2025 HCI C2 Prelim H2 Physics P2 Ans

1 HCi H2 Physics 9749 2025 C2 Preliminary Examination Paper 2 Suggested Solutions Q1 (a) ( ) forcePressure at depth in the fluid area of base of container weight of fluid column area mass of fluid column acceleration of free fall density volume Fph A W A mg A Vg A Ahg A gh    = = = = = = (b)(i) Let the volume of the object be V. Resultant force F = weight of object mg – upthrust U ( ) 1 1 F mg U Vd g V g Vdg d mg d    =− =− =−  =−  (b)(ii) At equilibrium, taking moments about the pivot, Clockwise moment due to sample = anticlockwise moment due to standard mass (Resultant force on sample) x = (Resultant force on standard mass) x ( ) 11 11 1.29 1.290.17851 1 1 8493 940.0 0.17873 kg (5 d.p.) s s s s mg m gdd mm dd   − = −    = − −     = − −        =

2 Q2 (a)(i) The mass passes through the equilibrium position P twice in a cycle, meaning that there are 100 complete cycles per minute. period, T = 1/f = 60/100 = 0.600 s (a)(ii) At equilibrium position x = 0, the kinetic energy of the mass Ek is at its maximum.  = 2 T  = 2 0.600  = 10.47 rad s-1 Ek = ( ) 2 2 2 211 (0.42) 10.4722 oom x x = Amplitude 2 2(0.500) 0.147 m = 14.7 cm = 15 cm (2 s.f.)0.42(10.47) ox == (a)(iii) Ek = 2 -121 2(0.500) =1.54 m s2 0.42 kEmv v m = = • Sinusoidal waveform • Maximum velocity of 1.54 m s-1 and time period of 0.60 s *No information is given whether mass starts oscillating from its equilibrium position or at its amplitude – sine or cosine graph (b)(i) No change (b)(ii) Frequency multiplied by 2 .

3 Q3 (a) Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point (without a change in kinetic energy) (b)(i) To just reach the neutral point from the Earth, gain in gravitational potential energy GPE = Loss in kinetic energy ( ) 8 7 -1 7 neutral point earth surface 7 neutral point earth surface 76 neutral point 6 -1 neutral point 10.0 6.10 10 J 6.10 10 J kg 6.10 10 6.10 10 6.10 10 62.3 10 1.3 10 J kg (2 s.f.) U m KE        =  =−  =   =  − =  =  + =  + −  =−  OR ( ) earth surface earth surface neutral point n eutral point 86 neutral point 6 -1 neutral point total energy at Earth surface = total energy at neutral point 6.10 10 10 62.3 10 0 10.0 1.30 10 J kg (3 s KE U KE U   + = +  + −  = −  =−  .f.) (b)(ii) The rock from the Moon must have enough energy to go past the neutral point, then resultant gravitational force of the Earth-mass system on the mass will accelerate it to the Earth. gravitational potential difference between the Moon’s surface and the neutral point ( ) neutral point moon surface 66 61 1.30 10 3.90 10 2.60 10 J kg    −  = − =−  − −  = Hence, the minimum kinetic energy needed to send a 1.4 kg rock from the Moon to the neutral point is m  = ( )( ) 61.4 2.60 10 = 3.64  106 J or 3.6  106 J (2 s.f.) (c)(i) The gravitational force exerted on one star by the other star provides the centripetal force for each orbit. This pair of forces is an action-reaction pair (Newton’s 3rd law), always equal in magnitude (and opposite in direction). OR The gravitational force between stars provides the centripetal force for each orbit. By Newton’s Law of Gravitation, the gravitation force between the stars 2 (2 ) (3 ) GM M R= , thus the two stars experience the same magnitude of centripetal force.

4 (c)(ii) 5 5 22 1.837 103.42 10T  −= = =   rad s-1 Consider the star of mass M, The gravitational force due to star of mass 2M provides the centripetal force for the orbit of M. ( ) ( ) 2 2 3 2 11 30 3 25 9 (2 ) (2 )(3 ) 9 6.67 10 3.14 10 9 1.837 10 4.10 10 m GcF Ma GM M MRR GMR   − − = = =  =  =

5 Q4 (a) Let the distance from the source to the first area be r. Using geometry, the distance from the point source is 7.5 times more for the second area. Method I 2 2 1 ArI . Hence, amplitude is inversely proportional to the distance away from a point source, 1A r [M1] Since the distance from the point source is 7.5 times more for the second area, the amplitude would be 7.5 times less. Hence, the new amplitude would be 0 20 0.137.5 AAA== [A1] Method II Since the source is a point source, the intensity of the wave is inversely proportional to the square of the distance that the waves travel, 2 1 rI . Hence the intensity at the second area, I2 2 2 2 22 2 (7.5 ) () r r rr=I I 2 27.5= II [M1] Intensity is directly proportional to the square of the amplitude. Hence, the amplitude: 2 220 2 202 2 2 A AAA → = II II 0 20 0.137.5 AAA== [A1] Method III Intensity = Power / Area Intensity  (1 / Area)  (amplitude)2 22 1.6 1.6 12 12 S S      = = → =        I III [M1] 00 1.6 0.1312A A A== [A1]

6 (b)(i) 1 & 2 Wavelength = 1.2 m. Two full wavelengths of stationary waves must be drawn with • Nodes at 0.0 m, 0.6 m, 1.2 m, 1.8 m and 2.4 m. • Antinode peaks at 0.3 m and 1.5 m or at 0.9 m and 2.1 m [B1] Maximum amplitude for Y at 12.5 ms ( ) 25.0sin 5.0sin 25.0sin 12.5 3.5 mm20 y t t T y   ==  = =−  [B1] Maximum amplitude for Z at 5.0 ms ( ) 25.0sin 5.0sin 5.0 20 5.0 mm Z Z yt y  ==  = [B1] The antinode peaks for Z should coincide with antinode troughs for Y and vice versa [B1] (b)(ii) 180° or π rad Z Y

7 Q5 (a)(i) Direction of the electric field at x = 25.0 cm is towards the right. Electric field is always directed towards lower potential. OR Direction of the electric field at x = 25.0 cm is towards the right. The potential gradient is negative at this value of x as can be observed by drawing a tangent to the curve at x = 25.0 cm. Since E is (the negative) of the potential gradient, E takes on a positive value, which means the electric field is directed towards the right. (a)(ii) Draw a tangent to the curve at x = 25.0 cm Potential gradient 545 200 8.12 (3 s.f.)0.0 42.5 dV dx −= = −− 18.12 V cm (acceptable range: 7.3 9.7)Electric field dVE dx −=− = − (a)(iii) Mobile charge carriers within a conductor will always re -distribute until a certain equilibrium state where the electric field within it is zero. Zero electric field means that there is zero potential gradient , and hence constant potential. (b) The ion will accelerate to the right until x = 35.0 cm , then the ion will decelerate and momentarily come to rest at x = 42.0 cm, and accelerate back to the left and momentarily stops at x = 25.0 cm before accelerating to the right again, repeating the motion. 900 800 700 600 500 400 300 200 V / V x / cm 10 0 20 30 40 50

8 OR The ion will oscillate between the points x = 25.0 cm and x = 42.0 cm.

9 Q6 (a) The motion of electrons is random, so there is no net flow of electrons in any direction. OR On average, as much mobile electrons move in one direction as in the opposite direction, Thus there is no net transfer / movement of electric charge in a particular direction. (b) Arrow to the right. (c)(i) I −−==   28 19 3 2 2.0 8.5 10 (1.60 10 ) (0.40 10 ) v neA = 2.9 x 10-4 m s-1 t = L/v = 0.50/ (2.9 x 10-4) = 1700 s (or 1710 s to 3 s.f.) Accept answer up to 3 significant figures. (c)(ii) All mobile electrons in the circuit start drifting at the same time as the electric field is established in the wire almost instantaneously. The lamp lights up as soon as the mobile electrons already in the lamp filament begin to move which is