2024 DHS Prelim H1 Phy P1 Ans

2024 DHS H1 Physics Prelim Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 D A A B B C A C D A 11 12 13 14 15 16 17 18 19 20 B C A C D A B B D C 21 22 23 24 25 26 27 28 29 30 B D D A D C D A C B Worked Solutions & Explanations: 1 D Units of v = m s-1 Units of ඥߣ݃= ඥm s-2 × m = ඥm2 s-2 = m s-1 2 A For a pair of forces of magnitude X and Y, Minimum Fnet = |X – Y| = 6 – 4 = 2 N Maximum Fnet = |X + Y| = 6 + 4 = 10 N Hence, the resultant force cannot be 1 N. 3 A average length measured = 891.5 mm difference from true value = 895 mm – 891.5 mm = 3.5 mm (results are not accurate to within 1 mm) ∆L = Lmax – Lavg = 0.5 mm (results are precise to within 1mm)

4 B vy2 = uy2 + 2as At max height, vertical component of velocity = 0 vy2 = 0 + 2(9.81)(13.0) = 255.06 m s-1 horizontal component of velocity is a constant, vx = 22 cos 30° v = ටvy 2 + vx 2 = ඥ225.06+222 cos2 30° = 24.9 m s-1 5 D The total area under the acceleration-time graph is the change in velocity. The maximum magnitude of area occurs at point K. Hence magnitude of the change in velocity (and hence speed) is the largest at point K. The total area under the graph from t = 0 s onwards is also strictly negative. This indicates that the velocity of the object is strictly negative. (car is always moving in the opposite direction to the defined positive from t = 0 to point M). Hence, the maximum displacement is at M. 6 C The 6.0 kg mass will accelerate upwards (F net upwards), while the 10 kg mass will accelerate downwards (Fnet downwards) By considering the free body diagrams of the two masses, From the 6.0 kg mass: T – 6g = 6a -------------(1) From the 10 kg mass: 10g – T = 10a ------------------(2) (1) + (2): 4g = 16a a =0.25 g Therefore, T = 6g + 6(0.25g) = 7.5g = 74 N

7 A By Newton’s Second Law, the rate of change in momentum is directly proportional to the resultant force. Hence, the resultant force can be determined from the gradient of a momentum-time graph of a body. At terminal velocity, the momentum of the object is a constant (due to constant v) and non- zero. (graph should be a horizontal line with a non-zero value) After the time of impact P, the deceleration is a constant and hence net force is a constant. The momentum -time graph should therefore have a constant negative gradient (linear graph sloping downwards) after P. 8 C During a collision, momentum is always conserved. However kinetic energy is only conserved when the collision is elastic. 9 A For an object to be in equilibrium, the three forces must be concurrent (i.e. the line of actions of the 3 forces must intersect at a point). Only option A fits this description. 10 A F = kx F = (500) [(90−60) × 10-3] = 15 N 11 B For a pair of equal and opposite forces, torque = F × perpendicular distance between the two forces = F (L sinߠ) 12 C Power input = rate of loss of GPE = m t gh = (6.0 × 1000)(9.81)(80) = 4.7088 MW Power output = 0.6 × 4.7088 MW = 2.8 MW

13 A EPE = area of shaded triangle = 1 2 × 6 × (30 × 10-3) = 0.09 J 14 C At 10 m s-1, F – 5.0 × 104 = ma Driving force, F = 5.0 × 104 + (3.0 × 105)(0.50) = 2.0 × 105 N Power delivered by locomotive = Fv = 2.0 × 105 (10) = 2.0 × 106 W Hence, max speed = P Fresistive = 2.0×106 5.0×104 = 40 m s-1 15 D For the ball undergoing vertical circular motion, the kinetic energy of the ball is the greatest at the bottom of the loop and decreases as it moves towards the highest point as kinetic energy is converted into gravitational potential energy. Hence the speed of the ball is the greatest at the bottom of the loop. Since v r  , the ball will have the greatest angular velocity at the bottom of the loop. 16 A Triangle OPQ is an equilateral triangle. Hence, angular displacement = ∠POQ = 60° = గ 3 rad

17 B For two identical objects lying on the same surface, the maximum frictional force is the same. For the objects to undergo uniform circular motion, the frictional force F must provide the centripetal force. Hence, F = mrω2 Therefore, for the same angular velocity experienced by P and Q (due to them being on the same rotating disc) , the frictional force experienced by the object is directly proportional to its distance from the centre of the disc. As ω increases, F will increase until it reaches a maximum value. When ω is increased even further than the maximum frictional force value, the object will start to slip. Since F ∝ r and rP < rQ, object Q will reach and exceed the maximum frictional force value before object P, and hence will start to slide first 18 B Effective resistance in circuit = 1000 + ቀ ଵ ଵ଴଴଴ + ଵ ଵ଴଴଴ቁ ିଵ = 1500 Ω Current through X = ଵ ଶ ቀ V Reff ቁ = ଵ ଶ ቀ ଵ.ଶ ଵହ଴଴ቁ = 0.0004 A Number of electrons passing through per minute = It e = 0.0004 × 60 1.6 × 10-19 = 1.5 × 1017 19 D The resistance of a NTC thermistor decreases non -linearly with respect to temperature. (Options A and B are incorrect) The resistance of a NTC thermistor is a finite value at 0°C, hence Option C is also incorrect.

20 C For the same potential difference and current to be tripled when connected in parallel with a 500 Ω resistor, Reff, parallel = 1 3 R ൬ 1 500 + 1 ܴ൰ ିଵ = 1 3ܴ 500ܴ 500 +ܴ= 1 3ܴ 500 = 1 3 (500 +ܴ) ܴ= 1500 − 500 = 1000 Ω 21 B Potential difference across 200 kΩ = ଶ଴଴଴଴଴ ଶ଴଴଴଴଴ାଵ଴଴଴଴଴଴ × [3 − (−15)] = 3.0 V Hence, potential at X = 3.0 – 3.0 = 0 V 22 D Maximum reading on voltmeter occurs when rheostat resistance is zero. Hence, maximum potential difference across the 15 kΩ resistor is 6.0 V Minimum reading on voltmeter occurs when rheostat resistance is a maximum at 75 kΩ. By the potential divider principle, Vmin = ଵହ ଵହା଻ହ × 6.0 = 1.0 V 23 D The magnitude of the magnetic force is given by the product of the magnetic flux density and the component of the length of a current carrying wire perpendicular to the magnetic field lines. F = B⊥Il = BIl cos 60 = BIl sin 30

24 A Since the reading on the top-pan balance increases, the wire exerts a downward magnetic force on the top -pan balance. By Newton’s third law, the wire experiences an equal and opposite magnetic force. Hence the force on the wire XY is upwards . By Fleming’s left-hand rule, pole P is a north pole. By considering the forces on the magnet, ∆mg = BIL (2.3 × 10-3)(9.81) = B(2.6)(4.4 × 10-2) ∴ B = 0.20 T 25 D Adjacent loops of a spring can be taken to be a pair of current carrying wires. When switch S is closed, the current in adjacent loops will carry current in the same direction with respect to each other. Hence, regardless of the direction of current of the input signal, the force between loops carrying the same direction of current will be strictly attractive, resulting in a compression in the spring. 26 C Since the mass of P is greater than mass of Q for the same number of nucleons, this implies that the mass defect of Q is greater than the mass defect of P. With a greater mass defect, the binding energy of Q is greater than P. Hence, nuclei Q is more stable compared to P and would require more energy to be separated into its individual nucleons. 27 D Nuclear fission results in a release of energy. (Hence, options A and B are eliminated) reactants 2(8.45)(118) 7.65(236) 1750 MeV released productsE BE BE     Hence, the energy released per reaction is approximately 1800 MeV

28 A The diagram shows that most of the alpha particles passed through the gold foil undeflected or deflected by small angles. This indicates that the size of the nucleus relative to the size of the atom is small, as most alpha particles interacting with the gold atoms were not close enough to the nucleus to experience significant electrical repulson and hence, were only deflected by small angles or remain undeflected. 29 C Alpha particles are positively charged and will m