YIJC 2023 H2Phy Prelim P2 Solution

2023 YIJC JC2 H2 Preliminary Examination Paper 2 Suggested Solutions 1 (a) The gradient of the graph represents the acceleration of the ball, which is changing with speed. This suggests that the resultant force acting on the ball (which is proportional to the acceleration by Newton’s second law) is changing with speed. When the ball is rising, the resultant force is given by weight and viscous force. Since its weight is constant, a changing resultant force implies that air resistance acting on the ball must vary with speed. OR when the ball is falling, the resultant force = weight – viscous force. A decreasing resultant force will imply that the viscous force is increasing with increase in speed. B1 B1 Marker’s comment Students should start the explanation by connecting the gradient of v -t graph to acceleration. This will allow the examiner to better follow the discussion. Students tend to be unclear in which part of the motion they were describing. The students did not seem to recognise that while the gradient is decreasing and hence the acceleration is decreasing, the implication is different when the ball is rising compared to the case when the ball is falling. Some answers are rather convoluted and answering directly to the question. Students should state clearly (b) (i) time = 1.75 s A1 Marker’s comment Some students did not understand at the highest point, the resultant force is the weight and thus this is the time at which the acceleration of the ball is g. There were several students who gave the answers as 4.2 s.

2 (ii) [B1] – Straight line graph starting from (0, 25) with negative gradient. [B1] – Cuts time axis at t = 2.55 s (so that the gradient of graph is –9.81 m s−2) B2 Marker’s comment Students need to plot the graph to half a small square. It should be shown clearly that the line cuts in between 2.5 and 2.6 s. Students do not gain credit if the graph is not plotted to half a small square. This part was poorly done as students did not show understanding that the gradient of graph needs to be 9.81 m s−2. Most students understood that the ball starts with a speed of 25. m s−1 -15 -10 -5 0 5 10 15 20 25 0 1 2 3 4 5 v / m s−1 t / s

3 (c) For the upward motion, energy lost = loss in KE - gain in GPE energy lost = ( )  −   21 0.350 25 0.350 9.81 192 = 44.14 J For the downward motion, energy lost = loss in GPE - gain in KE energy lost =   −    210.350 9.81 19 0.350 12.52 = 37.89 J Therefore, ratio 44.14 37.89= 1.16 C1 C1 A1 Marker’s comment Not well done. Students did not make sense of the given data from the graph. Also, some students tried to find the acceleration from the graph which would only be useful if the question assesses on the concepts of dynamics. Students need to connect the graph to the physical motion of the ball to better solve the question.

4 2 (a) Moment of a force is the product of the force and the perpendicular distance to a pivot. B1 Marker’s comment Students need to learn the definition and follow the definition closely. Simply saying that it is the product of force and perpendicular distance is not sufficient. Students need to state clearly the perpendicular distance is from the line of action of the force to the pivot. Saying “perpendicular distance f rom the pivot” is not clear enough. There was also some students who mentioned “perpendicular distance from the point the force acts to the pivot” which was incorrect. (b) (i) Taking moments about A, T(2.8) = 500(1.4 sin30°) + 4000(2.8 sin30°) T = 2125 N T = 2.1 kN (shown) M1 A0 Marker’s comment Most students were able to show the tension by taking moments about point A. There were a handful who were unable to find moment of forces correctly. (ii) Tv = T cos60° = 2100 cos60° = 1050 N A1 Marker’s comment Most students were able to resolve the tension correctly. (iii) There is an upward vertical component of force acting on the beam at A such that upward force at A + Tv = sum of downward forces of weight of beam and load. B1 Marker’s comment Not well done. The key idea that there is an “upward vertical component of the force acting on the beam” was not clearly brought across. Also, there were several students who incorrectly used the term “normal contact force” from the wall. Normal contact force acts perpendicular to the vertical wall and thus would have been acting horizontally and hence not awarded credit.

5 (iv) Vertical equilibrium: Av + Tv = 500 + 4000 Av = 4500 – 1050 = 3450 N Horizontal equilibrium: AH = T sin60 = 2100 sin60 = 1820 N Hence magnitude of force at A = (18202 + 34502)1/2 = 3900 N tan θ = 3450 / 1820 θ = 62 Direction is 62 above the positive horizontal. C1 C1 A1 A1 Marker’s comment Poorly done. Many students were not able to recall the steps to solve this kind of question. The angle was poorly expressed. Students should draw diagram to show clearly the angle they were determining. Use of scale diagram to solve this question is not encouraged. A AH AV θ

6 3 (a) Work done is the product of the force and the displacement moved in the direction of the force B1 Marker’s comment Many students left out the “in the direction of the force”. (b) (i) Displacement = Area of v – t graph = ½ (1.0 + 4.0) × 2.4 = 6.0 m M1 A1 Marker’s comment Many students left the answer in 1 sf. Students should know that the readings should be read to half of the smallest division, and hence to 1 d.p. Therefore, the displacement should be calculated to 2 s.f. BOD is given. The marking of this question is quite lenient. Students who showed workings to find area under the graph, 1 mark is given, although the range could be incorrect. (ii) Increase in GPE = mgΔh = 13000 × 6.0 = 7.8 × 104 J Increase in KE = ½mΔv2 ( )=   − 221 13000 2.4 02 9.81 = 3817 J Work done against friction = F × s = 2000 × 6.0 = 1.2 × 104 Work done by motor = Increase in energy of the lift + Work done against friction = ΔGPE + ΔKE + W = 78000 + 3817 + 12000 = 9.38 × 104 ≈ 9.4 × 104 J C1 C1 C1 A1 Marker’s comment Very poorly done. Many students attempted to use the approach of determining the work done by the forces instead of energy approach. However, many considered work done by the net force rather than work done by the tension. For those who considered work done by tension, many did not realise that the tension from t=0 to t=3 is different from t=3 to t=4.

7 (iii) Output power = F v = 1.6 × 104 × 2.0 = 3.2 × 104 W Output PowerEfficiency Input Power= 43.2 100.67 Input Power = Input Power = 4.78 × 104 ≈ 4.8 × 104 W C1 C1 A1 Marker’s comment Very poorly done as well. Some students found the work done from t=0 to t=2.5s and then divided by the time. These students have used the concept of average power instead of instantaneous power. The way to obtain instantaneous power is to use P = Fv where F and v are the force and velocity at that point in time. A few students thought that efficiency is input/output power instead.

8 4 (a) (i) Resultant force acts perpendicular to the velocity of the object. / Centripetal force is constant in magnitude B1 Marker’s comment Generally well done. Some students were vague when referring to “the centre” without specifying whether it is the centre of the circular motion or the centre of the object. (ii) The velocity of the object is always changing as the direction of the motion is always changing. Thus the object experiences a rate of change