ASRJC Prelim H2P1 MS final

1 9749/01/ASRJC/2023Prelim [Turn Over Anderson Serangoon Junior College 2023 H2 Physics Prelim P1 Exam Solution Paper 1 (30 marks) E – Easy, A – Average, D – Difficult 1 2 3 4 5 6 7 8 9 10 D A B B B B C A C C 11 12 13 14 15 16 17 18 19 20 C B C C A D C D D A 21 22 23 24 25 26 27 28 29 30 D C C D A D A B C A Easy: 6, Average: 15, Difficult: 9 1 D II= = =W Q Fs masV tt Base units of V = -2 2 -3 -1kg×ms ×m = kgm s AA×s E 2 A . .I. 1 200 250 48= = = VR Reading in ammeter is recorded to 2 dp. For division, the s.f. of calculated data is recorded to the lowest s.f. of the raw data A 3 B Using s = ut + ½ t2 9 = u(2) + ½ a(2)2 --- (1) 21 = u(4) + ½ a(4)2 --- (2) Solving (1) & (2), a = 0.75 m s–2 D 4 B Initial vertical velocities for both stones, uy = 0 sy = uyt + ½ gt2 = ½ gt2 Both stones will take the same time to fall through the same height since they experience the same gravitational acceleration g. E 5 B Change in momentum = 2(5.0 – 30) = -50 Area under F-t graph = Change in momentum -F x (4 + 6) / 2 = -50 F = 10 N A

2 9749/01/ASRJC/2023Prelim 6 B Due to smaller weight, a smaller air resistance is needed by the foam ball to achieve zero net for to reach terminal velocity. Hence time required to reach terminal velocity is shorter. So option A and C are eliminated. In addition. since air resistance depends on speed, a smaller air resistance will require a smaller terminal velocity. H ence option D is eliminated. D 7 C Tension in string is 3.0W (2.0 W + 1.0W) Let distance between string and CG be d Taking moments abouts the CG (which is at 0.5x) (2.0 W) × (0.3x) = (3.0 W) × d d = 0.2x When load is moved, it will be 0.3x away from the CG (but on the other side) Hence, the string needs to be placed 0.2x from the CG (but on the other side) for rod to be in equilibrium. Distance moved by string = 0.2x + 0.2x = 0.4x A 8 A Option A: WD to change speed of object from 2 m s-1 to 3 m s-1 = ½ m (32 – 22) = ½ m (5) WD to change speed of object from 1 m s-1 to 2 m s-1 = ½ m (22 – 12) = ½ m (3) So option A is correct. Option B: When an object slows down, there has to be negative work done by a force (opposing the motion of the object). Hence option B is not correct. Option C: When lifting an object 1 m at constant velocity, WD = mgh = mg(1) When moving an object 1 m across a smooth horizontal surface at constant speed, there is no need for a force to keep it at constant velocity. Therefore, WD is zero. Hence option C is not correct. Option D: When an object is dropped, the force of gravity is in the direction of the motion of the object. There is WD by the force of gravity and is given by Wd. Hence option D is not correct. A 9 C Distance moved in 60 = 0.40 × ( )60 2360  = 0.4189 m Since both masses were displaced vertically by 0.4187, the net increase in total GPE of the weights = m g h = (20 − 15) / 9.81 × 9.81 × 0.4189 = 2.1 J D

3 9749/01/ASRJC/2023Prelim [Turn Over 10 C If the roller coaster have lost contact, the normal contact force of track on roller coaster must be zero. Centripetal force will be provided by component of weight pointing towards O, if it just remains in contact. At position P, Component of weight towards O = mg cos 30° To maintain contact, 2 -1 cos30 cos30 1.0(9.81)cos30 2.91 m s mvmg r v rg = = = = So, option C will be the best option. D 11 C gN = gE and 2 GMg R= NE 22 NE NN EE GG 17 4.1 = = = = MM RR RM MR A 12 B p2 T2 = p1 T1 p2 50 + 273 = 200000 + 100000 25 + 273 p2 = 325000 Pa Hence the gauge reading is 225 kPa. Percentage increase is (225 – 200) / 200 x 100% = 12.5% A 13 C 1st Law of Thermodynamics ΔU = Q + W Since no work is done by or done on the blocks, W = 0 QUΔ = Energy exchange with surroundings is negligible, so A

4 9749/01/ASRJC/2023Prelim Heat gained by copper = Heat loss by iron ironcopper QQ −= ironcopper UΔUΔ −= Also, coppercoppercopper θΔmcQ = ironironiron θΔmcQ = Since ironcopper cc  , ironcopper θΔθΔ  14 C v = 22 xxo − 2 1 22 0 2 0 2 0 22 0 22 0 2030 20 30 1300 25 67 5 0.86630 90020 0.866 3 ms2 m06 − −= − = −= = = − =  = x x x xv x v A 15 A 2 2 4 4 1 4 16 1 16 I I I II  == = X x x r x x 2 4 2 44 2 4 4 4 1Since 16 1 16 1 44 x xx x x x A A A A A AA AA  = = = =  = I I I, I I A 16 D For zero resultant amplitude, two waves of the same type must meet in antiphase and have the same amplitude. When that happens, destructive interference occurs, and the resultant amplitude is the difference between the individual amplitude of the two waves. When two waves are emitted from their sources with the same intensity, they can travel different distances to meet at a point. Since intensity and hence amplitude decreases with distance from the source, they will meet with different amplitude. So Option A is incorrect. Two waves must be in antiphase with each other when they meet, so Option B is incorrect. Two waves do not need to travel in opposite directions for resultant amplitude to be zero. They can travel in any directions, as long as they fulfil the conditions stated in the first paragraph, to give zero resultant amplitude. So Option C is incorrect. E

5 9749/01/ASRJC/2023Prelim [Turn Over 17 C Since loudspeakers emit waves that are in antiphase, the condition for waves to meet in phase is path difference = (n + ½ )  For Options A and C, since OY < XY, observer O is in between X and Y as shown. For Option A, OX = XY – OY = 3.70 – 1.25 = 2.45 m Path difference of the waves from their respective sources to the observer = OX – OY = 2.45 – 1.25 = 1.20 m = 2 , so not correct. For Option C, OX = XY – OY = 2.50 – 2.00 = 0.50 m Path difference of the waves from their respective sources to the observer = OY – OX = 2.00 – 0.50 = 1.50 m = 2.5 , so correct. For Options B and D, since OY > XY, observer is to the left of X as shown in original diagram. Path difference of the waves from their respective sources to the observer = XY Option B: Path difference = 1.20 = 2 , so incorrect. Option D: Path difference = 1.60 m = 2.7 , so incorrect. D 18 D y =½at2 ………..(1) x = ut ………….(2) from (1) & (2) 2 2= 2 axy u , i.e y  x2 p: q: r: s = x2: (2x)2: (3x)2: (4x)2 = 1: 4: 9: 16 D 19 D I = nevA v = I /neA since I , n, e are constants v  1/A  1/d2 2 2 2 2 (2 ) 4 1 yx yx x y dv vd vd vd = == E

6 9749/01/ASRJC/2023Prelim 20 A Rate of arrival of electrons per unit area = number of electrons per unit time per unit area = n tA , where n is the number of electrons, t is time and A is area Since current, ne n t t e=  = II Therefore, rate of arrival of electrons per unit area = eA I E 21 D Volume, V =  d2 L When V is constant, L  1/d2 2 2 2 1 4 L LR d LLRL d  =    A 22 C Option A: Resistance of thermistor increases. By potential divider, p.d across thermistor increases. Hence voltmeter reading is negative. Option B: By potential divider, p.d across fixed resistor will decrease. Voltmeter reading will be negative. Option C: Resistance of LDR increases. By potential divider, p.d across thermistor decreases. Hence voltmeter reading is positive. Option D: same effect as Option B. D 23 C Since the net magnetic field at d from wire Y is zero, IX and IY flow in opposite directions. BI and 1B r At d