Part A Worked solutions ■■■■■■■ Unit Area of study 1: Motion in one and two dimensions 3 ■■■■■■■■■ Forces in action 1. u = 0, x = 10 m, t = 4.0 s, a = ? x = ut + 12 at 2 1 2 Part A — WORKED SOLUTIONS 4. The stationary car is pushed forward by the other vehicle. As a result, the seat pushes the body of an occupant forward. This happens almost instantaneously. However, without a headrest, there is nothing to push the occupant’s head forward quickly. The head remains at rest until pulled forward by the spine (Newton’s First Law of Motion). The head applies an equal and opposite force to the spine, (Newton’s Third Law of Motion) potentially causing serious injuries. 5. To say that the passenger is thrown forward implies that a force accelerates the passenger. The car slows down rapidly in most collisions as a result of a large external force. The passenger continues to move at the original speed of the car while the car slows down. 6. (a) (b) Chapter 1 ⇒ 10 = Jacaranda Physics 2, 3rd Edition TSK × a × 16 ⇒ 10 = 8a ⇒ a = 1.25 m s –2 Over first 5.0 m u = 0, x = 5.0 m, a = 1.25 m s−2 v=? v2 = u2 + 2ax = 2 × 1.25 × 5.0 ⇒ v = 3.5 m s−1 2. 10 m s−2 down 3. (c) 7. (a) F (down) (b) C (perpendicular to surface) (c) X (Constant velocity must be the result of a zero net force.) 8. (a) Δv = v − u ⇒ Δv = 50 km h−1 30 tan θ = 40 ⇒ θ = 37° Δv a= Δt 50 km h S 37°E 2.0 s = 25 km h−1 s−1 S 37°E 25 km h –1 a= S 37°E 1s (25 ÷ 3.6) m s –1 = S 37°E 1s = 6.9 m s−2 S 37°E = (b) (a) W = mg = 4.0 × 10 = 40 N (b) TH = 12 cos 30° = 10 N (c) Net vertical force = 0 ⇒ N + T sin 30° = W ⇒ N + 12 sin 30° = 40 ⇒ N = 40 − 12 sin 30° = 34 N 9. The matching reaction to the gravitational pull of Earth on you is the gravitational pull of you on the Earth. –1 Jacaranda Physics 2 TSK 2 © John Wiley & Sons Australia, Ltd 2009 10. The net force must be in the direction in which the elevator is speeding up. (a) (i) Normal reaction force is equal to your weight and the net force is zero. (ii) Normal reaction force is equal to your weight. The speed is not changing so the net force must be zero. (iii) Normal reaction force is greater than your weight because the speed of the elevator is increasing in an upwards direction. (iv) Normal reaction force is less than your weight because the net force must be down to make the elevator decrease its upward speed. (b) You feel the sensation of weight only if there is an upward push on you by an object like the ground, a floor, a chair or a bed. The apparent weight that you feel is the size of this upward push. If the elevator is accelerating upwards, the normal reaction force is greater than your weight and you feel heavier. If the elevator is accelerating downwards, the normal reaction force must be less than your weight and you feel lighter. Δv 11. (a) a = t –2.0 – 5.0 = 0.20 = −35 m s−2 = 35 m s−2 opposite to the initial direction of the dodgem car. (b) Fnet = ma = 200 × 35 = 7.0 × 103 N 12. (a) Braking distance = area under graph over last 20 s = 12 × 20 s × 20 m s –1 (d) a = Δv Δt ⇒ Forward force = 24 000 N = 2.4 × 104 N –20 m s –1 20 s = – 1.0 m s –2 = 13. (a) Fnet = ma = −4.0 × 104 × 1.0 = −40 000 N Additional frictional force = 40 000 − 8000 = 32 000 N = 3.2 × 104 N (b) N = W cos 20° = mg cos 20° = 1500 × 10 × cos 20° = 1.4 × 104 N (c) Zero (Car is at rest.) (d) Fnet = 0 ⇒ mg sin 20° − R = 0 ⇒ R = mg sin 20° = 1500 × 10 × sin 20° = 5.1 × 103 N 14. (a) No direction. The net force is zero. (b) = 200 m (b) Total distance travelled by train (and cyclist) = area under graph = 1 2 × 50 s × 20 m s –1 + 50 s × 20 m s –1 + 200 m = 1700 m distance travelled time taken 1700 m = 120 s = 14 m s –1 constant speed = (c) a = 15. Δv Δt ⇒ mg sin 30° − (air resistance + friction) = 0 air resistance + friction = mg sin 30° = 60 × 10 × sin 30° = 300 N 20 m s –1 50 s = 0.4 m s –2 = Fnet = ma = 4.0 × 104 × 0.40 = 16 000 N Total frictional forces = 8000 N ⇒ Forward force − 8000 N = 16 000 N Jacaranda Physics 2 TSK 3 © John Wiley & Sons Australia, Ltd 2009 16. (a) Fnet = ma = 70 × 2.0 = 140 N north ⇒ tension − frictional forces = 140 ⇒ tension − 240 = 140 ⇒ tension = 380 N north (b) Fnet = ma = 350 × 2.0 = 700 N north ⇒ thrust − tension − frictional forces = 700 ⇒ thrust − 380 − 600 = 700 ⇒ thrust = 1680 N north (1.7 × 103 N north) 18. (a) Fnet = m −2.7 0.10 = 5.4 N opposite to the original direction of motion Impulse on billiard ball = mΔv = 0.200 × −2.7 = 0.54 N s opposite to the original direction of motion The table doesn’t move because the net force acting on it is zero. It is zero because friction applied to the table by the floor is enough to match the force applied by the ball. The net force on the bullet is large because the forces acting on it, other than the one exerted by the rifle, are small. The net force on the rifle is zero if the shooter exerts a force on it to balance its weight and the force applied by the bullet. Therefore, the rifle does not accelerate. pf = pi ⇒ 4.0 kg × vrifle + 0.020 kg × 300 m s−1 = 0 ⇒ 4.0vrifle = −6.0 ⇒ vrifle = −1.5 m s−1 −1 ⇒ recoil speed is 1.5 m s Δp = impulse = area under graph = 10 N × 6 s = 60 kg m s−1 Δp = impulse = area under graph = 10 N × 6 s + 5 N × 6 s = 90 kg m s−1 ⇒ pf − pi = 90 kg m s−1 ⇒ pf = 90 kg m s−1 = 0.200 × (b) (c) 19. (a) (b) The net force on the magpie is zero. Resolving vertically: 2T sin θ = mg mg ⇒T = 2 sin θ 0.04 tan θ = 2.0 ⇒ θ = 1.146° 4.0 × 10 ⇒T = 2 × sin 1.146° = 1000 N This answer is based on the assumption that the wire has zero mass and is perfectly flexible. 20. (a) (b) 17. (c) Let the mass of the globe be m and assume that the mass of the wire is negligible. Resolving vertically: T cos θ = mg ⇒ T cos θ = 10m Resolving horizontally: Fnet = ma ⇒ T sin θ = ma ⇒ T sin θ = 1.5m Divide equation (2) by equation (1). sin θ 1.5 ⇒ = cos θ 10 ⇒ tan θ = 0.15 21. In each of cases (a), (b) and (c), the change in momentum is fixed. An increase in the time interval during which the momentum changes results in a smaller force applied to the occupants or cyclist as Δp = FΔt. 22. (a) Assigning original direction of motion of car as positive: (i) ⇒ Impulse = mΔv = 70 (0 − 14) = −980 N s (ii) Impulse = mΔv = 70 (0 − 14) = −980 N s ⇒ θ = 8.5° Jacaranda Physics 2 TSK Δv Δt 4 © John Wiley & Sons Australia, Ltd 2009 x = 0.50 m, u = 14 m s−1, v = 0, a = ? v2 = u2 + 2ax ⇒ 0 = 196 + 2a × 0.50 ⇒ a = −196 m s−2 (iv) x = 0.025 m, u = 14 m s−1, v = 0, a = ? v2 = u2 + 2ax ⇒ 0 = 196 + 2a × 0.025 –196 ⇒a = 0.05 = −3920 m s−2 (b) g = 10 m s−2 acceleration of driver = −196 m s−2 196 = g 10 = 19.6 g acceleration of passenger’s head = −3920 m s−2 3920 = g 10 = 392 g (c) The seatbelts allow the change in momentum of the occupant to take place over a longer time interval, thus reducing the force applied to the occupant. Without a seatbelt, the occupant continues to move forward, colliding with the interior of the car. This ‘secondary’ collision causes the change in momentum of the occupant to take place in a much smaller time interval than would be the case with a seatbelt. The force applied to the occupant is therefore much greater. (b) The vertically downward momentum of the coal decreases to zero because there is an upward net force acting on it when it strikes the cart. The total momentum of the Earth−coal system has not changed. (c) pi = 750 × 2.0 = 1500 kg m s−1 south When the coal falls from the cart it has a horizontal velocity of 2.0 m s−1 south. ⇒ 500 v + 250 × 2.0 = 1500 kg m s−1 south ⇒ 500 v = (1500 − 500) kg m s−1 south 1000 ⇒v= kg m s –1 south 500 = 2.0 m s−1 south 5. (a) p = mv = 70 × 2.0 east = 140 kg m s−1 east (b) Three seconds before impact, Dean is 6 m from Melita because he is gliding at 2.0 m s−1 east. Taking Dean’s position as the origin: m x + m2 x2 xcm = 1 1 m1 + m2 (iii) 70 × 0 + 50 × 6 120 = 2.5 m The centre of mass is 2.5 m east of Dean. (c) Before the collision, the centre of mass travels 3.5 m (from Dean to Melita) in 3.0 s. Δx v= Δt 3.5 m east = 3.0 s = (d) Chapter 2 Collisions (e) 1. Impulse is equal to the change in momentum or words to that effect. 2. To increase the time interval during which the momentum of the vehicle’s occupants changes during a collision. 3. No. The system of the two cars is not isolated. There are unbalanced frictional forces acting on the cars during and immediately after the collison. 4. (a) The system of the coal and railway cart can be considered to be isolated if the horizontal motion only is considered. pi = 500 × 3.0 + 0 (initial horizontal momentum of the coal is zero) = 1500 kg m s−1 south pf = 750 v where v = final velocity of cart pf = p i ⇒ 750 v = 1500 south ⇒ v = 2.0 m s−1 south Jacaranda Physics 2 TSK 6. (a) (b) (c) 5 = 1.2 m s –1 east The momentum of the centre of mass remains constant. ⇒ The common velocity of Melita and Dean after the collision is 1.2 m s−1 east. Impulse on Melita = her change in momentum = pf − pi = 70 kg × 1.2 m s−1 east − 0 = 60 N s east Assigning east as positive pi = 1500 × −20 + 2000 × 20 = 10 000 kg m s−1 pf = pi ⇒ 3500v = 10 000 ⇒ v = 2.9 m s−1 east (2.86 m s−1 east) Impulse on truck = its change in momentum = pf − pi = 2000 × 2.86 − 2000 × 20 = −3.4 × 104 = 3.4 × 104 N s west Δvcar = v − u = 2.9 − (−20) = 22.9 m s−1 Δvtruck = v − u = 2.9 − 20 = −17.1 m s−1 © John Wiley & Sons Australia, Ltd 2009 ⇒ The car experiences the greatest (in magnitude) change of velocity. (d) Δpcar = −Δptruck since the total change in momentum is zero. This can be verified. Δpcar = pf − pi = 2000 × 2.86 − 2000 × 20 = −3.4 × 104 = 3.4 × 104 N s west (e) Each vehicle experiences the same force (in magnitude) (Newton’s Third Law of Motion). 7. Student responses will vary. In most situations, an occupant in a larger car is safer. However, the design of the particular car, the nature of the collision and several other factors influence the likely effect of an accident on occupants. The following points should be made. • The forces applied to each car by the other are equal in magnitude and opposite in direction. • The change in velocity of each car is dependent on the mass of the car. Assuming that the sum of the forces other than that applied by the other car is zero, the change in velocity is inversely proportional to the mass of the car. • Assuming that you are properly restrained and that the collision is head-on, your change in velocity (and therefore the deceleration you are subjected to) is less if you are in a heavier car. • The body continues to move in the original direction and at the original speed of your car until an unbalanced force acts on you. If you are not restrained, the unbalanced force will be provided by the windscreen or part of the interior of the car, which has already slowed down. A smaller car will have slowed down more, so the impulse applied to you (mΔv) will be greater. 8. (a) Ek = 12 mv 2 = 1 2 W = ΔEk = (b) 2 Fav x = 9722 9722 0.70 = 1.4 × 104 N (c) An estimate of the depression of the dashboard by the driver needs to be made. If the dashboard is depressed by 5 mm, then 9722 Fav = 0.005 ⇒ Fav = = 2 × 106 N (approx.) 10. (a) Fspring + weight = 0 ⇒ Fspring = −mg = 1.0 × 10 = 10 N up F (b) k = x = gradient of F vs x graph = 15 N 0.40 m = 38 N m –1 (37.5 N m –1 ) (c) A — the spring with the greatest spring constant (d) W = mg = 0.500 × 10 = 5.0 N x = 0.50 m (from graph) Work done = area under F vs x graph = 12 × 0.50 × 5.0 = 1.3 J (1.25 J) (e) Greatest strain energy occurs in the spring with the F vs x graph of greatest area. At maximum extension Area A = 12 × 0.20 × 25 = 2.5 J = 1.8 × 105 J (b) 1.8 × 105 J (c) x = 0.40 m, u = 20 m s−1, v = 0 m s−1 v 2 = u 2 + 2as Area B = 2 v −u 2x 0 − 202 = 2 × 0.4 –400 = 0.8 = 500 m s –2 a= ⎛ 60 ⎞ × 70 × ⎜ ⎟ ⎝ 3.6 ⎠ = 9.7 × 103 J × 900 × 202 2 1 2 1 2 × 0.40 × 15 = 3.0 J Area C = 1.25 J (from (d)) Spring B has greatest strain energy. 11. (a) 2.3 × 104 N (b) 4.0 × 103 N 12. (a) F = ma = 900 × 500 = 4.5 × 105 N opposite to the initial direction of motion of the car 9. (a) Assume that the sum of the forces other than that of the seatbelt applied to the driver is zero. Jacaranda Physics 2 TSK 6 © John Wiley & Sons Australia, Ltd 2009 (b) ΔEgp = Area under graph of force vs height = m × area under graph in 12(a) = 150 × 12 = 1.8 × 103 J (c) Work = ΔEgp = 1.8 × 103 J 13. Let Egp = 0 at ground Air resistance can be assumed to be negligible. initial Egp (car) (a) initial Egp (cricket ball) = mgh (car) mgh (cricket ball) = M car M cricket ball = 1600 kg 0.160 kg ⇒ v 2 = 250 ⇒ v = 16 m s −1 (c) WN = 0 since there is no displacement in the direction of the force. (d) Work done by net force = change in kinetic energy ⇒ Fnet x = ΔEk ⇒ (mg sin 30° − friction) × x = 12 mv 2 ⇒ (300 − friction) × 25 = ΔEgp (car) Ek (car) (b) = Ek (cricket ball) ΔEgp (cricket ball) (b) mg Δh (car) mg Δh (cricket ball) 1600 kg = 0.160 kg = 10 000 = 1 mv 2 (car) 2 2 1 mv 2 ⇒ (cricket ball) 2 1600 vcar 2 0.160 vcricket ball = 2.475 × 2 2.5 = 1.4 m s−1 (c) F = kx, where x = compression When x = 10 cm, F = 22 N F k= x 22 N = 0.10 m Ek (car) Ek (cricket ball) = 10000 = 220 N m –1 2 vcar = 1 2 × 450 × (2.0) 2 900 × 2 3.0 × 107 = 7.7 × 10−3 m (b) Work done = strain energy = 900 J (c) 2.0 m s−1. If the rubber bumper obeys Hooke’s Law as it expands to its original shape, all of the energy will be returned to the kinetic energy of the dodgem car. = 7500 J ΔEk = Wgravity = Wgravity (since initial Ek = 0) ⇒ 12 × 60v 2 = 7500 ⇒ 30v 2 = 7500 Jacaranda Physics 2 TSK mv 2 ⇒x = = 60 × 10 × 25 sin 30° ⇒ 1 2 k = gradient of F vs compression graph 300 kN = 0.01 m = 3.0 × 107 N m−1 ⇒ 12 × 3.0 × 107 × x 2 = 900 (a) W = Fx where x = displacement in direction of force ⇒ Wgravity = mg Δh 1 mv 2 2 16. (a) ΔEk = = 900 J ⇒ Maximum strain energy = 900 J ⇒ 12 kx 2 = 900 J 14. (b) = 2.5 J (2.475 J) ΔEk = 2.4755 Ek = 2.475 J (since initial Ek = 0) 1 ⇒ 2 mv2 = 2.475 J ⇒v = =1 2 vcricket ball vcar ⇒ =1 vcricket ball ⇒ × 60 × (7.2)2 ⇒ 300 − friction = 62 ⇒ friction = 238 N 15. (a) Energy stored = energy under force vs compression graph = area under force vs length graph as length changes from 20 cm to 5 cm = 12 × 0.15 m × 33 N = 10000 (c) 1 2 7 © John Wiley & Sons Australia, Ltd 2009 20. (a) Assign east as positive. pi = mu + m × (−20) pf = 2m × 5 ⇒ mu − 20m = 10m ⇒ u − 20 = 10 ⇒ u = 30 m s−1 = 30 m s−1 east (b) Eki = 12 m(30) 2 + 12 m(20) 2 17. (a) The springs are inside each other as shown in the diagram below. Ekf = 1 2 × 2 m × (5) 2 Ekf = Eki 1 2 m × 25 m(900 + 400) ⎛ 1 ⎞ = 0.038 ⎜ ⎟ ⎝ 26 ⎠ pi = mv − mv =0 ⇒ pf = 0 ⇒v =0 (b) As the collision is perfectly elastic, the kinetic energy of each of the two cars has not changed. As the force on each car in the collision is equal (by Newton’s Third Law) and their masses are equal, they have both accelerated equally in the collision. Their speeds must then be equal and unchanged: that is, 60 km h−1. 22. (a) Ek = ΔEgp 21. (a) (b) The longest spring is the only one compressed for the first 30 cm of compression. 20 N F k = gradient = = x 0.30 m –1 = 67 N m (c) When the springs are compressed between 30 cm and 40 cm, the longest spring and the middle spring are compressed. Let the spring constants be kl and km respectively. F = kl x + km x ⇒ F = (kl + km) x gradient = kl + km 10 N = 0.10 m kl + km = 100 N m−1 When springs are compressed between 40 cm and 50 cm, all three springs are compressed. Let the spring constant of the shortest spring be ks. F = kl + km x + ks x ⇒ F = (kl + km + ks) x gradient = kl + km + ks 20 N = 0.10 m ⇒ kl + km + ks = 200 N m−1 but kl + km = 100 N m−1 ⇒ ks + 100 = 200 ⇒ ks = 100 N m−1 18. (a) No. The total kinetic energy of the ball and the ground is less after the collision than it was before the collision. (b) Sound, along with some heating of the ball, provide evidence that some of the ball’s initial kinetic energy is transformed. (c) Yes, assuming that the ball−ground system is an isolated system 19. (a) Noise, heating and any damage to the car indicate that some of the initial kinetic energy is transformed. (b) Yes. On an icy road, friction is small enough for the system of the two cars to be considered isolated. Jacaranda Physics 2 TSK = 60 × 10 × 30 = 1.8 × 104 J (b) At the instant that her head touches the water her kinetic energy is zero. Therefore, all of the gravitational potential energy lost has been transferred as strain energy in the cord. Strain energy = mgΔh = 60 × 10 × (50 − 1.70) = 28 980 J ⇒ strain energy = 2.9 × 104 J 23. (a) Assign north as positive. pi = 0.200 × 2.0 + 0 = 0.400 kg m s−1 pf = 0.200vw + 0.200 × 1.7 pf = pi ⇒ 0.200vw + 0.34 = 0.400 ⇒ vw = 0.30 m s−1 = 0.30 m s−1 north (b) Eki = 1 2 × 0.200 × (2.0)2 + 0 = 0.40 J Ekf = 1 2 × 0.200 × (1.7) 2 + 1 2 × 0.200 × (0.30) 2 = 0.298 J 0.298 100% × 0.40 1 = 75% Percentage returned = (c) (i) pi = 0.200 × 2.0 = 0.40 kg m s−1 8 © John Wiley & Sons Australia, Ltd 2009 pf = 0.200 × −0.5 + 0.200 × 2.5 = 0.40 kg m s−1 ⇒ momentum is conserved. (ii) The player’s claim suggests that the total kinetic energy of balls after the collision is greater than the total kinetic energy of the balls before the collision. This is not possible unless energy is transferred to the system other than by the cue. Eki = 12 × 0.200 × (2.0)2 Chapter 3 Projectile and circular motion 1. = 0.40 J Ekf = 12 × 0.200 × (0.5) 2 + 12 × 0.200 × (2.5)2 = 0.65 J 24. 2. When a basketball falls from rest, there is initially no air resistance. As it accelerates downwards due to the Earth’s gravitational pull, the air resistance increases. The magnitude of the net force, and subsequently its acceleration, decreases. The air resistance continues to increase as the acceleration continues downwards. Because the air resistance is small compared to the basketball’s weight, the basketball will not reach its terminal velocity (unless dropped from an aeroplane or helicopter in flight!). 3. (a) pi = pf ⇒ m1u1 − m2u2 = − m1v1 + m2 v2 ⇒ m1u1 + m1v1 = m2 v2 + m2u2 ⇒ m1 (u1 + v1 ) = m2 (v2 + u2 ) ⇒ ⇒ ⇒ ⇒ m1 v2 + u2 = m2 u1 + v1 Eki = Ekf 1 + 2 m2u22 = 12 m1v12 + 12 m2 v22 m1u12 − m1v12 = m2 v22 − m2u22 m1 (u12 − v12 ) = m2 (v22 − u22 ) m v2 − u 2 ⇒ 1 = 22 22 m2 u1 − v1 1 2 (1) m1u12 The velocity of the projectile at any point is found by the gradient of the x vs t graph. (b) (2) Combining (1) and (2) v2 + u2 v22 − u22 = u1 + v1 u12 − v12 ⇒ (v − u2 )(v2 + u2 ) v2 + u2 = 2 (u1 − v1 )(u1 + v1 ) u1 + v1 The acceleration of the projectile at any point is found by the gradient of the v vs t graph. ⇒ u1 − v1 = v2 − u2 ⇒ u2 − u1 = v2 − v1 (c) This graph shows the velocity is constant at −9.8 m s−1. The negative sign indicates the acceleration is towards the Earth. Jacaranda Physics 2 TSK 9 © John Wiley & Sons Australia, Ltd 2009 4. (a) Vertical component: vv = 20 sin 50° = 15 m s−1 Horizontal component: vh = 20 cos 50° = 13 m s−1 (b) Vertical component: vv = 11 cos 23° = 10 m s−1 Horizontal component: vh = 11 sin 23° = 4.3 m s−1 (c) Vertical component: vv = 5 m s−1 Horizontal component: vh = 5 sin 0° = 0 m s−1 (d) Vertical component: vv = 10 sin 0° = 0 km h−1 Horizontal component: vh = 10 km h−1 (e) Vertical component: vv = 33 cos 60° or 33 sin 30° = 17 m s−1 Horizontal component: vh = 33 sin 60° or 33 cos 30° = 29 m s−1 x = ut + 12 at 2 150 = 0t + 12 (10)t 2 150 = 5t 2 30 = t 2 ⇒ t = 30 = 5.5 s (take the positive square root) (b) v = ? v 2 = u 2 + 2ax v 2 = 02 + 2(10) (150) v 2 = 3000 v = 3000 = 55 m s −1 9. u = 18 m s−1 a = −10 m s−2 (consider up as positive) (a) Method 1: (consider whole motion) v = −18 m s−1 (due to symmetry) t=? v = u + at −18 = 18 − 10t 10t = 36 t = 3.6 s Method 2: (consider half motion to top of flight) v=0 t=? v = u + at 0 = 18 − 10t 10t = 18 t = 1.8 s ∴ for whole motion t = 1.8 × 2 t = 3.6 s (b) Consider 1st half of motion: v = 0 m s −1 5. As no forces are acting in the horizontal direction, there can be no horizontal acceleration. Therefore, the horizontal component of velocity must remain constant. 6. The time of a projectile’s flight is the time it takes to hit the ground. Therefore, the projectile cannot take longer to complete one part of its motion than the other. Time is the only useful variable that is a scalar and is the same in both the vertical and horizontal directions. 7. (a) (i) Air resistance will cause the horizontal component of the motion to decrease. As the horizontal component decreases, so will the horizontal component of air resistance. (ii) Vertical air resistance will increase as the parcel speeds up due to gravity. Its acceleration will decrease while the vertical component of air resistance continues to increase until it reaches the magnitude of the parcel’s weight. The vertical component of velocity will then remain constant until the parcel reaches the ground. (b) Air resistance increases as the speed of the parcel increases. (i) The horizontal component. During the first 2 seconds of its fall, the horizontal speed is greater than the vertical speed. The vertical speed is less than 20 m s−1 for at least the first 2 seconds. (ii) The vertical component. The horizontal component of speed is always 20 m s−1 or less. After about 2 seconds, the vertical component is greater. x=? 2 v = u 2 + 2ax 02 = 182 + 2(−10) x 20 x = 182 x= 324 = 16 m 20 (c) 10. m = 500 kg x = 10 m a = 10 m s−2 u = 0 m s−1 8. x = 150 m (consider down as positive) u = 0 m s−1 a = 10 m s−2 (due to gravity) Jacaranda Physics 2 TSK t =? (a) 10 © John Wiley & Sons Australia, Ltd 2009 (consider down as positive) (a) v=? v2 = u2 + 2ax v2 = 02 + 2(10)(10) v2 = 200 v = 14 m s−1 (b) t=? 11. (a) The car stopped moving because the brakes caused a force to be exerted on the car which opposed its motion. However, no such force was exerted on the tissue box, so it continued to move in a straight line with constant speed as stated in Newton’s first law. −1 (b) u = 100 km h (as car is travelling at the same speed) x = ut + 12 at 2 1000 m s −1 3600 = 28 m s −1 = 100 × 10 = 0t + 12 (10)t 2 10 = 5t 2 (c) 10 =2 5 t = 1.4 s (c) horizontally: u = 0.5 m s−1 a = 0 m s−2 (no forces in this direction) t = 1.4 s (linking factor between horizontal and vertical components) x=? t2 = Vertical Horizontal u = 28 m s−1 u=0 −2 a = 10 m s x = 2.5 m s−1 x=? Must find time, as it is the linking factor. Using horizontal component: x t= u 2.5 t= 28 x = ut + 12 at 2 t = 8.9 × 10−2 s (or 9 × 10−2 s if used exact value of u) Using this with vertical information: x = ut + 12 at 2 x = 0.5 (1.4) + 12 (0) (1.4)2 x = 0.7 m (d) x = 0 (8.9 × 10−2 ) + 12 (10) (8.9 × 10−2 ) 2 x = 0.040 m (or 0.041 m if using t = 9 × 10−2 s). (d) During sudden accelerations, objects could become projectiles, moving through the interior of the car, which could injure the occupants of the car. 12. Horizontal u = 7.0 sin 45° = 4.9 m s−1 a = −10 m s−2 v = −4.9 m s−1 (due to symmetry) t=? u = 7.0 cos 45° = 4.9 m s−1 t = 0.98 s (from vertical) x=? x = ut x = 4.9 × 0.98 x = 4.8 m v = u + at (or 4.9 if maintaining more than 2 sig. fig. through working) −4.9 = 4.9 − 10t Set up means you will be at a 10t = 9.8 distance of t = 0.98 s 0.5 × 11 = 5.5 m ⇒ your friend will not make the jump and you will be squashed! 13. Let v = initial velocity and θ be the angle of projection of the javelin. Then range = v cos θ × t. The time is given by the vertical components, vv = −v sin θ, uv = v sin θ, a = − g. v 2 = vv2 + vh2 v 2 = 142 + 0.52 v = 142 + 0.52 v = 14 m s −1 14 tan θ = 0.5 ⇒ θ = 88° (e) (i) While attached to magnet, a = 0 m s−2. ⇒ Fnet = ma Fnet = 500 × 0 Fnet = 0 N (ii) While falling, a = 10 m s−2 downwards. ⇒ Fnet = ma Fnet = 500 × 10 Fnet = 5000 N downwards Jacaranda Physics 2 TSK Vertical 11 © John Wiley & Sons Australia, Ltd 2009 v = u + at −v sin θ = v sin θ − gt t= = 6.9 m s−1 x = ut x = 6.9 × 0.69 x = 4.8 m (c) Horizontal component: xtop to goal = 7.0 m − 4.8 m = 2.2 m u = 6.9 m s−1 t=? x t= u 2.2 t= 6.9 t = 0.32 s (d) Vertical component: u = 0 m s−1 t = 0.32 s a = 10 m s−2 x=? x = ut + 12 at 2 2 v sin θ g Now range = v cos θ × 2 v sin θ g 2 v2 sin θ cos θ g but 2 sin θ cos θ = sin 2θ so = v2 sin 2θ g v and g are constant and sin 2θ is a maximum when 2θ = 90°. So the maximum range is given when θ = 45°. 14. (a) x = 2 m t = 0.42 s v=? x 2.0 = 4.8 m s –1 (4.76) v= = t 0.42 (b) range = x = 0 (0.32) + 12 (10) (0.32)2 x = 0.51 m ⇒ Final height = 2.4 − 0.51 = 1.9 m Therefore the ball goes into the net. 16. (a) Vertical component: u = 50 sin 35° km h−1 1000 = 29 × = 8.0 m s −1 3600 a = −10 m s−2 v = 0 m s−1 t=? v = u + at 0 = 8.0 − 10t t = 0.80 s (b) Vertically: x=? 4.76 5 ⇒ θ = 18° (c) Vertical: (consider up as positive) u = 5 sin 18° = 1.545 m s−1 a = −10 m s−2 v = 0 m s−1 x=? v2 = u2 + 2ax 0 = (1.545)2 + 2(−10)x x = 0.12 m 15. (a) Vertical component: u = 9.8 sin 45° = 6.9 m s−1 a = −10 m s−1 v = 0 m s−1 (at top of flight) t=? v = u + at 0 = 6.9 − 10t 10t = 6.9 t = 0.69 s x =? (b) Vertically: v2 = u2 + 2ax 02 = 6.92 + 2( −10) x 20x = 6.92 x = 2.4 m Horizontally: x=? u = 9.8 cos 45° cos θ = Jacaranda Physics 2 TSK v 2 = u 2 + 2ax 02 = 8.02 + 2(–10) x 20 x = 8.02 x = 3.2 m Horizontally: x = ? u = 50 cos 35° × 1000 3600 = 11 m s −1 t = 0.80 s x = ut x = 11 × 0.80 x = 8.8 m (equals 9.1 m if use more than 2 sig. fig. in calculations) 12 © John Wiley & Sons Australia, Ltd 2009 v = u + at = 8.2 m s−1 Final velocity: (c) Vertical component: x = 3.2 + 0.8 = 4.0 m a = 10 m s –2 u = 0 m s –1 t=? x = ut + 12 at 2 4.0 = 0t + 12 (10)t 2 v = (12) 2 + (8.2)2 4.0 = 5t 2 = 15 m s −1 t = 0.89 s (0.90 s if use more than 2 sig. fig. for x) (d) Horizontal component: u = 11 m s−1 ttotal = 0.80 + 0.89 = 1.7 s x =? x = ut x = 11 × 1.7 x = 19 m 17. Making use of the hint provided. Before highest point: Initial speed = 13.9 m s−1 (50 km h−1) Vertically, u = 13.9 sin 30° = 6.95 m s−1 v=0 8.2 12 ⇒ θ = 34° Final velocity = 15 m s−1 at 34° to the horizontal. tan θ = 18. Horizontal u = u sin 28° a = −10 m s t= x = 2.5 m u = u cos 28° t=? –2 2.5 2u cos 28° t= –1 x 2.5 = u 2u cos 28° v=0ms Apply the equation v = u + at to the vertical component of the first half of the gymnast’s motion. ⎛ ⎞ 2.5 0 = u sin 28° + (−10) ⎜ ⎟ ⎝ 2u cos 28° ⎠ a = −10 v = u + at ⇒ 0 = 6.95 −10t ⇒ t = 0.695 s Vertically, v2 = u2 + 2ax where x = maximum height above ramp edge ⇒ 0 = 48.3 −20x ⇒ x = 2.4 m After highest point: Vertically, x = 2.4 m + 1.7 m = 4.1 m u=0 a = 10 x = ut + 12 at2 ⎛ ⎞ 25 0 = u sin 28° − ⎜ ⎟ ⎝ 2u cos 28° ⎠ ⎛ ⎞ 25 ⎜ ⎟ = u sin 28° ⎝ 2u cos 28° ⎠ 25 = 2u 2 sin 28° cos 28° u= 12.5 sin 28° cos 28° u = 5.5 m s –1 19. 2 ⇒ 4.1 = 5t ⇒ t = 0.82 (a) Range = ut where u = horizontal velocity (constant) and t = 0.695 + 0.82 = 1.5 s ⇒ Range = 13.9 cos 30° × 1.5 = 18 m (b) Final horizontal velocity = 13.9 cos 30° = 12 m s−1 Vertically, (use downward flight only): u=0 a = 10 t = 0.82 Jacaranda Physics 2 TSK Vertical Vertical Horizontal u = 7 sin θ m s –1 x = 3.0 m a = −10 m s –2 u = 7 cos θ m s –1 t =? t= 3.0 14 cos θ t= x 3.0 = u 7 cos θ v = 0 m s –1 where θ = angle to horizontal v = u + at ⎛ 3.0 ⎞ 0 = 7 sin θ + (−10) ⎜ ⎟ ⎝ 14 cos θ ⎠ 30 0 = 7 sin θ − 14 cos θ 13 © John Wiley & Sons Australia, Ltd 2009 m = 35 kg a = 0.05 m s−2 Fnet = Fc = ma = 35 × 0.05 = 1.75 N towards the centre of the circle (1.7 N if using more than 2 sig. fig. for a) (c) m = 1500 kg a = 0.05 m s−2 Fnet = Fc = ma = 1500 × 0.05 = 75 N towards the centre of the circle (d) To move along the same path, the child and the train require the same acceleration. As the mass of the child and the train are different, different forces are needed to produce identical accelerations. 23. (a) r = 65 cm = 0.65 m m = 0.12 kg (b) 30 = 7 sin θ 14 cos θ 30 = sin θ cos θ 14 × 7 0.31 = sin θ cos θ ⇒ 2 sin θ cos θ = 0.62 ⇒ sin 2θ = 0.62 θ = 19° 20. (a) r = 120 m v = 6.0 km h−1 = 1.7 m s−1 a=? v 2 (1.7) 2 a= = r 120 (0.024) a = 0.02 m s−2 towards the centre of the circle (b) a = 0.024 m s−2 m = 65 kg F=? Fc = Fnet = ma = 65 × 0.024 = 1.6 N towards the centre of the circle 21. T = 5.2 s a=? 4π 2 r a= 2 T 4π 2 (0.65) = (5.2)2 = 0.95 m s –2 towards the centre of the circle (b) a = 0.95 m s−2 m = 0.120 kg F=? Fnet = Fc = ma = 0.12 × 0.95 = 0.11 N towards the centre of the circle (c) T = 35 s rN = 2.5 m rL = 3.2 m aN = = 4π 2 r T2 4π 2 (2.5) (35) 2 = 0.08 m s –2 aL = = 4π 2 r T2 4π 2 (3.2) (35) 2 = 0.10 m s –2 ∴ Lucy experiences the greatest centripetal acceleration. 22. r = 350 m v = 15 km h −1 × 1000 3600 = 4.2 m s –1 v 2 (4.2) 2 = = 0.050 m s –2 350 r towards the centre of the circle (a) a = Jacaranda Physics 2 TSK 14 © John Wiley & Sons Australia, Ltd 2009 24. Newton’s first law states that an object will continue to move in a straight line with constant speed unless an unbalanced force acts on it. Therefore, the mass will continue to move forwards without a propelling force, once in motion. The centripetal force acts to change the direction of the mass, not its speed. 27. r = 2.0 m T = 12 N (a) Fc = T cos 10° = 12 N (11.8 N) (b) m = 0.200 kg mv 2 = Fc = 11.8 r 0.200 v 2 ⇒ = 11.8 2.0 25. To go around a bend, a motorcyclist needs a horizontal force acting on the bike towards the centre of the curve of the bend. This is provided by the road acting on the tyres. The force of the road on the tyres needs to act through the centre of mass of the cyclist, otherwise the force will act to tip the bike over. As the horizontal component of this force is acting towards the centre of the curve, the motorcyclist must lean into the curve to avoid falling off. ⇒v = 11.8 × 2.0 0.200 = 11 m s −1 (10.9 m s −1 ) 26. (c) m = 0.0500 kg T = 0.800 s r = 1.50 cos 6.03° m = T2 ⎛ v2 ⎞ ⎜⎜ or ⎟⎟ r ⎠ ⎝ r = 4.5 m mv 2 r (90) (4.2)2 = 4.5 = 3.5 × 102 N toward the centre of the circle (b) 3.5 × 102 N toward the centre of the circle (as frictional forces are causing centripetal motion) (c) v = 4.2 m s −1 F = 350 × 90% = 315 N r =? F= 2 4π × 1.50 cos 6.03° (0.800) 2 = 92.0 m s −2 (c) Fc = = towards the centre of the circle 4π 2 r T2 4 × 0.0500 × π 2 × 1.50 cos 6.03° (0.800) 2 = 4.60 N towards the centre of the circle (d) = Fnet = ma mv 2 r 90 (4.2) 2 315 = r 90 (4.2)2 r= 315 r = 5.0 m The radius will increase to 5.0 m. = 0.0500 × 92.0 = 4.60 N towards the centre of the circle F= Fc cos 6.03° 4.60 = cos 6.03° = 4.63 N (e) T = Jacaranda Physics 2 TSK T2 0.200 × 4π 2 × 2.0 ⇒11.8 = T2 0.200 × 4π 2 × 2.0 11.8 = 1.2 s 28. (a) m = 90 kg 1000 v = 15 km h −1 × 3600 −1 = 4.2 m s 2πr T 2π × 1.50 cos 6.03° = 0.800 = 11.7 m s −1 4π 2 r m4 π 2 r ⇒T = (a) v = (b) a = Fc = 15 © John Wiley & Sons Australia, Ltd 2009 31. (a) 29. (b) mv 2 r = 360 + R sin 20° Fnet = Fnet mv 2 = 360 + R sin 20° r but R cos 20° = mg 10m ⇒R = cos 20° mv 2 10m sin 20° ⇒ = 360 + r cos 20° = 360 + 10 m tan 20° ⇒ (i) Fc = Fnet = W − N As N = 0: Fc = W = mg Fc = 800 × 10 Fc = 8 × 103 N downwards mv 2 Fc = (ii) r Fc r 2 ⇒v = m v= 8 × 103 × 4.0 800 v = 6.3 m s −1 32. (a) Using KEB = PEA: 1 2 v = g Δh 2 m (9.0) 2 = 360 + 3.64 m 10 ⇒ 8.1m = 360 + 3.64 m ⇒ 4.46 m = 360 ⇒ m = 81 kg The mass of the bicycle is 20 kg. ∴ The mass of the cyclist is 61 kg. 30. v = 30 m s−1, r = 12 m v2 a= r 302 = 12 = 75 m s −2 The horizontal component of the normal reaction force must therefore provide a centripetal acceleration of 75 m s−2. N sin θ = 75 m v 2 = 2 g Δh ⇒ v 2 = 2 × 10 × 10 v = 4.5 m s −1 mv 2 r 65 × 4.52 Fc = 4.0 (b) Fc = Fc = 3.3 × 102 N upwards (c) Fnet = Fc = T − W W = mg = 65 × 10 = 650 N T = Fc + W = 3.3 × 102 + 65 × 102 T = 9.8 × 102 N N cos θ = mg 75 So by dividing, tan θ = 9 ⇒ θ = 82° The problem with this road is that you would have to drive at 30 m s−1 to avoid slipping off! But Jacaranda Physics 2 TSK 16 © John Wiley & Sons Australia, Ltd 2009 33. (a) m = 10 kg g = 9.70 N kg−1 W=? W = mg = 10 × 9.70 = 97 N downwards (c) rVenus = 6.05 × 106 m (d) 35. m =? G = 6.67 × 10−11 N m 2 kg −2 kg mperson = 70 kg M = G = 6.67 × 10−11 N m 2 kg −2 g=? W =? 36. GM r2 6.67 × 10−11 × 5.98 × 1024 = (6.38 × 106 )2 M sun = 1.98 × 1030 kg rEarth’s orbit = 1.50 × 1011 m G = 6.67 × 10−11 N m 2 kg −2 F =? −1 F = 2 = 6.9 × 10 N = M Mars = 6.42 × 1023 kg rMars = 3.40 × 106 m GMm r2 6.67 × 10−11 × 1.98 × 1030 × 5.98 × 1024 (1.50 × 1011 ) 2 = 3.5 × 1022 N GM 37. r = rEarth + h = 6.38 × 106 + 3.55 × 105 m = 6.74 × 106 m T = 92 min = 5520 s r2 g = 3.70 N kg −1 W = mg W = 2.6 × 102 N Jacaranda Physics 2 TSK 6.67 × 10−11 M Earth = 5.98 × 1024 kg W = mg = 70 × 9.80 (ii) 4.3(5.0 × 105 ) 2 M = 1.6 × 1022 kg g= g= GM r2 gr 2 ⇒M = G rEarth = 6.38 × 10 m (i) r = 5.0 × 105 m g = 4.3 N kg −1 6 (b) W = mg W = 5 × 101 N g= = 9.80 N kg GM r2 g = 0.7 N kg −1 (ii) r = 6.41 × 10 m (ii) g= (i) 6 (i) M Pluto = 1 × 1022 kg rPluto = 1 × 106 m 6.67 × 10−11 × 5.98 × 1024 9.70 M Earth = 5.98 × 10 W = mg W = 6.2 × 102 N g = 9.70 N kg−1 r=? MEarth = 5.98 × 1024 kg G = 6.67 × 10−11 N m2 kg−2 GM g= 2 r GM r= g 24 GM r2 g = 8.87 N kg −1 (ii) (c) 34. (a) g= (i) (b) Consider up to be positive. Fnet = W + Fup ⇒ 0 = W + Fup ⇒ Fup = −W = −(−97 N) = 97 N r= M Venus = 4.87 × 1024 kg 17 © John Wiley & Sons Australia, Ltd 2009 (a) a = ? a= = 39. One possible answer. 4π 2 r T2 4π 2 (6.74 × 106 ) (5520)2 = 8.73 m s −2 (b) g = ? GM g= 2 r 6.67 × 10−11 × 5.98 × 1024 = (6.74 × 106 ) 2 a = 10 m s −2 r = 200 m = 8.78 N kg −1 (c) (i) The centripetal acceleration of the space station is caused by acceleration due to gravity (i.e. g). Therefore, the two answers should be the same. (ii) Discrepancies between the numbers are due to the rounding off of data. (d) Mss = 1200 tonnes = 1.2 × 106 kg g = 8.78 N kg−1 W=? W = mg = 1.2 × 106 × 8.78 = 1.1 × 107 N (e) r = 10 m M astro = 270 kg T =? a= T= F =? GMm F= 2 r 6.67 × 10−11 × 1.2 × 106 × 270 = (10) 2 TVenus = 1.94 × 107 s TSaturn = 9.30 × 108 s r3 = 2.2 × 10−4 N T = 24 h T = 8.64 × 104 s 2 = ⇒r = r = 6.38 × 106 m a=? = 4π 2 r a 4π 2 (200) 10 = 28 s On Earth, the ground pushes up on us. In the space station, the outer wall pushes the person in (centripetal force). This means that the person would feel as though the wall is the ground, and the direction opposite to the centripetal force is down. 40. Gravity holds solar systems together. As gravity is the only force acting on a planet, it is the net force. It is directed toward the sun. Whenever the net force on an object is toward the centre of a circle, the object will experience centripetal motion. 41. M sun = 1.98 × 1030 kg G = 6.67 × 10−11 N m 2 kg −2 a= T2 = M ss = 1.2 × 106 kg 38. 4π 2 r GM 4π 2 3 GMT 2 4π 2 2 2 ⎛T ⎞ 3 ⎛ 9.30 × 108 ⎞ 3 r ∴ Saturn = ⎜ Saturn ⎟ = ⎜⎜ 7 ⎟ ⎟ rVenus ⎝ TVenus ⎠ ⎝ 1.94 × 10 ⎠ = 13 4π 2 r T2 4π 2 (6.38 × 106 ) 42. (8.64 × 104 )2 = 0.034 m s −2 In Victoria, the radius of the circular path would be smaller. If r decreases, acceleration also decreases. Therefore, in Victoria, we experience less centripetal acceleration due to the Earth’s motion than people on the Equator. Jacaranda Physics 2 TSK 18 © John Wiley & Sons Australia, Ltd 2009 M Moon = 7.35 × 1022 kg 24 M Earth = 5.98 × 10 45. A geostationary satellite cannot remain above Melbourne as the net force on the satellite (the gravitational force) is towards the centre of the Earth, not towards the centre of the cirle mapped out by Melbourne as it rotates. kg rfrom Earth = x rfrom Moon = 3.3 × 108 − x F : m Area under a g vs x graph has the same units as F W x= (i.e. energy per kg). m m 46. As W = Fx, and g = FMoon = FEarth GM Moon m rm2 7.35 × 1022 (3.3 × 108 − x)2 = = GM Earth m rE2 5.98 × 1024 47. (a) Work/kg = area under graph x2 from 400 km to 600 km ≈ 12 (8.7 + 8.2) 2 × 105 (distance in metres) = 1.7 × 106 J kg−1 ⇒ Work = 1.7 × 106 × 800 = 1.4 × 109 J (approx.) (b) r = 6.0 × 105 + 6.38 × 106 (7.35 × 1922 ) x2 = (5.98 × 1024 ) (3.38 × 108 − x)2 (7.35 × 1022 ) x2 = (5.98 × 1024 ) ((3.38 × 108 )2 − 2(3.38 × 108 ) x + x2 ) x2 = 81.36 (1.14 × 1017 − (6.76 × 108 ) x + x 2 ) = 6.98 × 106 m 0 = 80.36 x2 − (5.5 × 1010 ) x + 9.28 × 1018 x= M Earth = 5.98 × 1024 kg 2 −b ± b − 4ac 2a T =? r3 5.5 × 1010 ± (5.5 × 1010 )2 − 4(80.36) (9.28 × 1018 ) = 2(80.36) T x = 3.02 × 108 m is correct because the spacecraft must be between the Earth and the Moon. T= 43. According to Newton’s first law, an object will move in a straight line, with constant speed, unless an unbalanced force is acting on it. As gravity acts at right angles to the satellite’s velocity, it does not change the speed of the satellite; rather it changes its direction. This causes the satellite to move around the Earth. Because the force of gravity and the speed of the satellite remain constant, so mv 2 must its radius as r = ; therefore it cannot move F closer to the Earth. 48. (a) T r= 3 GMm , G = 6.67 × 10−11 N m 2 kg −2 , r2 = 5.98 × 1024 kg, M satellite = 2400 kg F= F= 6.67 × 10−11 × 5.98 × 1024 × 2400 (8.38 × 106 ) 2 = 1.4 × 104 N Answers will vary if F = mg is used, taking the value of g from the graph. (b) Loss of GPE = area under graph from 8.38 × 106 m to 7.18 × 106 m. This is approximately a trapezium. A = 12 (a + b) h GMT 2 4π 2 6.67 × 10−11 × 5.98 × 1024 × (8.68 × 104 )2 = 12 (5.25 + 6.75) × 1.2 × 106 4π 2 = 1.7 × 1010 J. r = 4.2 × 107 m Jacaranda Physics 2 TSK 6.67 × 10−11 × 5.98 × 1024 = 8.38 × 106 4π 2 3 (6.98 × 106 )3 4π 2 r = 6.38 × 106 + 2 × 106 GM r= r 3 4π 2 GM r3 M Earth = 8.64 × 104 s r=? = 4π 2 = constant for any satellite of the Earth. T2 (d) (a) would be halved and (b) would remain the same. (c) T = 24 h 2 GM T = 5.8 × 103 s M Earth = 5.98 × 1024 kg r3 = T= = 3.83 × 108 , 3.02 × 108 44. 2 19 © John Wiley & Sons Australia, Ltd 2009 (c) It has gained 1.7 × 1010 J of kinetic energy. EK = 12 mv 2 Initially 50. (a) r = rEarth + h = 6.38 × 106 + 3.60 × 105 = 6.74 × 106 m GM g= 2 r 6.67 × 10−11 × 5.98 × 1024 = (6.74 × 106 )2 = 12 × 2400 × 69002 = 5.7 × 1010 After falling EK = 5.7 × 1010 + 1.7 × 1010 = 7.4 × 1010 J 1 2 1 mv 2 2 2 = 8.78 N kg −1 = 7.4 × 1010 (b) W = mg = 70 × 8.78 = 615 N (c) Zero, because both the space shuttle and the astronaut are in orbit around the Earth and have the same acceleration (centripetal acceleration). There is no normal reaction acting on the astronaut. × 2400 × v = 7.4 × 1010 v = 7.9 × 103 m s −1 49. Weightlessness can only be experienced by an object if it has no mass or is in a zero gravitational field (i.e. g = 0) since W = mg. Apparent weightlessness is experienced by an object if it is in free fall, that is, there is no normal reaction force acting on it. Any object in orbit around the Earth is in free fall and experiences apparent weightlessness. But g ≠ 0. Jacaranda Physics 2 TSK 20 © John Wiley & Sons Australia, Ltd 2009 Unit Area of study 2: Electronics and photonics 3 Jacaranda Physics 2, 3rd Edition TSK Part A — WORKED SOLUTIONS ■■■■■■■■■ V = Electronics 1. (i) (a) 1 = Reff = = 1 30 + = 1 20 (d) Reff = 12 Ω Reff = R1 + R2 = 12 + 12 = 24 Ω (b) Reff = 6 + 12 = 18 Ω 1 1 1 = + (c) Reff 30 + 10 40 = = (iv) (a) 2 40 1 20 (b) Reff = 20 Ω (d) Reff = 5 + 12 + 3 = 20 Ω V (ii) (a) I = R = 12 24 (d) 6Ω 0.50 A 3Ω V I= R = 93 = 0.25 A V (d) I = R 60 = 20 (b) Jacaranda Physics 2 TSK = 2. (a) 30 × 5 V = 5.0 V 30 + 10 15 4 = 3.7 V 3Ω 10 Ω V = IR V = IR =3×3 =3×5 = 9.0 V = 15 V 20 Ω 30 Ω V = 60 − 15 − 9 V = 36 V = 36 V 12 Ω 20 Ω 0.50 A V I = R 6 = 20 = 0.50 A V (c) I = R 5 = 20 (iii) (a) V = 5 4 10 Ω I = 0.13 A (c) = 0.50 A V (b) I = R = 189 = 3A 12 Ω V = IR = 0.5 × 12 = 6.0 V 6Ω V = IR = 0.5 × 6 = 3.0 V 10 ×5 30 + 10 = 1.3 V 5 60 1 12 40 Ω 30 Ω 10 Ω (c) Chapter 4 Vout 30 Ω V I= R 6 = 30 = 0.30 A 20 Ω V I= R 6 = 20 = 0.20 A 30 Ω V I= R 6 = 30 = 0.30 A 30 Ω I = 0.13 A = 0.20 A 40 Ω V I= R 5 = 40 10 Ω V I= R 15 = 10 20 Ω V I= R = 36 20 = 3.0 A = 1.5 A R2 = Vin R1 + R2 = 1.8 A = 0.13 A 30 Ω V I= R = 36 30 = 1.2 A 2.2 × 6.0 2.2 + 2.2 = 3.0 V 1 1 1 = + Reff 4.4 4.4 = 30 Ω V = 12 − 6 = 6.0 V 20 Ω V = 12 − 6 = 6.0 V 20 Ω V=9−3 = 6.0 V 30 Ω V=9−3 = 6.0 V (b) 2 4.4 1 = 2.2 = 2.2 kΩ = Reff so Vout = 3.0 V 21 © John Wiley & Sons Australia, Ltd 2009 Vout 2.0 = = Vin 6.0 3. so ⇒ R2 = R1 + R2 1 3 1 3 R2 =1 3.0 + R2 3 ⇒ 3R2 = 3.0 + R2 2R2 = 3.0 3.0 R2 = = 1.5 kΩ 2 (b) 4. (a) (c) R2 Vin R1 + R2 Vin = 10 V R1 = 60 Ω R2 = 40 Ω 40 Ω × 10V Vout = (60 Ω + 40 Ω) = 4.0 V R2 Vin (b) Vout = R1 + R2 + R3 Vout = (d) 7. (a) (b) Vin = 9 V R1 = 20 Ω R2 = 30 Ω R3 = 40 Ω 30 Ω × 9V ⇒ Vout = 90 Ω = 3.0 V 4 R × 20 V (c) Vout = R + 4R 8. (a) Vout = R2 Vin R1 + R2 2×9V 4+2 = 3.0 V (b) If the temperature is 80°C, R2 = 1 kΩ. R1 = ? Vin = 9 V Vout = 6.0 V 1× 9 V ⇒ 6.0 V = R1 + 1 ⇒ 6R1 + 6 = 9 ⇒ 6R1 = 3 ⇒ R1 = 0.5 kΩ = 500 Ω 9. (a) 70°C from graph (b) 8.0 kΩ from graph = ⎛ 4R ⎞ =⎜ ⎟ 20 V ⎝ 5R ⎠ = 0.8 × 20 V = 16 V 5. As the resistance decreases, so, too, does the voltage drop across the variable resistor. R V 6. (a) Vout = 2 in R1 + R2 Vout = 4.0 V Vin = 6.0 V R1 = 3.0 kΩ Jacaranda Physics 2 TSK R2 = R R kΩ × 6.0V ⇒ 4.0 V = (3.0 + R) kΩ ⇒ 12 + 4R = 6R ⇒ 2R = 12 ⇒R=6 R = 6.0 kΩ R V Vout = 2 in R1 + R2 12 R ⇒ 6.0 = R + 660 ⇒ 6R + 6 × 660 = 12R ⇒ 6R + 3960 = 12R ⇒ 6R = 3960 ⇒ R = 660 Ω R V Vout = 2 in R1 + R2 10 R ⇒ 2.5 = 10 + R ⇒ 25 + 2.5R = 10R ⇒ 7.5R = 25 ⇒ R = 3.3 kΩ R V Vout = 2 in R1 + R2 5×9 ⇒6 = R+5 ⇒ 6R + 30 = 45 ⇒ 6R = 15 ⇒ R = 2.5 kΩ (i) R = 5.0 kΩ (by reading graph) (ii) R = 1.0 kΩ (by reading graph) (i) 25°C (by reading graph) (ii) 50°C (by reading graph) If the temperature is 40°C, R2 = 2 kΩ R1 = 4 kΩ 22 © John Wiley & Sons Australia, Ltd 2009 T = 90°C Rthermistor = 8.0 kΩ (from (b)) (c) (c) VR = IR I = 4.0 × 10−3 A Vout Rthermistor = Vin Rvariable + Rthermistor (d) R= 3 8.0 = 9 Rvar + 8.0 4.0 × 10−3 = 1337.5 Ω 3Rvar = 48 Rvar = 16 kΩ (e) The output voltage drops. This is because the proportion of the resistance in the circuit taken by the resistor decreases, so the voltage drop decreases. 13. 10. (a) 5000 Ω (from graph) (b) 200°C (from graph) 14. (c) T = 200°C, Rthermistor = 400 Ω (from (b)) 15. Rthermistor = × Vin Rthermistor + Rvariable ⇒ Vout ( Rthermistor + Rvariable ) = Rthermistor Vin ⇒ ⎛ Vout V ⎞ Rvariable = Rthermistor ⎜1 − out ⎟ Vin Vin ⎠ ⎝ 16. = 1.3 × 103 Ω or 1.3 kΩ (d) −6.0 V as the diode is now reverse biased (e) I = 0 (f) VR = 0 The distortion is due to clipping. The flat batteries reduce the maximum value for Vout. In Jasper’s case the maximum value has fallen below the value produced by the amplification of a strong radio signal. Output voltage divided by input voltage. Gradient of linear section of the transfer characteristic. The output voltage of an inverting amplifier increases as the input voltage decreases, as long as the input voltage is in the linear range. Characteristic (a) (i) Gain = gradient of linear section of graph 5 −1 = 1 − (−1) = ⎛V ⎞ Rvariable = Rthermistor ⎜ in − 1⎟ ⎝ Vout ⎠ ⎛ 12 ⎞ = 400 ⎜ − 1⎟ ⎝8 ⎠ = 200 Ω 4 2 =2 (ii) Maximum output voltage = 5.0 V (from graph) Minimum output voltage = 1.0 V (from graph) (iii) Non-inverting (positive gradient in linear section) Characteristic (b) 0.5 − 5.5 (i) Gain = [3.0 − (−3.0)] × 10−3 11. VR = 12 − Vdiode = 12 − 0.7 = 11.3 V V I = R 11.3 = 1.0 × 103 = −5.0 6.0 × 10−3 = 830 (take positive value for gain) (ii) Maximum output voltage = 5.5 V Minimum output voltage = 0.5 V (iii) Inverting (negative gradient) Characteristic (c) 1−5 −4 (i) = (4 − 1) × 10−3 3 × 10−3 = 11.3 × 10−3 A = 11.3 mA = 1.3 × 103 (take positive value for gain) (ii) Maximum output voltage = 5.0 V Minimum output voltage = 1.0 V (iii) Inverting (negative gradient) Δ Vout 17. (a) Gain = Δ Vin or 1.13 × 10−2 A. 12. (a) Read value from graph. Answer is approximately 0.65 V. Accept 0.6 V to 0.7 V. (b) VR = E − VD = 6.0 V − 0.65 V = 5.35 V Jacaranda Physics 2 TSK 5.35 V = 3Rvar + 24 = 72 Vout V I 8 0.4 = 20 = 23 © John Wiley & Sons Australia, Ltd 2009 (b) Non-inverting. An increase in Vin leads to an increase in Vout. (c) R2 Vin R1 + R2 (b) Vout = (i) = 3000 ×6 3000 + 2000 = 3 5 ×6 = 3.6 V (c) As illumination increases, the resistance of the LDR decreases, resulting in an increase in Vout. 5. (a) (ii) Relative light LDR Current LDR Voltage LDR Resistance intensity IL (μA) (V) (kΩ) Log RL Log IL 0.6 350 38.7 111 2.04 −0.22 0.4 220 39.0 177 2.25 −0.40 0.2 90 39.1 434 2.64 −0.70 0.1 40 39.0 975 2.99 −1.00 0.05 20 39.0 1950 3.29 −1.30 (b) (iii) (c) log RL = 1.17 log IL + 1.80 RL = IL1.17 × 101.80 RL = 63 IL1.17 (iv) V is not constant. R 6. (a) LEDs are semiconductor diodes designed to emit light when they conduct a current. They contain a semiconductor chip with p-type material connected to the anode by a fine wire and n-type material connected to the cathode through contact with a metal base. (b) A diode is forward biased when the p-type material — the anode — is connected to a higher voltage than the cathode. In this situation, current will flow if the voltage difference is greater than a minimum value, which is determined by the nature of the semiconducting material and, hence, the colour. A diode is reverse biased, and blocks the flow of current if the anode is connected to a lower voltage than the cathode. (c) (d) An LDR is not ohmic because Chapter 5 Introductory photonics 1. A transducer is a device that transforms energy from one form to another. 2. The resistance (in an LDR) is a function of the intensity of light shining on the LDR — the more light, the less resistance. 3. Some cameras contain light meters that use voltage dividers containing an LDR. The intensity of the light affects the resistance of the LDR, which changes the voltage to the light meter. 4. (a) 0.4 lux (from graph) Jacaranda Physics 2 TSK 7. Indicator lights on electronic equipment — they need to last for the life of the equipment, be compact and energy efficient. 24 © John Wiley & Sons Australia, Ltd 2009 Some clock and cash register displays for the same reasons. Stop lights in cars. (b) VR = 12 − 1.7 = 10.3 V V R= R I 10.3 = 0.02 = 515 Ω 8. LEDs emit light according to the choice of impurities in the semiconductor. All diodes emit some light but the choice of impurities and the structure of the diode maximise the light produced. = 5.2 × 102 Ω 9. Limiting resistors are needed because a forward-biased diode has a very low resistance. This can result in a current large enough to destroy the diode. (c) VR = 24 − 1.7 = 22.3 V V R= R I 22.3 = 0.02 = 1115 Ω 10. (a) (i) 1.6 V (from graph) (ii) VR = 6.0 − 1.6 = 4.4 V V (iii) R = I 4.4 = 0.02 = 220 Ω (b) (i) 6.0 V (reverse-biased diode with 6.0 V supply) (ii) 0 A (iii) 0 V (V = IR) = 1.1 k Ω (d) VR = 50 − 1.7 = 48.3 V V I 48.3 = 0.02 = 2415 R= 11. (a) 1.7 V (or 1.8 V) (from graph) (b) V = IR = 0.06 × 500 = 30 V (c) emf = 30 + 1.7 (or 30 + 1.8) = 31.7 V (or 31.8 V) = 2.4 × 103 Ω 15. (a) 12. (a) The diode is reverse biased and therefore nonconducting. (b) 2.5 V — the voltage must drop across the diode as the voltage across the resistor is given by V = IR (= 0). Resistor has 15 − 2 = 13 V across it. V 13 I = = 100 = 0.13 A R (b) The LED might burn out because the current is too high. 13. (a) V = 9.0 − 1.8 = 7.2 V V (b) R = I 7.2 = 0.04 = 180 Ω 16. A photodetector converts light signals into electrical signals. 17. (a) A photodiode is one that is sensitive to light. It can be used to measure light intensity. If reverse biased, a photodiode begins to conduct when light shines on it. Alternatively, it can be used as a source of electric current when light shines on it. (b) Phototransistors operate like photodiodes but they amplify the collector–base current, making them more sensitive than photodiodes. (c) Photodiodes have a faster response time. (d) Phototransistors are more sensitive. (e) Photodiodes respond more quickly than phototransistors, increasing the bandwidth of photonic systems that use them instead of phototransistors. 14. (a) VR = 9.0 − 1.7 = 7.3 V V R= R I 7.3 = 0.02 = 365 Ω = 3.7 × 102 Ω Jacaranda Physics 2 TSK 25 © John Wiley & Sons Australia, Ltd 2009 (d) Metal wires have a higher density of charge carriers (free electrons) near the surface when carrying an alternating current. This makes the resistance of the wire greater for alternating current than direct current, causing loss of signal power. (e) The skin effect is the result of electromagnetic effects. 20. (a) Optical intensity modulation occurs when the intensity of a light source is affected by the varying electrical signal from a transducer. (b) The signal from a microphone is a current of changing magnitude. This signal can be used to modulate the light emitted from a suitably biased LED. 21. (a) Modulation: changing the intensity of the carrier wave to replicate the amplitude variation of the sound signal. Demodulation: Separating the sound signal from the modulated carrier wave. (b) LED. (c) LDR, photodiode or phototransistor. 18. (a) Phototransistors have a base current that is proportional to the intensity of the light incident on it, so the collector current is also proportional to the intensity of the light. In a normal transistor, the base current is produced by a voltage divider and a signal from an external device. Phototransistors usually do not have a base terminal as transistors do. (b) The base current in a phototransistor is produced by light shining on the collector–base junction giving sufficient energy for some electrons in the depletion region to escape their atoms and form a small current. 19. (a) Bandwidth is the amount of data that can be transmitted in a fixed amount of time. (b) 15 000 Hz — the system bandwidth is equal to the maximum frequency it can carry (c) Optical fibres can carry up to 1 terabps (= 1 × 1016 binary digits per second). Telephone wires can carry up to 100 Mbps (= 1 × 108 binary digits per second). 1016 Optical fibre can carry = 108 times as much 8 10 data per second as metal wire. Jacaranda Physics 2 TSK 26 © John Wiley & Sons Australia, Ltd 2009 Unit 3 Detailed study 3.1: Einstein’s special relativity Part A — WORKED SOLUTIONS ■■■■■■■■■ 14. Question 13 gives an example of where a head-on collision can make a crash in a car travelling at 50 km h−1 as dangerous as one travelling at 100 km h−1. 15. (a) The Earth’s direction changes by only about 1° each day so, from our point of view, it is travelling in a straight line with constant speed. The acceleration due to the revolution of the Earth around the Sun is about 6 × 10−2 m s−2 which is tiny. According to the principle of relativity, if we are not accelerating, we cannot feel the movement of the Earth no matter what our speed is. (b) Daily rotation about its axis and precession of the equinoxes (the motion of the Earth’s axis), which has a period of about 22 000 years (c) The accelerations are so tiny that they are difficult to detect. Therefore, for most purposes, the Earth can be considered to be an inertial reference frame. 16. (a) The motion of the ball when viewed from the deck of the ship would depend on the speed of the ship. However, if the ship is an inertial reference frame, the speed of the reference frame should not affect objects within that frame. (b) At the base of the mast. If the ship is stationary the ball would hit the deck at the bottom of the mast. According to the principle of relativity, the ball must fall the same regardless of the speed of the ship. Δv 17. (a) a = t 100 = 10 = 10 km h −1s −1 Δv (b) a = t ( − 100 − −200) = 10 = 10 km h −1 s −1 (c) It is invariant because it is independent of the reference frame in which it is measured. 18. (a) 2.9979 × 108 m s−1 (b) v = 1.00 × 105 + 2.9979 × 108 v = 2.9989 × 108 m s−1 19. Yes, because it would increase the time difference between the two journeys, making it easier to detect. 20. Parallel to the Earth’s motion: 2L t= ⎛ v2 ⎞ c ⎜⎜1 − 2 ⎟⎟ c ⎠ ⎝ Chapter 6 Relativity before Einstein 1. An oscillating electric field induces an oscillating magnetic field, which in turn induces an electric field. These fields move at c. No medium is required for the electromagnetic wave. 2. Maxwell’s theory of electromagnetic waves produced a model consistent with the properties of light. Young’s double-slit experiment; the speed of light is slower in water than in air; light is polarised. 3. When Maxwell’s equations are solved for the velocity of an electromagnetic wave, the answer is a constant, equal to 3 × 108 m s−1. Apparent wind is the effect of air on an observer moving through air. It is the result of the motion of the air relative to the Earth (wind) plus the motion of the observer relative to the Earth. 4. (a) 6.24 p.m. and 30 seconds (b) d = vt = 9.9 × 1010 m = 9.9 × 107 km 5. A. Those at rest in the aether would measure the speed of light to be c. Those moving away from the light source would measure c − v, those approaching the light source c + v. 6. A frame of reference is the environment from which motion is observed and measured. For the purpose of measurement, it consists of a system of clocks and rulers, all moving with a constant velocity. 7. The value of the velocity depends on the reference frame in which it is measured. 8. Both clocks read the same time at any instant. When a long distance separates them, they do not appear to tell the same time but they do when the time required for light from both clocks to reach the observer is taken into account. 9. Inertial reference frames are those that are not subject to acceleration. Non-inertial frames are accelerating. 10. The principle of relativity states that all inertial reference frames are equal and that all laws of physics will be the same regardless of the reference frame. If the speed of light was relative to the aether, the laws of electromagnetism in the reference frame of the aether would be different from other reference frames. 11. There are no indications of movement with constant velocity. While accelerating, however, a force is required and you feel this as a push from your seat. 12. Each passenger is measuring the velocity relative to different reference frames: one in motion with the car; the other moving at 50 km h−1 relative to the car. 13. v = 50 − −50 = 100 km h−1 Jacaranda Physics 2 TSK Jacaranda Physics 2, 3rd Edition TSK = 2× ⎛ 1 − (3.0 × 104 )2 ⎞ 100 × 108 ⎜⎜ 8 2 ⎟ ⎟ 3.0 ⎝ (3.0 × 10 ) ⎠ = 6.666 666 773 × 10−7 s 27 © John Wiley & Sons Australia, Ltd 2009 Perpendicular to the Earth’s motion: t= 4. The principle of relativity did not appear to hold for the theory of light and electromagnetism. Light and electromagnetic waves were thought to be disturbances in a stationary aether, suggesting that the speed of the Earth could be measured relative to the aether. This is a violation of the principle of relativity. 5. It would seem to be impossible. An object travelling at a speed of 0.99c toward a stationary observer emits light in all directions. The light travels at speed c. According to Newtonian physics, the observer would measure the speed of the light from the approaching object to be 1.99c. However, the observer actually measures the speed of light to be c. 6. c 7. Einstein’s second postulate states that the speed of light is the same for all observers. The Michelson–Morley experiment worked on the basis that the speed of light was relative to the aether, which the Earth was moving through. If this were the case then the light would take the two paths in different times, however, as the speed of light is the same for all observers in all directions, it does not matter which path the light takes, it will take the same time, if the paths have equal length. This would mean that the interferometer would detect no changes in the interference pattern as it was rotated. 8. Events that are simultaneous in one reference frame are not simultaneous in all reference frames. 9. Time dilation is the slowing of time for objects in motion relative to the observer. This is significant for only very great speeds. However, it can be detected with very sensitive instruments when comparing times on board aeroplanes with the time on the ground. 10. Depth. Only the length in the direction of the motion is contracted. 11. The clock in motion relative to you. 12. False. By the principle of relativity, the situation must be symmetrical so any two inertial reference frames are indistinguishable. Both observers see the other frame as moving and, therefore, both see the other’s time running slow. 13. Time stops for a photon; it does not age and, therefore, it does not decay or change in any way. 14. Three examples of time dilation effects are seen: when fast muons fall to earth, when atomic clocks travel around the world on commercial airliners, and in GPS satellite navigation systems, which would be incorrect unless time dilation was accounted for. 15. The length of the returning twin is not contracted. As shown in the ‘parking spot’ paradox on page 160, the length contraction is due to a disagreement regarding what is simultaneous. Measuring lengths involves measuring where the two ends of an object are at one point in time. When the twin returns to the same reference frame as his brother, they again agree on simultaneity and therefore on length. 16. Refer to diagram in text, p. 154. 17. The principle of relativity states that the laws of physics are the same for observers in all inertial reference frames. If all clocks in a reference frame at rest relative to the observer ticked at the same rate, they must all tick at the 2 v 2L 1− 2 c c = 2× ⎛ 1 − (3.0 × 104 ) 2 ⎞ 100 × 108 ⎜⎜ ⎟ 8 2 ⎟ 3.0 ⎝ (3.0 × 10 ) ⎠ = 6.666 666 7 × 10−7 s Assuming that c and v are determined precisely, the difference is 7.3 × 10−8 s. Despite the large size of this apparatus, the time difference is still tiny. 2L . 21. The time of the parallel journey is t = ⎛ v2 ⎞ c ⎜⎜1 − 2 ⎟⎟ ⎝ c ⎠ 2L v2 1− 2 . c c Dividing the parallel time by the perpendicular time gives 1 . Therefore, if the parallel journey was ⎛ v2 ⎞ ⎜⎜1 − 2 ⎟⎟ ⎝ c ⎠ The time of the perpendicular journey is t = ⎛ v2 ⎞ ⎜⎜1 − 2 ⎟⎟ , as long as the perpendicular journey and light ⎝ c ⎠ travelled at the same speed in both directions, the time for the two light journeys would be the same. 22. The speed of sound is relative to whatever medium it is passing through. Without the medium there is no sound. The sound must then be relative to its medium and no laws of physics will change if the speed of sound is measured by different observers. Light, however, is dependent on the speed of electric and magnetic fields, which travel at the speed of light. If the speed of light was relative then the laws determining the size of the electric and magnetic fields would be different for different observers. 23. Absolute means that the speed of light was not relative but had the same value in all cases. Chapter 7 A new relativity 1. (a) The laws of physics are the same in all inertial reference frames and light speed is constant in vacuum, for all observers. (b) In previous physics, the laws of electromagnetism were not the same in all inertial reference frames — light speed depended on the motion of the source and the receiver, so was not the same for all observers. 2. Einstein rejected the luminiferous aether altogether. 3. (a) Newton’s Laws are an excellent approximation at speeds lower than light speed. (b) Newton’s Laws are accurate enough for most purposes. They are easier to use and learn. Jacaranda Physics 2 TSK 28 © John Wiley & Sons Australia, Ltd 2009 same rate if the reference frame is set in motion. The length of the tick is not absolute but the idea of time passing at a certain rate within a reference frame must be maintained or the laws of physics will be broken. 18. See pages 152−54 and 157−58. 19. t0 is the proper time; that is, the time as measured in the reference frame of the event. t is the time as measured in a different inertial reference frame. 1 20. γ = ⎛ v2 ⎞ − 1 ⎟ ⎜⎜ c 2 ⎟⎠ ⎝ 24. (a) 5 minutes (b) 23. (a) (b) ⎛ v2 ⎞ ⎜⎜1 − 2 ⎟⎟ ⎝ c ⎠ = 0.714 L0 = Lγ = 4.2 × γ = 3.0 light-years d v 3.0c = 0.7c = 4.3 years (b) t = d v 4.2c = 0.7c = 6 years (c) t = L0 γ 80 = 1.40 = 57 m = 0.01 × 1.048 28 = 0.010 482 8 minutes 1 f′= t = 95.4 beats per minute. (It beats more slowly.) Unchanged It would be contracted to 60% of its proper length. c 1 γ= ⎛ v2 ⎞ ⎜⎜1 − 2 ⎟⎟ ⎝ c ⎠ = 7.0888 t t0 = γ 20 = 7.0888 = 2.82 s L L= 0 γ 5 = 7.0888 = 0.7053 m Jacaranda Physics 2 TSK ⎛ v2 ⎞ ⎜⎜1 − 2 ⎟⎟ ⎝ c ⎠ 1 25. (a) γ = 2⎞ ⎛ ⎛ 30 5⎞ ⎜ 1 − ⎜ × 10 ⎟ ⎟ ⎜ ⎝ 3 ⎠ ⎟⎠ ⎝ = 1.000 000 5 1 21. (a) γ = ⎛ v2 ⎞ ⎜⎜1 − 2 ⎟⎟ ⎝ c ⎠ 1 = ⎛ ⎛ 0.3c ⎞2 ⎞ ⎜1 − ⎜ ⎟ ⎜ ⎝ c ⎟⎠ ⎟ ⎝ ⎠ = 1.048 28 1 (b) T = f = 1 100 min t = t0 γ 22. (a) (b) (c) t0 ⎛ ⎛ ⎛ t ⎞2 ⎞ ⎞ ∴ v = ⎜ c2 ⎜1 − ⎜ 0 ⎟ ⎟ ⎟ ⎜ ⎜ ⎝ t ⎠ ⎟⎟ ⎠⎠ ⎝ ⎝ = 0.78c 1 = t = 26. (a) L = (b) d = ct , L + vt = ct L ∴t= c−v cL d = c−v = 190 m (c) L + vt = ct t= L (c − v) = 6.3 × 10−7 s (d) 29 L0 c 80 = × 108 3 = 2.67 × 10−7 s t= © John Wiley & Sons Australia, Ltd 2009 32. C. They would require more force to accelerate due to the increase in their inertial mass. 33. Energy and mass are equivalent. We cannot talk about an increase in energy without an increase in mass, and vice versa. 34. (a) m = m0γ = 70 × 1.67 = 117 kg (b) E = m0γc2 − m0c2 = 4.2 × 1018 J (It takes a lot of energy to reach this speed.) 35. E = mc2 = 6.0 × 1024 × (3 × 108)2 = 5.4 × 1041 J 36. m = m0(γ − 1) = 5.98 × 1024 × 0.000 000 5 = 2.99 × 1018 kg 37. Ek = (γ − 1)m0 c2 L0 γ 100 = 2.29 = 43.6 m 28. No matter how fast an object is moving it can send out a light beam that travels at c. or If v is greater than c, γ involves the square root of a negative number. At the speed of light, m becomes infinite, time stops and the length becomes zero. Therefore, matter cannot travel at light speed because it would take up zero volume and have infinite mass. 29. Ek = (γ − 1)m0c2 ⎛ ⎞ 1 − 1⎟ × 1000 × (3 × 108 )2 (a) = ⎜ ⎜ 1 − 0.12 ⎟ ⎝ ⎠ 27. L = = 4.5 × 1017 J ⎛ ⎞ 1 − 1⎟ × 1000 × (3 × 108 )2 (b) = ⎜ ⎜ 1 − 0.52 ⎟ ⎝ ⎠ ⎛ ⎞ 1 =⎜ − 1⎟ × 10 000 × (3 × 108 ) 2 ⎜ 1 − 0.62 ⎟ ⎝ ⎠ = 1.4 × 1019 J = 2.3 × 1020 J ⎛ ⎞ 1 (c) = ⎜ − 1⎟ × 1000 × (3 × 108 )2 ⎜ 1 − 0.82 ⎟ ⎝ ⎠ Ek = (γ − 1)m0c2 so γ = 38. 1 = 6.0 × 1019 J 2 ⎛ ⎛v⎞ ⎞ ⎜1 − ⎜ ⎟ ⎟ ⎜ ⎝ c ⎠ ⎟⎠ ⎝ ⎛ ⎞ 1 − 1⎟ × 1000 × (3 × 108 ) 2 (d) = ⎜ ⎜ 1 − 0.92 ⎟ ⎝ ⎠ = Ek +1 m0c2 Ek +1 m0c2 2 ⎛ 1 ⎛v⎞ ⎞ ⎜1 − ⎜ ⎟ ⎟ = 2 ⎜ ⎟ ⎝c⎠ ⎠ ⎛ E ⎞ ⎝ k + 1 ⎜⎜ ⎟⎟ 2 ⎝ m0c ⎠ 20 = 1.2 × 10 J 30. Sketch a graph of energy versus speed using your answers to the previous question. ⎛ ⎜ ⎜ v = ⎜1 − ⎜ ⎜⎜ ⎝ ⎞ ⎟ ⎟ 1 ×c 2⎟ ⎛ Ek ⎞ ⎟ + 1⎟⎟ ⎟ ⎜⎜ 2 ⎟ ⎝ m0c ⎠ ⎠ = 1.0 × 104 m s −1 39. E = mc2 = 0.25 × (3 × 108)2 = 2.25 × 1016J. 40. All of the chemical products of combustion must add up to less than the mass of the initial coal because energy has been released. 41. It would take a lot of energy to separate the Moon from the Earth. This would increase the mass of the Moon. 31. As seen in the previous question, even accelerating 1000 kg to 0.9c involves enormous amounts of energy. This is not currently feasible. Jacaranda Physics 2 TSK 30 © John Wiley & Sons Australia, Ltd 2009 Unit Detailed study 3.2: Materials and their use in structures 3 ■■■■■■■■■ Part A — WORKED SOLUTIONS (b) B (c) Toughness is given by the area under the σ vs ε graph. B is therefore the tougher material. (d) B exhibits more plastic behaviour and is therefore more ductile. (e) A Chapter 8 Investigating materials 1. (a) Reading from the graph, Δl = 0.60 mm (b) Young’s modulus is determined from the slope of the σ vs ε graph. However, in this case use the relationship σ = Y × ε: σ F×L Y = = ε A × Δl = Jacaranda Physics 2, 3rd Edition TSK = = 3.2 MPa 50 × 10−6 × 0.6 × 10−3 (b) ε = = 500 × 106 Pa = 500 MPa π(0.100) 2 8. From: F σ= A =Y ×ε = 1.3 × 105 Pa Y × Δl L F×L ⇒ Δl = A×Y = The column causes the larger stress. 3. (a) Strain = ε Δl = L 100 = 10 000 = 0.01 = 15 × 103 × 5 π × (0.001)2 × 200 × 109 = 0.119 m ≅ 120 mm F (b) Stress = σ = A 50 × 103 = π × (0.004) 2 F , A where A = 12.7 × 12.7 × 10−6 m2 ≅ 161 × 10−6 m2, Δl and strain = ε = = , where L = 5.08 × 10−2 m L 9. (a) Using stress = σ = = 9.9 × 108 Pa σ ε 80 MPa = 0.002 = 4 × 104 MPa 4. Y = = 40 GPa 5. (a) Young’s modulus is given by the slope of the σ vs ε graph. A has the larger Young’s modulus as it is the steepest. Jacaranda Physics 2 TSK Δl 0.05 × 10−3 = = 1 × 10−5 5 L 7. σ = Y × ε = 110 × 109 × 3 × 10−4 = 3.3 × 107 Pa = 33 MPa 2. Estimate the area of your feet. Taking a mass of 60 kg supported on two feet, each approximately 25 cm × 6 cm, the stress would be 60 × 10 = 2 × 104 Pa. 2 × 0.25 × 0.06 4000 π × 0.1002 = 3.2 × 106 Pa 150 × 0.100 The stress in the column is F A 100 × 103 6. (a) Stress = σ = 31 Stress (MPa) Strain 0 0 449 1.97 × 10−3 673 2.95 × 10−3 898 3.94 × 10−3 999 4.92 × 10−3 1170 5.91 × 10−3 © John Wiley & Sons Australia, Ltd 2009 12. (a) (b) Y = (b) It is ductile because of its noticeable plastic behaviour. (c) From the slope of the linear portion of the graph 24 MPa = 200 MPa. before yielding, Y = 12 × 10−2 (d) When stretched to twice its original length, ΔL = L; L that is, ε = = 1 = 100%. L The stored energy is found from the area under the σ vs ε graph. Stored energy ≅ 20 × 106 × 1 = 2 × 107 J m−3 13. The stiffness of fishing line (a) is constant but the stiffness of (b) and (c) changes. When the stress–strain graph is steeper, the fishing line is stiffer. In each case the fisherman will feel the same increasing force, assuming that the stress in each case is applied at a constant rate. 14. Energy = 12 σ × ε 680 MPa 3 × 10−3 = 227 × 103 MPa = 227 GPa (c) A strain of 0.5% = 0.005 is beyond the elastic limit. Therefore use the graph rather than the relationship σ = Y × ε. From the graph σ ≅ 1030 MPa. 10. (a) 54 × 106 × 2 × 10−3 5 −3 4 = 1.1 × 10 J m F 15. (a) Using stress = σ = , A = 1 2 × 2 ⎛ 12.5 × 10−3 ⎞ 2 2 −6 where A = π ⎜⎜ ⎟⎟ m ≅ 122 × 10 m , 2 ⎝ ⎠ Δl , where L = 50.00 × 10−3 m and strain = ε = L Stress (MPa) Area under curve A = area under curve B. (b) As A and B are equally tough, the areas under the graphs must be equal. This shows that A is stiffer by a factor of 2. 11. σ = Y × ε = 110 × 103 MPa × 5 × 10−4 = 55 MPa Jacaranda Physics 2 TSK 32 Strain 0.0 0.0000 36.7 0.0004 109 0.0014 182 0.0026 254 0.0036 272 0.0150 287 0.0400 291 0.0600 291 0.0800 275 0.1030 © John Wiley & Sons Australia, Ltd 2009 (iv) (v) (vi) (vii) (viii) (ix) (x) (b) (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (b) and (c) (i) σY = 31.2 × 103 = 254 MPa 122 × 10−6 (ii) The ultimate tensile strength is given by the largest stress = 291 MPa. (iii) Breaking strength = stress at fracture = 275 MPa 16. (a) The minimum force to cause plastic deformation will be just larger than the force at yield. That is, F = σY × A = 55 × 106 × π × (1.5 × 10−3)2 ≅ 3.9 × 102 N (b) Δl = ε × L σ ×L = Y Y 55 × 106 × 40 × 10−3 = 76 × 109 (d) Compression Tension Tension Tension Compression Tension Compression Nylon — tough, elastic, stiff Steel — strong, hard, stiff Skin — elastic (decreases with age), soft, tough Plastic — tough, stiff Rubber — elastic, tough Plastic — tough Nylon — elastic Rubber — tough, elastic Plastic — plastic, tough Putty or silicon — depends on material and purpose 20. (a) ΣF = 0 Στ ≠ 0 21. = 2.9 × 10−5 m 22. = 2.9 × 10−2 mm (c) F = σ × A = 125 × 106 × π × 0.00152 = 884 N σ 90 × 106 17. ε = × = 1.2 × 10−3 Y 75 × 109 Δl = ε × L = 1.2 × 10−3 × 1 = 1.2 × 10−3 m = 1.2 mm F 6 × 103 = = 850 MPa 18. (a) σ = A π × (1.5 × 10−3 )2 23. Δl 0.4 = = 8 × 10−4 L 500 σ 850 × 106 = 1.1 × 1012 Pa (c) Y = = ε 8 × 10−4 (d) From the area under the graph of force vs extension, strain energy = 12 × 6.0 × 103 × 0.4 × 10−3 (b) ε = 24. ΣF = 0 (b) No. Rotational equlibrium is not satisfied. Because of the lever arm, Vince will get maximum effect when the pedals are horizontal and he pushes vertically down. The force he applies will have no effect when the pedals are vertical because the lever arm is zero and consequently the torque is zero. To pick the rod up, a force equal to 500 + 300 + 800 N must be applied. If the rod is not going to rotate it must be picked up at the CM where rotational equilibrium would be satisfied i.e. Στ = 0. Assuming the CM is x cm from the bucket on the left and clockwise torques are positive: Στ = 0:300 × 1.5 − 800 × x = 0 1.5 x = 300 × = 0.56 m 800 As Sam moves beyond the left of the fulcrum, the seesaw will rotate anticlockwise. This will happen when the torque from Sam about the fulcrum is larger than the torque of the bag. (a) The reaction from the left abutment decreases and the reaction from the right abutment increases. (b) = 1.2 J (e) Total energy = 1 2 × 6.0 × 103 × 0.4 × 10−3 = 1.2 J 19. (a) (i) Tension (ii) Shear (iii) Tension Jacaranda Physics 2 TSK ΣF = 0: RL + RR = 12 tonne (1) Στ = 0: taking torques about the left hand support 12 × 16 − 20RR = 0 (2) 33 © John Wiley & Sons Australia, Ltd 2009 28. This question once again relies on applying the equations of equilibrium. ΣF = 0: 2000 + 800 + 600 + 200 − RL − RR = 0 that is, RL + RR = 3600 N (1) Στ = 0: taking torques about the left hand support 800 × 2.0 + 2000 × 3.0 + 600 × 4.0 + 200 × 5.0 − RR × 6.0 = 0 (2) 16 = 9.6 tonne 20 Substituting in (1), RL = 2.4 tonne. 25. (a) The wall resists the loads from the balcony and the person with an upwards force and an anticlockwise torque. (b) As the person moves towards the wall, the total reaction remains constant. However, the torque on the wall decreases. RR = 12 × 29. Pirate Bill will tip into the water when the overturning torque caused by his weight is greater than the stabilising torque of the plank’s own weight. This will occur when the pirate is x m past the edge of the boat. The CM of the plank is 1 m inside the edge of the boat. Στ = 0: −500 × +800 × 1 = 0 800 x= = 1.6 m 500 26. 30. If the tension in the cable is T, the force can be found by considering vertical equilibrium. 2T sin 8° = 4 kN T = l4.4 kN Equilibrium must be satisfied at each joint if the tent is to stay up. The forces can be simply resolved into their vertical and horizontal components. Let the force in the guy = T and the force in the pole = P: (a) ΣFhorizontal : 500 − T cos 60° = 0 31. No, the wire did not return to its original length. The force–extension graph shows that the wire was loaded beyond its elastic limit. When it was unloaded, some of the energy had been used to permanently deform the wire. 500 cos 60° T = 1000 N tension (b) ΣFvertical : T sin 60° + P = 0 T = 1000 sin 60° + P = 0 P = −866 N i.e. compression The guys create a downward force on the poles. (c) The poles would sink into the ground until the ground can resist the downward force from the poles. 27. Taking torques about the left support: 1.8 W − 20(5 − 1.8) = 0 72 1.8 W = 35.6 tonnes ≅ 36 tonnes W = Jacaranda Physics 2 TSK 34 © John Wiley & Sons Australia, Ltd 2009 Unit Detailed study 3.3: Further electronics 3 Jacaranda Physics 2, 3rd Edition TSK Part A — WORKED SOLUTIONS ■■■■■■■■■ Chapter 9 (iv) VRMS = Electronics at work = (b) Vp − p = max value − min value (b) = 5 V − (−5 V) = 10.0 V = 6.0 cm × 2 ms cm −1 = 12 ms 2 5V (ii) f = 2 = 3.5 V (iii) Vpeak = 1.5 cm on screen = 1.5 cm × 5 mV cm = 7.5 mV 1 = (iv) VRMS = 4.0 × 10−3 s = 250 Hz = V = RMS R 3.5 V = 100 Ω 3. (a) (i) T = time to complete one cycle = 20 ms (ii) f = (i) T = time to complete 1 cycle = 6.0 cm on screen = −1 = 6.0 s = 1 (ii) f = T 1 20 ms 1 2 × 10−2 s (iii) Vpeak = 1.5 cm × 1 V cm −1 1 6.0 s = 0.17 Hz = 1.5 V (iv) VRMS = (iii) Vpeak = 1.5 cm on screen = 1.5 cm × 1 V cm 1 T = 50 Hz = −1 = = 1.5 V Jacaranda Physics 2 TSK 2 7.5 mV = 4 cm × 5 ms cm −1 = 35 mA = 6.0 cm × 1 s cm Vpeak 2 = 5.3 mV = 3.5 × 10−2 A 2. (a) 1 1.2 × 10−2 s = 83 Hz 1 T (f) I RMS 1 T = (d) T = time to complete 1 cycle = 4.0 ms (e) f = (i) T = time to complete 1 cycle = 6.0 cm on screen Vpeak = 2 1.5 V 2 = 1.1 V 1. (a) Vpeak = 5.0 V (c) VRMS = Vpeak Vpeak 2 1.5 V 2 = 1.1 V 35 © John Wiley & Sons Australia, Ltd 2009 (b) R = 33 kΩ = 3.3 × 104 Ω C = 100 μF = 10−4 F τ = 3.3 × 104 Ω × 10−4 F = 3.3 s (c) R = 68 kΩ = 6.8 × 104 Ω C = 10 μF = 10−5 F τ = 6.8 × 104 Ω × 10−5 F = 6.8 × 10−1 s = 0.68 s (d) R = 470 kΩ = 4.7 × 105 Ω 4. (a) Connect the probes to the terminals marked ‘Com’ and ‘V’. Set the dial on the multimeter to an appropriate voltage scale and connect the probes in parallel across the part of the circuit being measured. (b) Never measure mains voltage with a multimeter because it could destroy the multimeter and kill or harm the person holding it. 5. (a) 0.1 μF = 0.1 × 10−6 F C = 0.47 μF = 4.7 × 10−7 F τ = 4.7 × 105 Ω × 4.7 × 10−7 F = 0.22 s 8. (a) 1.8 V ± 0.2 V (by reading graph) (b) The time constant for an RC circuit is the time it takes the capacitor to reach 0.63 of its final voltage when charging. Assume that the final voltage is approximately 10 V. Time constant is the time at which the voltage is 6.3 V. τ = 5.0 ms (by reading graph) (c) τ = RC τ ⇒C = R 5.0 × 10−3 s = 100 Ω = 1.0 × 10−7 F (b) 220 μF = 220 × 10−6 F = 2.20 × 102 × 10−6 F = 2.2 × 10−4 F (c) 100 pF = 1.0 × 102 × 10−12 F = 1.0 × 10−10 F (d) 140 nF = 1.4 × 102 × 10−9 F = 1.4 × 10−7 F 6. (a) Time constant for an RC circuit is the time it takes the capacitor to reach 0.63 of its final voltage when charging, or 0.37 of its initial voltage when discharging. (b) After one time constant, the voltage will fall to 0.37 of its initial voltage. 0.37 × 10 V = 3.7 V 7. (a) τ = RC R = 390 Ω = 5.0 × 10−5 F (or 50 μF) τ (d) R = C 5.0 × 10−3 s = 1.0 × 10−7 F = 5.0 × 104 Ω (or 50 kΩ) 9. Rectification is the conversion of an AC voltage signal into a DC voltage signal. 10. Only the positive or the negative part of the signal will be passed — not both. 11. (a) A diode is an electronic device that allows current to flow through it in one direction. (b) About 0.7 V when forward biased (c) One = 3.9 × 102 Ω C = 100 μF = 100 × 10−6 F = 10−4 F τ = 3.9 × 102 Ω × 10−4 F = 3.9 × 10−2 s (τ = 39 ms) Jacaranda Physics 2 TSK 36 © John Wiley & Sons Australia, Ltd 2009 17. (a) τ = RC 12. (a) It smooths the output voltage of the half-wave rectifier. (b) When the diode ‘blocks’ the voltage, the capacitor discharges through the load resistor. (c) It becomes smoother. 13. (a) Bridge rectifier = 1 × 106 Ω × 1 × 10−7 F = 1 × 10−1 s = 0.1 s 1.0 V is approximately 0.63 of 1.5 V; that is, it takes approximately one time constant for the voltage across the capacitor to reach 1.0 V. (b) The capacitor is considered to be fully charged after five time constants = 0.5 s. (c) When fully charged, the voltage across the capacitor equals the emf of the cell = 1.5 V. (d) Q = CV = 0.1 × 10−6 F × 1.5 V = 1.5 × 10−7 C Centre-tap rectifier (e) (b) Bridge rectifier, Centre-tap rectifier, ⇒ VR = 1.5 V − 1.0 V = 0.5 V 18. (a) The peak output voltage equals the peak input voltage minus the voltage drop across the diode. The maximum voltage drop across a silicon diode is approximately 0.7 V. Vpeak output = Vpeak input − Vdiode Vout = 12.0 − 1.4 = 10.6 V Vout = 12.0 − 0.7 = 11.3 V = 4.0 V − 0.7 V = 3.3 V 14. Vr(p − p) = 6.3 − 5.8 = 0.5 V V T 15. Vr(p − p) = max RC 1 12.5 × 100 = RC 0.125 = RC (a) Vr(p − p) = (b) (c) (i) If C = 10 μF τ = RC 0.125 3 10 × 10 × 100 × 10 −6 = 0.13 V (c) Vr(p − p) = 1 × 103 Ω × 1 × 10−5 F 0.125 (b) Vr(p − p) = 3 = 1 × 10−2 s The capacitor therefore has a little more than one time constant to discharge before the next pulse arrives. The voltage will fall to slightly less than 0.37 × 3.3 V = 1.2 V. −6 3.3 × 10 × 100 × 10 = 0.38 V 0.125 = 3 10 × 10 × 5.0 × 10−6 = 2.5 V (d) Vr(p − p) = E = VR + VC 0.125 25 × 10 × 20 × 10−6 3 = 0.25 V (e) Vr(p − p) = 0.125 100 × 103 × 25 × 10−6 = 0.050 V 16. 4.2 V AC Jacaranda Physics 2 TSK 37 © John Wiley & Sons Australia, Ltd 2009 (ii) If C = 100 μF τ = RC (e) τ = RC = 500 Ω × 2.0 × 10−5 F = 1 × 103 Ω × 1 × 10−4 F = 0.010 s = 10 ms (f) The capacitor discharges for slightly more than one time constant between peaks, so Vout will fall to 0.37 of 12.0 V or 4.4 V. −1 = 1 × 10 s The capacitor has approximately 0.1 of a time constant to discharge, so the voltage will fall very little. (g) Use a larger value resistor. Use a larger value capacitor. 20. (a) τ = RC 19. (a) Vpeak = 2 VRMS = 2.0 × 104 Ω × 1.0 × 10−5 F = 2 × 9.0 V = 0.20 s = 13 V (= 200 ms) (b) Yes. A capacitor is considered to be fully charged (or discharged) after five time constants. The input voltage is at a high or low value for one second each, which is five time constants. (c) (b) f = 50 Hz (input and output have same frequency) 1 ⇒T = f 1 = 50 Hz = 0.02 s = 20 ms (c) Maximum output voltage = 12.7 V − 0.7 V = 12.0 V (d) (d) VD max = 0.7 V When reverse biased, VD is the same as Vin. Vin = VC + VR ⇒ VR = Vin − VC 21. • The AC supply to your house • A battery going flat • Variations in the load resistance 22. (a) Voltage regulation is the maintenance of a steady voltage supply. (b) Most electronic devices require a steady voltage supply. Jacaranda Physics 2 TSK 38 © John Wiley & Sons Australia, Ltd 2009 (c) Between the power supply and the load (d) The output voltage is a little less than the input voltage. 23. Zener diodes are designed to ‘breakdown’ in a reliable and non-destructive way so that they can be used in reverse to maintain a fixed voltage across their terminals. 24. The breakdown voltage of a zener diode is the reverse bias voltage at which the junction breaks down and can conduct large currents. 25. The zener diode is in series with a limiting resistor (to limit the current through the diode) and a voltage supply. It is placed in parallel with the load to provide a regulated (steady) voltage across the load. 26. (a) 20 V (b) 0.20 A (c) 20 V (d) 0.40 A (e) 0.20 A 27. (a) 40 V to 70 V (b) Imax = 1.75 × 10−2 A (17.5 mA), Imin = 0.010 A (10 mA) (c) I = 3.0 × 10−3 A (3.0 mA) (d) Imax = 1.45 × 10−2 A (14.5 mA), Imin = 0.007 A (7 mA) 28. (a) 12 V (b) 8 V (c) 108 mA (d) 120 Ω (e) 74 Ω Jacaranda Physics 2 TSK 29. (a) Read value from graph. Answer is approximately 0.65 V. Accept 0.6 V to 0.7 V. (b) VR = E − VD = 6.0 V − 0.65 V = 5.35 V (c) VR = IR I = 4.0 × 10−3 A R= = V I 5.35 V 4.0 × 10−3 = 1337.5 Ω (e) (f) (g) 30. (a) (b) 39 = 1.3 × 103 Ω or 1.3 kΩ − 6.0 V as the diode is now reverse biased I=0 VR = 0 Transistors and diodes Heat sinks conduct, convect or radiate the thermal energy away from the device. © John Wiley & Sons Australia, Ltd 2009 Unit 4 Area of study 1: Electric power Jacaranda Physics 2, 3rd Edition TSK Part A — WORKED SOLUTIONS ■■■■■■■■■ Chapter 10 Magnetism – fields and forces 8. 1. If the magnet attracts the piece of metal, the metal must be a magnetic material, which may or may not be a magnet. If the magnet repels the piece of metal from some angles, the metal must be a magnet with similar strength to the known magnet. Otherwise, if the piece of metal is a magnet, either much stronger or weaker than the known magnet, the two magnets will attract but the strength of attraction will depend on the part of the metal the pole of the magnet is touching. 2. As the material cools, the magnetic domains are aligned by the Earth’s magnetic field. 3. There is a magnetic field at both ends which induces a field of the same orientation in the iron nail. Opposite poles attract and so the nail moves to the magnet. 4. The magnetic pole in the southern hemisphere is a north magnetic pole. 5. Each iron rod is magnetised in the same orientation. Adjacent north ends repel each other. 6. (a) 9. 10. 11. (a) The direction of the magnetic field beneath the wire is north ⇒ No change. (b) The direction of the magnetic field beneath the wire is south ⇒ Opposite direction. (a) Out of page (b) Out of page (c) Out of page (d) Up the page (e) To the right (a) Up the page (b) to the left (c) Down to the right (d) Into the page (e) Out of page (a) W: Out of page; X, Y, Z: Into the page (b) W, X: To the right; Y, Z: To the left Use the left-hand rule or the right-hand-slap rule. (a) Into page (b) Into page (c) South (d) Out of page (e) North (f) East (g) South-east (h) South 12. (a) (b) (c) (b) (d) Yes, by Newton’s third law, the force by A on B is equal and opposite to the force by B on A. 13. F = nIlB = 1 × 4.5 × 0.05 × 0.30 = 0.07 N 14. F = nIlB = 1 × 2.4 × 0.30 × 0.25 = 0.18 N F = nIlB 15. 7. Use the right-hand-grip rule. Jacaranda Physics 2 TSK 40 but l = 2πr ⇒ F = 500 × 15 × 10−3 × (2π × 1.5 × 10−2 ) × 2.0 = 1.4 N © John Wiley & Sons Australia, Ltd 2009 16. F = nIlB (b) Yes, just reverse the direction of the current. (c) Yes, just increase the voltage of the DC battery, which increases the size of the current. 27. (a) The commutator would counter the change in voltage and consistent movement would not be possible. (b) The motor would turn if it were turning at the same rate as the AC frequency and if the voltage changed direction when the coil was perpendicular to the field. This would be very unlikely. 28. (a) The two ways are series and parallel. (b) A series-wound motor produces a very large turning effect when it is started. This makes this sort of motor usual for trains and trams. Parallel-wound or shuntwound motors maintain a steady speed with different loads and are used to drive such machines as lathes. = 1 × 1.8 × 8.0 × 10−2 × 40 × 10−3 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. = 5.8 × 10−3 N Use several coils of wire wrapped around a matchbox floating in a dish of water. Supply a DC current. Fire an electron into a space. The direction it deflects can be used to determine the direction of the magnetic field, except where the electron is undeflected. The direction of the deflection is the direction of the force, the initial direction of the electron gives the opposite direction of the conventional current, and the hand rule (left-hand rule or right-hand-slap rule) gives the direction of the magnetic field. If the electron is undeflected, the magnetic field is either parallel or anti-parallel to the electron direction. By travelling parallel or anti-parallel to the magnetic field. An east–west power line would experience an up–down force because the current reverses direction at 50 Hz. For most lines, this would be too high a frequency to produce noticeable movement in the line. But there would be a particular length for which 50 Hz would be the resonant frequency. No, not if the electron is at rest relative to the magnet. (a) As the electrons approach the wire they experience a downwards magnetic field. They initially experience a force to the right. (b) The force on the electron is away from the wire. (a) The electron experiences a force in a westerly direction and moves in a circle. (b) The electron moves in a circle and moves up, at the same time producing a spiral path. A DC current flows through a coil in a magnetic field. The field exerts forces of the same size, but of opposite direction on the opposite sides of the coil. These forces initially make the coil rotate, but the direction of the current needs to be reversed if the coil is to continue to rotate. This reversal is done by a commutator twice each cycle when the coil is at right angles to the field. (a) The magnet provides a magnetic force that interacts with the electric current to produce a force. (b) The brushes prevent tangling of wires by allowing current to flow from the stationary wires coming from the battery to the rotating coils of wires. (c) The commutator reverses the current, twice every cycle when the coil is at right angles to the field. (d) Each turn experiences the magnetic force, so the more turns the larger the force and the faster the rotation. (a) If the coil is in a position at right angles to that in the figure on page 252, then either there will be no electrical connection between the brushes and the coil and so no current, or possibly the commutator touches both brushes allowing the current to bypass the coil, that is, to short out and possibly burn out the brushes and the commutator. This can be overcome by using extra sets of coils at angles to the first, each with it own set of sections on the commutator. The figure on page 253 has three sets of coils. Jacaranda Physics 2 TSK Chapter 11 Generating electricity 1. Magnetic flux is the amount of magnetic field passing through an area. 2. He used these to increase the amount of magnetic flux passing through the iron core. 3. (a) Φ = Bperpendicular × A = 3.0 T × 0.050 m 2 = 0.15 Wb (b) Φ = 0.4 T × 4.5 cm 2 = 0.4 T × 4.5 × 10−4 m 2 = 1.8 × 10−4 Wb (c) Φ = 0.025 T × 12 × 10−4 m 2 × 50 = 1.5 × 10−3 Wb 4. 5. The loss in gravitational potential energy as the metal rod fell was not fully converted into kinetic energy. Some of it was converted into electrical potential energy of the separated charges. 6. The falling magnet induces circular currents in the metal tube whose magnetic fields produce repulsion from below and attraction from above. This is an effective air resistance as the size of the induced currents depends on the speed of the falling magnet — a condition for terminal velocity. 41 © John Wiley & Sons Australia, Ltd 2009 (c) ΔΦ = − 35 − 60 = − 95 Wb 7. (a) As the rod falls through the magnetic field, the falling electrons are pushed to the left end of the rod, making the right end positive. This establishes a difference in voltage between the ends. The rod will continue to fall, gaining speed. The electrons will have a greater vertical speed and will experience a greater sideways force. This will make the difference in voltage greater. As long as the rod accelerates, the voltage difference will increase. (b) This process differs from the charging of a capacitor in that the force needed to separate the charges is from within the rod, rather than from an external battery, and the voltage difference will continue to increase as the rod speeds up until the rod reaches a terminal velocity. The rate at which the voltage increases will also depend on the acceleration and is likely to be constant. In a capacitor, the initial increase is rapid, followed by a levelling off. ΔΦ Δt −95 = −1 × 2.5 = 38 V 13. (a) Circumference of loop = 10 cm C r= = 1.59 cm 2π ΔΦ = − 0.60 T × (π × 0.01592 ) m 2 ε = −N ε = −N = −1 × 9. (a) Anticlockwise current seen from above (b) No (c) Yes, clockwise current 10. (a) Anticlockwise current as seen from the left (b) Clockwise current produced (b) ΔΦ = −1.0 T × (0.26 m)2 = 0.0676 Wb 11. (a) Yes; as the coil starts to leave the magnetic field an anticlockwise current will be induced in order to produce a magnetic field out of the page. (b) Same as (a) (c) No, there is no change in flux. (d) Same as (c) ΔΦ Δt 0.0676 = −1 × 0.5 = − 0.1352 ε=−N A = πr 2 −2 2 = π × (5.0 × 10 ) m ε R 0.1352 = 2.5 = 0.054 A I = 2 = 7.895 × 10−3 m 2 ΔΦ = Φ final − Φ initial = 0 − BA (c) ΔA = π(0.04)2 − π(0.08)2 −3 = 0 − 0.40 × 7.85 × 10 ΔΦ ε=−N Δt −1 × − 0.40 × 7.85 × 10−3 = 0.2 = 0.016 V (b) ΔΦ = 35 − 60 = − 25 Wb = 1.51 × 10−2 m 2 ΔΦ = 2.0 × 1.51 × 10−2 Wb ΔΦ ε=−N Δt 2.0 × 1.51 × 10−2 = −1 × 0.8 = − 0.037 75 V ε R 0.037 75 = 0.2 = 0.19 A ΔΦ Δt −25 = −1 × 1.5 = 17 V I = ε=−N Jacaranda Physics 2 TSK −0.60 × π × 0.01592 0.3 = 1.59 × 10−3 V ε I = R 1.59 × 10−3 = 0.4 = 0.004 A 8. Clockwise current to produce a magnetic field into the page 12. (a) ΔΦ Δt 42 © John Wiley & Sons Australia, Ltd 2009 14. (a) f = 50 Hz T = 0.02 s (c) ΔΦ = BA = 0.1 × 10−3 T × 3.0 × 107 Time for a quarter of a turn = 0.005 Φinitial = 0 = 3000 Wb ΔΦ ε=−N Δt 3000 = −1 × 1 = − 3000 V ε (magnitude) = 3000 V Φ final = 2.5 T × 0.042 = 0.10 Wb ΔΦ ε=−N Δt = − 100 × 0.10 0.005 18. (a) = 2000 V (b) ε = −N Φ initial = 0 Φ final = 0 ⇒ ΔΦ = 0 ⇒ ε=0 ΔΦ Δt (b) 15. The rotating coil of the motor is in the magnetic field and so produces an induced EMF. If the motor is connected to a battery, this induced EMF will oppose the battery and is called a back EMF. If the motor is disconnected from the battery and forced to turn, an alternating EMF will be produced which will be rectified by the commutator. (c) The loss in gravitational potential energy as the magnet fell was not fully converted into kinetic energy — some went into electrical energy. The magnet fell slower than it would have without the coil. (d) The change in magnetic flux is zero, so the two areas under the graph cancel each other out. (e) If the magnet is accelerating, the current below the coil, where it is travelling faster, will produce a larger current, but for a shorter time. ΔΦ 19. ε = −N Δt ΔΦ ⇒ IR = (magnitude only) Δt Q ΔΦ ⇒ R= where Q = charge Δt Δt ΔΦ Q= ⇒ R 16. (a) Simplify the induced EMF equation to: BA ε Δt ε= , rearranging B = , Δt A volt × sec so tesla = metre2 (b) Voltage = Current × Resistance (V = IR), and Charge Q QR Current = ( I = ), so ε = . Time Δt Δt Substituting into the expression from coulomb × ohm QRΔt QR B= , so tesla = = Δt × A A metre 2 (c) Magnetic flux = Magnetic field × Area (Φ = BA) QR Φ = BA = × A = QR, so weber = A coulomb × ohm. = 3.0 T × 1.6 × 10−3 m 2 0.2 Ω = 0.024 C 20. In both cases, there is an induced EMF; however, plastic is an insulator, so the electrons are not free to move as in the case of a conductor. In the wire there is a current, but there is no current in the plastic. 17. (a) v = 6000 m s−1 t = 1.0 s distance = vt = 6000 m 21. If the approaching north end of the magnet to a coil produced a magnetic field with the south end towards the magnet, the magnet would be attracted and speed up, and so induce an even stronger field whose south end would make the magnet travel even faster. Thus, kinetic energy would be created from nothing. (b) Area = L × W = 6000 m × 5000 m = 3.0 × 107 m Jacaranda Physics 2 TSK (i) Anticlockwise from above (ii) Zero current (iii) Clockwise current from above 43 © John Wiley & Sons Australia, Ltd 2009 22. (a) The faster the motor is turning, the quicker the coil is cutting the magnetic field in the motor, so the larger the back EMF. (b) At slow speeds there is a small back EMF, so the current will be higher than when the motor is spinning faster. (c) The low resistance of the coils will produce a very large current which will overheat the motor. 23. Distance covered by axle in one second = vt 3. (a) N prim = × Nsec 240 × 50 12 = 1000 turns P V 20 × 5 = 12 = 8.3 A (b) I = (c) Pprim = Vprim I prim = 20 × 5 W 20 × 5 240 = 0.42 A 4. A transformer does not work with constant DC voltage, such as that supplied by a battery or a regulated power supply, because there is no change in the magnetic flux. I prim = = 50 m 2 ΔΦ ε=−N Δt 40 × 10−6 T × 50 m 2 = −1 × 1 −3 = 2.0 × 10 V 5. To enable the direction of the magnetic field to change as quickly as the current changes direction. 24. 8.9 V 25. (a) 28 m s (b) 35 Hz (c) 50 mV (d) 100 mV (e) 35 mV 26. 6.4 V 6. (a) Step down transformer (b) 48 turns 7. (a) 125 V (b) 37.5 W (c) 0.625 A 8. (a) Voltage across generator = 250 VRMS P I = V 50 000 = 250 = 200 A Chapter 12 Transmission of power Vsec N = sec Vprim N prim (b) Ploss = I 2 R 2000 × 240 100 = 4800 V N = sec × Vprim N prim = (200)2 × 0.3 ⇒ Vsec = 2. (a) Vsec Vsec = = 120 km h −1 × 1 s 120 = 3.6 = 33.33 m Area covered = L × W = 33.33 × 1.5 1. Vprim = 12 kW (c) V = IR = 200 × 0.3 = 60 V (d) V = 250 − 60 = 190 V (e) 6 × 240 300 = 4.8 V N prim × Vsec (b) Vprim = Nsec = 300 ×9 6 = 450 V = Jacaranda Physics 2 TSK 44 © John Wiley & Sons Australia, Ltd 2009 P V 50 000 = 250 × 20 = 10 A V2 P (240)2 = 20 000 = 2.88 Ω Effective resistance of workshop and house (in parallel): 1 1 1 = + Reff 2.88 6.0 I = (b) ⇒ R= Ploss = I 2 R = (10) 2 × 0.3 = 30 W (c) V = IR ⇒ Reff = 1.95 Ω (c) Total resistance = 1.95 + 0.20 = 2.15 Ω = (10)2 × 0.3 = 3.0 V (d) V = 250 − 3 1.95 × 240 2.15 = 218 V V = = 247 V V 240 + = 6.0 Ω 40 I (b) R = 6.0 + 0.2 = 6.2 Ω (c) V = 6.2 × 240 = 232 V (d) (i) Calculate resistance of workshop: P = VI 9. (a) R = = Jacaranda Physics 2 TSK (e) (f) 10. (a) (b) (c) (d) 11. (a) (b) V2 R 45 Yes Increase 667 A 178 kW 267 V 330 kV, 220 mW 25% 0.25% © John Wiley & Sons Australia, Ltd 2009 Unit 4 ■■■■■■■■■ Area of study 2: Interactions of light and matter Part A — WORKED SOLUTIONS (b) Two waves are exactly out of phase when one wave is exactly half a cycle ahead of another; for example, a wave crest from one source coincides with a wave trough from another source. Destructive interference happens when the addition of waves results in zero disturbance, commonly as a result of two waves arriving at a point in space exactly out of phase, with crests meeting troughs. A node is a point, or a line, where destructive interference occurs, that is, where there is no wave disturbance. Chapter 13 Light – waves and particles 1. (a) (b) 5. (a) A: antinode, B: node, C: antinode (b) A: bobbing up and down B: stationary C: bobbing up and down (c) A: 0, A is on central antinode at an equal distance from both sources 3λ , B is on second nodal line B: 2 C: λ, C is on the first antinodal line (d) A: bright, B: dark, C: bright The wavelets at the edge of the slits form a wavefront which represents the bending of diffracting waves around the edge of the slit. 2. (a) 6. (a) The first minima occur where the path difference for light rays travelling from different places on the λ opening reaches . This will be a region where the 2 intensity of the light will be diminished due to destructive interference. (b) (b) The same number of wavefronts pass X and Y each minute, since wavefronts cannot disappear nor appear between those points. The wave frequency is the same in both materials. (c) Wave speed is greater in material B. Since the wave frequency is the same in materials A and B, the greater speed will be seen as a greater wavelength. This occurs in material B. 3. Superposition means the adding together of the effects of waves coinciding at a particular location. For example, the bright and dark bands in the pattern produced when light passes through two slits to a screen are caused by the addition of the effects of waves travelling from each of the two slits to each point on the screen. 4. (a) Two waves are in phase when corresponding points on the waves are synchronised; for example, wave crests leaving two slits at the same time, or wave troughs arriving at a certain place at the same time. Waves experience constructive interference when they are in phase at a point in space, and continue to be in phase as the waves pass through that point. The crests arrive together and the troughs arrive together. This point is said to be on an antinode, which is a point, or a line, where maximum wave disturbance occurs. Jacaranda Physics 2 TSK Jacaranda Physics 2, 3rd Edition TSK (c) 7. (a) Diffraction depends on wavelength or colour of light. White light is a mixture of the different colours in the spectrum, so when it diffracts when passing through a small slit, the colours that make up the white light form slightly different diffraction patterns. 46 © John Wiley & Sons Australia, Ltd 2009 (b) The positions of the minima are given by mλ . The red end of the spectrum has the sin θ = d longest wavelengths, therefore θ for any particular minimum occurs at a greater angle than for blue light and other parts of the visible spectrum. This means that the bright fringes in an interference pattern will be reddened on their outer edges. 8. See table at bottom of page. 9. (a) Ek = 12 mv 2 ⇒v= = 10. Ephoton = = = = 6.21 eV − 5.1 eV = 1.11 eV To stop an electron with 1.11 eV of kinetic energy requires a stopping voltage of 1.11 V. 11. (a) Minimum energy required to raise the electron energy to zero is 4.5 eV. (b) In each case Ek of electron is given by Ephoton + initial electron energy (i) Ek = 5.9 eV − 4.7 eV = 1.2 eV Electron is ejected with kinetic energy of 1.2 eV. (ii) Ek = 5.9 eV − 5.3 eV = 0.6 eV Electron is ejected with kinetic energy of 0.6 eV. (iii) Ek = 5.9 eV − 5.9 eV = 0.0 eV Electron is ejected with zero kinetic energy. (iv) Energy of photon not sufficient to raise electron energy to zero, so electron is not ejected. 12. (a) The maximum current occurs when the accelerating voltage causes all ejected electrons to be collected at the anode. The voltage required for this is greater than zero because some electrons leave at an angle and their parabolic path may miss the anode at lower voltages. When the voltage opposes the motion towards the cathode, electrons travelling towards the anode slow down. When the magnitude of the voltage is large enough, the electrons reverse direction and so do not contribute to the current. At a high enough retarding voltage, all electrons turn around before reaching the anode so the current is zero. (b) Increasing the intensity without changing the frequency would increase the number of photons per second reaching the cathode but not change their energy. As a result, the photoelectrons will have the same energy spread, and therefore the same stopping voltage, but there will be more electrons ejected per second, resulting in a higher photocurrent. 9.1 × 10−31 kg = 5.3 × 105 m s −1 (b) (c) (d) Infra-red from CO2 laser Red helium–neon laser Yellow sodium lamp UV from eximer laser X-rays from aluminium Jacaranda Physics 2 TSK 10.6 μm 633 nm 589 nm 0.193 μm 0.988 nm 9.932 × 10−19 J 1.60 × 10−19 J eV −1 = 6.21 eV Maximum Ek, photoelectron = Ephoton − W 2 × 0.8 × 1.60 × 10−19 J eV −1 Wavelength 200 × 10−9 m = 9.932 × 10−19 J 2 Ek m Source hc λ 6.6262 × 10−34 J s × 2.9979 × 108 m s −1 Frequency 13 2.83 × 10 Hz 4.74 × 1014 Hz 5.09 × 1014 Hz 1.55 × 1015 Hz 3.03 × 1017 Hz 47 Energy −20 1.87 × 10 J, 0.117 eV 3.14 × 10−19 J, 1.96 eV 3.37 × 10−19 J, 2.11 eV 1.03 × 10−18 J, 6.42 eV 2.01 × 10−16 J, 1.25 keV Momentum 6.25 × 10−29 kg m s−1 1.05 × 10−27 kg m s−1 1.125 × 10−27 kg m s−1 3.43 × 10−27 kg m s−1 6.69 × 10−25 kg m s−1 © John Wiley & Sons Australia, Ltd 2009 the effect of a single photon. The pattern of arrangement of these spots demonstrates the wave characteristics of the light. 15. Neon atoms have well defined energy levels, and the wavelengths of light emitted by neon atoms correspond to atoms jumping from higher to lower energy levels. The sharpness of the wavelengths in the spectrum indicates the sharpness of the possible energies of the atoms. 16. The magnitude of the energy change of the electrons is equal to the photon energy, hf. Ephoton = 6.6262 × 10−34 J s × 4.59 × 1014 Hz (c) Increasing the frequency of the light would increase the energy of each photon, resulting in higher energy photoelectrons and a greater stopping voltage. If the intensity is unchanged, the energy per second reaching the cathode is not changed, but since each photon has a greater energy, this means that there are fewer photons per second reaching the cathode and the photocurrent will be reduced. = 3.04 × 10−19 J = 3.04 × 10−19 J 1.60 × 10−19 J eV −1 = 1.90 eV ΔEelectron = −3.04 × 10−19 J = − 1.90 eV 17. (d) There will be the same number of photons per second and they have the same energy but it requires a different energy to eject an electron. This will change the maximum Ek of the electrons and therefore the stopping voltage but since the same number of photons reach the cathode each second, the maximum photocurrent will be unchanged. Two answers are shown. One is for a material with a greater work function (lower maximum Ek and therefore lower stopping voltage) and the other is for a material with a smaller work function (higher maximum Ek and therefore higher stopping voltage). Each wavefront forms a set of centres for circular wavelets. The wavelets closest to the boundary will reach the material where they will travel at higher speed first. These wavelets will form the new wavefront in this material as shown in the diagram, at a different direction to the incoming wavefront. 18. (a) λ = = c f 3.00 × 108 m s −1 4.59 × 1014 Hz = 6.53 × 10−7 m (b) 4.35 × 10−7 m 19. Since violet light experiences the greater change in direction, its change in speed must also be greater. The speed in air . Since refractive index is equal to the ratio speed in flint this ratio is greater for violet light, violet light travels more slowly than red light in flint. 13. The frequency of the light determines the photon energy, and so determines energy delivered to each electron and the maximum kinetic energy of the electrons. The maximum kinetic energy also depends on the surface because this determines the energy supplied to allow the electron to escape. The wave model predicts that electrons would be ejected below this threshold frequency. It would just take a longer time for the energy to accumulate for electrons to escape. 14. The individual points showing where a chemical reaction of the film emulsion has occurred, would be the sign of Jacaranda Physics 2 TSK 20. Constructive: nλ, n = 0, 1, 2, . . ., 0, 1.06 μm, 2.12 μm, 3.18 μm . . . Destructive: ( n + 12 )λ, n = 0, 1, 2, . . ., 0.53 μm, 1.59 μm, 2.65 μm . . . 48 © John Wiley & Sons Australia, Ltd 2009 21. (a) (i) The bright band corresponds to constructive interference where crests from the two slits arrive together, and troughs in the light waves arrive together. (ii) The dark band is where destructive interference occurs. At all times, the sum of the waves from the two slits is zero, including crests from one slit coinciding with troughs from the other. (b) (i) 0 λ (ii) = 317 nm 2 (iii) 2λ = 1266 nm (c) The wavelength is smaller so the bright fringes in the interference pattern would be closer together. As path difference for the bright fringes equal nλ, a smaller wavelength means that each bright fringe would be moved closer to the central bright band. (d) The increase in distance between the slits would result in the bright bands being closer together. (e) Moving the screen away would result in the interference pattern spreading out, increasing the distance between the bright bands. 22. Circular wavefronts can be used when the slit width is extremely narrow in comparison with the wavelength of the light. Then diffraction produces effectively circular wavefronts. 23. Each stripe on the soap film corresponds to a certain path difference for light reflected from the front and back of the film, resulting in destructive interference for a particular set of wavelengths. The stripes are approximately horizontal because the part of the film that has a certain thickness is a horizontal slice through the film. As the soap film drains, the part of the film with this thickness moves downwards, so each stripe does too. 24. The centre of the pattern consists of red and blue light, hence the pattern would appear magenta coloured. Red light diffracts more than blue light and so there would be a point of the centre where there would be red light but no blue light as it would be at a minimum in intensity. Thus the pattern would be magenta in the middle with a red fringe on either side. 25. For green light at, say, 515 nm: hc Ephoton = λ 6.6262 × 10−34 J s × 2.9979 × 108 m s −1 = 515 × 10−9 m ⇒ Eblue photon rate of emission of red photons = rate of emission of blue photons Ered photon = λ red photon λ blue photon = 1.33 27. Electron energy = 4.0 × 10−19 J = 4.0 × 10−19 J 1.60 × 10−19 J eV −1 = 2.5 eV (a) A retarding voltage of 1.0 V would reduce the kinetic energy by 1.0 eV to 1.5 eV, or 2.4 × 10−19 J. (b) A retarding voltage of 2.5 V is required to completely transform the 2.5 eV of kinetic energy into electric potential energy, that is, to stop the electron. (c) A stopping voltage of 4.3 V means that the highest kinetic energy of electrons is 4.3 eV, or 6.9 × 10−19 J. (d) 28. (a) Maximum Ek = Ephoton − W hc Ephoton = λ 6.6262 × 10−34 J s × 2.9979 × 108 m s −1 = 254 × 10−9 nm = 7.82 × 10−19 J = 4.88 eV Maximum Ek = 4.89 eV − 2.30 eV = 2.59 eV, or 4.14 × 10−19 J (b) To transform this kinetic energy into electric potential energy would require 2.59 V. (c) and (d) = 3.86 × 10−19 J So the detection limit of 2 × 10−17 J is equivalent to: 2 × 10−17 J = 52 photons. 3.86 × 10−19 J P = energy per second 26. = rate of photon emission × energy of one photon Pblue = Pred ⇒ rate of emission of blue photons × Eblue photon = rate of emission of red photons × Ered photon Jacaranda Physics 2 TSK 49 © John Wiley & Sons Australia, Ltd 2009 4. Atoms and molecules in the outer part of the Sun absorb at characteristic wavelengths, reducing the intensity of those colours in the light which reaches us. 5. Emission lines are produced when electrons return from an excited state to a lower energy state. The energy is released in the form of photons of particular frequencies. Absorption lines are produced when light from a continuous spectrum passes through a gas. This light excites some of the electrons in the atoms making up the gas, so photons with the energies allowed by the atoms will be removed from the continuous spectrum. As the energy required to raise an electron to a more excited state is equal to the energy released when the electrons drop back to the lower state, the emission lines and absorption lines for a particular element will be the same. 6. Possible answers include refracting the light through a prism. Spectral yellow will remain yellow whereas a mixture of green and red light will separate into two beams. h 7. (a) λ = p h = mv 6.63 × 10−34 = 1.67 × 10−27 × 3.0 × 107 29. (a) Threshold frequency is where the graph crosses the frequency axis, that is, where the maximum Ek of the electrons, and the stopping voltage, is zero: f0 = 4.6 × 1014 Hz c (b) λ = f = (c) (d) 30. (a) (b) 2.9979 × 108 m s −1 4.6 × 1014 Hz = 6.5 × 10−7 m W = hf0 = 6.6262 × 10−34 J s × 4.6 × 1014 Hz = 3.0 × 10−19 J = 1.9 eV or, use the intercept with the stopping voltage graph. Planck’s constant is the gradient of the graph: 6.6 × 10−34 J s, or 4.1 × 10−15 eV s Red light is diffracted more as it passes through a narrow slit because it has a longer wavelength. As a result its diffraction pattern will be broader. The edge of the pattern will be reached by white light, but minus the colour which experiences destructive interference there. The violet light has the narrower interference pattern, so its first node occurs closest to the centre of the pattern. The remaining light will therefore have a yellow tinge. = 1.3 × 10−14 m h (b) λ = p h = 2mE = 2 × 9.11 × 10−31 × 1.6 × 10−19 × 54 = 1.7 × 10−10 m h (c) λ = p h = mv Chapter 14 Matter – particles and waves = 6.63 × 10−34 0.2 × 50 = 6.6 × 10−35 m 8. (a) Ek = − ΔEp 1. The cathode rays are deflected by both magnetic and electric fields, and have mass and charge, unlike electromagnetic radiation. 2. There is an electric field between the plates. When a negative particle enters the electric field, it is attracted to the positive plate and repelled from the negative plate. 3. A fire glows with a continuous range of wavelengths, and in a red fire, the red wavelengths have the greatest intensity. Neon in a discharge tube glows red because the electrons in the neon atoms are excited to particular energies. When the electrons return to the ground state they produce light of a few fixed wavelengths, mainly in the red part of the spectrum. Jacaranda Physics 2 TSK 6.63 × 10−34 = − Vqelectron = − (5 kV × − 1 e) = 5 keV = 5 keV × 1.60 × 10 −16 J keV −1 = 8.0 × 10−16 J (b) Ephoton = 8.0 × 10−16 J = 50 hc λ © John Wiley & Sons Australia, Ltd 2009 ⇒ λ = = hc Ephoton 13. (a) λ = 6.63× 10−34 J s × 3 × 108 m s −1 = = 9.11 × 10−31 × 2.5 × 106 = = 2.3 × 10−24 kg m s −1 h λ= p 6.63 × 10−34 2.3 × 10−24 λ ≈ 1 . As the d wavelength is a little greater than the atomic spacing, diffraction will be significant. 11. (a) 3000 eV, 4.8 × 10−16 J (b) 2.96 × 10−23 Ns, 2.2 × 10−11 m (c) λ/w = 0.045, thus the scientist would not expect to observe any diffraction effects since the wavelength of the electrons is too small. (d) The scientist should make the accelerating voltage smaller. This would reduce the energy of the electrons and hence reduce the momentum and hence increase the wavelength. (e) 2.2 × 10−11 m, 2.96 × 10−23 Ns. To obtain the same diffraction pattern the wavelength and hence the momentum of the photons much be the same as the electrons used in the previous experiment. (f) 8.88 × 10−15 J, 5.55 × 104 eV c = fλ 12. c ⇒λ = f (b) Significant diffraction occurs when = 14. 2 × 9.11 × 10−31 × 1000 × 1.60 × 10−19 6.63 × 10−34 2 × 1.67 × 10−27 × 1000 × 1.60 × 10−19 = 9.1 × 10−13 m h λ = p h = 2mE h = 2mqV ⇒V = = h2 2mqλ2 (6.63 × 10−34 )2 2 × 9.1 × 10−31 × 1.60 × 10−19 × (2.0 × 10−10 )2 = 38 V 15. For a 10 eV electron: h λ= p h = 2mE 3.0 × 108 4.5 × 1014 = 6.7 × 10−7 m h p = λ h ⇒v = λm 6.63 × 10−34 = 6.7 × 10−7 × 9.1 × 10−31 = 6.63 × 10−34 2 × 9.1 × 10−31 × 10 × 1.60 × 10−19 = 3.9 × 10−10 m For a 10 eV photon: hc E= λ hc ⇒ λ= E = 1.1 × 103 m s −1 Jacaranda Physics 2 TSK 6.63 × 10−34 = 3.9 × 10−11 m for the electron Using m = 1.67 × 10−27 kg for the proton, h λ= p h = 2mE = 2.9 × 10−10 m = h 2mE Since E and h are constant, the larger the value of m, the smaller the wavelength. So the proton will have the shorter wavelength. h (b) λ = p h = 2mE 8.0 × 10−16 J = 2.5 × 10−10 m 9. Light exhibits both wave and particle behaviour, depending on the experiment you are performing at the time. For further discussion of this issue argue with your friends and teacher! 10. (a) p = mv = h p 51 © John Wiley & Sons Australia, Ltd 2009 = 6.63 × 10−34 × 3.0 × 108 19. (a) and (b) 10 × 1.60 × 10−19 = 1.2 × 10−7 m The electron has the smaller wavelength. 16. The magnitude of the energy change of the electrons is equal to the photon energy, hf. Ephoton = 6.6262 × 10−34 J s × 4.59 × 1014 Hz = 3.04 × 10−19 J = 3.04 × 10−19 J 1.60 × 10−19 J eV −1 = 1.90 eV ΔEelectron = −3.04 × 10−19 J = −1.90 eV 17. (a) Light of this wavelength corresponds to a photon energy of hc 6.63 × 10−34 × 3.0 × 108 E= = = 4.7 × 10−19 J λ 420 × 10−9 Electrons in helium absorb these photons indicating that there is an energy level within the helium atom 4.7 × 10−19 J above the ground state of the atom. (b) The increase in energy of the electron is equal to the energy of the absorbed photon: ΔE = 4.7 × 10−19 J = 2.94 eV (c) E1 = 12.8 − 12.1 = 0.7 eV E2 = 12.8 − 10.2 = 2.6 eV E3 = 12.8 eV E4 = 12.1 − 10.2 = 1.9 eV E5 = 12.1 eV E6 = 10.2 eV 18. See table at bottom of page. For red light: Convert E (eV) to E (J) by multiplying by 1.60 × 10−19. E (J) Calculate p = . c E (J) Calculate f = . h c Calculate λ = . f For electron: Convert E (eV) to E (J) by multiplying by 1.60 × 10−19. Calculate p = 2mE . h Calculate λ = . p Follow the above in reverse for the blue light and electron. (d) The least energy is 0.7 eV. The photon with this energy has a wavelength of hc λ= E 6.63 × 10−34 × 3 × 108 = 1.60 × 10−19 × 0.7 = 1.78 × 10−6 m (1.8 × 10−6 m) The greatest energy photon has energy 12.8 eV: hc λ= E 6.63 × 10−34 × 3 × 108 = 1.60 × 10−19 × 12.8 = 9.71 × 10−8 m (9.7 × 10−8 m) λ (nm) f (Hz) E (J) E (eV) p (Ns) Red light 633 4.73 × 1014 3.14 × 10−19 1.96 1.05 × 10−27 Electron 0.877 — 3.14 × 10−19 1.96 7.56 × 10−25 Blue light 405 7.41 × 1014 4.91 × 10−19 3.07 1.64 × 10−27 Electron 405 — 1.47 × 10−24 9.19 × 10−6 1.64 × 10−27 Jacaranda Physics 2 TSK 52 © John Wiley & Sons Australia, Ltd 2009 20. (a) and (b) Third excited state: Ephoton = − ΔEatom = − 0.77 eV − ( −5.12 eV) = 4.35 eV = 6.96 × 10−19 J λ= = = − 3.01 eV − (−5.12 eV) = 2.11 eV = 3.38 × 10−19 J = hc Ephoton 6.63 × 10−34 J s × 3 × 108 m s −1 3.38 × 10−19 J = 5.88 × 10−7 m Second excited state: Ephoton = − ΔEatom = − 1.37 eV − (−5.12 eV) = 3.75 eV = 6.0 × 10 hc λ= Ephoton = −19 Ephoton 6.63 × 10−34 J s × 3 × 108 m s −1 6.96 × 10−19 J = 2.86 × 10−7 m The energy change from the first excited state is responsible for the yellow glow. A comparison with the answers to question 10 helps here. Yellow light is between red and blue light in the spectrum, that is between 405 nm and 633 nm. The first excited state is the only one that produces a wavelength in this range. 21. At room temperature, virtually all hydrogen atoms are in their ground state. As a result, the absorption spectrum shows only lines corresponding to absorption by atoms in the ground state. The emission spectrum is formed when electrons which are in excited states due, for example, to the energy supplied by a flame or an electron in an electric discharge emit light as they drop to lower energy levels. Since there is no restriction on which levels they can fall to, series of spectra are seen which correspond to any final state of the atom, not just the ground state. The 0.0122 nm UV radiation corresponds to the transition between the ground and first excited state, so it will appear in both emission and absorption spectra. The 656 nm light corresponds to a transition between the first and second excited states of hydrogen. It will appear in the emission spectrum but not the absorption spectrum. 22. There is no wave equation for matter of the type v = fλ since the speed of matter, unlike light, does not depend on the medium through which it moves. For photons, the energy is ‘locked-up’ in the frequency variable f: E = hf. For matter, the energy is ‘locked-up’ in the speed, which 2 unlike light is variable: E = 12 mv . It serves no purpose (c) Ephoton = −ΔEatom = Eatom initial − Eatom final First excited state: Ephoton = − ΔEatom λ= hc to develop a frequency for matter. However, the de Broglie wavelength is useful as it determines the extent to which both light and matter behave like a wave and determines the extent to which both matter and light behaves like a particle. J 6.3 × 10−34 J s × 3 × 108 m s −1 6.0 × 10−19 J = 3.32 × 10−7 m Jacaranda Physics 2 TSK 53 © John Wiley & Sons Australia, Ltd 2009 Unit Detailed study 3.1: Synchrotron and its applications 4 ■■■■■■■■■ Jacaranda Physics 2, 3rd Edition TSK Part A — WORKED SOLUTIONS V d 480 = 0.12 = 4.0 × 103 V m−1 from the bottom plate to the top plate (b) F = Eq = 4.0 × 103 × 1.6 × 10−19 = 6.4 × 10−16 N (c) W = qV = 480 eV = 7.7 × 10−17 J (d) Increase in kinetic energy = work done = 7.7 × 10−17 J Chapter 15 5. (a) E = The Australian synchrotron 1. Similarities: Both accelerate electrons to high speeds in a straight line using a potential difference. Differences: The cathode ray tube uses a single pair of charged plates, while a linear accelerator uses a series of charged plates or tubes whose polarities are reversed as the electron passes from one pair of plates, or one tube, to the next. This enables the linac to give the electron energy that would be impractical with a cathode ray tube due to the high voltages that would be required. 2. The filament is a wire that is heated with an electric current. This heating gives some of the electrons in the filament sufficient energy to escape the wire with the assistance of the potential difference between the charged plates. 3. A voltage can be used to accelerate a charged particle because where there is a voltage there is an electric field. An electric field exerts a force on charged particles and forces result in acceleration of the particles. A particle that is neutral, however, is not influenced by the electric field. 4. (a) 1 2 6. (a) (b) (c) (b) (d) = 7.7 × 10−17 J × v = 7.7 × 10−17 J ⇒ v = 1.3 × 107 m s−1 V E= d 480 = 0.24 = 2.0 × 103 V m−1 from the bottom plate to the top plate F = Eq = 2.0 × 103 × 1.6 × 10−19 = 3.2 × 10−16 N W = qV = 480 eV = 7.7 × 10−17 J Increase in kinetic energy = work done = 7.7 × 10−17 J F = Eq 1 2 Vq = d 200 × 1.60 × 10−19 = 0.10 = 3.2 × 10−16 N × 9.1 × 10 1 mv 2 2 −31 2 × 9.1 × 10 1 mv 2 2 −31 2 = 7.7 × 10−17 J × v = 7.7 × 10−17 J v = 1.3 × 107 m s−1 ⇒ 7. (a) For an electron: (i) V = 1000 V E = Vq = 1000 × 1.6 × 10−19 (c) = 1.6 × 10−16 J 1 2 ⇒ 2 × 3.2 × 10−16 = 6.4 × 10−16 N Jacaranda Physics 2 TSK × 9.1 × 10 54 1 2 −31 mv 2 = 1.6 × 10−16 J × v 2 = 1.6 × 10−16 J v = 1.9 × 107 m s−1 © John Wiley & Sons Australia, Ltd 2009 (ii) V = 10 000 V (ii) V = 10 000 V E = Vq = 10000 × 1.6 × 10 E = Vq −19 = 10000 × 3.2 × 10−19 = 1.6 × 10−15 J 1 2 × 9.1 × 10 1 mv 2 2 −31 2 = 3.2 × 10−15 J = 1.6 × 10−15 J × v = 1.6 × 10−15 J 1 2 v = 5.9 × 107 m s−1 ⇒ (iii) V = 100 000 V v = 9.8 × 105 m s−1 E = Vq −19 = 1 000 000 × 3.2 × 10−19 = 1.6 × 10−14 J 1 2 × 9.1 × 10 1 2 −31 = 3.2 × 10−14 J mv 2 = 1.6 × 10−14 J × v 2 = 1.6 × 10−14 J 1 2 v = 1.9 × 108 m s−1 ⇒ (b) For a proton: (i) V = 1000 V V = = 1000 × 1.6 × 10−19 = 1.6 × 10−16 J 1 2 × 1.67 × 10 1 2 × 1.67 × 10 ⇒ (iii) V = 100 000 V J × v 2 = 1.6 × 10−15 J v = 1.4 × 106 m s−1 E = Vq = 1.6 × 10−14 J × 1.67 × 10 = 1.6 × 10−14 J × v = 1.6 × 10−14 J ⇒ v = 4.4 × 106 m s−1 (c) For an alpha particle: (i) V = 1000 V E = Vq = 3.2 × 10−16 J 1 2 × 6.64 × 10 ⇒ Jacaranda Physics 2 TSK × 9.1 × 10−31 × (3.0 × 107 ) 2 1.6 × 10−19 1 2 × 1.67 × 10−27 × (3.0 × 107 ) 2 1.6 × 10−19 1 2 × 6.64 × 10−27 × (3.0 × 107 )2 3.2 × 10−19 = 5.1 × 1016 m s −2 10. (a) F = Bqv = 0.25 × 1.60 × 10−19 × 5.0 × 106 = 2.0 × 10−13 N F (b) a = (the electron is travelling at speeds where m relativistic effects are not important) = 1000 × 3.2 × 10−19 1 mv 2 2 −27 2 1 2 = 9.3 × 106 V 9. (a) F = Bqv = 2.4 × 1.60 × 10−19 × 1.2 × 105 = 4.6 × 10−14 N (b) Assuming the electron to be in a synchrotron with the magnetic field coming from above, the electron will be deflected towards the south (using the right-handslap rule or equivalent), resulting in a circular arc for the period that the electron is in the magnetic field. F (the electron is travelling at speeds where (c) a = m relativistic effects are not important) 4.6 × 10−14 N = 9.1 × 10−31 = 100 000 × 1.6 × 10−19 1 2 v = 3.1 × 106 m s−1 q (c) V = mv 2 = 1.6 × 10−15 J 1 mv 2 2 −27 2 × v = 3.2 × 10−14 J = 4.7 × 106 V = 10000 × 1.6 × 10−19 1 2 −27 = 3.2 × 10−14 J mv 2 (b) V = E = Vq = 1.6 × 10 1 mv 2 2 −2 2 = 2.6 × 103 V v = 4.4 × 105 m s−1 −15 1 2 (a) V = = 1.6 × 10−16 J × v = 1.6 × 10−16 J ⇒ (ii) V = 10 000 V × 6.64 × 10 ⇒ 2 1 8. 2 mv = qV E = Vq 1 mv 2 2 −27 2 = 3.2 × 10−15 J × v = 3.2 × 10−15 J ⇒ (iii) V = 100 000 V E = Vq = 100 000 × 1.6 × 10 × 6.64 × 10 1 mv 2 2 −2 2 = 3.2 × 10−16 J × v = 3.2 × 10−16 J v = 3.1 × 105 m s−1 55 © John Wiley & Sons Australia, Ltd 2009 = 2.0 × 10−13 N 9.1 × 10 16. B = −31 = 2.2 × 10−17 m s −2 F (c) a = (the proton is travelling at speeds where m relativistic effects are not important) 2.0 × 10−14 N = 1.67 × 10−27 = 17. 18. = 1.2 × 1013 m s −2 11. (a) Down the page (b) Out of the page (c) Down the page perpendicular to the velocity of the proton 19. 12. 13. Out of the page 20. mv 2 14. F = r mv 2 Bqv = r mv r= Bq (a) r = 22. 9.1 × 10−31 × 3.0 × 107 4.0 × 1.6 × 10−19 = 4.3 × 10−5 m (b) r = 1.67 × 10−27 × 3.0 × 107 4.0 × 1.6 × 10−19 = 7.8 × 10−2 m (c) r = 6.64 × 10−27 × 3.0 × 107 4.0 × 3.2 × 10−19 = 0.16 m 15. mv 2 r mv 2 Bqv = r mv B= qr F= = 23. 9.1 × 10−31 × 3.0 × 107 1.6 × 10−19 × 0.10 = 1.7 × 10−3 T perpendicular to the velocity of the electron Jacaranda Physics 2 TSK 56 p qr 1.0 × 10−18 1.6 × 10−19 × 1000 = 6.3 × 10−3 T p = Bqr = 2.0 × 1.6 × 10−19 × 34.4 = 1.1 × 10−17 kg m s−1 Synchrotrons are designed to produce intense beams of electromagnetic radiation by accelerating charged particles (synchrotron radiation). In a synchrotron, electrons are accelerated to near light speeds and their circular motion ensures they are accelerating and emitting synchrotron radiation. Linac: accelerates electrons to 99.995% of the speed of light. Circular booster: increases the energy of the electrons greatly so that they travel at 99.999994% of the speed of light. Storage ring: uses magnets in straight-line sections joined by curved sections to deflect electrons into a circular path and emit synchrotron radiation at a tangent to the path. Beamlines: where the synchrotron radiation is directed onto an experiment. An electron gains speed in the electron gun, the linac, the circular booster and the straight sections of the storage ring. It loses speed in the curved magnets of the storage ring. As these photons have energy less than 100 keV, they will undergo Thompson scattering. This results in Bragg diffraction. nλ = 2d sin θ nλ ⇒ sin θ = 2d 0.125 × 10−9 × n = 2 × 0.35 × 10−9 = 0.179n When n = 1, θ = 10.3°; n = 2, θ = 21.0°; n = 3, θ = 32.5°; n = 4, θ = 45.7°; n = 5, θ = 63.5°. Higher values of n mean that 0.179n will be greater than 1. There is no angle whose sine is greater than 1, so photons are only detected at these five angles. (a) nλ = 2d sin θ nλ ⇒ = sin θ 2d ⇒ sin 15.4° = 0.266 nλ = 0.266 when n = 1. So 2d nλ = 0.531 ⇒ θ = 32°. For n = 2, 2d For n = 3, θ = 53°. For n = 4 and above, there is no solution. © John Wiley & Sons Australia, Ltd 2009 (b) nλ = 2d sin θ, n = 1, θ = 22.5° (half the detector angle), λ = 0.154 nm 1 × 0.154 × 10−9 = 2 d sin 22.5° d = 2.01 × 10−10 m nλ = 2d sin θ, n = 2, θ = 50° (half the detector angle), λ = 0.154 nm 2 × 0.154 × 10−9 = 2 d sin 50° d = 2.01 × 10−10 m (iii) (2.0 ± 0.1) × 10−10 m. Add up the five values, divide by 5. Use the range of values to estimate the uncertainty. (c) For RbCl: (i) 20°, 24°, 46°, 53°, 87° The angles 20° and 46° correspond to the shorter wavelength photon as the angles are smaller than the corresponding order of peaks for the other series of peaks in accordance with Bragg’s Law. (ii) nλ = 2d sin θ, n = 1, θ = 10° (half the detector angle), λ = 0.138 nm 1 × 0.138 × 10−9 = 2 d sin 10° d = 3.97 × 10−10 m nλ = 2d sin θ, n = 2, θ = 23° (half the detector angle), λ = 0.138 nm 2 × 0.138 × 10−9 = 2 d sin 23° d = 3.53 × 10−10 m nλ = 2d sin θ, n = 1, θ = 12° (half the detector angle), λ = 0.154 nm 1 × 0.154 × 10−9 = 2 d sin 12° d = 3.70 × 10−10 m nλ = 2d sin θ, n = 2, θ = 26.5° (half the detector angle), λ = 0.154 nm 2 × 0.154 × 10−9 = 2 d sin 26.5° d = 3.45 × 10−10 m nλ = 2d sin θ, n = 3, θ = 43.5° (half the detector angle), λ = 0.154 nm 3 × 0.154 × 10−9 = 2 d sin 43.5° d = 3.36 × 10−10 m (iii) (3.6 ± 0.4) × 10−10 m Add up the five values, divide by 5. Use the range of values to estimate the uncertainty. 27. (a) To calculate the angles for the three principal planes of each crystal, use Bragg’s Law, nλ = 2d sin θ. nλ ⇒ sin θ = 2d ⎛ nλ ⎞ ⇒ θ = sin −1 = ⎜ ⎟ ⎝ 2d ⎠ Calculate θ for each crystal spacing for n = 1, 2, 3 . . . for valid values of sin θ (that is, where sin θ ≤ 1). (b) To construct the diffraction pattern, the intensity of peaks relative to the first peak can be plotted against the detector angle for each crystal. This can be done by setting the first peak for D1 at 100. The subsequent heights are found by multiplying the number in brackets by 100. λ = 0.266 2d ⇒ λ = 1.2 × 10−10 m 24. As nλ = 2d sin θ and d and λ are unknown, we have one equation with two unknowns. There are many possible wavelengths and crystal spacings that would produce these diffraction angles. 25. nλ = 2d sin θ For n = 1, λ = 2 × 543.09 nm × sin 10.00° = 188.6 nm 26. (a) For NaCl: (i) 28°, 34°, 60°, 67°, 112° The angles 28° and 60° correspond to the shorter wavelength photon as the angles are smaller than the corresponding order of peaks for the other series of peaks in accordance with Bragg’s Law. (ii) nλ = 2d sin θ, n = 1, θ = 14° (half the detector angle), λ = 0.138 nm 1 × 0.138 × 10−9 = 2 d sin 14° d = 2.85 × 10−10 m nλ = 2d sin θ, n = 2, θ = 30° (half the detector angle), λ = 0.138 nm 2 × 0.138 × 10−9 = 2 d sin 30° d = 2.76 × 10−10 m nλ = 2d sin θ, n = 1, θ = 17° (half the detector angle), λ = 0.154 nm 1 × 0.154 × 10−9 = 2 d sin 17° d = 2.63 × 10−10 m nλ = 2d sin θ, n = 2, θ = 33.5° (half the detector angle), λ = 0.154 nm 2 × 0.154 × 10−9 = 2 d sin 33.5° d = 2.79 × 10−10 m nλ = 2d sin θ, n = 3, θ = 56° (half the detector angle), λ = 0.154 nm 3 × 0.154 × 10−9 = 2 d sin 56° d = 2.79 × 10−10 m (iii) (2.8 ± 0.2) × 10−10 m. Add up the five values, divide by 5. Use the range of values to estimate the uncertainty. (b) For LiCl: (i) 41°, 45°, 87°, 100° The angles 41° and 87° correspond to the shorter wavelength photon as the angles are smaller than the corresponding order of peaks for the other series of peaks in accordance with Bragg’s Law. (ii) nλ = 2d sin θ, n = 1, θ = 20.5° (half the detector angle), λ = 0.138 nm 1 × 0.138 × 10−9 = 2 d sin 20.5° d = 1.97 × 10−10 m nλ = 2d sin θ, n = 2, θ = 43.5° (half the detector angle), λ = 0.138 nm 2 × 0.138 × 10−9 = 2 d sin 43.5° d = 2.00 × 10−10 m Jacaranda Physics 2 TSK 57 © John Wiley & Sons Australia, Ltd 2009 For example, for Algodonite, the number in brackets for D2 is 0.4, so set the first peak for D2 to 40. For D3, the number in brackets is 0.2, so set the first peak for D3 to 20. The detector angle is twice the Bragg angle. The graph can be drawn freehand with the height values on the Y-axis and the detector angles on the X-axis, with the graph coming back down to the X-axis after each point. Alternatively, the graph can be drawn in a spreadsheet program such as Microsoft Excel. For Excel you need to enter dummy data points either side of each proper point to bring the graph down to the X-axis, for example for data point (38.6, 100), include two extra points (38,0) and (39,0). When finished, sort your data, do an ‘X−Y scatter’ graph and select the graph type that has straight lines between the points. (b) Aluminium: Detector Angles D1 Height D2 Height D3 Height 32.4 100 65.0 90 37.6 70 67.7 50 80.3 35 113.4 25 150.4 17 (a) Algodonite: Bragg Angles (degrees) D1 D2 D3 19.3 18.2 17.0 41.4 38.6 35.8 82.6 69.2 61.2 (a) Aurostibite: Bragg Angles (degrees) (b) Algodonite: Detector Angles D1 Height D2 Height D3 Height 38.6 100 36.3 40 34.0 20 82.8 50 77.1 20 71.5 10 165.2 25 138.4 10 122.5 5 D1 D2 D3 19.1 12.7 11.4 40.9 26.2 23.3 78.9 41.4 36.3 62.0 52.2 80.8 (b) Aurostibite: Detector Angles D1 Height D2 Height D3 Height 38.2 100 25.5 75 22.8 70 81.7 50 52.4 37 46.5 35 157.8 25 82.9 18.0 72.6 17 123.9 9 104.3 8 161.7 4 (a) Aluminium: Bragg Angles (degrees) D1 D2 D3 16.2 32.5 18.8 33.9 40.1 56.7 75.2 Jacaranda Physics 2 TSK 58 © John Wiley & Sons Australia, Ltd 2009 (a) Bilibinskite: Bragg Angles (degrees) D1 D2 D3 16.1 32.3 18.7 33.7 39.9 56.3 74.2 (b) (a) Bilibinskite : Detector Angles D1 Height D2 Height D3 Height 32.2 100 64.5 75 37.4 70 D1 D2 D3 67.4 50 79.8 35 18.6 31.4 37.7 112.7 25 148.4 17 39.7 Diamond: Bragg Angles (degrees) 73.2 (b) Diamond: Detector Angles D1 Height D2 Height D3 Height 37.2 100 62.9 25 75.4 16 79.3 50 146.5 25 (a) Copper: Bragg Angles (degrees) D1 D2 D3 18.4 21.3 31.0 39.0 46.7 70.9 28. X-rays travel in straight lines and are not charged. They diffract as though they had a wavelength less than that of visible light. 29. (a) 1.2 × 10−10 = 0.12 × 10−9 = 0.12 nm c (b) f = λ 3.0 × 108 = 1.2 × 10−10 (b) Copper: Detector Angles D1 Height D2 Height D3 Height 36.7 100 42.7 46 61.9 20 78.1 50 93.3 37 141.7 25 Jacaranda Physics 2 TSK = 2.5 × 1018 Hz 59 © John Wiley & Sons Australia, Ltd 2009 (c) E = hf = 6.63 × 10−34 × 2.5 × 1018 = 1.7 × 10−15 J = 10 keV h (d) p = λ 6.63 × 10−34 = 1.2 × 10−10 (ii) λ = (iii) λ = 31. = 2.0 × 1018 Hz 3.00 × 108 32. 23 × 10−9 = 1.3 × 1016 Hz (iii) f = 3.0 × 108 57 × 10−9 33. = 5.3 × 1015 Hz (b) λ = c f (i) λ = 34. 3.0 × 108 3.2 × 1018 = 9.4 × 10−11 m Jacaranda Physics 2 TSK 7.5 × 1019 = 4.0 × 10−12 m = 5.5 × 10−24 kg m s −1 c 30. (a) f = λ 3.0 × 108 (i) f = 0.15 × 10−9 (ii) f = 3.0 × 108 60 3.0 × 108 5.7 × 1017 = 5.3 × 10−10 m Compton scattering occurs when a beam of photons collides with electrons. When a beam of photons is shone through a material, some of the photons are scattered by electrons in the material. These photons have longer wavelength and lower frequency than the photons in the beam, indicating that they have lost energy. The scattering can be treated like a collision between two particles, where momentum and energy are conserved. Compton scattering requires high energy (>100 keV) photons. Eincoming photon = Escattered photon + Ek scattered electron Ek scattered electron = hfincoming photon − hfscattered photon = 6.63 × 10−34 (2.2 − 1.4) × 1018 = 5.3 × 10−16 J = 3.3 keV Thompson scattering occurs with photons with energy less than 100 keV. The photons are scattered without any change in energy, unlike the loss of energy that occurs in Compton scattering. Reflection of X-rays is really an interference effect. Wavelets of X-rays are scattered by each atom in the crystal plane. Wavelets from adjacent atoms overlap and reinforce at an angle equal to the incident angle. © John Wiley & Sons Australia, Ltd 2009 Unit 4 Detailed study 3.2: Photonics Jacaranda Physics 2, 3rd Edition TSK Part A — WORKED SOLUTIONS ■■■■■■■■■ Chapter 16 Light and lightemitting devices 6. 1. A photon 2. (a) Electricity heating the filament — the electrons gain sufficient energy from the hot surroundings that they emit visible light when returning to the ground state. (b) The high temperatures produced in the burning of the candle provide electrons in the gas surrounding the candle with sufficient energy that they emit visible light when returning to ground state. (c) Fusion reactions in the core produce very high temperatures, causing the Sun to behave as an incandescent light source. (d) The atoms are vibrating sufficiently due to the thermal energy that collisions between them result in the emission of photons across the visible spectrum. (e) A current of electrons is passed through a low density gas. Atoms that are hit by the electrons are excited. When the electrons in these atoms return to lower energy states they emit photons. Fluorescent tubes produce light in the UV wavelengths. A phosphor coating on the tube absorbs the UV light and emits photons in the visible wavelengths, producing a whitish light. (f) LEDs emit light when they are forward biased because as conducting electrons fall into holes in the p-type material, they release energy in the form of photons. 3. Lasers produce coherent light. The input power to the laser excites a high percentage of the atoms of an optical resonator to a high energy state. This is known as a population inversion. The atoms are then stimulated to release their energy (in the form of photons) by interacting with photons of the same energy. This is called stimulated emission. The process is also called amplification because it takes one photon and produces many identical photons. 4. Remote sensing is the analysis of photons that have come from a distant place in order to gain an understanding of that place. (a) By analysing the infra-red radiation coming from the surface of a planet. The wavelength with the peak intensity indicates the average surface temperature. (b) Water in the atmosphere of a planet reduces the intensity of some wavelengths of electromagnetic radiation coming from the surface of the planet. (c) The absorption lines in the spectra from stars reveal the elements present in the stars’ atmospheres. 5. (a) Mercury (b) A current of electrons is passed through a low density gas. Atoms that are hit by the electrons are excited. Jacaranda Physics 2 TSK 7. 8. 9. 10. 11. 12. 13. 14. 61 When the electrons in these atoms return to lower energy states they emit photons. A phosphor coating on the tube absorbs the UV light and emits photons in the visible wavelengths, producing a whitish light. Not all diodes emit visible light because the energy released goes into increasing the temperature of the crystal lattice. (a) Because the energy levels of atoms split when they come close to other atoms, as is the case with the atoms of a solid. The many splittings that occur in the energy levels due to an atom being in a solid result in an energy band of many energy levels close together, rather than a specific energy level. (b) The valence band in a solid is the highest natural energy band of a material and contains the valence electrons. (c) The conduction band is above the valence band. The energy levels of different atoms join to form a common band in which it is easy for an electron to pass between atoms — conduction. (d) The difference in energy between the conduction and valence bands (the gap energy) is greater for insulators than for semiconductors. (a) The band gap is the difference in energy between the conduction and valence bands. (b) The band gap determines the wavelength of photon hc . emitted by an LED by E = λ The wavelength and frequency of a released photon is related to the gap energy of the LED by the formula Eg = hf = hc/λ. The materials used in constructing the LED are manipulated to produce a desired wavelength of photon. 1 Eg ∝ λ Blue, green, red 2.14 eV 6.9 × 10−7 m hc E= λ hc ⇒λ= E 4.14 × 10−15 × 3.00 × 108 = 2.09 −7 = 5.94 × 10 m = 594 nm hc E= λ 4.14 × 10−15 × 3.00 × 108 = 930 × 10−9 = 1.34 eV © John Wiley & Sons Australia, Ltd 2009 E= 15. = hc λ 4.14 × 10−15 × 3.00 × 108 (c) 1500 × 10−9 = 0.828 eV 16. (a) Coherent photons have the same wavelength and are in phase. (b) Light bulbs produce photons that are out of phase because their emission is spontaneous and therefore at random. 17. (a) Light amplification by stimulated emission of radiation (b) One photon results in the emission of multiple identical photons. (c) The atoms in a gas are excited by the power supply of the laser and are stimulated to emit a photon by a photon of the correct energy. The photons that are travelling along the length of the laser resonator are reflected back and forward between two mirrors, resulting in the stimulated emission of many coherent photons. One of the mirrors allows some of the photons to pass through, forming the coherent laser beam. The laser light is parallel because the photons travelling in other directions escape the resonator without being amplified. 18. • The beam does not spread out as much. • There is a smaller range of wavelengths produced. • The light has a greater intensity. 5. (a) An example of 2 modes in a optic fibre (b) A low-order mode makes few internal reflections; a high-order mode makes many internal reflections. 6. (a) A single-mode step-index optical fibre is a very thin core with thick cladding that only has one possible mode for light travelling along it. (b) A single-mode step-index optical fibre is a very thin core with thick cladding; a multimode step-index optical fibre has a thicker core and thinner cladding. (c) A multimode step-index optical fibre has a core of constant refractive index; a graded-index optical fibre has a refractive index that decreases from the centre to the edge of the core. 7. (a) Dispersion is the spreading out of a light signal or pulse as it travels down an optical fibre. (b) Modal dispersion occurs when there is more than one possible path (mode) for the light to take. Some of the modes are longer than others, resulting in the spreading of the signal over the length of the fibre. (c) As dispersion results in the pulse spreading out, fewer pulses per second can be sent, which means a lower bandwidth. (d) Use a smaller diameter fibre; use a graded-index fibre; use a single-mode fibre. 8. (a) A graded-index optical fibre has a refractive index that decreases from the centre to the edge of the core. (b) Chapter 17 Fibre-optic waveguides and systems 1. The core and the cladding that surrounds it 2. 500 MHz 3. A core surrounded by cladding of lower refractive index. This is covered by a protective layer called a jacket. n1 sin θc = n2 sin 90° 4. (a) ⇒ (b) Jacaranda Physics 2 TSK ⎛n ⎞ θc = sin −1 ⎜ 2 ⎟ ⎝ n1 ⎠ ⎛ 1.53 ⎞ = sin −1 ⎜ ⎟ ⎝ 1.58 ⎠ = 75.5° ⎛n ⎞ θc = sin −1 ⎜ 2 ⎟ ⎝ n1 ⎠ ⎛ 1.46 ⎞ = sin −1 ⎜ ⎟ ⎝ 1.48 ⎠ = 80.6° (c) Modes that take the light on a longer path and hence closer to the edge of the fibre spend more time in lower refractive index fibre where the light travels faster. Those that take the light closer to the centre of the core and therefore on a shorter path travel more slowly, reducing the spreading due to modal effects. 9. (a) If there is only one mode, then all light signals will pass through the fibre in the same time. (b) Material dispersion (c) Material dispersion spreads the pulse so that the number of pulses per second that can be sent is reduced — lower bandwidth. (d) The modal dispersion is a much larger effect. ⎛n ⎞ θc = sin −1 ⎜ 2 ⎟ ⎝ n1 ⎠ ⎛ 1.46 ⎞ = sin −1 ⎜ ⎟ ⎝ 1.47 ⎠ = 83.3° 62 © John Wiley & Sons Australia, Ltd 2009 Attenuation (2) length of fibre Pin and Pout can be calculated by rearranging formula 1 to Pout = 10− attenuation/10 Pin 10. (a) NA = n12 − n22 and Fibre attenuation = = 1.482 − 1.462 = 0.242 α = sin −1 NA = 14° (b) NA = n12 − n22 Fibre attenuation per km Attenuation (dB km−1) (dB) P in (mW) Pout (mW) Length (km) = 0.171 7.5 1.5 20 α = sin −1 NA 4.0 0.15 0.5 14 28 5.0 0.005 2.0 30 15 0.51 0.41 0.98 0.92 2.3 2.0 0.046 0.55 2 = 1.47 − 1.46 2 = 9.9° (c) NA = n12 − n22 = 1.582 − 1.532 = 0.394 0.35 0.94 12 17. n1 sin θ c = n2 sin 90° α = sin −1 NA ⎛n ⎞ θ c = sin −1 ⎜ 2 ⎟ ⎝ n1 ⎠ ⎛ 1.33 ⎞ = sin −1 ⎜ ⎟ ⎝ 1.50 ⎠ = 62.5° = 23° 11. (a) The loss of optical power per kilometre along the fibre (b) Scattering, due to imperfections in the fibre resulting in some of the light travelling in different directions; absorption, when imperfections in the fibre absorb some of the light, turning it into thermal energy. (c) The unit of attenuation is the dB km−1. 12. (a) Rayleigh scattering is the partial reflection of light when it passes from one refractive index to another. (b) Variations in composition and structure of the fibre (c) It increases with frequency. 13. (a) Microphone: changes sound waves into an electrical signal (b) LED: changes electrical signal into a light signal (c) Modulator: imposes the electrical signal onto the light carrier (d) Optical fibre: carries the light signal from one place to another (e) Photodiode: changes the light signal into an electrical signal (f) Multiplexer: sends a number of signals along the fibre at one time 14. Pulse coded modulation involves converting the analogue electrical signal into a pulse modulated electrical signal. To do this, the voltage of the signal is measured (sampled) thousands of times per second. These sample voltages are then converted into binary form as a series of pulses. 15. Optical fibre sensors can be embedded in structures and skins to carry information from across the structure to central monitoring equipment. 16. The missing values in the table are calculated using the formulae P Attenuation = − 10 log10 out (1) Pin Jacaranda Physics 2 TSK 7.0 18. (a) NA = n12 − n22 = 1.502 − 1.462 (b) 19. (a) (b) (c) = 0.344 α = sin−1 NA = 20° 0.96 μm (read directly from the graph) 1.7 dB km−1 (read directly from the graph) There would be less optical loss due to Rayleigh scattering. 20. (a) NA = n12 − n22 = 1.52 − 1.42 (b) (c) 21. (a) (b) 22. (a) (b) (c) 63 = 0.539 α = sin−1 NA = 33° Use a fibre with a smaller diameter. As it is a stepindex fibre, the option of a graded-index fibre is ruled out. It is a short-distance fibre, so a multimode fibre would probably be used. 2 dB km−1 (read directly from the graph) Approximately 1300 nm or >1500 nm, where the fibre attenuation is least The optical fibre represented in figure 17.22 is better because it has a smaller minimum attenuation than the one described in question 21. 1.0–1.7 µm (read directly from the graph) The increase in attenuation at a wavelength of about 1.4 μm (1400 nm) in both materials is due to the presence of hydroxyl (OH−) ions. The difference in © John Wiley & Sons Australia, Ltd 2009 25. (a) Infra-red (b) 1.4458 to 1.4470 (c) c v= n attenuation between the materials at this wavelength occurs because they contain different amounts of hydroxyl ions. P 23. (a) Attenuation = − 10 log out Pin = 0.020 = − 10 log 0.50 = 14 dB (b) Attenuation per kilometre = = 2.0718 × 108 m s −1 v= 14 2.0 = 7.0 dB km −1 24. (a) Attenuation = − 10 log = − 10 log = Pout Pin 2.9979 × 108 1.4458 (d) The range of wavelengths produced by the LED will be transmitted through the optical fibre with a range of speeds, spreading the pulses and hence causing dispersion. (e) This is material dispersion. (f) Material dispersion is reduced by minimising the range of frequencies in the optical signal. 0.048 2.5 17 = 14 km 1.2 Jacaranda Physics 2 TSK c n = 2.0735 = 108 m s −1 = 17 dB (b) 2.9979 × 108 1.4470 64 © John Wiley & Sons Australia, Ltd 2009 Unit Detailed study 3.3: Sound 4 Jacaranda Physics 2, 3rd Edition TSK Part A — WORKED SOLUTIONS ■■■■■■■■■ 10. v = 340 m s−1 f = 256 Hz v = fλ (a) Chapter 18 Background sound distance travelled time taken 996 m = 3.0 s ⇒λ = 1. speed = = v f 340 m s −1 256 Hz = 1.33 m v (b) λ = f = 332 m s −1 2. A wave travels a distance of one wavelength in a time of one period. 3. No, generally it is only the medium that determines the speed of the wave. 4. The band may appear to be ‘out of step’ with the music because the observer sees the band before the corresponding sound is heard. This is because light travels at 3.0 × 108 m s−1 and sound travels at approximately 340 m s−1. 5. High frequency sounds have the same speed as low frequency sounds. 6. f = 256 Hz 1 T = f 1 = 256 Hz 1.50 × 103 m s −1 256 Hz = 5.86 m = 11. −1 f (Hz) λ (m) 335 500 0.67 300 12 1500 5000 60 24 340 1000 0.34 260 440 0.59 v (m s ) = 3.91 × 10−3 s 200 wavelengths take 200 periods to pass a point. ⇒ t = 200T = 0.78 s 7. Assume that the flash is seen almost instantly. d = vt 25 0.30 2.5 12. = 335 m s −1 × 5.0 s = 1675 m = 1.7 × 103 m λ 8. v = T ⇒ λ = vT 13. (a) = 340 m s −1 × 3.0 × 10−3 s = 1.02 m λ 9. v = T 1.32 m = 4.0 × 10−3 s (b) = 330 m s −1 Jacaranda Physics 2 TSK 65 © John Wiley & Sons Australia, Ltd 2009 (c) (b) f = 1 T 1 = 4 × 10−3 s = 250 Hz v (c) λ = f (d) 330 m s −1 250 Hz = 1.32 m 16. (a) T = time of one complete cycle = 14. (a) wavelength = length of one complete cycle on the graph = 0.80 m (b) T = ? = 8 cm × 2 m s cm −1 = 16 m s (1.6 × 10−2 s) 1 (b) f = T 1 = 1.6 × 10−2 s = 62.5 Hz (c) v = f λ = 62.5 Hz × 5.28 m v = 340 m s −1 λ = 0.80 m v f = λ 340 m s −1 = 0.80 m = 425 Hz T = = (d) 1 f 1 425 Hz = 330 m s −1 (i) If the frequency is doubled, the period is halved. If the loudness stays the same, assume that the amplitude remains constant. = 2.35 × 10−3 s (c) (ii) If the loudness is increased, the amplitude increases. Explanation When a particle is at the centre of a compression (or rarefaction), it is at its ‘mean’ or ‘rest’ position. As the wave moves forwards, the particle will move forward from its rest position. It will return to its rest position after half a period. (d) 17. power = energy 2.0 J = time 100s = 2.0 × 10−2 W 18. intensity = = 15. (a) Period is given by the ‘length’ of one complete cycle on the graph. T = 4 ms Jacaranda Physics 2 TSK power area 4.0 × 10−8 W 0.080 m 2 = 5.0 × 10−7 W m −2 66 © John Wiley & Sons Australia, Ltd 2009 19. power = intensity × area = 4.5 × 10−5 W m−2 × 2.0 m2 = 9.0 × 10−5 W 20. I0 = 3.0 × 10−3 W m−2 r = 10 m Itotal = 5 × I0 = 5 × 3.0 × 10−3 W m−2 = 1.5 × 10−2 W m−2 21. (a) 1.6 × 10−5 W m−2 (b) 4.0 × 10−6 W m−2 (c) 2.5 × 10−7 W m−2 (d) 1.0 × 10−18 W m−2 I 22. (a) L = 10 log10 I0 = 10 log10 L = 10 log10 ⇒ ΔL = 10 log10 = 10 log10 0.5 = − 3.0 dB 25. A = 5.0 × 10−5 m 2 E=? t = 3 min = 180 s I = 1.0 × 10−2 W m −2 P A ⇒ P = IA E = Pt I = = IAt = 1.0 × 10−2 W m −2 × 5.0 × 10−5 m 2 × 180 s I 1.0 × 10−17 W m −2 26. 5.0 × 10−10 W m −2 1.0 × 10 −17 Wm −2 27. = 27 d B (b) L = 10 log10 3.2 × 10−7 W m −2 1.0 × 10−12 W m −2 = 55 d B (c) L = 10 log10 4.9 × 10−3 W m −2 28. 1.0 × 10−12 W m −2 = 97 d B (d) L = 10 log10 0.5 I1 I1 1.8 × 10−9 W m −2 1.0 × 10−12 W m −2 29. = 9.0 × 10−5 J The threshold of hearing is the lowest sound intensity that can be detected by an ear at a particular frequency. Refer to the figure on page 438. (a) (i) 36 dB (ii) 2 dB (iii) −3 dB (iv) 10 dB (b) approx. 3000 Hz (c) approx. −6 dB Refer to the figure on page 438. (a) approx. 300 Hz–10 000 Hz (b) approx. 6000 Hz (c) approx. 250 Hz–400 Hz (a) = 33 d B 23. (a) I ⎛L ⎞ − 12 ⎟ ⎜ ⎠ = 10⎝ 10 (b) ⎛ 7.0 ⎞ − 12 ⎟ ⎜ ⎠ = 10⎝ 10 = 5.0 × 10−12 W m −2 (b) I (c) ⎛ 25 ⎞ − 12 ⎟ ⎜ ⎠ = 10⎝ 10 = 3.2 × 10−10 W m −2 (c) I 30. λ = 2 × nodal separation = 2 × 0.75 m = 1.50 m 31. ⎛ 54 ⎞ − 12 ⎟ ⎜ ⎠ = 10⎝ 10 = 2.5 × 10−7 W m −2 (d) I ⎛ 115 ⎞ − 12 ⎟ ⎜ ⎠ = 10⎝ 10 = 0.32 W m 24. ΔL = 10 log10 (a) −2 (b) I2 I1 (c) If I 2 = 0.5 I1 Jacaranda Physics 2 TSK (d) 67 © John Wiley & Sons Australia, Ltd 2009 32. (a) f = 330 Hz The regions of maximum sound intensity are caused by constructive interference of the waves produced by the two speakers. (b) λ = 2 × distance between adjacent nodes or antinodes. ⇒ λ = 2 × (4.0 m − 3.5 m) = 1.0 m (c) v = fλ = 330 Hz × 1.0 m 36. = 330 m s −1 String Nodes Anti-nodes 33. (a) f = 4.0 Hz λ = 1.2 m amplitude = 10 cm v=fλ = 4.0 Hz × 1.2 m = 4.8 m s−1 (b) nodal separation = λ ÷ 2 = 1.2 m ÷ 2 = 0.60 m (c) The maximum displacement of the standing wave = 2 × the amplitude of the individual waves = 2 × 10 cm = 20 cm (d) The wavelength of a standing wave is equal to the wavelength of the individual waves. λ = 1.2 m (e) The string is straight 2 times per period, that is, it will be straight 2f times per second. 2 × 4 = 8 times The string is straight 8 times per second. 34. (a) v = 250 m s−1 length of string = 1.0 m For the longest standing wave in a string fixed at both ends, λ = 2L = 2 × 1.0 m = 2.0 m (b) v = f λ v f = λ 250 m s −1 = 2.0 m λ Tone Harmonic B 2 1 2L f0 First D 5 4 f3 = 4 f0 Fourth C 3 2 L f1 = 2 f0 Second A 4 3 2L f2 = 3 f 0 3 Third 37. (a) For strings, f1 = 2 f 0 f1 2 500 Hz = 2 = 250 Hz (b) The second harmonic = 2f0 ⇒ f0 = 516 Hz ÷ 2 = 258 Hz 38. 810 Hz = 3f0 (third harmonic) ⇒ f0 = 810 Hz ÷ 3 = 270 Hz f 4 = 1400 Hz 39. (a) ⇒ f0 = f4 = 5 f0 1400 Hz 5 = 280 Hz (b) 2nd harmonic frequency = 2f0 = 2 × 280 Hz = 560 Hz (c) f2 = 3f0 = 3 × 280 Hz = 840 Hz 40. (a) (i) V = 340 m s−1 λ = 2L = 2 × 40 cm = 80 cm = 0.80 m v (ii) f 0 = λ 340 m s −1 h = 0.80 m ⇒ f0 = = 125 Hz (c) For strings, f1 = 2f0 = 2 × 125 Hz = 250 Hz (d) The second resonant frequency above the fundamental corresponds to the third harmonic, since f2 = 3f0: f2 = 3 × 125 = 375 Hz 35. (a) f0 = 240 Hz f1 = 2f0 = 2 × 240 Hz = 480 Hz Jacaranda Physics 2 TSK (b) f3 = 4f0 = 4 × 240 Hz = 960 Hz (c) The third harmonic = 3f0 = 3 × 240 Hz = 720 Hz (d) The 22nd harmonic = 22f0 = 22 × 240 Hz = 5280 Hz = 425 Hz 68 © John Wiley & Sons Australia, Ltd 2009 (b) (i) λ = 2L = 2 × 60 cm = 1.20 m v (ii) f 0 = λ 340 m s −1 = 1.20 m = 283 Hz (c) (i) λ = 2L = 2 × 1.21 m = 2.42 m v (ii) f 0 = λ 340 m s −1 = 2.42 m = 140 Hz (d) (i) λ = 2L = 2 × 1.00 m = 2.00 m v (ii) f 0 = λ 340 m s −1 = 2.00 m = 170 Hz (e) (i) λ = 4L = 4 × 0.50 m = 2.0 m v (ii) f 0 = λ 340 m s −1 = 2.0 m = 170 Hz (f) (i) λ = 4L = 4 × 0.25 m = 1.0 m v (ii) f 0 = λ 340 m s −1 = 1.0 m = 340 Hz (g) (i) λ = 4L = 4 × 0.125 m = 0.50 m v (ii) f 0 = λ 340 m s −1 = 0.50 m = 680 Hz (h) (i) λ = 4L = 4 × 0.17 m = 0.68 m Jacaranda Physics 2 TSK v λ 340 m s −1 = 0.68 m = 500 Hz (ii) f 0 = 41. Pipe Nodes B D 3 2 Antinodes 2 1 A 4 3 C 5 4 Pipe Nodes Antinodes λ L 2L 2L 3 L 2 Resonant frequency f1 = 2 f 0 Harmonic f0 Second First f2 = 3 f 0 Third f3 = 4 f 0 Fourth Resonant frequency Harmonic 42. λ 4L Fifth f2 = 5 f 0 5 C 1 1 4L f0 First 4L A 2 2 Third f1 = 3 f 0 3 4L B 4 4 Seventh f3 = 7 f 0 7 43. (a) The pipe is closed at one end as there is a pressure antinode at the closed end of a pipe when it is sustaining a standing wave. (b) 0.70 m. There is a pressure node at the open end of a pipe. Beyond the open end there will be little variation in pressure. (c) λ = 0.40 m (one complete cycle on the graph) v (d) f = λ 340 m s −1 = 0.40 m = 850 Hz (e) For this pipe, the longest wavelength is given by: λ = 4L = 4 × 0.7 m D 3 3 = 2.8 m v f0 = λ 340 m s −1 = 2.8 m = 121 Hz The frequency is seven times the fundamental, so it is the seventh harmonic. Only odd numbered harmonics occur with a pipe closed at one end, so this is the third resonant frequency above the fundamental. (f) See previous answer. f0 = 121 Hz 69 © John Wiley & Sons Australia, Ltd 2009 (g) This is the first resonant frequency above the fundamental, that is, the third harmonic. Variation from Normal air pressure (b) The wavelength is given by four times the distance between neighbouring nodes and antinodes: 4L 3 4 × 0.50 m = 3 = 0.67 m λ= 44. (a) Draw pressure variation diagram. v λ 340 m s −1 = 0.67 m (c) f = Three antinodes suggests the third harmonic. (b) λ = 2 × distance between adjacent nodes. L distance between adjacent nodes = 3 2L ⇒ λ= 3 2 × 0.85 m = 3 = 0.57 m (0.567 m) v (c) f = λ 340 m s −1 = 0.567 m = 600 Hz (d) = 5.1 × 102 Hz (d) f 0 = 5.1 × 102 Hz 3 = 1.7 × 102 Hz (e) The third resonant frequency above the fundamental is the seventh harmonic as only odd-numbered harmonics are possible. f3 = 7 f0 = 7 × 1.7 × 102 Hz = 1.2 × 103 Hz (f) (e) 46. An example of 2 modes in a optic fibre (f) T = = 1 f 1 600 Hz = 1.7 × 10−3 s (g) This is the third harmonic, therefore f = 600 Hz. 45. (a) λ = 2 × distance between adjacent antinodes = 2 × (0.506 m − 0.170 m) = 2 × 0.336 m = 0.672 m f = 500 Hz v = fλ = 500 Hz × 0.672 m = 336 m s −1 Jacaranda Physics 2 TSK 70 © John Wiley & Sons Australia, Ltd 2009 Chapter 19 Sounding good 1. A transducer is a device that transforms energy from one form to another. 2. The resistance of carbon dust inside the microphone changes with the pressure exerted on it. The pressure changes when the diaphragm responds to sound waves, producing a signal. 10. (a) A baffle is a cabinet or partition that reduces the interference of the sound waves produced from the front and back of the speaker. (b) It improves the fidelity of the sound production. (c) The sound waves from the back of the speaker cone have no access to those in front of the speaker cone. They have an infinitely long journey to take before interference can occur. 3. They need a power supply. 4. A crystal microphone uses the piezoelectric property of crystals. They can produce a small electric current that is proportional to the pressure applied to them. 5. The diaphragm is connected to a coil that is wrapped around a permanent magnet. As the diaphragm responds to pressure changes in the air, it moves the coil back and forth over the magnet, generating an electric current by electromagnetic induction. 11. The 10 000 Hz sound will be diffracted less than the 200 Hz sound because it has a smaller wavelength. For the 10 000 Hz sound: v λ= f 340 = 10000 = 0.034 m The first minimum from the 10 000 Hz sound will occur at 1.22λ sin θ = w 1.22 × 0.034 ⇒ θ = sin −1 0.05 = 56° from the normal to the speaker. For the 200 Hz sound: v λ= f 340 = 200 = 1.7 m The first minimum from the 200 Hz sound would occur at 1.22λ sin θ = w 1.22 × 1.7 ⇒ θ = sin −1 0.05 which has no solution. Therefore there is no minimum in front of the speaker. 6. (a) This is the range of human hearing, so frequencies outside of this are irrelevant for most uses. (b) The decibel (dB) is a unit for sound level intensity, which is what the microphone is designed to respond to. (c) Approximately 10 000 Hz (read directly from the graph) (d) 30 Hz and 60 Hz correspond to natural frequencies of the diaphragm of the microphone. (e) When f = 35 Hz, the frequency response is −12 dB. When f = 60 Hz, the frequency response is −4 dB. The difference is 8 dB. 7. (a) A (b) B 8. A loudspeaker converts electrical energy into sound energy and is, therefore, a transducer. 9. Dynamic loudspeakers involve a cone attached to a moveable voice coil wrapped around the central pole of a circular magnet. The signal from the amplifier passes through the coil, which results in a force between it and the magnet proportional to the current. The changing current results in a changing force, causing the speaker to move in time and magnitude with the electrical signal. Jacaranda Physics 2 TSK 71 © John Wiley & Sons Australia, Ltd 2009 v f 340 = 10000 = 3.4 cm 12. (a) λ = (b) The first minimum from the 10 000 Hz sound will occur at 1.22λ sin θ = w 1.22 × 0.034 ⇒ θ = sin −1 0.05 = 56° from the normal to the speaker. 13. (a) The lower curve belongs to a tweeter. Tweeters respond best to high frequencies as shown in the lower curve. (b) The top curve belongs to a woofer. Woofers respond best to low frequency sounds as shown in the top curve. 14. (a) Diffraction is the spreading out of waves as they pass through an opening or around an obstacle. (b) High frequency sounds are diffracted less as they leave a loudspeaker, so will be more audible from directly in front of the speaker. Low frequency sounds will diffract much more when leaving the loudspeaker. v f 340 = 200 = 1.7 m 15. (a) λ = 18. (b) There won’t be much difference as λ is greater than the gap width and significant diffraction occurs. (c) The drill will be much louder at A than at B as it produces sound with a much shorter wavelength and little diffraction occurs. 19. v f 340 = 1250 = 0.272 m 20. 16. λ = 21. 1.22λ w 1.22λ w= sin θ 1.22 × 0.272 = sin 65° = 0.366 m = 0.37 m sin θ = ⇒ 17. (a) For 1500 Hz: Jacaranda Physics 2 TSK v f 343 = 1500 = 0.229 m For 8500 Hz: v λ= f 343 = 8500 = 0.040 m (b) For 1500 Hz: 1.22 λ sin θ = w ⎛ 1.22 × 0.229 ⎞ ⇒ θ = sin −1 ⎜ ⎟ 0.30 ⎝ ⎠ = 68.6° For 8500 Hz: 1.22 λ sin θ = w ⎛ 1.22 × 0.040 ⎞ ⇒ θ = sin −1 ⎜ ⎟ 0.30 ⎝ ⎠ = 9.36° 1.22 λ (c) w = sin θ 1.22 × 0.040 = sin 68.6° = 0.052 m (a) Less diffraction (b) More diffraction (c) More diffraction (d) Less diffraction The fan shape distributes the high frequency sounds more effectively, because they spread less than the low frequencies due to diffraction. Anne is correct because the diffraction of sound from the speaker increases as the diameter of the speaker decreases. (a) Ferric oxide, chromium dioxide (b) A recording head is an electromagnet formed from a ring of ferromagnetic material wrapped in a coil of wire that the signal passes through. A small gap in the ring produces the north and south poles of the magnet, to which the magnetic tape is exposed. (c) The signal must be amplified before going to the recording head to provide sufficient current to magnetise the tape. (d) As the voltage of the signal changes, so does the strength of the magnet. (e) The erase head has a wider gap and receives a constant signal of high frequency alternating current to jumble the magnetic fields stored on the tape. λ= 72 © John Wiley & Sons Australia, Ltd 2009 22. Guitar pickups contain a permanent magnet wrapped in a coil of wire. When the ferromagnetic strings vibrate, they affect the magnetic field passing through the coil, inducing an electric current in the coil. (f) Without the erase head, a recording would be a combination of any previous recordings and the new recording. (g) As the tape moves, the changing magnetic field induces a current in the playback head. (h) The playback head produces an electrical signal that is too weak to drive a loudspeaker, so it is amplified. Jacaranda Physics 2 TSK 73 © John Wiley & Sons Australia, Ltd 2009