2020 YIJC H1 Physics Prelims P2 Answers

2020 JC2 H1 Physics Preliminary Examination Paper 2 Suggested Solutions (SECTION A) 1 (a) (i)  (Contact) force on plank by ground – correct direction and labelling  (Contact) force on plank by man – correct direction and labelling  Weight of plank (force on plank by Earth) – correct direction and labelling *Do not accept symbols. B1 B1 B1 Marker’s comment: Poorly attempted. Many students could only identify weight of the plank. Many wrongly labelled the contact force on plank by man as weight of man. (ii) The moment of a force about a point is defined as the product of the force and the perpendicular distance from the line of action of the force to that point. B1 Marker’s comment: Mostly well done. Some did not get any credit as their phrasing of “perpendicular distance” is vague and did not reflect the essence of definition of moment. (iii) Taking moment about A (anticlockwise moments) = (clockwise moments) FB × 5.0 = 880 × 0.5 + 200 × 2.5 FB = (440 + 500) / 5.0 = 188 N M1 A1 Marker’s comment: Mostly well done. Common mistakes include not factoring in the moment due to weight of man or weight of plank. (b) The clockwise moment will increase due to the increase in perpendicular distance (between CG & point A), B1

2 thus the anticlockwise moment due to FB will have to increase to maintain equilibrium as well and magnitude of FB will increase. B1 A0 Marker’s comment: Poorly attempted. While many could correctly state that the CW moment will increase, they did not explain the reason for that increase. Similarly, they did not mention the corresponding increase in ACW to maintain rotational equilibrium and just drew link between increase in CW moment and hence increase in FB. 2 (a) Area under F-x graph is work done in stretching bow. More potential energy is stored Arrows gain more kinetic energy, has less deviation from intended flight path / more accurate OR Arrows gain more kinetic energy, has further range OR Arrow hits target faster B1 B1 Mostly well done. Most could identify that in stretching the bow more potential energy is stored and in turn identify reasonable advantage of that. (b) (i) By counting squares under F-x graph, Elastic potential energy = area under F-x graph = 1.25 J × 65 = 81 J Acceptable to +/- 2 squares M1 A1 Marker’s comment: Mediocre attempt. While some students were able to correctly identify that the EPE is given by area under F-x graph, they were unable to give an accurate estimate as they did not take into consideration the scale of both axes. Some students also used the wrong method – using formula (EPE = ½ kx2) to calculate without realising that F is not a constant. (b) (ii) By Conservation of Energy, Loss in Ep = Gain in Ek for arrow   22181 2 3.5 10 v  v = 68.0 m s-1 M1 A1 Marker’s comment: Mostly well done. Ecf was given and most students were able to apply Principle of COE to calculate the speed of arrow . Students who were not awarded the full credit mostly had calculation error. 3 (a) For a body moving in uniform circular motion, there is a change in velocity as the direction of motion is changing. The rate of change of velocity gives rise to an acceleration. Since the change in velocity is directed towards the centre of the circular motion, the acceleration is also towards the centre. (acceleration is perpendicular to velocity) *Resultant force (centripetal force) B1 B1

3 Marker’s comment: Mediocre attempt. Some students did not get any credit as they were merely describing circular motion and not explaining how the force allowed it to move in a circular path. (b) (i) 𝑎 = 𝑟𝜔2 = (5.0 cos 60˚) (2𝜋 3.5) 2 = 8.06 𝑚 𝑠−2 M1 A1 Marker’s comment: Mediocre attempt. Some students did not know the correct formula for acceleration, other did not know that 3.5 s is the period and its role in the ccalculation of acceleration. (ii) 𝑇 cos 60˚ = 𝑚𝑎 𝑇 = 0.40 × 8.06 cos 60˚ = 0.40 × 8.06 0.5 = 6.45 N M1 A1 Marker’s comment: Mediocre attempt. Some students were unable to get the correct tension because they did not resolve it to equate the horizontal component to the centripetal force, instead they just equated the tension to centripetal force reflecting their lack of understanding of how vectors are computed. (iii) 𝐿 = 𝑚𝑔 + 𝑇 sin 60˚ = (0.40 × 9.81) + 6.45 sin 60˚ = 9.51 N M1 A1 Marker’s comment: Mediocre attempt. Many did not include the vertical component of tension in the downward vertical force, hence they only balanced lift force with weight of the toy plane. (iv) The tension in the wire is unchanged, since centripetal force is unchanged. The vertical component of tension together with the new lift force now balance the weight of the toy plane. Hence new lift force will be lesser than before. M1 A1 Marker’s comment: Poorly attempted. Many thought that the since the magnitude of tension and centripetal force remain unchanged, the lift force will remain unchanged too . For students who were able to correctly deduce that lift force will be lesser than before, they could n ot explain the reason convincingly. 4 (a) (i) I = 1.8 + 0.60 = 2.4 A A1 Marker’s comment: Generally well done. (ii) VX = VY + VZ

4 0.60 × 8.0 = 1.8 (2.0 + RZ) RZ = 0.67 Ω C1 A1 Marker’s comment: Most are able to find the voltage and the current flowing through the resistor Z, thus enabling them to find Z resistance. Some wrongly equate the current in the two parallel branches to be the same (the voltage across the two parallel branches is the same). (iii) Either E = Ir + IXRX E = 2.4 × 1.5 + 0.60 × 8.0 Or E = Ir + IYRY + IRZ E = 2.4×1.5 + 1.8(2.0+0.67) Or 1 1 1 8 0 2 0 0 67effR . . .  8 0 2 672 4 1 5 8 0 2 67 ..E . . ..    E = 8.4 V C1 A1 Marker’s comment: Student are generally able to find the total resistance of the circuit and then the e.m.f. required. (b) lR A  XY YX RA RA 80 20 Y X .A .A 0 25X Y A .A  C1 A1 Marker’s comment: Students are generally able to compute the ratio of the area from the ratio of the resistance. Some are unable to manipulate the relationship between the resistance and the area while other wrongly thought that the resistance is inversely proportional to the square of the area. (c) Either output powerEfficiency input power IV IE 0 60 8 0 84 .. .  Or 1 1 1 8 0 2 0 0 67effR . . .  Reff = 2.00 Ω M1

5 output powerEfficiency input power IV IE   2 2 eff eff IR I R r    2 2 20 2 0 1 5 I. I . .   Efficiency = 57.1% A1 Marker’s comment: Not well done. It is surprising that many did not use power to compute efficiency. Some who tries to use voltage to compute efficiency should illustrate how the current in the output and input power is the same. 5 (a) The direction of the magnetic force acting on the electron is always normal to its velocity. The magnitude of the magnetic force is constant. Hence the magnetic force provides for the centripetal force, leading to the electron following a circular path. M1 M1 A0 Marker’s comment: Poorly done. Most are not aware that the question is asking how this scenario fulfil the conditions of circular motion. Many wasted their efforts to find the direction of the magnetic force and then simply state that this force acts as a centripetal force, without explaining why the force lead to circular motion. (b) Increase in KE = Decrease in PE ½ mv2 = qV Since p = mv, 21 2 mmv qVm 2 2 p qVm   2p mqV M1 M1 A0 Marker’s comment: Poorly done with many not attempting it. Students did not understand that the decrease in potential energy leads to increase in kinetic energy (and also moment