MI 2024 PU3 H2 PHYSICS PRELIM P2 ANS

2024 PU3 H2 Physics Prelim Paper 2 Suggested Answers 1 (a)(i) The vertical acceleration of the ball is constant in magnitude (9.81 ms-2) and directed downwards throughout the path. Hence as the ball moves from the ground to the highest point, vertical velocity decreases at a constant rate till vertical velocity reaches zero. As the ball moves from the highest point towards the ground, vertical velocity increases at constant rate. B1 B1 B1 (a)(ii) The horizontal acceleration of the ball is zero throughout the path. Hence the horizontal velocity is constant in magnitude. B1 B1 (b)(i) Considering the ball just after it leaves the ground to the point just before it hits the ground, 𝑣𝑦 = 𝑢𝑦 + 𝑎𝑦𝑡 15 sin 20𝑜 = ( −15 sin 20𝑜) + (9.81)𝑡 𝑡 = 30 sin 20𝑜 9.81 𝑠𝑥 = 𝑢𝑥𝑡 = (15 cos 20𝑜) 30 sin 20𝑜 9.81 = 14.7 m C1 C1 A1 (b)(ii) Considering the ball just as it drops down the cliff to just before it hits the ground. 𝑠𝑦 = 𝑢𝑦𝑡 + 1 2 𝑎𝑦𝑡2 70 = −15 sin 20𝑜𝑡 + 1 2 (9.81)𝑡2 t = - 3.29 (NA) s or 4.34 s 𝑠𝑥 = 𝑢𝑥𝑡 − 14.7 = (15 cos 20𝑜) (4.34) − 14.7 = 46.4 𝑚 C1 C1 C1 A1

2 (a) Gravitational field strength is defined as the force per unit mass acting on a small mass placed at a point in the gravitational field. A1 (b) For the same difference in gravitation potential, the distance between the potential lines is increasing as the distance from the surface of Mars, show that gravitation field strength is decreasing. A1 (c) Loss in gravitational potential energy = gain in kinetic energy of the satellite (90)(-6.0 –(-8.0)) x 106 = (1/2)(90)v2 v = 2000 ms-1 C1 A1 (d)(i) Centripetal force of the satellite is provided by the gravitation force. 𝑚𝑟(2𝜋 𝑇 )2 = 𝐺𝑀𝑚 𝑟2 𝑇 = √4𝜋2𝑟3 𝐺𝑀 = √ 4𝜋2(3.4 ×106+1.7 ×107)3 𝐺(6.4 ×1023) = 88607 s = 24.6 h C1 C1 (correct radius) A1 (d)(ii) KE = 1 2 𝑚𝑟2𝜔2 = 1 2 (90)(3.4 × 106 + 1.7 × 107)2 ( 2𝜋 88607) 2 = 9.42 x 107 J Alternative KE = 𝐺𝑀𝑚 2𝑟 = 𝐺(6.4 ×1023)(90) 2(3.4 ×106+1.7 ×107) = 9.42 x 107 J C1 A1 (d)(iii) Due to work done against friction, total energy of the satellite decreases, causing the satellite to fall towards Mars (or radius of orbit decreases). As the satellite fall towards Mars, it experiences a loss in gravitational potential energy and an increase in kinetic energy. (Or Since KE = GMm/2r as radius decreases, KE increases) B1 B1

3 (a) Despite both being at 100 C , steam transfers more (heat or thermal) energy to the skin than boiling water. This is because steam releases (heat) energy proportional to the latent heat of vaporisation during condensation in addition to the heat energy released on cooling of condensed water on skin. This is much higher than the (heat) energy released by boiling water on contact with the skin. B1 B1 (b)(i) Use state A: 𝑛 = 𝑃𝐴𝑉𝐴 𝑅𝑇𝐴 = 20×105×1.2×10−4 8.31×800 0.0361 moln = If state D is used, 0.0362 moln = . C1 A1 (b)(ii)1. 5 -420 10 5.2 10 8.31 0.0361 BB B B PVT Rn   ==  3467 KBT = If 0.0362 moln = is used, 3457 KBT = . OR 𝑉𝐴 𝑇𝐴 = 𝑉𝐵 𝑇𝐵 1.28 800 = 5.2 𝑇𝐵 𝑇𝐵 = 3467 K A1 (b)(ii)2. 3 2U nR T =  3 2 0.0361 8.31 (3467 800)U =    − 1200 JU = (allow ecf) Accept 3 2 1200 JpV= . C1 A1 (b)(ii)3. 5420 10 (5.20 1.20) 10W P V −= −  = −   −  800 JW =− Award 1 mark for +800 J. C1 A1 (b)(ii)4. From the first law of thermodynamics, U Q W = + Q U W= − ( )1200 800 2000 JQ= − − = (allow ecf) A1 (b)(iii) From D to A, since Q is zero, DA DAWU = ( )3 DA 2 0.0361 8.31 800 226 258 JW =    − = or 259 J if used n 0.0362 net by gas 800 390 0 259 932 JW = + + − = or 931 J is used n=0.0362 Allow ecf. Students can also use 𝑊DA = 3 2 (𝑃𝐴𝑉𝐴 − 𝑃𝐵𝑉𝐵) = 258 J C1 C1 A1

4 (a) In a polarised wave, the vibrations of wave particles are limited to only one axis; whereas an unpolarised wave is not (i.e. no specific axis of vibrations or many different axis of vibrations). B1 (b)(i) 0 as no light will be able to pass through the polariser given it is perpendicular. A1 (b)(ii)1. A3 = A2 cos(62°-23°) A2 = A1 cos(23°) thus A3 = A1 cos(23°) cos(62°-23°) = 0.715 A1 M1 M1 A0 (b)(ii)2. let the initial amplitude (unpolarised light) = A0 1 2 = ( 𝐴1 𝐴0 )2 𝐴0= √2𝐴1 = 1.41 A1 𝐼3 𝐼0 = ( 𝐴3 𝐴0 )2 = ( 0.715𝐴1 √2𝐴1 )2 = 0.256 Percentage of intensity reduced = (1 - 0.256) x 100% = 74.4% C1 C1 A1

5 (a)(i) RR = 105Ω RTh = 50 Ω A1 A1 (a)(ii) 𝑉 = 50 105+50 × 12 = 3.87 V C1 A1 (a)(iii) The p.d across resistor and thermistor is equal so this implie s that the resistance across each component is equal. Hence the temperature is 50 °C. C1 A1 (b)(i) p.d across length x = 𝑥 𝑙 E Since the appliance is parallel to length of x, p.d is the same. Thus p.d across the appliance is x = 𝑥 𝑙 E M1 A1 (b)(ii) p.d across appliance = 20 / 100 x 5.0 = 1.0 V current = 1/10.0 = 0.10 A A1 (b)(iii) ε = 45.0 / 100 x 5.0 = 2.25 V A1

6 (a) A magnetic field is a region of space in which a magnetic force is experienced by moving charges or current -carrying conductors or permanent magnets placed in this field. A1 (b)(i) As the positive ion passes into the region of magnetic field, by Flemming’s left hand rule, there is a magnetic force on the ion towards the positive plate. Due to the electric field, there will be a electric force on the ion towards the negative plate. If the magnitude of these two forces are equal, the ion will be able to pass through without deviation. (If student state “Electric force is opposite in direction to magnetic force” award only 1 mark out of the first 2 B1 marks). B1 B1 B1 (b)(ii)1. In the region with B -field only there exist a magnetic force (of constant magnitude) perpendicular to direction of motion causes ions to undergo centripetal acceleration / provides the centripetal force. A1 (b)(ii)2. 𝐵′𝑞𝑣 = 𝑚𝑣2 𝑟 𝑚 𝑟 = 𝐵′𝑞 𝑣 = constant 𝑀 26.2 = 12𝑢 22.4 𝑀 = 14𝑢 C1 A1

7 (a)(i) Photoelectric effect is the ejection/emission of an electron from a metal surface when the surface is irradiated with electromagnetic radiation of a high enough frequency. A1 (a)(ii) Allow 350 nm to 400 nm A1 (a)(iii) Threshold frequency is determined when there is no responsivity from photocathode/ no release of photoelectrons. From Fig. 7.2, threshold wavelength is approximately 640 mm. Threshold frequency is found by using speed of light divided by threshold wavelength (or 630 nm) B1 B1 (a)(iv) Using hc/λ = work function + Ek, max, hc/(450 x 10-9) = h (4.76 x 1014) + 0.5(9.11 x 10-31)v2 v = 5.27 x 105 m s-1* C1 C1 A1 (a)(v) Using a low work function photocathode will allow electrons to be emitted easily with photons of larger wavelength. A1 (b)(i) Current = nq/t 7.8 x 10-4 = (n/t)*(1.6 x 10-19) n/t = 4.88 x 1015 s 1 C1 A1 (b)(ii) Number of photons per unit time = 4.88 x 1015 /(0.01) = 4.88 x 1017 A1 (b)(iii) 4.8 mA of current is detected from photocathode when 1 W of power of EM radiation is incident on it. A1 (c)(i) Use wavelength = 225 nm, R = 9.0 mA/W and wavelength = 540 nm (or 550 nm), R = 22 mA/W to correctly determine η at 5 %. B1 B1 (c)(ii) Correct computation of values with maximum values of 2sf To deduct one mark for one wrong calculation. η / % λ / nm R / mA W-1 3 300 7.2 3 400 9.7 3 500 12 3 600 14