of the cup is larger than the upthrust acting on the cup, an upwards external force F is required to keep the cup stationary. Since the cup is held in equilibrium, the net force acting on the cup is zero. Hence, F + U = Wcup F = Wcup – U = 1.4676 – 0.66708 = 0.800496 = 0.80 N M1 M1 A0 (a)(iii) Since the cup was already fully submerged, there i