NJC_2024_H2_Physics_Prelim_P1_Ans

NJC Preliminary Examination 2024 H2 Physics Paper 1 Solutions 1 2 3 4 5 6 7 8 9 10 C C A B A A C A B B 11 12 13 14 15 16 17 18 19 20 B A B B B D C B C D 21 22 23 24 25 26 27 28 29 30 C D A A C A B C B D

1 C 1 per second, 3600 per hour 2 C by definition 3 A Slope has constant gradient. v2 = u2 + 2as. So KE is proportional to the displacement with a being negative. 4 B Let time interval between each drop be t. First drop took 3t to reach the ground and travelled 9 m 9 = 0 (4t) + ½ (g)(3t)2 t2 = 2 𝑔 distance travelled by 2nd drop = 0(2t) + ½ g (t)2 distance travelled by 3rd drop = 0 (3t) + ½ g (2t)2 distance between 2nd and 3rd drop = ½ g (2t)2 - ½ g (t)2 = 3.0 m alternative solution: v t 4th drop start to fall 1st drop touch the ground distance between 1st drop and 4th drop = 9 triangles = 9m distance between 2nd and 3rd drop = 3 triangles = 3m 5 A F = 70 = (20+6) a a = 2.692 ms-2 f on B = 6 a = 16.15 N (to right) N3L, f on A = f on B = 16.15 N = 16 N (2sf) (to left) 6 A Taking moments. A has the greatest moments. 7 C Kinetic energy is conserved only for elastic collision, momentum is conserved for both elastic and inelastic 2nd drop 3rd drop

8 A GPE = KE 0.3 (9.81) (2) = 5.886 = ½ (0.3) v2 v = 6.2641 m s-1 Pi = Pf 0.3 (6.2641) = (0.3 + 0.5) V V = 2.3491 m s-1 KE after collision = ½ (0.8) 2.34912 = 2.20725 J therefore EPE = 5.886 – 2.20725 = ½ k (0.20)2 k = 184 N m-1 (3 s.f.) 9 B At B, resultant force towards centre of dome Fnet = mg cos 48.2o Centripetal acceleration = Fnet / m = g cos 48.2o = 6.54 m s–2 10 B Geostationary orbit is always at the same radius, have the same period and mass does not affect the orbit 11 B Uniform g field has the same g field strength at all points 12 A Mean speed = (2+10+11)𝑢 3 = 7.67𝑢 Mean square speed = (4+100+121)𝑢2 3 = 75𝑢2 Root mean square speed = 𝑢√75 = 8.66𝑢 13 B 𝑝𝑉 = 𝑛𝑅𝑇 = 𝑚 𝑀 𝑅𝑇 At constant pressure and volume, 𝑚𝑇 = constant. 𝑚2𝑇2 = 𝑚1𝑇1  𝑚2 = 𝑚1𝑇1 𝑇2 = (273+10) (273+30) × 15 = 14 kg 14 B Q = Pt = mc𝛥 P = mc(𝛥 /t) P and m is constant, 𝛥 /t is bigger for liquid than solid, therefore specific heat capacity c is smaller for liquid than solid. (A is wrong) Q = mL 2000 x 3 = 1 x L 🡺 L = 6000 J kg-1 (B is correct) The substance melts after an increase in temperature of 3K from room temperature. The melting temperature is not 3K. (C is wrong) Unless the graph becomes horizontal again, we are unable to determine when the substance starts to become gaseous. And only after it stops being horizontal again will it be completely gaseous. (D is wrong)

15 B Taking the initial position at 650 mm mark, 𝑥 = 𝑥0 𝑠𝑖𝑛 𝑠𝑖𝑛 𝜔 𝑡 = 𝑥0 𝑠𝑖𝑛 𝑠𝑖𝑛 [(2𝜋 𝑇 ) 𝑡] 25 = 50 𝑠𝑖𝑛 𝑠𝑖𝑛 [(2𝜋 2 ) 𝑡] 𝑡 = 0.167 𝑠 𝑥 = 𝑥0 𝑠𝑖𝑛 𝑠𝑖𝑛 𝜔 𝑡 = 𝑥0 𝑠𝑖𝑛 𝑠𝑖𝑛 [(2𝜋 𝑇 ) 𝑡] 10 = 50 𝑠𝑖𝑛 𝑠𝑖𝑛 [(2𝜋 2 ) 𝑡] 𝑡 = 0.064 𝑠 tTotal = 0.167 + 0.064 = 0.231 = 0.23 s 16 D