(VJC) 2024 H2 Prelim P2 Soln

2024 Physics Prelim Exam H2 Paper 2 suggested solutions 1(a) Moment of a force: The moment of a force about an axis is the product of the force and the perpendicular distance [1] from the line of action of the force to the axis [1]. Minus 1 mark if no diagram given or diagram does not show corresponding d and F correctly. Torque of a couple: Torque of couple is defined as the product of one force F [1] and the perpendicular distance 𝑑⊥ between the two forces [1]. Minus 1 mark if no diagram given or diagram does not show corresponding d and F correctly. (b)(i) F d F d F T: Tension T: Tension W: Weight of truck N: Normal reaction force

(b)(ii) Considering the truck in equilibrium: sin40 2Tm g T = [1] Taking moments about the centre of the pulley, Sum of clockwise moments = sum of anti-clockwise moments 3mg r T r =  sin403 2 Tmgmg r r  =  [1m for correct clockwise expression, 1m for correct anti - clockwise expression] 1500sin40 2(3)m = m = 161 kg [1] 2(a) (i) 1m for correct graph P 1m for correct graph K (ii) 1m for correct graph F Explanation: Some potential energy lost will be converted to do work against friction [1], hence the gain in kinetic energy will be less as compared to the situation without friction. [1] Energy/ J h / m P K F

3(a) momentum after elastic collision = - mu [1] time between collisions = 2L/u [1] number of collisions per unit time = u/2L [1] rate of change of momentum = - mu2/L [1] average force = mu2/L [1] (b)(i) Gas molecules collide with one another elastically. [1] (ii) Since collisions are elastic, the average speed of gas molecules incident on the wall remains the same. [1] So frequency of collision between each wall and the molecule also remains unchanged. [1] Therefore the pressure exerted on the wall is not affected. (c) Given in (b) pV = 1/3 Nm<c2> - - - - (1) By Ideal Gas Law: pV = NkT - - - - (2) [1] Equating (1) = (2) 1/3 Nm<c2> = NkT => ½ m<c2> = 3/2 kT [1] Hence average ke, ½ m<c2> = 3/2 kT  Ave KE  T (because k is a constant) (d)(i)1. Shade the area under p-V graph [1]

2. Since the process A -> B is such that U = 0 (isothermal), and U = 3 2 pV then pAVA = pBVB. [1] Hence there is no difference in the product pAVA and pBVB. [1] (ii) If the change takes place very quickly, there is no time for heat transfer with the surrounding. So it is an adiabatic change. i.e. Q = 0. [1] Fig. 3.2 shows a compression. i.e. W is positive. So U = Q + W is also positive. i.e. U increases. [1] Since U  crms2 of the molecules, it means the mean square speed of the molecules increases. [1] , 4(a) Since the tube is in equilibrium, Weight of (tube and ball bearings = Upthrust [1] (m + M)g = AHg ( )MmH A += [1] (b) When the tube is displaced downwards by y, the upthrust will increase to A(H+y)g while the weight would remain the same as (m+M)g. Hence the net force = (m+M)g - A(H+y)g = (m+M)g - AHg - Ayg = -Agy since (m + M)g = AHg (negative means that the net force is opposite in direction to the displacement y) [1] By N2L ( ) m M a A gy Agay mM   + =− =− + [1]

(c) From (b) since Agay Mm =−  + , thus a  -y. Hence motion is S.H.M. Comparing with defining equation: a = - 2 y, then: 2 34 2 2 0.012 0.0252 1.00 10 6.0 10 9.81 0.4982 0.50 s (Shown) Ag Mm Ag Mm Ag T M m MmT Ag       − = + = + = + += +=     = = (d) As the test-tube oscillates, it experiences drag force exerted by the water . This results in light damping and energy is gradually lost as heat. [1] (e)(i) [1] [1] [1] Sinusoidal graph [1] Label period 3.3 s [1] Intensity 0 90 180 270 360 Angle / o

(ii) The amplitude of oscillation is small because the frequency of the driving force (the waves) is too low (0.30 Hz) compared to the natural frequency of the test-tube (2 Hz). The amplitude of the oscillations can be increased by adding ball bearings to the test- tube to decrease the natural frequency so that it is closer to the frequency of the driving force. 6(a)(i) From Fig 6.1 it can be observed that at specific wavelengths there were dips in the intensity meaning that the energy at these wavelengths were being absorbed. [1] So this is an absorption spectrum. [1] (ii) Transition from n = 3 to 2 should corresponds to the least energetic transition i.e. transition giving rise to 656 nm line. ( )( ) 34 8 656 9 19 6.63 10 3.0 10 656 10 3.03 10 J − − −  =  = nmE 5(a) The direction of the induced emf is such as to produce an effect that opposes the change causing it [1] (b)(i) X = 0.85 A [1] Y = 2/ 0.040 [1] = 160 rad s–1 [1] (b)(ii) two cycles of a sinusoidal curve with a period of 0.040 s [1] correct phase (i.e. V2 max / min at t = 0, 0.02, 0.04, 0.06 and 0.08 s, and V2 zero at t = 0.01, 0.03, 0.05, 0.07 s) [1] maximum / minimum V2 shown (consistently) at ± 6.5 V [1] (b)(iii) (magnitude of) V2 is proportional to rate of change of (magnetic) flux [1] • V2 is proportional to gradient of I1–t curve • V2 has maximum magnitude when I1–t curve is steepest • V2 is zero when I1–t curve is horizontal / a maximum or minimum • V2 changes sign when sign of gradient of I1–t curve changes Any 2, give [1] [1] [1]

3 2 656 19 19 19 5.44 10 3.03 10 2.41 10 J = 1.51eV −− − = + =−  +  =−  − nmE E E (b)(i) If light behave as waves, then light waves should continuously transfer energy to the electrons to overcome the work function. [1] Even if lower frequency waves are used, the electrons can accumulate energy over time to gain enough energy to overcome the work function. [1] Hence no minimum frequency should exist. (ii) From graph, stopping potential Vs = 2.7 V [1] ( )( ) ( ) ( )( ) 34 8 19 9 19 6.63 10 3.0 10 1.60 10 2.7 184 10 6.49 10 J   − − − − =−  = −   = s hc eV (iii) 7(a) Longitudinal wave – oscillations of wave particles parallel to direction of transfer of energy of wave [1] Transverse wave – oscillation of wave particles perpendicular to direction of transfer of energy of waves [1] (b) S-waves cannot pass through the liquid outer core. [1] [1] [1] I / nA 10 5 0 15 - 4.0 - 2.0 2.0 4.0 6.0 0 V / V Graph with lower stopping potential [1]

(c)(i) Akita – 30 mm Pusan – 56 s, 540-550 km, Tokyo – 425-435 km, 210 mm All amplitudes correct [1] Time interval for Pusan correct [1] Distance for Pusan (allow ecf) and Tokyo read correctly [1] (ii) Understand must draw 3 circles center at stations [1] All circles drawn with compass and correct scale [1] Epicentre at KOBE [1] (iii)1. -Distance from epicenter -Scattering at boundary of different materials, cracks etc. -Absorption due to rocks in Earth Any other suitable answers [1] Magnitude = 6.8 [2] (d)(i) 1. Ratio = 109/106.8 = 158 [1] The distance between the epicentre and the station. [1] 2. The above shows that a change in magnitude of 2.2 correspond to a change in intensity of about 160 times. Hence using a log scale allows us to compress the scale to more manageable numbers [1] (ii) Ratio = 10(1.5)(2.2) = 1995 (2000) [1] (e) - Population density of the affected area might be low. - The affected area might have stricter building codes that require buildings to have earthquake-proof features.