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 p