2025 HCI C2 Prelim H2 Physics P3 Ans

HCI H2 Physics 9749 2025 C2 Preliminary Examination Paper 3 Suggested Solutions Q1 (a) (i) Vertical component of the ball’s initial velocity, uy = u sin θ = 25 sin 30° = 12.5 m s-1 = 13 m s-1 (ii) Considering the vertical component motion, At maximum height, the vertical component of the ball’s velocity will decrease to zero. Using vy2 = uy2 + 2aysy and taking upwards direction as positive, 0 = (12.5)2 + 2(-9.81)sy Vertical displacement (i.e. max height reached), sy = 7.96 = 8.0 m (iii) Initial total energy = 21 2K mu= At maximum height of 8.0 m, ball’s vertical component velocity, vy = 0. Its velocity will be its horizontal component velocity, vx = u cos 30 = 3 2 u kinetic energy at 8.0 m = 2 21 3 3 1 0.752 2 4 2m u mu K ==   Gain in potential energy of system = Loss in kinetic energy of ball Potential energy at 8.0 m = 3 0.254K K K−= OR: Initial total energy 2 2 12 2 KK mu m u=  = at 8.0 m, ( )( ) ( ) ( ) ( ) 22 2 9.81 8.02 0.25 2 s.f. 25 0.25 0.75 2 d.p. KPE mgh gh K Ku KE K PE K K K = = = = = − = − =

2 (b) (i) Correct graph Ep parabolic curve starts from zero, peaks at K/4 at the middle, ends at zero. Ek parabolic curve starts from K, decreases to 3K/4 at the middle, ends at K. Correct labels, clear markings, symmetry ` 0 x energy 0.5x Fig. 1.2 Ek Ep K parabolic horizontal distance

3 Q2 (a) ( )( )( ) 2 2 1 108 10001.20 0.30 2.502 60 60 405 N 1 2 ddF C Av   = = =  ( ) ( ) 2 220.05 0.02 0.05 405 1.20 0.30 2.50 108 70 N 1s.f. dd dd d d FC Av F C A v F F      = + + +  = + + + = Fd   Fd = 410  70 N (b) velocity of car A relative to car B, AB A Bv v v=− By Cosine Rule, ( )( ) 22 -1 40.0 50.0 2 40.0 50.0 cos135 83.2 km h ABv = + −  = By Sine Rule, sin sin135 50.0 83.2  = , hence 25.1 = direction: bearing = 90.0° + 25.1° = 115.1° 135° 40.0 km h-1 50.0 km h-1 θ

4 Q3 (a) Let mass of Planet Z be M, mass of argon molecules be m, Consider an argon molecule escapes from the planet’s surface to infinity. By conservation of energy, Loss in KE = Gain in GPE 2 3 2 1 002 2 42 3 8 ()3 surface surfaceKE KE U U GMmmv r GMv r Gr r G r shown   − = − − = − − =  = = (b) 2 -11 3 2 -1 8 3 8 (6.67 10 ) (413 10 ) (5500)3 724 ms v G r   = =   = (c) Assume that escape velocity v is equal to the root-mean-square speed crms of the argon molecules. By Kinetic Theory, average KE of one argon molecule = 3/2 kT Average KE of one mole of monatomic gas is therefore 3/2 RT. ( ) 2 max 2 max 2-3 13 where is the molar mass of argon.22 3 (40 10 ) 724 3 8.31 841K r rms r r rms M c RT M McT R = = =  = (d) The root-mean-square speed of argon molecules will be less than escape velocity when temperature decreases. However, due to the random distribution of speeds, some molecules will have speeds greater than the escape velocity and would be able to escape.

5 Q4 (a) From Fig. 4.2, 5 mm corresponds to 3 fringe separations. (b) Let  be the angle made by 1st order minimum of the single slit diffraction envelope from the principle axis. From Fig. 4.1 and Fig 4.2, 3 min 10.0 10tan 0.0102 0.980 −== ( ) min 9 5 sin Applying small angle approximationsin 633 10 0.0102 6.2 10 m b b    − − = = = = ; Alternative Method From Fig. 4.2, the first minimum for single slit diffraction envelope coincides with the 6 th maximum for double slit interference (missing order). Single slit diffraction: Double slit interference: Since , 4 5 1 6 3.7 10 66 6.2 10 m b d db − − = == = (c)(i) Increase slit separation d. From Ly d = , a smaller d makes the fringe separation y wider, spreading the interference pattern. (c)(ii) Make the slits narrower. Narrower slits produce a much wider single -slit diffraction envelope with lower overall intensity, so the bright fringes vary less in intensity. 3 5.0 5.0 3 1.67 mmyy =   =  = 9 3 4 0.980(633 10 )1.67 10 3.7 10 m Ly d d d  − − − = = = 1sin 1b = 6sin 6d = 16=

6 Q5 (a)(i) The component of velocity perpendicular to the magnetic field results in a magnetic force on the electron, which is always perpendicular to the motion . Hence, the electron rotates in a circular path. The component of velocity parallel to the magnetic field remains constant. H ence, the electron travels at constant velocity along the direction of magnetic field. The combination of both components of velocity of the electron results in a helical path. (a)(ii) 14 19 6 -1 sin20 4.3 10 sin20 0.088(1.60 10 )sin20 8.93 10 m s B B F Bqv Bqv Fv Bq v ⊥ − − = =  ==   = (a)(iii) Magnetic force provides centripetal force to the electron’s component of velocity perpendicular to the magnetic field. Find period T: 2 2 2 2 2 B mv BqrF Bqv v rm v r r T r rm mT v Bqr Bq     ⊥ ⊥⊥ ⊥ ⊥ = =  = == = = = Pitch:` 6 31 19 3 2cos20 (8.93 10 )cos20 (2 )(9.11 10 ) (0.088)(1.60 10 ) 3.4 10 m p v T mv Bq   − − − = =    =  = P (b) The pair of forces generate a torque. The maximum output torque is F x d 3 Torque ( ) 395 (1200)(96)(6.1 10 ) 0.562 T Fd NB d NB A B NA B   − = = = = ==  = Il I I

7 Q6 (a)(i) For any given metal, electrons are emitted only when the frequency of incident light is above some minimum value. This frequency is known as the threshold frequency. (a)(ii) At threshold frequency, energy of photon matches the work function. Hence, o hc = 11 MAX oo hc hcE hf hc     = − = − = −   (b)(i) Extend line to the x-axis. The x-intercept is 1 o 61 2.3 10 o = 74.35 10 mo −= (b)(ii) Gradient of graph = hc ( ) ( ) ( ) 19 8 6 3.1 1.0 10 3.00 103.85 2.8 10 h −− =− 346.7 10 Jsh −= (c) Same gradient Higher y-intercept, x-intercept more to the left. (d) As electrons escape the surface, the sphere becomes more positively charged and the potential at the surface increases. It comes to a point when the electrons with maximum kinetic energy lose all the energy at almost infinite distance and they are attracted back to the surface due to the electric field. The system is finally in dynamic equilibrium. The rate of return is equal to the rate of emission, keeping the total charge constant at the surface. Thus, electric potential energy gained by an electron at infinity = EMAX lost ( ) 110 4 4 11 surface MAX oo o o U U E Qe hcr hc rQ e       −= − − = −   =−  

8 Q7 (a) The binding energy E of a nucleus is the energy required to separate the nucleus into its constituent free neutrons and protons. It is related to the mass defect ∆m by the formula, 2E mc= . (b)(i) The energy equivalent of 1.00 u is given by, ( )( ) ( )( ) ( )( ) 2 227 8 227 8 19 6 1.00 1.66 10 3.00 10 1.66 10 3.00 10 MeV 1.60 10 10 933.75 MeV 934 MeV J uc − − − = =    =  = = (b)(ii) For Nitrogen-14, Binding energy ( ) ( )( ) 27 1.007 276 7 1.008665 14.003074E uc= + − ( )0.108513 934 MeV== 101.35 MeV Binding energy per nucleon 101.35 7.2414== MeV per nucleon (c)(i) The energy released in the reaction = final binding energy – initial binding energy ( ) ( ) ( )17 7.530 14 7.24 4 6.836 128.01 128.70 0.694 MeV = − +  =− =− Hence there is actually a gain of mass in this reaction and 0.694 MeV of energy needs to be provided. (c)(ii) We need at least 0.694 MeV of energy from the kinetic energy of the alpha particles as the oxygen nucleus and proton may also have kinetic energy