DHS 2026 H3 Physics Prelim (Ans)

H3 Physics Prelim Answers 1 (a) time of flight, cosRt u = A1 (b) tan (taking upward direction as posit ive)hR =− A1 (c) ( ) ( ) ( ) ( ) 2 2 22 taking upward direction as positive, for constant vertical acceleration, we have 11sin tan sin2 cos 2 cos tan t an 2cos RRh u t g t R u g uu gR u       = + − − = −  − = − ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 tan tan 1 cos2 sin1 cos2 1 cos2 tancos 2sin cos 1 cos2 tan sin2 1 cos2 tan gR u uR g u g u g             − = − +  = + + + = + + = + + M1 M1 A1 (d) ( ) ( ) ( ) 2 22 sin2 1 cos2 tan 22 cos2 tan sin2 0 cos2 tan sin2 1 tan tan2 uR g dR u u d g g            = + + = + − =  = = tan tan 2 2 2 2 22       = −   = −  + = M1 A1

2(a) From diagram, 1sin 2 = , hence 𝜃 = 30° and 3cos 2 = Let v be the velocity of m1 after free falling a distance of L cos𝜃, 2 cosv gL = When the string becomes taut, initial velocity of the mass when it undergoes the motion in stage 2 = vtan, the tangential velocity of the mass. tan sin 1 2 cos2 vv gL   = = By conservation of energy, ( ) 22 1 1 1 tan 1 22 1 tan 2 11 cos22 2 (1 cos ) 1 2 cos 2 (1 cos )2 2 cos 2 (1 cos )4 32 cos2 332 4 m u m v m g L L u v gL gL gL gL gL gL gL       = + − = + − = + − = + − =−  =−  Therefore, 1 332 4u gL =−  M1 M1 C1 A0 0.5 L L 𝜃 v vtan The motion happens in 2 stages: 1. Free fall a distance of L cos𝜃. 2. Vertical circular arc after stage 1.

2(b)(i) 1 1 339.81 (1.0) 2 4 2.6223 m s u − =−  = 1 1 2 2 12 1 1 0.100(2.6223) 0.400 1 4 0.65558 0.66 m s ZMF m u m uv mm u − += + = = = = C1 A1 2(b)(ii) Let S denote the laboratory reference system and S’ denote the zero momentum frame. Taking leftwards to be positive, u sing a Galilean transformation, the balls’ velocity in S’ before collision are 21 10 4ZMFv v u= − =− After the collision, the velocity of m2 is 1 1 4 u in S’. Hence in S, 21 1 1 1 4 1 2 1 (2.6223)2 1.31115 m s ZMFV u v u − =+ = = = By Conservation of Energy, 2 2 2 2 2 2 1 2 2 0.088 m m gh m V Vh g = = = M1 C1 A1

3 (a) 0 0 2 0 2 0 2 0 3 0 4 0 4 0 2 0 2 2 2 4 2 1 2 m r r mm V rL m Lr dm dV Lrdr I r dm Lr dr rL Lr mr          == = == = = =  = =   M1 M1 A1 (b) 2 2 2 2 00 2 2 2 0 0 0 0 22 0 2 0 1 1 12 2 2 2 Taking pivot at point of contact on the ground, By parallel axis theorem, 1(2 ) (3 ) 2 () () () 1(3 ) 2 (3 ) cm cm cm I MR mr MR mr I I M m R M m R mr I F R r F R r I F R r M m R mr aR R Rr Mm    = + = + = + + = + + = = − −= −= ++ = −= + 22 012 FR mr+ M1 M1 A1

(c) ( ) 2 0 22 0 2 2 2 00 22 0 22 00 22 0 0 Suppose f is directed to the left, 2 (2 )( ) 1(3 ) 2 1(3 ) (2 ) (2 ) 2 1(3 ) 2 1 (2 )2 1(3 ) 2 (2 ) Since f >0, it is cm cm F f M m a M m R Rrf F F M m R mr M m R mr M m R M m Rr F M m R mr MR mr M m Rr F M m R mr I M m Rr FI − = + +−=− ++ + + − + + + = ++ + + + = ++ ++= directed to the left. M1 A1 A1 (d) A1

4 (a) Curve connecting 1 and 2 with correct arrow Value of P1 and label V1 and V2 Horizontal line connecting 2 and 3 with correct arrow Value of P3 and label V3 B1 B1 B1 B1 (b) Curve connecting 3 and 1 B1 (c) 1 1 3 3 1 31 3 33 31 3 1 1 1 3 3 1 1 Since at same temperature, Density of gas at 3, Since pressure remains constant, Mass of air entering the cylinder, 1.02 cyl cyl cyl PV P V PVV P PPmm V V P P mV P VgP    = = = = = = == M1 A1 (d) 6 0 0 0 1.23 575 10 0.70725 % increase 100% 44% cyl mg mm m −=   = −=  = A1

5 (a) The decay 238 238 0 92 93 1U Np −→+ cannot occur spontaneously because the mass of the products (237.99979 u + 0.00055 u = 238.00034 u) is greater than the mass of the parent nucleus (238.00019 u) so energy must be supplied instead, making the process energetically forbidden. B1 B1 (b) (i) ( ) ( ) ( )( ) ( )( )( ) 2 227 8 227 8 13 13 19 6 energy released rest mass of reactants res t mass of products 238.00019 233.99409 4.00150 1.66 10 3.00 10 0.00460 1.66 10 3.00 10 6.8724 10 J 6.8724 10 eV1.60 10 4.29525 10 eV 4.3 c − − − − − =− = − +   =   = =  = = MeV (to 2 s.f.) M1 M1 A1 (ii) The decay products emerge with total kinetic energy 4.3 MeV. Momentum is conserved; since the parent U-238 nucleus is at rest, the daughter Th-234 nucleus and the α-particle have momenta of equal magnitude p but opposite direction. Kinetic energy is K = p2/2m: Since p is the same for the two particles, the kinetic energy divides inversely as their masses. Hence the α- particle gets 234 / (234 + 4) of the total, or 4.3 MeV. ( ) ( ) ( ) 6 19 71 27 234kinetic energy of -particle, 4.3 MeV 4.23 MeV234 4 2 4.23 10 1.60 10 J2speed of -particle 1.43 10 m s 4 1.66 10 kg K K m   − − − == +    = = =   C1 M1 A1 (c) (i) The rate at which thorium-234 nuclei are produced is given by the activity A of the uranium-238 sample. The intermediate daughters (Th-234 and Pa-234) decay very quickly compared to U-238, so the production rate of U-234 is essentially the decay rate of U-238. ( )1 2 U-238 U-238 U-238 U-238 U-238 23 91 41 massln2production rate of N-234 molar mass ln2 2.38 g 6.02 104.47 10 365 24 3600 s 238 g mol 2.96 10 s ANN t − − = =    =        = M1 A1 (ii) Because U-238 decays so slowly, its activity is almost constant over the U-234’s timescale. Eventually, U -234’s activity (decays per second) equals the parent’s activity, so the two appear to be decaying at the same rate.

( )1 2 U-238 U-234 U-234 U-234 4 U-234 U-234 23U-234 51 4 U-234 massln22.96 10 234 massln2 6.02 102.48 10 365 24 3600 s 234 g mol mass 1.30 10 g A AA N N t  − − = =  =   =        = M1 A1 6 (a) 1. The laws of physics are the same in all inertial frames of reference. 2. The speed of light in free space is the same regardless of the motion of the light source or observer. B1 B1 (b) Consider two frames S and S’. Frame S’ moves at velocity v relative to S. Use inverse Lorentz transformation: ( ) 2''t t vx c=+ . For a clock at rest in S’ (the source of proper time), the interval is measured at the same position, so 0x'= . Therefore, ( )( ) 20 t t' v c t t' =  +  =  . M1 A1 (c) (i) Take noon = t = t′ = 0 at the event “ship passes Earth”. Ship speed v = 0.8c. Ship proper time 12:30 is t′ = 0.5 h =1800 s. ( ) 22 2 21 1 1 1 0.8 1 0.6 5 3v c c c = − = − = = The event “ship passes the station” has ship-frame coordinates x’ = 0, t’ = 1800 s. Lorentz transform to Earth time: ( ) ( ) 2 5 3 1800 3000 s = 50 mint t' vx' c t'= + = =  = So, Earth time = noon + 50 min = 12:50 p.m. M1 A1 (ii) Earth-frame position ( )x x' vt' vt'= + = (since x’ = 0): 115 3 0.8 1800 s 7.2 10 mx vt' = c=   =  So, the station is 7.2 × 1011 m from Earth. (Notice also 110.8 3000 s 7.2 10 mx vt c= =  =  .) M1 A1

(iii) Emission event: 113000 s, 7.2 10 m.emit emittx = =  Light travels back to Earth (at x = 0) at speed c. Travel time = emitxc . 11 8 7.2 10 m 2400 s 40 min3.0 10 mt  = = = So, Earth receives at t = 3000 + 2400 = 5400 s = 1.5 h after noon = 1:30 p.m. M1 A1 (i