YIJC 2024 JC2 PRELIM H1 Phy P1 Solutions

©YIJC 9749/01/YIJC/24 2024 YIJC JC2 Prelim Exam H1 Physics Paper 1 Solution Question Answer Question Answer Question Answer 1 B 11 B 21 B 2 C 12 C 22 A 3 C 13 B 23 C 4 B 14 D 24 B 5 A 15 D 25 B 6 B 16 B 26 B 7 C 17 A 27 C 8 B 18 B 28 B 9 B 19 C 29 C 10 B 20 D 30 C MCQs Solutions Qn Answer Explanation 1 B Option A: container ship can go up to about 40km h–1 = 11 m s–1 Option B: Usain bolt complete 100 m at 9.63 s Olympic sprinter can go up to about 10 m s–1 = 0.01 km s–1 Option C: The modern F1 car can reach speeds of roughly 220mph = 100 m s–1 = 10000 cm s–1 Option D: The top speed of a snail is about 0.04 km h–1 2 C  = 64o 3 C Taking average can only reduce random error. For systematic error, the error has to be known and then to be added/subtracted from all measurements. Vrain -VBus Vrain relative to bus  360 10 1tan 3600 8 

©YIJC 9749/01/YIJC/24 2 4 B The slope of the v-t graph is equals to the acceleration. Downward velocity is taken to be positive. So acceleration downward is positive too. In region OP, decreasing (positive slope) acceleration with downwards direction. In region PQ, terminal velocity is reached, acceleration (slope) is zero. In region QR, increasing (negative slope) acceleration with upwards direction. In region RS, new reduced terminal velocity, acceleration (slope) is zero. 5 A As angle θ increases from zero, the range increases. R reaches a maximum value, then starts to drop to zero when angle θ is 90°, where the projectile is shot straight upwards. Taking upwards positive, sin sin 2 sin V V gt Vt g        Taking rightward positive,   2 2 2 sin coscos sin 2 VR V t g V g       6 B Taking downwards as positive, 21 2 y y ys u t a t  212.0 9.812 t   0.639t  s x xs u t 2.5 0.639xu  3.92xu  ms−1 7 C Impulse = area under the F-t graph. Since the graph shown is a-t, we can multiply the y- axis value by the mass to obtain the F-t graph.         1Area under graph (1.2) areaunder graph (1.2) (3.0) 2.0 3.02 9.0 N s F t a t OR The area of a-t = change of velocity = 7.5 m s-1 Impulse = m v = 1.2 x 7.5 = 9.0 N s

©YIJC 9749/01/YIJC/24 3 8 B Assuming the 3.0 kg mass accelerates downwards, the 5.0 kg mass will accelerate rightwards at the same rate. Let the acceleration be a.            2 for the 5.0 kg mass, taking right as 20 5.0 (1) for the 3.0 kg mass, taking down as 3.0 3.0 (2) solving simultaneously, (1) (2) 3.0 20 8.0 1.2 m s ve T a ve g T a g a a 9 B A – tension is the force the rope acts on the two different boys. Since there are three systems (rope and 2 boys), the tension acting on the two boys cannot be an action- reaction pair. C – the centripetal force is a net force (sum of all forces) and can never be one of the forces in an action-reaction pair. D – the buoyancy force acts upwards on the boat, while the thrust pushes the boat forward (horizontal