2025 HCI C2 Prelim H2 Physics P1 Ans

2025 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 © Hwa Chong Institution 1 2025 C2 H2 Physics Prelim Exams Paper 1 Suggested Solutions 1 C 6 D 11 C 16 A 21 A 26 D 2 B 7 D 12 A 17 B 22 C 27 B 3 D 8 C 13 B 18 B 23 A 28 C 4 B 9 D 14 B 19 D 24 D 29 A 5 C 10 D 15 C 20 B 25 B 30 D 1 C 3 3 344volume (12) 7200 cm33 r= = = 2 B Upwards direction is taken as negative. Sandbag when released will continue to move upwards but with acceleration downwards. Sandbag will move vertically upwards till it slows down to zero velocity and then it will accelerate downwards. 3 D Since trolley A is faster than B, and they have the same mass, the net momentum before collision is towards the right and non-zero. Hence Option A is true. Option B is true by Newton’s Third Law. Option C is true. If B is moving to the left then so must be A (as A cannot pass through B) and net momentum will be towards the left which is not possible if momentum is conserved. Option D is false. Whether a collision is elastic or not depends on whether kinetic energy is conserved and not on the duration of the collision. 4 B Taking moments about the left end of the ruler, 2 (0.250)(0.200) (0.500)(0.0500) 0.075k x g g g= + = ------ (1) Taking moments about the right end of the ruler, 1 (0.750)(0.200) (0.500)(0.0500) 0.175k x g g g= + = ------ (2) Dividing (1) by (2): 2 1 0.075 3 0.430.175 7 k k = = = Distractors (remove from suggested solution for students): A: 1/3 → forget to account for mass of ruler B: 3/7 → Correct C: 7/3 → flipped B D: 3/1 → flipped A 5 C By the Principle of Floatation (or by considering the net force on the cube), the upthrust on the cube is equal to the weight of the cube. 3 00weight of cube mg Vg L g= = = 6 D The shelf is light, so its weight can be neglected. The centre of gravity of the system is thus at the centre of the book.

2 © Hwa Chong Institution 2 When the system is in rotational equilibrium, the lines of action of the forces must meet at a point: The three forces must form a closed vector triangle, so that net force is zero. 7 D ( ) 2 2 d kxdUF kx dx dx= − = − = − The negative sign indicates that direction of force is towards the reference point O. 8 C ( ) ( ) : sin : cos (0.8 ) tan 1.25 51.3 cc o N mg N F ma m g  = → = = = = = Option A: shift cos (0.8) Option B: swap the angles in resolving Option D: shift sin (0.8) 9 D The rotation of the Earth results in the acceleration of free fall being smaller than the gravitational field strength at the Earth’s surface, except at the poles where they are equal. 10 D As the orbital radius r increases, GMmGPE r=− becomes less negative (i.e. increases). Kinetic energy of the satellite decreases since the orbital speed drops as the gravitational pull provides less centripetal acceleration; or by 2 GMmKE r= , KE decreases. book shelf wall A B C D mg

3 © Hwa Chong Institution 3 11 C From kinetic theory of gas, p = 21 3 c Since the oxygen gas is 16 times as dense as hydrogen gas, the corresponding root-mean-square speed of its molecules must be 4 times as slow, meaning that the root-mean-square speed of the hydrogen molecules are 4 times as fast. 12 A By First law of thermodynamics, besides cooling, the internal energy can also be decreased by mechanical work through expanding the gas, as it does work on its surroundings. Option D is true only for ideal gas, which has zero microscopic PE. Option B is true, except during change of state. E.g. when ice melts, its internal energy is increased because of the increase in its microscopic PE. But there is no change in its microscopic KE, and thus no temperature change. Option C Temperature is a measure of the average (microscopic) KE (not KE+PE). 13 B By Newton’s 2nd law, Fnet on piston = ma Since gas exerts an upward force Fnet = Fgas – mg where mg is the piston’s weight Hence Fnet = Fgas – mg = ma Fgas = m(g + a) = 0.75 (9.81 + 3.5) = 9.983 N Work done by the gas = Fgas  s = 9.983  0.12 = 1.20 J 14 B Using v0 = w x0, 3 = 2p 12 x0 giving x0 = 5.730 cm Using a0 = w 2x0, a0 = 2p 12 æ èç ö ø÷ 2 ´ 5.730 = 1.571 cm s-1 Hence maximum restoring force = 3.0 x 1.571 ´10-2 = 24.7 10 − N Option C: wrong T = 15/1.5 = 10 s Option D: forgot to square omega, and mass multiply by max v = 3.0 x 0.03 15 C Since the wave pulse is travelling to the right, the last point on the displacement time graph will be the first point on the displacement position graph. Correspondingly, the first point on the displacement time graph will be the last point on the displac ement position graph. The sequence follows through for the points in between. 16 A As the light passes through the first sheet half of the intensity of the light is lost. The second sheet will need to reduce the intensity by another half in order for the emerging light to be 0.25Io. So,

4 © Hwa Chong Institution 4 2 2 1 cos24 1cos 2 1cos 2 1cos 45 2 oo     − = = = = =  II Either sheet could be moved as both will reduce the intensity by half each time the light passes through. 17 B Using small angle approximation, min between the stars from telescope from telescope between the stars 9 16 2 10 (630 10 )(5.9 10 ) 45 10 8.3 10 m d Db D d b    − −  = = =  = 18 B The distance from the node to the end of the tube is 4  . Since 0.17 0.68 m4  = → = . The frequency to generate this stationary wave is 340 500 Hz0.68 vf = = = The wavelengths of 357, , , ....4 4 4 4  are equal to 0.17 m to obtain a next node at the same position. The corresponding resonant frequencies would then be 500 Hz, 1500 Hz, 2500 Hz, 3500 Hz, 4500 Hz, 5500 Hz… etc. 19 D We cannot use Coulomb’s Law because that is for point charges (or charge distributions with spherical symmetry), but we have metal plates. Uniform electric field strength between the metal plates, 12 3VVE d r r == + aperture stars α θmin d 5.96 x 106 m

5 © Hwa Chong Institution 5 Electric force on charged oil drop 12 3() VqE q rr== + Distractors (remove from suggested solution for students): A → formula for electric potential energy if the plates were point charges B → formula for electric force if the plates were point charges C → calculated electric field strength due to each plate separately D → correct 20 B work done by external force U q V= =  The net potential at a point is the scalar sum of the potentials due to the first two point charges. U is independent of the path. Since the charges have opposite polarity, • the electric potential is positive near +2q and negative near –q. • the point where the electric potential is zero is between the two charges, and nearer to –q. A: always equidistant from +2q, starts and ends about the same distance from –q. 0V C: starts and ends about equidistant from both +2q and –q. 0V D: initialV is a large negative number, and finalV is a smaller negative number. V is positive. B: initialV is a large negative number, and finalV is a large positive number. V is the greatest. 21 A R