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) Relative speed of approach = relative speed of separation, AB BAu u VV−=− ( )2.0 2.0 BAVV−− = −

DUNMAN HIGH SCHOOL 2019 YEAR 5 H2 PHYSICS 3 4.0 BAVV= − ------------------ (2) Solving equations (1) and (2), VA = -1.0 m s-1 VB = 3.0 m s-1 Marker’s comments: Many students did not adopt a sign convention in this part of question. As a result, the answers did not reflect accurately the direction of motion of the boxes. Students were encouraged to explain clearly how the boxes moved after collision. (c) (i) ( )AA BB A B fmu mu m m v+= + ( )( ) ( )( ) ( ) 0.50 2.0 0.30 2.0 0.50 0.30 fv +−= + = 0.5 m s-1 Marker’s comments: Some students wrongly assumed the collision to be elastic and attempted to solve by applying the law of conservation of kinetic energy. (ii) During the collision, force on box A and box B is the same but box A has a greater mass. Hence, acceleration of box A is smaller. Marker’s comments: An acceptable alternative method is to consider the change in velocity of both boxes and deduce the acceleration based on the rate of change of velocity of each box. (iii) ( )( ) ( )( ) 22110.50 2 0.30 2 1.6 22 iKE J= + −= ( )( ) 21 0.50 0.30 0.5 0.1 2 fKE J= += 0.1 0.0631.6 f i KE KE = = Marker’s comments: Some students substituted -1.0 m s-1 and 3.0 m s-1 as the final velocities of box A and box B instead. There were also students who found i f KE KE .

DUNMAN HIGH SCHOOL 2019 YEAR 5 H2 PHYSICS 4 3 (a) weight = mass x g = density x volume x g = ρ ( πd2 4 h)g = 2700 π(0.024)2 4 (0.05)(9.81) = 0.599 (must be seen) = 0.60 N (must end with value asked to show) Marker’s comments: For “show” kind of question, all explanations, including substitutions of intermediate and final values in the calculations must be seen. (b) (i) At equilibrium, the sum of forces in any direction is zero. The sum of moments about any point is zero. Marker’s comments: There is no reason not to obtain full marks for this kind of equation. (ii) clockwise moment about pivot = anti-clockwise moment about pivot 12W = (0.25 x 8) + (0.60 x 38) W = 2.1 N Marker’s comments: There were some careless mistakes in obtaining the correct distance to calculate the anti-clockwise moments. (c) (i) pressure on bottom of greater than pressure on top or force on bottom greater than force on top Marker’s comments: Many students did not understand the meaning of the word “origin”. To attribute upthrust to difference in pressure is not specific enough. (ii) anti-clockwise moment reduced and reducing weight of X redu ces clockwise moment Marker’s comments: It is wrong to explain that upthrust contribute clockwise moments. Upthrust acts on the cylinder which is in equilibrium. However, the question asked how about equilibrium of the bar. Some students even mistakenly thought upthrust reduces the weight of cylinder! (iii) clockwise moment about pivot = anti-clockwise moment about pivot 12W = (0.25 x 8) + [38 x (0.60 − π(0.024)2 4 (0.05)(1260)(9.81))] W = 1.2 N Marker’s comments: Some students did not using the density of liquid to calculate the upthrust. Some answers did not take the resultant downward force at A to calculate the anti -clockwise moment.

DUNMAN HIGH SCHOOL 2019 YEAR 5 H2 PHYSICS 5 4 (a) (i) any three from o ( speed is constant but) direction is continuously changing (towards centre) o (velocity is changing) with time ( so body accelerates) o by Newton’s 2nd law o a force is required (for acceleration towards centre) (ii) a = v2/r (b) R – mg = mv2/r or R = 200 + [(20/9.81)2 × 4.72]/2.8 = 200 + 161 R = 361 N (c) any three from o statement of Newton’s first law; o (hence) without car wall/restraint/friction at seat, the people in the car would move in a straight line/at a tangent to circle; o (hence) door/seat belt/seat exerts centripetal force; o (in frame of reference of the people) straight ahead movement is interpreted as “outwards” Comments: (a) (i) Many did not explain that the direction of motion is changing steadily with time. (ii) A few included the mass m when stating the expression for a. (b) The main error was to write the equation as R =