DHS 2019 JC1 Promos Physics (Solutions)

DUNMAN HIGH SCHOOL 2019 YEAR 5 H2 PHYSICS 1 Answers to P romo Exam MCQ Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 B A B D A D C B C D Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 C C C A D B B D A D Structured Q uestions 1 (a) Speed decreases/ stone decelerates to rest/zero at 1.25 s. Speed then increases/ stone accelerates in opposite direction. Marker’s comments: From the description, some students demonstrate poor understanding of the motion, mistakenly thinking that as the gradient is positive, velocity is increasing and velocity is decreasing when gradient is negative. There were also some mistakes in reading the time the stone reaches highest displacement. (b) (i) v = u + at = 0 + (9.81) (3.0 – 1.25) = 17.2 m s-1 Marker’s comments: There were many wrong answers that substitute u = 0 with t = 3 s in the given equation. In addition, it is not possible to calculate the velocity from the gradient as numerical values for the displacement are not given. (ii) s1 = 1 2at2 = 1 2 (9.81)(1.25)2 = 7.66 m s2 = 1 2 at2 = 1 2 (9.81)(1.75)2 = 15 m t otal distance = 7.66 + 15 = 22.7 m Marker’s comments: Another method is to find the initial upward speed at t = 0 = the downward speed at t = 2.5 s = 12.3 m s-1. A shorter method is to use the above. (iii) displacement = s2 − s1 = 7.34 m direction: downward Marker’s comments: Using the alternative method in (ii), the answer can be found by substituting u = 12.3 m s-1 and t = 0.5 s into the equation for displacement with uniform acceleration. (c) straight line with negative gradient, positive value of v from t = 0 to t = 1.25 s negative values of v from t = 1.25 s to t = 3.0 s, same gradient max height t = 0 t = 3 s t = 1.25 s s1 s2

DUNMAN HIGH SCHOOL 2019 YEAR 5 H2 PHYSICS 2 Note: acceptable range for horizontal intercept: 1.2 to 1.3 Marker’s comments: Some students overlooked that in (b), the question already mentioned air resistance is negligible. Hence there is only acceleration due to free fall which is constant. The direction of acceleration is downwards, hence the gradient of the line is negative. 2 (a) The total momentum of a system of bodies is constant, provided no external resultant force acts on the system. Marker’s comments: Some students attempted to explain the answer by using formula instead. In addition, there were also some students who provided the definition for principle of moments. (b) Taking rightwards as positive, By principle of conservation of momentum, AA BB AA BBmu mu mV mV+=+ ( )( ) ( )( ) ( ) ( )0.50 2.0 0.30 2.0 0.50 0.30 ABVV+ −= + ( ) ( )0.4 0.50 0.30 ABVV= + ------------------- (1)