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, vx2 = ux2 + 2 ax sx = 12.52 + 2(-2.5)(14) vx = 9.31 m s-1 Since the speed at the wall is not zero, car will collide with the wall. , Alternatively, Kinetic energy left after impact in (c) = ½ mu2 = ½ x 800 x 12.512 = 62600 J Work done needed to decelerate car to a stop = (ma) x s = (800 x 2.5) x (100 – 86) = 28000 J. Since the initial kinetic energy is more than the work done required to bring the car to a stop at the wall, the car will have some more energy to move beyond the wall. { Principle: Change in KE = Work Done by Net Force} [1, ecf] [1]

3 SAJC 2025 H2 Physics Timed Trial/ 9478 Examiner’s Comments: • Some students drew a v-t graph to find the time for braking and distance travelled during braking (area of graph). This is also allowed. • 2nd mark is only awarded if there was a valid/ logical reasoning based on correct physics. • Some students erroneously compared the ‘time needed for deceleration’ with the ‘time taken during constant speed motion’. 2(a) (i)1. By conservation of momentum, → : mAuA + mBuB = mAvA + mBvB 6.0 (5.0) + 10 (-3.0)= 6.0 (vA) + 10 (2.0) vA = - 3.33 m s −1 speed = 3.3 m s −1 Examiner’s Comments: • Some students forgot to consider account of the signs when subbing in the velocities as they forgot momentum is a vector. [1] [1] 2. KE before collision: KEi = ½ (6)(5)2 + ½ (10)(3)2 = 120 J KE after collision: KEf = ½ (6)(3.3)2 + ½ (10)(2)2 = 53 J Since there is a loss in kinetic energy during the collision, the collision is inelastic. Examiner’s Comments: • Some students use the relative speed of approach = relative speed of separation method which is acceptable if they did not mixed up the terms. However if student used this method, the signs of the velocities must be taken into account. • Students are advised to present their answers properly. They should not present in such a way that KE before and after are equal and then later has a line that shows the number does not tally. They should calculate KE before and after separately then compare the two. Such poor presentation may be penalised for the A levels. [1] [1]

4 SAJC 2025 H2 Physics Timed Trial/ 9478 (ii)1. Correct shape (starting and ending point of the slope, straight lines) – [1] Correct values – [1] Allows ECF from part (a) Examiner’s Comments: • Most students got this correct. Concept is that the total momentum before collision = total momentum after collision. 2 to 4 ms is the duration of the collision. • Some number of students forgot to that the final velocity is negative as (a)(i)1 ask for speed of A after collision. [2] 2. For B, Force = d(mv)/dt = 10(2 –(-3))/ (2.0 x 10-3) = 25 000 N For A, Force = d(mv)/dt = 6(- 3.33 - (-5))/ (2.0 x 10-3) = 24999 = 25 000 N Examiner’s Comments: • Many students only got 1 mark as they did not see that the time at the x-axis is in milliseconds. • However, apart from that, most students are able to get at least 1 mark. [1] [1]

5 SAJC 2025 H2 Physics Timed Trial/ 9478 (b)(i) When water goes into or exit the ballast tanks, the weight of the submarine will change. If the weight is bigger than upthrust, the submarine will sink, if the weight is smaller than upthrust, it will float. [1] (ii) Δ(mv) = m(Δv) = area under the graph 500 (Δv) = 0.5 (500 x 10-3 ) (30000) Δv = vf – vi = 15 vf = 15 + 2 = 17 m s −1 Examiner’s Comments: • A significant number of students forgot that the area of the graph gives change in velocity, not final velocity and forgot to add that to the initial velocity for the final velocity. [1] [1] Prelim 3 (a) Resultant/ net force is zero Resultant/ net moment about any axis/point is zero [1] [1] (b)(i) W is the weight of the rod, T is the tension acting on the rod, R is the force acting on the rod by the hinge/ contact force {do not accept: “normal contact force”}. [1] [1] (b)(ii) Take moments about the hinge, NT LTLo 1274.127 )2(30cos81.930 = = (LHS & RHS each 1m) {penalise 1m for scale diagram methods} [1,1] T • T and W correctly labelled with full name, T and W in correct direction. • All 3 forces intersect to meet at a common point (clearly shown with dotted lines!) and R correctly labelled & in correct direction W R

6 SAJC 2025 H2 Physics Timed Trial/ 9478 (b)(iii) NTR oo X 7.6330sin4.12730sin)( ===+ NTR o Y 18430cos81.930)( =−=+ NRRR YX 1957.1941847.63 2222 =+=+= ecf allow o9.707.63 184tan ==  above the horizontal as shown in Fig. 1 ecf allow {accept cosine rule method, 3m, with correct R direction} {award ec