2025 DHS H2 Prelim P2 Ans w MC

2025 DHS Prelim Paper 2 Suggested Solutions 1 (a) (i) mass/volume (ratio must be clear) B1 (ii) kg m-3 OR kg/m3 B1 (b) v has unit of m s-1 B1 p/ρ has unit of kg m-1 s-2/ kg m-3 (no e.c.f from (a)) M1 √𝑝/𝜌 has unit of m s-1 B1 LHS = RHS so 𝛾 has no unit A0 Marker’s Comments: Common mistakes made by equating quantities to units will not be accepted. 2 (a) Using v2 = u2 + 2as 1 0 2(5.0)(4.0) 6.32 m s v − =+ = A1 (b) ( ) 1 1 22 2 1 Horizontal component 6.32cos37.0 5.05 m s Initial vertical component of velocity 6 .32sin37.0 3.80 m s Using 2 Final vertical component of velocity 3.8 0 2 9.81 15.0 17.6 m s o o y y yv u gs − − − = = == =+ = +   = A1 A1 (c) =+ −== =+ −== = + = = Ball rolling down the roof: using 6.32 0 1.264 s1 5.00 Ball falling from edge of roof to ground: using 17.57 3.80 1.404 s2 9.81 Total time taken 1.264 1.404 2.668 2.67 s v u at t v u gtyy t M1 M1 A1 Marker’s Comments: Common mistakes made by using the wrong acceleration for calculation of the times.

(d) (i) B1: Shape B1: Correct labelled values (allow ecf) (ii) B1: Shape B1: Correct labelled values (allow ecf) Marker’s Comments: Common mistakes made is inappropriate spacing of the time, velocity and acceleration in the sketch. 3 (a) uniform electric and magnetic fields normal to each other AND charged particle enters region normal to both fields correct B direction w.r.t. E for zero deflection For no deflection, v = E/B B1 B1 B1 Marker’s Comments: Many did not state that for no deflection, v = E/B. (b) r = mv / BQ , so B proportional to m B = (0.680) × (87/86) = 0.6879 T ΔB = 7.9 × 10-3 T C1 A1 Marker’s Comments: Most were able to solve this part. Some did not calculate the change but instead gave the value of the calculated magnetic field as the answer. a t 9.81 5 1.26 2.67 v t 18.3 6.32 1.26 2.67

4 (a)   − − = =  = = 2 0 72 12 0 2 4 (3.20 10 ) 4 (8.85 10 )(2 0.50sin3.0 ) 0.33616 0.34 N (to 2sf) QqF r C1 A1 Marker’s Comment: Most students knew which formula to apply but often times made mathematical errors such as forgetting to square, or did not x 2 after resolving for the horizontal distance of the central line to the centre of one sphere. (b) By considering the forces on one sphere,  = =  = = tan 0.33616(9.81) tan3.0 0.65385 0.65 kg (to 2sf) EF W m m Note: Allow ecf. C1 A1 Marker’s Comment: Quite a number did not attempt. Those who attempted generally managed to resolve correctly either by finding the tension first or use of the tangent function directly. A small number just simply divided the electric force by g. (c) (i) The electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point. B1 Marker’s comment: Generally the responses were missing certain key words (e.g. “work done per unit charge…”) or used wrong key words. (e.g. “bring a small point charge …”)

(ii) B1: Correct Shape B1: Intercept the horizontal axis at the midway point. Marker’s comment: Varied responses of graphical shapes were seen. Most wrong answers gave the U-shaped graph above the horizontal axis. Many did not label the value of the x-intercept. (iii) Since the electric field strength E is related to the potential as dVE dx=− , and the graph in Fig. 4.2 is E against x. The potential difference between two points can be found by determining the area under the graph between the two points. B1 B1 Marker’s comment: Not very well done. Most responses suggested that gradient of the E – x graph can be used to find the potential difference. Those who identified correctly that area of the graph is to be used often did not explain how did they arrive at that method. x / m E / N C -1 0.026

(d) Each spheres will be deflected by the same, but smaller than before angle 𝜃 from the vertical. (𝜃 < 3.0°) Note: • Zero is given if diagram / explanation suggests unequal angle from centre vertical line. • BOD if diagram has at least 1 angle unlabelled but there is clear symmetry about the centre vertical line B1 B1 Marker’s Comment: Often times the angles in the diagram was not clearly labelled neither does their explanation often suggest that both spheres are deflected to the same extent. BOD was very often exercised as long as students drew a somewhat symmetrical diagram. 5 (a) Acceleration is directly proportional to the displacement. Acceleration is always in the opposite direction to its displacement B1 B1 (b) (i) 0.034 m B1 (ii) 0 0 1 0.60 0.034 17.647 18 rad s v x − = = = = C1 A1 Marker’s Comment: Parts (a) and (b) were generally well done. 𝜃 ° 𝜃°

(c) (i) Upwards is taken to be positive from the graph, 2 netF N W m x = − =− When the object loses contact, N = 0 (Concept must be correct, incorrect concept penalise this 1 mark) 2 2 2 9.81 17.647 0.0315 0.032 m mg m x gx   = = = = = B1 C1 A1 Marker’s comment: Many responses revealed misconceptions amongst students. These include: • Fnet = 0 (contradiction as this would mean acceleration = 0) • N = W = m𝜔2x (this would imply that W = 0 since N = 0) • mg = −m𝜔2x (implies W is opposite direction to Fnet) If students somehow managed to obtain the correct mathematical working despite the obvious conceptual error, marking was very lenient to award partial credit for these responses. Students should also refrain from forcefully removing “ −“ signs from their workings as this would violate equivalence in their workings. (ii) C marked at (0.032, 0.2) Note: e.c.f. only if marked in 1st quadrant. B1 Marker’s comment: Students found this challenging. Most students did not mark point C on the curve but rather on its x-value on the horizontal axis. Students who managed to get the right answer to c(i) often failed to consider the context of the oscillation (starting point and subsequent motion) and marked it in the 4th quadrant instead. 6 (a) (i) cannot predict when a particular nucleus will decay or cannot predict which nucleus will decay next B1 (ii) (decay is) not affected by external (environmental) factors B1

(b) fluctuations in (measured) count rate B1 (c) The average time taken for the initial number of nuclei (or activity) of that particular radioactive nuclide to reduce to half of its initial value. B1 Marker’s Comment: The word 'average' is crucial because individual nuclear decay is unpredictable. Half-life describes the statistical behaviour of many nuclei, not a precise countdown for each nucleus. Distinguish: 'nuclide' = a specific isotope type (e.g. C-14); 'nucleus' = individual atomic core. (d) (i) activity (of X at time t) B1 (ii) • Y is a stable isotope • total number of nuclei is constant • half-life (of X) is 13.6 s • decay constant (of X) is 0.051 s–1 1 2 12 2 13 6 s 0 051 sln t ln . . − == • amount (of X) at t = 0 is 0.066 mol 22 23 0 0 A 4 00 10 6 02 10 0 066 molen N N . . . = =   = • activity (of X) at t = 0 is 2.0 × 1021 Bq ( )( ) 22 21 00 0 051 4 00 10 2 0 10 BqA N . . . = =  =  Any three points, 1 mark each B3 Marker’s Comment: Incorrect to say Y's half -life is longer than X's. Y is stable (doesn't decay), so the concept of half-life doesn't apply to Y. (e) mass of 1 nucleus = (7.3 × 10–4) / (4.0 × 1022) nucleon number = mass of nucleus / (1.66 × 10–27) = (7.3 × 10–4) / (