Solutions to SQA examination

1995 Higher Grade Physics

Paper II Solutions

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1.a.	Vectors have magnitude and direction, whereas, scalars have magnitude only.

Use the cosine rule to calculate the magnitude of the displacement. a2 = b2 + c2 -2bcCosA a2 = 7002 + 7002 -2x700x700Cos45o a2 = 980000 - 692964.65 a2 = 287035.35 a = 535.7m As the triangle is isosceles angles B and C are equal in size. Angle C = (180 - 45)/2 = 67.5o Displacement (s) = 535.7m at a bearing of 067.5o b.ii. Velocity(v) = Displacement(s)/Time(t) vandy =535.7/t t1 = Total distance(d)/Speed t1 = 1400/3 t1 = 466.67s =>vAndy = 535.7/466.67 vAndy = 1.15m/s The velocity bearing is also 067.5o v = 1.15m/s (067.5o) b.iii.vPaul = 2.5s (067.5o) b.iv. t2 = Total distance(d)/Speed t2 = 535.7/2.5 t2 = 215.1s Including Paul's five minutes waiting time the total time(ttotal)) for Pauls journey, in seconds, is can be calculated. ttotal) = t2 + 215.1 ttotal) = 5x60 + 215.1 ttotal) = 515.1s Pauls total time = 515.1s Andy's journey time(t1) = 466.67s Andy reaches the checkpoint ahead of Paul by 48.4s 2.a. Sprinters P and Q have the same journey time. For sprinter P: s = 20m a = 1.6m/s/s u = 0m/s t = ? Use: s = ut + 1/2(at2) 20 = 0xt + 1/2(1.6xt2) t2 = 20/0.8 t2 = 25 t = 5s The time for both sprinters is 5s b. vP = u + at vP = 0 + 1.6x5 vP = 8m/s vQ = u + at vQ = 0 + 1.2x5 vQ = 6m/s c. To calculate the displacement of Q from their starting point to the finishing line use: s = ut + 1/2(at2). s = 0xt + 1/2(1.2x52) s = 15m The distance between starting points is therefore 5m 3.a.i. Initially the sand filled box is at rest and has no kinetic energy. The bullet embeds itself in the sand filled box and its kinetic energy is shared with the box. This constitutes an inelastic collision and kinetic energy is not conserved. However, after this event the kinetic energy of the bullet and box system can be calculated using the principle of conservation of energy. This means that the potential energy gain of the bullet and box, at their maximum height, is equal to the initial kinetic energy of the bullet and box. Note: the bullet and box have no kinetic energy at the maximum height. Epgain = Ekinitial mgh = 1/2(mv2) v2 = 2gh v2 = 2x9.8x0.1 v2 = 1.96 v = 1.4m/s a.ii. To solve this problem apply the law of conservation of linear momentum. Momentum before(Pbefore) = Momentum after(Pafter) mbulletubullet + mboxubox= (mbullet+mbox)v The initial velocity of the box is zero therefore the term with this value can be ignored. ubullet = (mbullet+mbox)v/mbullet ubullet = (10x1.4)/0.025 ubullet = 560m/s b. The change in momentum of the bullet in this collision will be greater than in the first experiment. To conserve momentum this means that the change in momentum of the box of sand will also be greater. The only way that this change in momentum can be increased is if the box moves off with a greater velocity. The box thus has a greater initial kinetic energy that will be transferred into potential energy resulting in the box reaching a greater height. 4.a. The tension force provided by the cable will be equal in size to the weight of the block, but act in the opposite direction.
W = ? w = mg m =5.0x103kg w = 5.0x103x9.8 g = 9.8N/kg w = 49000N b.i.
Weight(W) = Tension(T) + Buoyancy(B) T = W - B B = W - T B = 49000 - 29000 B = 20000N b.ii. The pressure acting on the lower surface of the block is greater than the pressure acting on the upper surface, because the lower surface is at a greater depth.
As the force on each surface is given by the equation F = PA, the upward force is greater than the downward force. This net upward force is called buoyancy or upthrust. c. The pressure difference will be the same at all depths. This will result in the upthrust being the same at any depth. 5.a. The circuit in the problem can be drawn in the more familiar form as shown in the diagram below.
R1 is in series with R2 R1+2 = 20W R3 is in series with R4 R3+4 = 20W R1+2, R3+4 and R5 are in parallel. 1/Rtotal = 1/R1+2 + 1/R1+2 + 1/R5 1/Rtotal = 1/20 +1/20 +1/10 1/Rtotal = 4/20 Rtotal = 20/4 = 5W b.i. The greatest current will flow through R5 thus this is where the greatest power will develop. b.ii. IR5 = V/R5 IR5 = 12/10 IR5 = 1.2A P = IR52R5 P = 1.22x10 P = 14.4W c.i. The electric potential is the same at points A and C. With no no potential difference across R5 no current will be flowing through it. c.ii. Resistors R1 and R2 are in series. The current flowing through branch BCD can be calculated using ohms law. IBCD = VBD/R2+3 IBCD = 12/20 IBCD = 0.6A Similarly the current flowing through branch BAD can be calculated. IBAD = VBD/R2+3 IBAD = 12/20 IBAD = 0.6A Ibattery = IBCD + IBAD Ibattery = 0.6 + 0.6 Ibattery = 1.2A 6.a. "Lost volts" = Ir "Lost volts" = 0.5x.2 "Lost volts" = 0.1V emf(E) = Terminal Potential Difference(Vtpd) + "Lost volts" Vtpd = V1 V1 = E - "Lost volts" V1 = 12.0 - 0.1 V1 = 11.9V 7.a. E = I(R+r) E/I = R+r R = E/I-r As required b.i. The above equation can be written as: R = E(1/I) - r Compare this to: y = mx + c The y-intercept(c) is equal to -r. The gradient(m) is equal to E. c = -2.5 =>r = 2.5W b.ii. m = (y2-y1)/(x2-x1) m = (6-0)/(0.5-0.15) m = 6/0.35 m = 17.1 =>E = 17.1V c. I = E/r I = 17.1/2.5 I = 6.8A 8.a. The minimum energy an electron needs to absorb to escape from an atom is called the work function. b.i. Ephoton = hf Ephoton = 6.63x10-34x6.1x1014 Ephoton = 4.04x10-19J Emax. kinetic = Ephoton - Ework Emax. kinetic = 4.04x10-19 - 3.04x10-19 Emax. kinetic = 1.00x10-19J b.ii. Ektotal = Emax. kinetic + Efield Note: Efield is the energy gained by the electron in the electric field. Efield = qV Efield = 1.6x10-19x0.8 Efield = 1.28x10-19 Ektotal = 1.00x10-19 + 1.28x10-19 Ektotal = 2.28x10-19J c. To stop the photoelectrons the wok done by the electric field must equal the maximum kinetic energy of the photoelectrons. Efield = Emax. kinetic qV = Emax. kinetic V = Emax. kinetic/q V = 1.00x10-19/1.6x10-19 V = 0.625V 9.a. qcrit = sin-1(1/n) qcrit = sin-1(1/1.33) qcrit = sin-1(0.752) qcrit = 48.7o
Note: Some internal reflection occurs for the light incident at the first and second boundary, however, for clarity these reflected rays have been omitted from the diagram. 9.b. To an observer the light appears to originate from where the broken rays cross. This produces a virtual image of the lamp at the apparently shallower depth.
a.ii. When forward biased the majority charge carriers in the n-type material, electrons, flow to the p-type material. This movement of electrons also makes it appear that holes in the p-type material move towards the n-type material. a.iii.When conduction band electrons in the n-type material pass into the p-type material they fall into lower energy holes and emit energy as as a visible photon, if the energy loss is equal to the energy of a visible photon. b.i Conduction starts when the applied voltage is 0.5V. b.ii. The resistance of the diode decreases as the applied voltage increases. This can be justified by using ohms law to calculate the resistance at different applied voltages. R = V/I R1 = V1/I1 R1 = 1.0/0.275 R1 = 3.6W R2 = V2/I2 R2 = 1.2/0.5 R2 = 2.4W 11.a. The reaction is an example of nuclear fusion. b. The total mass of the products from the fusion reaction is less than the mass of the initial reactants. The difference in mass is converted into energy. The amount of energy produced can be calculated using the equation: E = mc2, where m is the difference in mass and c is the speed of light. c. Mreactants = (5.00890x10-27+3.34441x10-27)kg Mreactants = 8.35331x10-27kg Mproducts = (6.64632x10-27+1.67490x10-27)kg Mproducts = 8.32122x10-27kg Mass defect = Mreactants - Mproducts Mass defect = 8.35331x10-27 - 8.32122x10-27 Mass defect = 0.03209x10-27kg E = Mdefectc2 E = 0.03209x10-27 x (3x108)2 E = 2.8881x10-12J d. Number of reactions = Total energy per second/energy per reaction Number of reactions = 25x106/2.8881x10-12 Number of reactions = 8.658x1018

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