SAJC_Final Exam (H2) JC1 2025_timed trial soln only

1 SAJC 2025 H2 Physics Timed Trial/ 9478 H2 Physics 2025 Timed Trial Solutions 1 (a) sin 50 = ℎ 458 h = 39.91 = 39.9 m Note: COE & kinematics equations are unaccepted. Examiner’s Comments: • Some students erroneously wrote vy2 = u y2 + 2a ysy, where 282 = 0 + 2(9.81)sy. This is totally inappropriate although the answer yielded happens to be 39.9 m. • Some erroneously applied the COE method, saying that the total initial energy = KE i + GPE i = 0 + 0, and total final energy = ½ (800)(28 2) + (800)(9.81)(h). Total initial energy is simply not equal to the total final energy in this case! There is unknown energy input from the car’s motor. [1] (b) Taking downwards as positive, from the cliff top to the ground, 𝑠𝑦 = 𝑢𝑦𝑡 + 1 2 𝑎𝑦𝑡2 39.91 = (-28 sin 5o) t + ½ (9.81) t2 (ecf for wrong h) t = 3.11 s or -2.61 s (NA) Examiner’s Comments: • Many made careless mistakes in substitution (esp negative signs) because they did not consider a certain direction as positive. • Some students neglected “sin 5o” in uy. • Some took a lengthy method to find the time taken for different parts of the flight path and then adding them up. But some students only calculated the time for part of the flight path. • Note that vy is not zero at the ground! [1, ecf] [1] (c) By conservation of energy, taking the ‘start’ point to be at the edge of cliff, 10% of total mech energy of car before impact = KE remaining after impact 0.1 (𝑚𝑔ℎ + 1 2 𝑚𝑣2) = 1 2 𝑚𝑢2 0.1 [(9.81)(39.91) + ½ (282)] = ½ u2 (ecf for wrong h) u = 12.51 = 12.5 m s-1 Alternatively, taking the ‘start’ point to be at just before touching the ground (hence GPE = 0), 10% of KE (just before touching ground) = KE remaining after impact 0.1 x ½ (m)(v2) = ½ (m)(u2) [1, ecf] [1]

2 SAJC 2025 H2 Physics Timed Trial/ 9478 where speed just before touching ground is v = √𝑣𝑥2 + 𝑣𝑦2 = 39.58 where vx = 28 cos 5o , and vy has to be solved using vy = uy + ayt = (-28 sin 5o) + (9.81)(3.11) = 28.09 m s-1 Examiner’s Comments: • There are different ways to solve this question, depending on where you see the ‘start’ point as. • Many students forgot to include either the initial KE or GPE at the ‘start’ point on the LHS of the eqn above. • Some erroneously wrote “90%” instead of “10%” in the eqn above. (d) Taking rightwards as positive, v2 = u2 + 2as 0 = 12.512 + 2 (-2.5) s s = 31.3 m Total distance = 86 m + 31.3 m = 117 m, which is greater than 100 m. The car will collide with the wall. Alternatively, 𝑣 = 𝑢 + 𝑎𝑡 0 = 12.51 + (-2.5)(t) 𝑡 = 12.51 2.5 = 5.004 s From 𝑠 = 𝑢𝑡 + 1 2 𝑎𝑡2 s = (12.51 x 5.004) + ½ (-2.5)(5.0042) (ecf for wrong u) = 31.3 m Alternatively, Taking rightwards as positive,