ANSWERS AND SOLUTIONS S N A P Student Notes and Problems PHYSICS 12 British Columbia ANSWERS AND SOLUTIONS Credits Every effort has been made to provide proper acknowledgement of the original source and to comply with copyright law. However, some attempts to establish original copyright ownership may have been unsuccessful. If copyright ownership can be identified, please notify Castle Rock Research Corp so that appropriate corrective action can be taken. Some images in this document are from www.clipart.com, © 2019 Clipart.com, a division of Vital Imagery Ltd. CASTLE ROCK RESEARCH ii Copyright Protected ANSWERS AND SOLUTIONS Lesson 2—Special Theory of Relativity MEASUREMENT OF MOTION Lesson 1—Frames of Reference PRACTICE EXERCISES ANSWERS AND SOLUTIONS PRACTICE EXERCISES ANSWERS AND SOLUTIONS 1. t t0 1 1. 2. 2.5 m/s south 1.0 m/s south 3.5 m/s south v2 c2 1.25 years 2 2.40 108 m/s 1 8 3.00 10 m/s 2.08 years 2.5 m/s north 1.0 m/s south 1.5 m/s north 3. 2. t t0 1 R (v1 ) 2 (v2 ) 2 2.5 m/s 1.0 m/s 2.7 m/s opposite tan adjacent 1.0 m/s 2.5 m/s 1.0 m/s tan 1 2.5 m/s 22 2 2 7.50 107 m/s 1 8 3.00 10 m/s 3.10 years 3. 5. v2 c2 1.50 h 44.4 m/s 1 8 3.00 10 m/s 1.50 h 4. t 2 t0 The spacecraft accelerates relative to the stars because of the force acting on it. However, there is no force acting on the floating object. By Newton’s first law, the object remains in its state of motion with the same velocity relative to the stars. To the astronaut, who is fixed to the spacecraft, it seems that the object is accelerating in the opposite direction to the spacecraft, i.e., toward him. 2 t0 1 v2 c2 v2 t0 t 1 2 c 2 2.80 108 m/s 50.0 years 1 8 3.00 10 m/s 18.0 years From the point of view (reference frame) of the shark, it appears that the fish is approaching it at (64–15) km/h. Therefore, the motion of the fish appears to be 49 km/h west. Not for Reproduction t 1 R 2.7 m/s 22 S of W 4. v2 c2 3.00 years 1 Physics 12 SNAP ANSWERS AND SOLUTIONS 5. t 8. t0 1 2 v c2 v2 t0 t 1 2 c 2 2.90 108 m/s 35.0 years 1 8 3.00 10 m/s 8.96 years 1 1 PRACTICE EXERCISES ANSWERS AND SOLUTIONS v2 1 2 c 10.0 years 18.0 years 1 Lesson 3—Length Contraction t0 t 6. v2 8 v2 3.00 10 v 8 m/s 2 2 3.00 10 8 m/s 2 v2 3.00 10 8 1. 3.00 10 m/s 2 m/s 2 10.0 years 18.0 years 10.0 18.0 1 60.0 years 1 8 m/s 8 m/s 2 2 18.4 m 2 10.0 1 18.0 2. 2 L L0 1 L0 0.691 v2 c2 L v2 c2 80.0 m 1 1 0.900 184 m v2 c2 15.0 years v2 3.00 10 v2 3.00 10 8 1 v2 c2 2 t0 t 1 L L0 1 2.55 10 35.0 m 1 3.00 10 v (0.691)(3.00 108 m/s) 2 2.49 108 m/s 7. The teacher should give 2.5 h according to the space bus clock. Not only would the space bus clock slow down as the space bus travels, but the biological clocks of all the passengers would slow down at the same rate. If you travel at a high speed through space, you are not aware that there is anything different. Writing an exam for 2.5 h on the space bus will seem the same to you as writing the same exam in your class room. v m/s 2 2 3.00 10 8 m/s 2 v2 3.00 10 8 m/s 2 8 3. L L0 1 m/s 2.85light years = 6.25light years 1 2 3.00 10 8 m/s 2 2.85 light years v2 1 6.25 light years (3.00 108 m/s) 2 v2 2 0.456 1 (3.00 108 m/s) 2 2 v 2 1 0.456 8 2 (3.00 10 m/s) 2 15.0 years 1 60.0 years 0.938 2 v (0.792)(3.00 108 m/s) 2 2.67 108 m/s v (0.938)(3.00 108 m/s) 2 2.90 108 m/s CASTLE ROCK RESEARCH v2 Note: You do not have to convert light years. 15.0 years 60.0 years 15.0 years 60.0 years v2 c2 2 Copyright Protected ANSWERS AND SOLUTIONS 4. L L0 1 v2 c2 2. m0 m 1 90.0 m 1 0.800 54.0 m NOTE: Height does not change. It is only the dimension parallel to the direction of travel that changes. 2 5. L L0 1 v2 m0 m 1 2 c 2 2.50 108 m/s 7.20 1020 kg 1 8 3.00 10 m/s 3.98 1020 kg v2 c2 3.10 10 1.20 km 1 3.00 10 8 m/s 8 m/s 2 3. 2 L L0 1 4. 2 1 (0.800) 2 kg v c m0 1 x2 2 m0 2 1.96 10 m/s 8 v2 c2 1.67 1027 kg 1 0.95 6.05 10 27 kg 1 —27 kg 8.00 10 0.654 v xc 0.654 3.00 108 m/s 1 10 m x 1 0 m PRACTICE EXERCISES ANSWERS AND SOLUTIONS m let x m Lesson 4—Mass Increase 1. v2 c2 v2 m0 1 2 c 5.50 10-10 kg 3.30 10 v2 c2 1.00 m 1 0.600 0.800 m m0 m 1 You will soon discover that the mathematics is impossible to do as one cannot find the square root of a negative value. According to Einstein’s special theory of relativity, nothing can travel faster than light (3.00 × 108 m/s). The salesman was not telling the truth. 6. v2 c2 5. 2 5.3 1027 kg let x m v c m0 1 x2 m x 1 0 m 2 2 1 1 3 0.943 v xc 0.943 3.00 108 m/s 2.83 10 m/s 8 Not for Reproduction 3 Physics 12 SNAP ANSWERS AND SOLUTIONS 6. let x m v c m0 Lesson 5—Relativistic Addition of Velocities 1 x2 m x 1 0 m 2 PRACTICE EXERCISES ANSWERS AND SOLUTIONS 2 1 1 4 0.968 v xc 0.968 3.00 108 m/s u 2. u 2.90 10 m/s 8 7. E mc 2 10.0 kg 3.00 108 m/s 2 9.00 1017 J 8. 9. E mc 2 E m 2 c 7.25 1016 J (3.00 108 m/s) 2 0.806 kg v u vu 1 2 c 0.80c 0.75c 0.80c 0.75c 1 c2 0.13c or 0.13c toward NOTE: The negative sign indicates that the object is moving toward the stationary observer. E mc 2 E m 2 c 5.3 1010 J (3.00 108 m/s) 2 5.9 10 7 kg NOTE: A grain of sand may have a mass approximately 2.00 × 10–6 kg. 5.9 × 10–7 kg is very small amount of gasoline, approximately onethird of the mass of a grain of sand. NOTE: in order to obtain 5.3 × 1010 J of energy by chemical means, you would burn approximately 1500 L of gasoline in your car. 16 000 km 1500 L 10.6 km/L CASTLE ROCK RESEARCH v u vu 1 2 c 0.80c 0.75c 0.80c 0.75c 1 c2 0.97c away 1. 4 v u vu 1 2 c 0.75c 0.60c 0.75c 0.60c 1 c2 0.93c toward 3. u 4. u v u vu 1 2 c 0.80c c 0.80c c 1 c2 c or c toward the observer Copyright Protected ANSWERS AND SOLUTIONS 5. v u vu 1 2 c 0.80c c 0.80c c 1 c2 c toward the observer 3. u b) Observer 2 will observe light from light 1 and 2 at precisely the same time because the light from both sources travels precisely the same distance. NOTE: It does not matter if the vehicle is travelling toward or away from the stationary observer. The velocity of light is the same. 6. u 4. v u vu 1 2 c 2.5 108 m/s 2.0 108 m/s 2.5 10 m/s 2.0 10 3.00 10 m/s 8 1 8 a) Observer 1 will observe the illumination of light 1 before light 2 because for observer 1, light 1 travels less distance than light 2. 8 a) Proper time is measured by the observer who is at rest with respect to the event. José is at rest with respect to the event. t b) m/s 18.0 years 2 v2 (3.00 108 ) 2 v 2.96 108 m/s 2.9 10 m/s away v u vu 1 2 c 2.5 m/s 6.0 m/s (2.5 m/s)(6.0 m/s) 1 (3.00 108 m/s) 2 8.5 m/s away u 5. Proper time is the time measured by an observer who is at rest with respect to the event. Proper time is also less than the time measured by an observer who is not at rest with respect to the event. Note that for non-relativistic speeds, the discrepancy from the proper time is extremely small. t0 t 2 v 1 c 6. Proper length is defined as the length as determined by the observer who is at rest with respect to the length determined. But, unlike proper time, proper length is greater than the length determined by the observer that is moving with respect to the length determined. NOTE: This is the same answer that you would get if you just added the two velocity vectors together as you added vectors earlier in this course. Relativistic effects are only observable at very high velocities. Practice Test ANSWERS AND SOLUTIONS 1. 2. v2 c2 where L0 is the proper length. Michelson and Morley designed an experiment to measure the speed of light relative to ether. Their result is known as the null result, because they did not detect any evidence of the ether. This made many scientists question its existence. L L0 1 7. An inertial frame of reference is any frame of reference that is at rest or is moving at a constant velocity. Not for Reproduction 2 v 1 c 3.00 years 1 8 7. t0 5 Einstein’s special theory of relativity has two postulates: 1. the laws of physics in all inertial frames of reference are the same 2. the speed of light in space is the same for all observers, regardless of the motion of the source or observer. Physics 12 SNAP ANSWERS AND SOLUTIONS 8. If Einstein’s postulates are correct, then space and time are not absolute. It also follows that mass and energy are equivalent. 9. The length contraction will only be in the direction of the motion. Therefore, side Y will remain at 50 m. There is length contraction of side Z so it will be shorter. Again Fg2 = 1.19Fg1 = (1.19)(7.74 N) = 9.21 N = 9.2 N 2. 10. An observer will have the same experiences in any inertial frame of reference. Pavel will not observe any change to his watch other than the time changing as he would normally notice. FORCES CAUSE MOTION Fx 0 FT – Fgx 0 Fgx FT 5000 N Lesson 1—The First Condition of Equilibrium Fgx Fg sin Fgx 5000 N Fg sin sin 25.0 1.18 10 4 N PRACTICE EXERCISES ANSWERS AND SOLUTIONS 1. 3. Fg1x Fg1cos1 Fg1cos40 0.766Fg1 Fg1 y Fg1sin1 Fg1sin40 0.643Fg1 Fg 2 x Fg2 cos 2 Fg2 cos50 0.643Fg2 Fg2 y Fg2 sin 2 Fg2 sin50 0.766Fg2 FTx FT cos FT cos63.0 0.454FT Fy 0 FTy Fg 0 FTy Fg 20.0 N Fx 0 Fg 2 x Fglx 0 0.643Fg2 0.766Fg1 0 0.766 Fg1 Fg 0.643 Fg 2 1.19Fg1 0 Fg1 y Fg2 y – Fg3 0 0.643Fg1 0.766Fg2 –12 N 0 1.55Fg1 12 N Fg1 7.74 N y Fapp 6 FTy 20.0 N 0.891 0.891 22.4 N F x0 Fapp FTx 0 0.454FT 0 Fapp 0.454 22.4 N 10.2 N FT F CASTLE ROCK RESEARCH FTy FT sin FT sin63.0 0.891FT Copyright Protected ANSWERS AND SOLUTIONS Fx 0 FTx F 0 FTx 39.7 N FTx 0.866FT F 39.7 N FT Tx 0.866 0.866 45.9 N Fy 0 FTy Fg 0 0.500 FT Fg Fg 0.500 45.9 N 22.9 N 4. FT1x FT1cos1 FT1cos22.0 0.927FT1 FT2x FT2 cos 2 FT2 cos80.0 0.174FT2 FT1y FT1sin1 FT1sin22.0 0.375FT1 FT2y FT2 sin 2 FT2 sin80.0 0.985FT2 Fg mg Fg 22.9 N m g 9.81 m/s 2 2.3 kg Fx 0 FT2x FT1x 0 0.174FT2 0.927FT1 0 0.927 FT1 FT2 0.174 5.33FT1 Fy FT2y FT1y – Fg 0.985FT2 0.375FT1 0.985 5.33FT1 – 0.375FT1 4.88FT1 FT1 6. 0 0 75.0 N 75.0 N 75N 15.4 N FTx FT cos FT cos78 0.208FT FT2 = 5.33 (15.4 N) = 82.1 N 5. FTy FT sin FT sin78 0.978 FT Fx 0 FTx F 0 F FTx 0.208FT F FTy FT sin FTx FT cos FT sin30 FT cos30 0.500 FT 0.866FT F Ffr μmgcos 0.27 15 kg 9.81 m/s cos 0 39.7 N Not for Reproduction y FTy Fg FTy 0.978FT FT 7 0 0 Fg 735 N 735 N 0.978 751 N 7.5 102 N Physics 12 SNAP ANSWERS AND SOLUTIONS FT1y FT1sin1 FT1x FT1cos1 7. FT1cos26.5 FT1sin26.5 0.895FT1 0..446FT1 FT2x FT2 cos 2 FT2 cos63.5 0..446FT2 FT1x FT1cos1 96 N cos 60 96 N sin 60 48 N 83 N FT2x FT2 cos 2 FT2 sin63.5 0.895FT2 Fx 0 FT1x FT2x 0 0.895FT1 0.446 FT2 0 0.446 FT2 FT1 0.895 0.498FT2 0.498FT2 FT1 FT2y FT2 sin 2 FT2 cos30 FT2 sin30 0.866FT2 0.500 FT2 Fx 0 FT2x FT1x 0 48 N 0.866 T2 48 N FT2 0.866 55.43 N FT2y FT2sin 2 FT1y FT1sin1 55 N 0.498 FT2 110.4 N or 1.1102 N FT2 55 N F y Fy 0 FT1y FT2y – Fg 0 83 N 0.500 FT2 Fg 83 N 0.500 55 N Fg 0 FT1y FT2y – Fg 0 .446FT1 0.895 FT 2 Fg 0.446 55 N 0.895 110 N Fg Fg =123 N or 1.2 102 N 8. 9. For the whole system, Fy 0 FT1 Fg1 0 FT1 Fg1 25 N 12 N 37 N For the lower block, Fy 0 FT2 Fg 0 FT2 Fg 12 N Find 1 op 3.0 m hyp 6.7 m 1 26.5 sin1 Find 2 op 6.0 m hyp 6.7 m 2 63.5 sin 2 CASTLE ROCK RESEARCH 8 Copyright Protected ANSWERS AND SOLUTIONS 2. 10. F 0 F y x | FTx F 0 a) 0 1 – 2 0 r1 F1 sin1 – r2 F2 sin 2 0 0 FTy Fg 0 FTx F FTx 410 N L FT sin 40 – 12 L Fg sin 90 0 0.643LFT – 200 N L 0 FTy Fg 675 N 0.643FT 200 N 200 N 0.643 311 N 3.1102 N FT b) op adj 410 N 675 N 31.3 tan Lesson 2—The Second Condition of Equilibrium FTx FT cos 311 N cos 40 238 N FTy FT sin 311 N sin 40 200 N PRACTICE EXERCISES ANSWERS AND SOLUTIONS 1. rF sin F r sin F x 45 N m (0.35 m)(sin 90°) 129 N or 1.3 102 N 0 FH – FTx 0 FH FTx FH 238 N F y 0 FTy Fg FV 0 200 N – 400 N FV 0 Fv 238 N Fv 200 N Not for Reproduction 9 Physics 12 SNAP ANSWERS AND SOLUTIONS Fx 0 FT2 – Fgb – Fgd FT1 0 3744 N – 400 N – 735 N FT1 0 3. FT1 –3744 N 400 N 736 N FT1 –2605 N or – 2.61103 N 5. 0 a) 1 – 2 3 0 r1 F1sin1 – r2 F2sin2 r3 F3sin3 0 5.00 m FTR sin90 – 2.50 m 250 N sin90 – 1.50 m 650 N sin90 0 5.00 m FTR – 625 N – 975 N 0 0 1 – 2 0 r1 F1 sin1 – r2 F2 sin 2 0 L FT sin 30 – 12 L Fgb sin 90 0 0.500 FT –100 N 0 0.500 FT 100 N 100 N FT 0.500 200 N =2.0 102 N (625 N m 975 N m) FTR 5.00 m FTR 320 N b) Fx 0 FTR FTL – Fgs – Fgb 0 320 N FTL – 650 N – 250 N 0 FTL 900N – 320 N FTL 580 N 6. 4. 0 1 – 2 0 r1 F1 sin1 – r2 F2 sin 2 0 point of rotation is arbitrary 0 1 – 2 3 0 r1 F1sin1 – r2 F2 sin 2 r3 F3sin3 0 L FT sin 50 – 12 L Fgb sin 40 0 0.766FT – 64.3 N 0 r1 FT2 – r2 Fgb – r3 Fgd 0 0.766FT 64.3 N 64.3 N 0.766 83.9 N L F L (400 N) 1 4 T2 FT 1 2 ( L)(75.0 kg 9.81 m/s 2 ) 0 1 FT2 – 200 N – 736 N 0 4 FT2 4 936 N 3744 N or 3.74 103 N CASTLE ROCK RESEARCH =84 N 10 Copyright Protected ANSWERS AND SOLUTIONS Lesson 3—Uniform Circular Motion 6. mv 2 r (925 kg)(25 m/s) 2 72 m 8.03 103 N Fc PRACTICE EXERCISES ANSWERS AND SOLUTIONS 1. 2. Here the frictional force is equal to the centripetal force. mv 2 r (925 kg)(22 m/s) 2 75 m 6.0 103 N Fc Again FN Fg mg 925 kg 9.81 m/s 2 9.07 103 N C 2 r (2 ) 3282 m 2.06 104 m v Ffr FN F fr FN 8.03 103 N 9.07 103 N 0.89 d t 2.06 104 m (2.0 min)(60 s/m) 172 m/s 2 v r (172 m/s) 2 3 282 m 9.0 m/s 2 ac 3. 4. 5. mv 2 Fc r (822 kg)(28.0 m/s) 2 105 m 6.14 103 N 2 r T 2 (2.8 m) (0.98 s) 18 m/s mv 2 Fc r 2.00 102 m 4.60 1014 N v2 (9.11 1031 kg) mv 2 r (2.7 03 kg)(4.7 103 m/s) 2 1.8 107 m 3 3.3 10 N 7. Fc 8. a) v b) Fc c) Find time to reach the ground using the vertical component v0 v a d t 1.2 m × v (4.60 1014 N)(2.00 102 m) 9.11 1031 kg 3.18 107 m/s v Not for Reproduction 2 r T 2 (0.90 m) 0.30 s 19 m/s mv 2 r (3.7 kg)(19 m/s) 2 0.90 m 1.5 103 N 0 11 × 9.81 m/s 2 Physics 12 SNAP ANSWERS AND SOLUTIONS 1 d y v0 t at 2 2 1 1.2 m (9.81 m/s 2 )(t 2 ) 2 2(1.2 m) t 9.81 m/s 2 0.495 s mv 2 r Fc r v m (135 N)(1.10 m) 2.00 kg 8.62 m/s 10. Fc Horizontally d vx x t d x vx t 19 m/s 0.495 s 9.3 m 9. 11. a) Ffr FN 0.95 9143 N 8.69 103 N Speed 2r Diameter 30.0 102 m r 2 15.0 102 m T FN mg 932 kg 9.81 m/s 2 9143 N mv 2 r (932 kg)(v 2 ) 3 8.69 10 N 82 m (8.69 103 N)(82 m) v 932 kg 28 m/s Fc 1 f 1m f 33 1 rev/minute 3 60 s 1 33 3 Hz 60 60 T 1 33 Hz 3 1.80 s 2 r v T 2 (15.0 102 m) 1.80 s 0.524 m/s b) Ffr FN 0.40 9143 N 3.66 103 N mv 2 r (932 kg)(v 2 ) 3 3.66 10 N 82 m (3.66 103 N)(82 m) v 932 kg 18 m/s Fc Acceleration v2 ac r (0.524 m/s) 2 15.0 10 2 m 1.83 m/s 2 CASTLE ROCK RESEARCH 12 Copyright Protected ANSWERS AND SOLUTIONS FT 0 Fc Fg mv 2 mg r v rg 4. Lesson 4—Vertical Circular Motion PRACTICE EXERCISES ANSWERS AND SOLUTIONS 1. (2.90 m)(9.81 m/s 2 ) 5.33 m/s FT 0 Fc Fg mv 2 mg r v rg Fg mg 5. 1.50 kg 9.81 m/s 2 14.7 N (5.00 m)(9.81 m/s 2 ) 7.00 m/s 2. Fc FT – Fg 186 N – 14.7 N 171 N mv 2 Fc r Fc r v m (171 N)(1.90 m) 1.50 kg 14.7 m/s FT 0 Fc Fg mv 2 mg r v rg (0.95 m)(9.81 m/s 2 ) 3.1 m/s 3. Fc FN – Fg 1.96 103 N – 655 N 1.305 103 N 6. Fg mg Fg m g 655 N 9.81 m/s 2 66.8 kg 4 2 rm T2 (4 2 )(1.0 m)(2.2 kg) (0.97 s) 2 92.3 N Fc Fg mg 2.2 kg 9.81 m/s 2 21.6 N Fc FT Fg FT Fc Fg 92.3 N – 21.6 N 71 N mv 2 r Fc r v m (1.305 103 N)(18.0 m) 66.8 kg 18.7 m/s Fc Not for Reproduction a) 13 Physics 12 SNAP ANSWERS AND SOLUTIONS Fc FT Fg FT Fc Fg 92.3 N 21.6 N 114 N b) b) Fg mg 2.5 kg 9.81 m/s 2 24.5 N mv 2 r (2.5 kg)(12 m/s) 2 0.75 m 4.8 10 2 N Fc i) The tension of the string at its low point is FT Fc Fg 480 N 24.5 N 5.0 102 N FT 0 Fc Fg mv 2 mg r v rg 7. ii) The tension of the string at its high point is FT Fc – Fg 480 N – 24.5 N 4.6 102 N (48 m)(9.81 m/s 2 ) 22.7 m/s or 22 m/s Practice Test FT 0 Fc Fg mv 2 mg r v rg 8. ANSWERS AND SOLUTIONS 1. (43 m)(9.81 m/s 2 ) 20.5 m/s or 21 m/s Force of friction depends on the surfaces (coefficient of friction, ) and the normal force (FN = mg cos). Ffr = mg cos 9. a) 2. 1 2 m1 gL1 m2 gL2 32 kg 1.5 m 42 kg d . d 1.14 m 1.1 m High Fc FT Fg FT Fc – Fg Low Fc FT Fg FT Fc Fg 3. Rotational motion is related to torque. Therefore, if a wooden beam has no rotational motion, then the torques acting on the beam must be balanced. It is clear from the above formulas that the tension in the string greater at its low point than at its high point CASTLE ROCK RESEARCH 14 Copyright Protected ANSWERS AND SOLUTIONS 4. 7. Fnet 0 Ffr Fg x Fg x Fg sin 2.55 N 9.81 m/s 2 sin 30 12.5 N F 0 FT1 – Fg 0 FT1 Fg mg 18 kg 9.81 m/s 2 5. 177 N or 1.8 102 N 8. 0 1 2 0 m1 gd – m2 g 1 m – d 0 3.00 kg d – 5.00 kg 1 m – d 0 3.00 kg d – 5.00 kg m 5.00 kg d 0 8.00 kg d 5.00 kg m d 0.625 m or 62.5 cm In circular motion, the velocity of a moving object is tangent to the path. 9. 6. In circular motion, the net force ( Fc ) is toward the centre of the path, and the acceleration of an object is always in the direction of the net force. 10. a) Fnet Fnet Ffr Fg x 0 Fg x – Ffr Fg x Fg sin The tension in the cord will be directed toward the centre of the circle wherever the mass is in its path in the vertical circle. The force due to gravity will be directed straight down to Earth’s surface wherever the mass is in its path in the vertical circle. Therefore, when the mass is precisely at the top of the vertical circle, both the force due to gravity and the tension are directed down through the centre of the circle. 3.0 N 9.81 m/s 2 sin 37 17.7 N Ffr mgsin Ffr mg cos (17.7 N) (3.0 kg)(9.81 m/s 2 )(cos 37) 0.75 Not for Reproduction 15 Physics 12 SNAP ANSWERS AND SOLUTIONS 10. b) At the top of the vertical circle, both the tension and the force of gravity act toward the centre of the circle. Therefore, the magnitude of the centripetal force at that moment is equal the sum of both forces. Fc Fg FT 11. The key to drawing this graph is recognizing that the object is twirled at a constant speed. If the speed remains constant, the tension in the cord will remain constant. This is represented in a graph that looks like this: Fg 3. Fg Fc Ffr mv 2 mgcos r v r g cos Fg 5. Fg . FORCES CAUSE MOTION m2 Lesson 1—Newton’s Law of Universal Gravitation Gm1m2 r2 Fg r 2 Gm1 (3.2 106 N)(2.1 101 m) 2 2 11 N m 1 6.67 10 (5.5 10 kg) kg 2 3.8 101 N PRACTICE EXERCISES ANSWERS AND SOLUTIONS Gm1m2 r2 2 11 N m (70.0 kg)(52.0 kg) 6.67×10 2 kg (1.50 m) 2 –7 1.08 10 N Fg CASTLE ROCK RESEARCH Gm1m2 r2 2 11 N m (5.98 1024 kg) 6.67 10 2 kg (6.50 102 kg) (4.15 106 m 6.37 106 m) 2 2.34 103 N 4. (40.0 m)(0.50)(9.81 m/s 2 ) 14 m/s 1. Gm1m2 r2 As both the objects have equal mass, Fg r 2 m2 G (2.30×10 8 N)(10.0 m) 2 N m2 6.67×10 11 kg 2 4 2 3.45 10 kg m 1.86 10 2 kg 12. Centripetal force is the net force that causes an object to travel in a circular path. 13. Gm1m2 r2 2 11 N m (5.98×1024 kg) 6.67×10 2 kg (7.35×1022 kg) (3.84×108 m) 2 1.99 1020 N 2. 6. Gm1m2 r2 Gm1m2 r Fg Fg 2 11 N m (2.0 10 2 kg) 6.67 10 2 kg (2.0 102 kg) 3.7 106 N 8.5 101 m 16 Copyright Protected ANSWERS AND SOLUTIONS 7. 8. Gm1m2 r2 2 11 N m (5.98 10 24 kg) 6.67 10 2 kg (70.0 kg) (6.38 106 m) 2 2 6.86 10 N Fg changes by a factor of Fg Fg 6.86 102 N 0.444 3.05 102 N 11. Gm1m2 r2 Fg m Fg Fg1 Fg2 , and since these forces are in opposite directions, the net force will be zero. Mass changes by a factor of: To 35.0 kg From 70.0 kg 0.500 12. Gm1m2 r2 2 11 N m 6.67 10 (10.0 kg)(10.0 kg) kg 2 (5.00 10 1 m) 2 2.67 108 N Fg1 Mass changes by a factor of 0.500 Fg will change by a factor of 0.500 Fg 6.86 102 N 0.500 3.43 102 N 9. Gm1m2 r2 1 Fg 2 r Fg Gm1m2 r2 2 11 N m 6.67 10 (10.0 kg)(15.0 kg) kg 2 (5.00 101 m) 2 –8 4.00 10 N Fg2 Distance changes by a factor of: to 1.276 107 m from 6.38 106 m 2.00 Fnet Fg2 Fg1 4.00 10 –8 N – 2.67 10 –8 N 1.33 10 –8 N Distance changes by a factor of 2.00 Fg changes by a factor of 0.250 = 1 = 0.444 (1.50) 2 1 (2.00) 2 Fg 6.86 102 N 0.250 13. Fg 1.72 102 N Fg Gm1m2 r2 m1m2 r 2 (2)(2) (3) 2 0.444 Gm1m2 r2 1 Fg 2 r 10. Fg Fg(new) 3.24 10–7 N 0.444 1.44 10–7 N Distance changes by a factor of: to 9.57 106 m from 6.38 106 m 1.50 Distance changes by a factor of 1.50 Not for Reproduction 17 Physics 12 SNAP ANSWERS AND SOLUTIONS Lesson 2—Field Explanation GM r2 GmY 2 r GmY r g g ANSWERS AND SOLUTIONS PRACTICE EXERCISES 1. 2. GM r2 Gm 2M r 2 11 N m 23 6.67 10 (6.37 10 kg) kg 2 (3.43 106 m) 2 3.61 N/kg 2 11 N m (4.83 1024 kg) 6.67 10 2 kg 1.67 N/kg 1.39 107 m g 5. GM r2 Gm 2E r GmE r g g g new 9.81 N/kg 0.25 2.45 N/kg 2 11 N m 24 6.67 10 (5.98 10 kg) kg 2 7.33 N/kg 6 7.377 10 m 6. From surface: d 7.377 106 m – 6.38 106 m 1.0 106 m 3. g GM r2 1 g 2 r 1 (2) 2 0.25 g GM r2 Gm 2B r 2 11 N m 22 6.67 10 (4.00 10 kg) kg 2 (6.0 105 m) 2 7.4 N/kg g Fg m 63.5 N 22.5 kg 2.82 N/kg 7. gB 4. g Fg gA gB m 50.0 N 30.0 kg 1.67 N/kg rA 2 rA rA CASTLE ROCK RESEARCH GM rA 2 GM 2 rB rB 2 2 rA g rB 2 B gA 1 rB ( 1.20 0.913 rB gA 18 g A 1.20 g B ) Copyright Protected ANSWERS AND SOLUTIONS 8. GM r2 Gm 2m r 2 11 N m (7.35 10 22 kg) 6.67 10 2 kg (1.74 106 m) 2 1.62 N/kg g g 4. Fg 1 1 Ep Gmmoon mmeteor ri rf 2 Nm 6.67 10 11 (7.35 10 22 kg)(1 750 kg) 2 kg 1 1 6 6 15 000 m 1.74 10 m 1.74 10 m 7 –4.2110 J Ek – Ep 4.21 107 J Ek Ek f – Ek i 1 1 mv 2 mvo 2 2 2 1 2 m v vo 2 2 m Fg mg 20.0 kg 1.62 N/kg 32.4 N Lesson 3—Gravitational Potential Energy 4.21 10 J 7 2 v 2 24 (5.98 10 9.90 106 m Nm kg 2 2 4.21 10 7 1750 kg 3 2 4.21 10 J 7 1750 kg J 1.00 10 m/s 2 3 2 2 2 kg)(5.00 10 kg) 3 5. W Ep 1 1 Ep GmE ms ri r f 6. N m2 (5.98 1024 kg)(5.00 103 kg) 6.67 1011 2 kg 1 1 6 6 6.38 10 m 9.90 10 m 1.11 1011 J 3. 3 2 v 1.02 103 m/s –2.011011 J 2. 2 (1750 kg) v 1.00 10 m/s v 1.00 10 m/s PRACTICE EXERCISES ANSWERS AND SOLUTIONS GmE ms 1. Ep r 11 6.67 10 1 GmE mo r 2 11 N m 24 6.67 10 (5.98 10 kg)(10.0 kg) kg 2 6.38 106 m –6.25 108 J Ep 1 1 Ep GmE ms ri r f 2 11 N m 24 3 6.67 10 (5.98 10 kg)(2.50 10 kg) kg 2 1 1 6 6 6.38 10 m 6.90 10 m 10 1.18 10 J W Ep The work done to lift the satellite into orbit was 1.11 1011 J , so the change in gravitational potential energy was 1.11 1011 J . Not for Reproduction 19 Physics 12 SNAP ANSWERS AND SOLUTIONS Lesson 4—Satellites: Natural & Artificial v 3. v 4. v PRACTICE EXERCISES ANSWERS AND SOLUTIONS 1. 2. 2GM r 2(6.67 1011 N m 2 /kg 2 )(7.35 1022 kg) (1.74 106 m) 3 2.37 10 m/s v 2GM r 2(6.67 10-11 N m 2 /kg 2 )(1.9 1027 kg) (7.18 107 m) 5.94 104 m/s GmE r 2 11 N m 24 6.67 10 (5.98 10 kg) kg 2 (4.4 105 m 6.38 106 m) 7.65 103 m/s 2. v GmJ r 2 11 N m 6.67 10 (1.90 10 27 kg) 2 kg (5.60 106 m 7.18 107 m) 4.06 104 m/s Gms r N m2 )(1.98 1030 kg) kg 2 1.50 1011 m 4 2.97 10 m/s (6.67 1011 3. 2GM r v2 r M 2G (9.0 103 m/s) 2 (7.2 106 m) 2(6.67 10 11 N m 2 /kg 2 ) 4.37 1024 kg v 3 5. T 2 r 2 Gm 3 (2 )(2.3 1011 m) 2 4. 2 11 N m (1.98 1030 kg) 6.67 10 2 kg 6.0 107 s or 1.9 Earth years 2GM r v2 r M 2G (2.92 103 m/s) 2 (2.57 106 m) 2(6.67 1011 N m/kg 2 ) 1.64 10 23 kg v Lesson 6—Electric Forces PRACTICE EXERCISES ANSWERS AND SOLUTIONS Lesson 5—Satellites in Orbit 1. PRACTICE EXERCISES ANSWERS AND SOLUTIONS 1. GmE r 2 11 N m (5.98 10 24 kg) 6.67 10 2 kg 3.84 108 m 1.02 103 m/s v CASTLE ROCK RESEARCH 20 Gm1m2 r2 2 11 N m (70.0 kg)(52.0 kg) 6.67 10 2 kg (1.50 m) 2 1.08 107 N Fg Copyright Protected ANSWERS AND SOLUTIONS 2. 3. 4. 5. 6. Gm1m2 r2 2 11 N m (5.98 1024 kg) 6.67 10 2 kg (7.35 10 22 kg) (3.84 108 m) 2 1.99 1020 N Fg 7. kq1q2 r2 q1q2 Fe 2 r (3)(3) (2) 2 2.25 Fe 4.5 10–3 N 2 25 Fe 1.0 10–2 N kq1q2 r2 2 9 Nm 6 9.00 10 (4.00 10 C) C2 (3.00 106 C) (2.00 102 m) 2 2 2.70 10 N Fe 8. kq1q2 r12 2 9 Nm 6 6 9.00 10 (2.0 10 C)(2.0 10 C) 2 C (0.40 m) 2 2.25 10 –1 N kq q F2 22 3 r2 2 9 Nm 6 6 9.00 10 (2.0 10 C)(3.0 10 C) C2 (0.40 m) 2 3.37 101 N Fnet F1 F2 2.25 101 N 3.37 101 N 1.110 –1 N 1.110 –1 N left F1 kq1q2 r2 F r2 q2 e k (3.40 102 N)(1.00 10 1 m) 2 N m2 9.00 109 C2 14 2 3.78 10 C q 1.94 107 C Fe kq1q2 r2 kq q r2 1 2 Fe 2 9 Nm 6 9.00 10 (2.0 10 C) 2 C (4.0 106 C) (5.6 101 N) 2 –1 1.28 10 m 2 r 3.6 10 –1 m Fe Therefore the magnitude of the net electric force is 1.1 10 –1 N 9. kq1q2 r2 1 Fe 2 r 1 2 (3) 0.111 Fe Fe 6.20 10–2 N 0.111 6.89 10 Not for Reproduction –3 kq1q2 r2 1 Fe 2 r Distance changes by a factor of: to 4.04 101 m from 3.11101 m 1.30 Distance increases by a factor of 1.30. 1 Fe (1.30) 2 0.592 Fe 5.2 10–4 N 0.592 Fe 3.110–4 N N 21 Physics 12 SNAP ANSWERS AND SOLUTIONS 10. kq1q2 r2 12. Fe 2 9 Nm 6 6 9.00 10 (2.00 10 C)(3.00 10 C) 2 C (3.50 101 m) 2 –1 4.40 10 N F1 kq1 q2 r12 Fnet ma F a net m 4.40 101 N 2.00 105 kg =2.20 104 m/s 2 2 9 Nm 6 6 9.00 10 (3.0 10 C)(4.0 10 C) 2 C (0.60 m) 2 3.00 10 –1 N kq q F2 22 3 r2 2 9 Nm 6 2 9.00 10 (4.0 10 C) 2 C (0.60 m) 2 4.00 10 –1 N q2 r12 q2 2 r2 r2 12 r2 13. FA on B FC on B FC on B FA on B 1.5 cm 2 4.5 cm 2 0.11 Lesson 7—Electric Fields Fnet ( F1 )2 ( F2 )2 PRACTICE EXERCISES ANSWERS AND SOLUTIONS (3.00 101 N) 2 (4.00 101 N) 2 5.0 10–1 N 11. kq1q2 r2 2 9 Nm 6 2 9.00 10 (1.50 10 C) 2 C (2.00 101 m) 2 5.06 10 –1 N F2 CASTLE ROCK RESEARCH 22 kq1 r2 2 9 Nm 6 9.00 10 (8.00 10 C) C2 (7.50 10 1 m) 2 1.28 105 N/C 1. E 2. g GM r2 2 11 N m (6.37 1023 kg) 6.67 10 2 kg (3.43 106 m) 2 3.61 N/kg Copyright Protected ANSWERS AND SOLUTIONS 3. 8. kq1 r2 kq r |E| E E1 E2 2 9 Nm 6 9.00 10 (4.60 10 C) 2 C 2.75 105 N/C 3.88 10 –1 m 2 9 Nm 6 9.00 10 (4.50 10 C) C2 (2.50 10 1 m) 2 5 6.48 10 N/C Fg 4. g 5. Fe q Enet E1 E2 6.48 105 N/C 6.48 105 N/C 1.296 106 N/C towards right Enet 1.296 106 N/C m 63.5 N 22.5 kg 2.82 N/kg 9. to 1.60 1019 C from 3.20 1019 C 0.500 Fe 0.500 0.250 N 0.125 N 6. 7. kq r2 kq1 r12 2 9 Nm 6 9.00 10 (3.0 10 C) 2 C (4.0 101 m) 2 1.688 105 N/C E1 Fe q 7.11103 N 5.20 106 C 1.37 103 N/C E kq2 r2 2 2 9 Nm 6 9.00 10 (6.0 10 C) 2 C (4.0 101 m) 2 3.375 105 N/C Enet E1 E2 1.688 105 N/C 3.375 105 N/C 1.7 105 N/C towards left Enet 1.7 105 N/C E2 Fe q 3.20 10 –19 C 7.60 104 N/C E 2.43 10 –14 N Fe ma F a e m 2.43 1014 N 6.65 1027 kg 3.65 1012 m/s 2 10. Since both the point charges have equal value and their distance are same on both side of the test charge E1 E2 Enet 0 Not for Reproduction 23 Physics 12 SNAP ANSWERS AND SOLUTIONS 11. Fe ma 9.1110–31 kg 7.50 1012 m/s 2 Lesson 8—Electric Potential due to a Point 6.83 10–18 N PRACTICE EXERCISES ANSWERS AND SOLUTIONS F E e q 6.83 1018 N 1.60 1019 C 4.27 101 N/C V 2. V 3. V1 kq r2 1 E 2 r 12. E Distance changes by a factor of: To 4.50 101 m From 3.00 101 m 1.50 kq1 r 2 9 Nm 6 9.00 10 (2.5 10 C) 2 C 6.0 102 m 5 3.8 10 V 1. Distance changes by a factor of 1.50. kq1 r 2 9 Nm 6 9.00 10 ( 2.5 10 C) C2 25 102 m 4 –9.0 10 V 2 1 E will change by a factor of 0.444 1.50 E 3.60 105 N/C 0.444 1.60 105 N/C 13. kq r2 1 E 2 r 1 2 r r E E kq1 r1 2 9 Nm 6 9.00 10 (5.0 10 C) C2 45 102 m 4 9.99 10 V kq1 r2 2 9 Nm 6 9.00 10 (5.0 10 C) C2 15 102 m 3.00 105 V V2 1 E Field is changed by a factor of: To 4.20 104 N/C From 2.10 106 N/C 2.00 V V2 – V1 3.00 105 V – 9.99 104 V 2.00 105 V W E qV 0.030 10 –6 C 2.00 105 V E changes by a factor of 2.00 r changes by a factor of 1 0.707 2.00 6.0 10 –3 J r 7.50 101 m 0.707 5.30 01 m CASTLE ROCK RESEARCH 24 Copyright Protected ANSWERS AND SOLUTIONS 4. VT V1 V2 V3 1.20 105 V –1.54 105 V –1.63 105 V kq1 r1 2 9 Nm 18 9.00 10 (6.4 10 C) 2 C 2.0 1011 m 2.88 103 V V1 –2.0 105 V 6. kq1 r2 2 9 Nm 18 9.00 10 (6.4 10 C) 2 C = 0.50 m 1.15 10–7 V V2 r2 r12 r32 (8.0 cm) 2 (6.0 cm) 2 1.0 101 cm W E qV 1.60 10 –19 C 2.88 103 J 4.60 10 –16 J kq2 r2 2 9 Nm 6 9.00 10 ( 3.0 10 C) 2 C 1.0 101 m –2.70 105 V V2 E E k 1 Ek mv 2 2 2 Ek v m 2(4.60 10 16 J) 1.67 10 27 kg =7.4 105 m/s 5. kq3 r3 2 9 Nm 6 9.00 10 (4.0 10 C) C2 6.0 102 m 5 5.99 10 V V3 kq1 r1 2 9 Nm 6 9.00 10 (1.0 10 C) 2 C 7.5 102 m 1.20 105 V V1 VT V1 V2 V3 1.12 105 V –2.70 105 V 5.99 105 V 4.4 105 V kq2 r2 2 9 Nm 6 9.00 10 ( 3.0 10 C) C2 17.5 102 m 5 –1.54 10 V 7. V2 Ek Ep Ek Ep 0 Ep Ek Ep kq1q2 1 kq q 1 m1 (v1 )2 m2 (v2 )2 1 2 r 2 2 r Both masses m1 and m2 are alpha particles, so they are equal in size. Therefore, the speeds v1 and v2 are also equal. Therefore, m1 (v1 ) 2 and m2 (v2 )2 are like terms. kq V3 3 r3 2 9 Nm 6 9.00 10 ( 5.0 10 C) 2 C 27.5 102 m –1.63 105 V Not for Reproduction kq1 r1 2 9 Nm 6 9.00 10 (1.0 10 C) 2 C 8.0 102 m 1.2 105 V V1 m1 m2 m v1 25 2 v2 v 2 2 Physics 12 SNAP ANSWERS AND SOLUTIONS kq1q2 1 kq q 1 m(v) 2 m(v) 2 1 2 r 2 2 r kq1q2 kq1q2 2 m(v) r r kq1q2 kq1q2 2 m(v) r r kq1q2 kq1q2 r r v m 2. 3. 2 9 Nm 19 2 9.00 10 (3.20 10 C) 2 C kq1q2 r 2.5 1012 m 3.69 1016 N m 2 9 Nm 19 2 9.00 10 (3.20 10 C) 2 C kq1q2 r 0.75 m 1.23 1027 N m V d V Ed E V 2.0 103 V/m 7.3 10 –2 m 1.5 102 V 4. Inserting the values into equation 1: 3.69 1016 N m 1.23 10 27 N m 6.65 1027 kg 5 2.4 10 m/s v 8. V d V d |E| 2.0 102 V 5.0 103 V/m 4.0 10 –2 m E 5. W E qV E V q 4.40 105 J 3.00 106 C 14.7 V Ek q 1.50 1015 J 3.20 1014 C 4.69 103 V V E k q Ek qV 1.60 10 –19 C 7.20 10 2 V V 1.15 10 –16 J 6. V Ep(e) q Ee(p) qV 3.20 10 –19 C 7.50 103 V Lesson 9—Electric Potential in a Uniform Electric Field 2.40 10 –15 J kinetic energy gained by the alpha particle Ek Ep(e) PRACTICE EXERCISES ANSWERS AND SOLUTIONS 1 2 mv 2 2 Ek v m 2(2.40 1015 J) 6.65 1027 kg 8.50 105 m/s Ek 1. E V d 12.0 V 9.00 102 m 1.33 102 V/m 7. CASTLE ROCK RESEARCH 26 Because there is no change in the potential energy of the proton, there is no work done. Copyright Protected ANSWERS AND SOLUTIONS 8. energy (kinetic) of the alpha particle when it hit the plate is: Ek 4.90 10–18 J 1.70 10–17 J 2.19 10–17 J 1 2 Ek m v 2 2 Ek v m 2(2.19 1017 J) 6.65 1027 kg 8.11 104 m/s V d 7.5 101 V 6.0 102 m 1.25 103 V/m E Fe q Fe q E E 1.60 10 –14 C 1.25 103 V/m 2.00 10 –16 N work Fd 2.00 10–16 N 3.0 10–2 m 6.0 10 9. –18 11. Find initial energy: 1 Ek1 mv12 2 1 (9.11 1031 kg)(5.0 105 m/s) 2 2 1.14 10 –19 J J Ek q V E V Find energy of electron at plate: 1 Ek2 mv2 2 2 1 (9.11 1031 kg)(1.0 105 m/s) 2 2 4.56 10 –21 J Energy is changed by a factor of: to 9.00 1017 J from 3.00 1017 J 3.00 The energy changes by a factor of 3.00. Therefore, the potential difference changes by a factor of 3.00. V 4.20 102 V 3.00 E 1.14 10–19 J – 4.56 10–21 J 1.09 10–19 J 1.26 103 V E q 1.09 1019 J 1.60 1019 C 6.8 10 –1 V V 10. Find initial kinetic energy first: 1 Ek mv 2 2 1 (6.65 10 27 kg)(7.15 10 4 m/s) 2 2 1.70 10 –17 J 12. Now find the change in energy using following steps: V E d V Ed 1.70 102 V/m 9.00 10 –2 m 1.53 101 V E V q E qV 3.20 10 –19 C 1.53 101 V V d 1 E d To 5.0 cm From 7.0 cm 0.714 E Distance changes by a factor of 0.714. the electric field will change by a factor of 1 1.40 0.714 E 9.3 102 V/m 1.40 4.90 10 –18 J 1.3 103 V/m Not for Reproduction 27 Physics 12 SNAP ANSWERS AND SOLUTIONS V d 3.00 10 2 V 5.00 102 m 6.00 103 V/m Fe q 5.30 1014 N 3.20 1019 C 1.66 105 N/C E 13. 16. E V d V Ed E 1 2 mv 2 1 (9.111031 kg)(6.00 106 m/s) 2 2 1.64 10–17 J Ek 14. 1.66 105 N/C 7.50 10 –2 m 1.24 104 V E 1.64 10–17 J 0 1.64 10–17 J 17. When a charged object is moved perpendicular to an electric field, no work is done. E q 1.64 1017 J 1.60 1019 C 1.02 102 V V 18. a) 15. Uniform accelerated motion: vi 0 t 2.50 10 –5 s d 4.00 10 –2 m b) 1 d vi t at 2 2 2d a 2 t 2(4.00 10 2 m) (2.50 10 5 s) 2 1.28 108 m/s 2 Ek Ep 0 1 2 1 mv mv0 2 qV 0 2 2 1 (1.67 1027 kg)(v 2 ) 2 (1.60 1019 C)(60.0 V) 0 v 1.07 105 m/s Ek Ep 0 1 2 1 mv mv0 2 qV 0 2 2 1 (9.111031 kg)(v 2 ) 2 (1.60 1019 C)(60.0 V) 0 v 4.59 106 m/s 1 2 mv 2 E qV 1 qV mv 2 2 v 2 V or V v 2 19. Ek Fnet Fe ma 6.65 10–27 kg 1.28 108 m/s 2 8.5110–19 N Fe q 8.51 1019 N 3.20 1019 C 2.66 N/C E speed changes by a factor of: To 7.25 104 m/s From 2.90 104 m/s 2.50 V changes by a factor of 2.50 6.25 2 V 6.25 2.50 105 V 1.56 106 V CASTLE ROCK RESEARCH 28 Copyright Protected ANSWERS AND SOLUTIONS E qV 1.60 10 –19 C 5.00 103 V 22. Draw a free-body diagram to identify the forces involved. 8.00 10 –16 J 20. E E k 0 Ek E 1 Ek mv 2 2 2 Ek v m 2(8.00 10 16 J) 9.11 10 31 kg 4.19 107 m/s E 21. a) Fg mg 0.500 kg 9.81 m/s 2 4.90 N V d Considering equilibrium condition: Fy 0 FTy – Fg 0 FT sin – Fg 0 FT sin 80.0 – 4.90 N 0 0.985FT 4.90 N 4.90 N FT 0.985 4.975 N Fx 0 Fe – FTx 0 q E – FT cos 0 1 500 N/C q FT cos80.0 1 500 N/C q 0.174FT 1 500 N/C q 0.174 4.975 N q 5.77 10 –4 C 12.0 V 2.00 102 m 6.00 102 V/m Fe q E 3.20 0 –19 C 6.00 102 V/m 1.92 10 –16 N b) Fg mg 6.65 10 –27 kg 9.81 N/kg 6.52 10 –26 N c) Fnet Fe – Fg 1.92 10 –16 N – 6.52 10 –26 N 1.92 10 –16 N d) Fnet ma F a net m 1.92 1016 N 6.65 1027 kg 2.89 1010 m/s 2 e) 23. Find how long the electron will be in the electric field: d v t d t v 14.0 102 m 8.70 106 m/s 1.61 10 –8 s Fe Fg Fe 6.52 1026 N Fe q 6.52 1026 N 3.20 1019 C 2.04 107 N/C E Find the electric force on the electron: Fe q E 1.60 10–19 C 1.32 103 N/C 2.1110–16 N V E d 2.04 107 N/C 2.00 102 m 4.07 109 V Not for Reproduction 29 Physics 12 SNAP ANSWERS AND SOLUTIONS Fnet ma F a net m 2.111016 N 9.111031 kg 2.32 10 –14 m/s 2 3. Now find the vertical component of displacement of the electron: vi 0 t 1.6110 –8 s a 2.32 10 –14 m/s 2 4. Lesson 11—Magnetic Forces on Current-Carrying Conductors Lesson 10—Magnetic Forces and Fields PRACTICE EXERCISES ANSWERS AND SOLUTIONS PRACTICE EXERCISES ANSWERS AND SOLUTIONS 2. B 0 In 800 (4 10 7 N/A 2 )(2.0 A) 0.30 m 6.7 10 –3 T Magnitude of displacement 1 d vi t at 2 2 1 (2.32 1014 m/s 2 )(1.61108 s) 2 2 3.00 10–2 m 1. 775 turns n 30.0 cm 100 cm n 2 583 B 0 In B I 0 n 0.100 T 7 T m 4 10 (2583) A 31 A 1. Fm qvB F B m qv 9.50 1014 N (1.60 1019 C)(2.10 105 m/s) 2.83 T B 0 In 7 T m 4 10 (1.25 A)(1800) A 25.0 102 m 2 1.13 10 T 2. B 0 In B n 0 I Fm B Il 7.50 10–4 T 0.960 A 0.222 m 1.60 10–4 N Fm 1.60 10–4 N east 2.1 103 T 7 T m 4 10 (0.72 A) A 2.32 103 / m 3 2.32 10 x 100 cm 25 cm x 580 3. Fm qvB 1.60 10 –19 C 3.52 105 m/s 2.80 10–1 T 1.58 10 –14 N Fm 1.58 10 –14 N west 4. Fm qvB 3.20 10–19 C 7.50 104 m/s 5.50 T 1.30 10–13 N Fm 1.30 10–13 N west CASTLE ROCK RESEARCH 30 Copyright Protected ANSWERS AND SOLUTIONS 5. Fm qvB F B m qv 10. Fm qvB F v m qB 1.70 1014 N (1.60 1019 C)(1.40 104 m/s) 5.59 T B 5.59 T up 6. 4.1 1013 N (1.60 1019 C)(7.20 101 T) 3.56 106 m/s 1 Ek mv 2 2 1 (9.11 10 31 kg)(3.56 106 m/s) 2 2 5.77 1018 J E qV E V q 5.77 1018 J 1.60 1019 C 3.6 101 V Fm qvB F B m qv 7.11014 N (1.60 1019 C)(2.7 105 m/s) 1.6 T B 1.6 T west 7. The alpha particle is moving parallel to the magnetic field. 11. Fm qvB 1.60 10 –19 C 6.20 105 m/s there is no force. Fm qvBsin 0 8. E qV 1.60 10 –19 C 1.70 103 V 2.72 10 –16 J 12. Fm Fg 1 2 mv 2 2 Ek v m 2(2.72 1016 ) 9.11 1031 kg 2.44 107 m/s Fm qvB 1.60 10 –19 C 2.44 107 m/s Ek Fg mg 1.76 102 kg 9.81 N/kg 1.72 101 N Fm 1.72 101 N Fm B Il F B m Il 2.50 10 –1 T 9.77 10 Not for Reproduction T 2.28 10 N F ma F a m F a m m 2.28 1014 N 9.111031 kg 2.50 1016 m/s 2 Fm B Il 5.00 10–1 T 3.60 A 2.50 10–1 m –13 2.30 10 –1 14 4.50 10–1 N 9. 1.72 101 N (6.00 A)(1.90 102 m) 1.51 T N 31 Physics 12 SNAP ANSWERS AND SOLUTIONS 6. Lesson 12—Cathode Rays PRACTICE EXERCISES ANSWERS AND SOLUTIONS 1. Fm Fc mv 2 r qB r v m (3.20 1019 C)(4.22 101 T)(1.50 103 m) 6.65 1027 kg 4 3.05 10 m/s qvB 2. Fm Fc mv 2 qvB r qB r v 2 m v qB r v m 3.20 1019 C 3.60 10 1 T 8.20 10 2 m 6.65 1027 kg 1.42 106 m/s 1 2 mv 2 2 1 6.65 1027 kg 1.42 106 m/s 2 6.7110–15 J Ek Fm Fc mv 2 qvB r mv B qr (1.67 1027 kg)(5.40 105 m/s) (1.60 1019 C)(7.20 103 m) 7.83 101 T 7. E q E qV 1.60 1019 C 2.50 103 V V 4.00 1016 J 1 2 mv 2 2 Ek v m 2(4.00 1016 J) 9.11 1031 kg 2.96 107 m/s Ek 3. Fm Fc mv 2 qvB r q v m B r (3.60 105 m/s) (6.10 101 T)(7.40 10 2 m) 7.98 106 C/kg 4. 8. Fe Fm q E qvB E vB 7.80 105 m/s 2.20 101 T 1.72 105 N/C 5. Fe Fm q E qvB 9. |E| B 2.10 105 N/C 6.50 101 T 3.23 105 m/s v CASTLE ROCK RESEARCH 32 1 2 mv 2 1 (9.111031 kg)(4.75 107 m/s) 2 2 1.03 10–15 J E V q 1.03 1015 J 1.60 1019 C 6.42 103 V Ek E q E qV (1.00 1014 C)(1.40 103 V) 2.24 1016 J V Copyright Protected ANSWERS AND SOLUTIONS 12. When charged particles travel parallel to a magnetic field there is no force. If there is no force, there is no deflection. 1 2 mv 2 2 Ek v m 2(2.24 1016 J ) 9.11 1031 kg 2.23 107 m/s Ek 13. Fm Fc mv 2 r qB r v m (1.60 1019 C)(0.75 T)(0.30 m) v 1.67 1027 kg 7 2.16 10 m/s qvB Fm Fc qvB mv 2 r mv qB (9.11 1031 kg)(2.23 107 m/s) (1.60 1019 C)(2.20 102 T) 5.74 103 m r 14. Fm Fc mv 2 r qB r v m (1.60 1019 C)(2.2 104 T)(0.77 m) 9.11 1031 kg 7 2.98 10 m/s 1 E mv 2 2 1 (9.11 10 31 kg)(2.98 107 m/s) 2 2 4.03 10 –16 J E V q 4.03 1016 J 1.60 1019 C 2.5 103 V qvB 10. Fe Fm q E qvB |E| B 4.20 105 N/C 1.40 102 T 3.00 107 m/s 1 Ek mv 2 2 1 (9.11 10 31 kg)(3.00 107 m/s) 2 2 4.10 10 16 J E V q 4.10 1016 J 1.6 1014 C 2.56 103 V v 15. Fm Fc mv 2 r qB r v m (1.0 104 C/kg)(9.10 101 T) 0.240 m 2.4 103 m/s d C 2 r 2 0.240 m 1.51 m d v t d t v 1.51 m 2.40 103 m/s 6.28 104 s qvB 11. Fm Fc qvB mv 2 r mv rB (8.4 1027 kg)(5.6 105 m/s) (3.5 102 m)(2.8 101 T) 4.8 10 –19 C 4.8 1019 C # e 1.6 1019 C 3 The particle carries an extra 3 electrons. q Not for Reproduction 33 Physics 12 SNAP ANSWERS AND SOLUTIONS Lesson 13—Electromagnetic Induction N 5. N PRACTICE EXERCISES ANSWERS AND SOLUTIONS 1. area r 2 (5.0 10–2 m) 2 7.85 103 m 2 BAcos 3.2 103 T 7.85 103 m 2 cos 0 2.5 105 Wb 2. 3. area r 2 (6.0 10 –2 m) 2 1.13 10 –2 m 2 f BAcos (3.6 104 T)(1.13 102 m 2 ) cos 90 0 0 BAcos (3.6 104 T)(1.13 102 m 2 ) cos 0 4.07 106 Wb f 0 0 – 4.07 106 Wb – 4.07 106 Wb N t 4.07 106 Wb (200) 0.015 s 5.4 10 –2 V t –50 15 Wb/s –7.5 102 V 4. t (t ) N (0.92 V)(0.12 s) 100 –1.10 103 Wb BAcos B A cos 1.10 103 Wb 4.0 10-3 m 2 2.8 101 T 6. area r 2 (12.5 102 m) 2 4.91102 m 2 0 BAcos (2.7 103 T)(4.91 102 m 2 ) cos 0 1.33 104 Wb f BAcos (2.7 103 T)(4.9110 2 m 2 ) cos 180 –1.33 104 Wb f 0 –1.33 104 Wb –1.33 104 Wb –2.66 10–4 Wb N t 2.66 10 4 Wb (10) 0.30 s 8.8 103 V f BAcos (2.2 10–2 T) 0 cos 90 0 0 BAcos (2.2 10 –2 T)(2.5 10 –3 m 2 ) cos 0 5.5 10 –5 Wb f 0 0 – 5.5 10 –5 Wb –5.5 10 –5 Wb N t 5.5 105 Wb 0.10 s –4 5.5 10 V 7. 0 BAcos 0.40 T (5.0 103 m 2 ) cos 0 2.0 103 Wb f BAcos 0 (5.0 103 m 2 ) cos 0 0 f 0 0 – 2.0 10 –3 Wb –2.0 103 Wb Direction: clockwise CASTLE ROCK RESEARCH 34 Copyright Protected ANSWERS AND SOLUTIONS t 2.0 103 Wb (25) 1.0 s 5.0 102 V N 8. a) IR I R 1.6 103 V 2.0 8.0 104 A 0 BAcos 1.6 T (7.2 103 m 2 ) cos 0 1.15 102 Wb b) counter-clockwise Lesson 14—Electric Generator f BAcos 0 (7.2 103 m 2 ) cos 0 0 f 0 0 –1.15 102 Wb –1.15 102 Wb N t 1.15 102 Wb 0.050 s 2.3 101 V b) PRACTICE QUESTIONS— ANSWERS AND SOLUTIONS 1. Vback – Ir 120 V – 12.0 A (6.0 ) 48 V 2. a) initial Vback – Ir Vback r I 120 V 0 r 40.0 A 120 V 40.0 A 3.00 IR I R 2.3 101 V 12.0 1.9 10 –2 A 9. c) clockwise a) area l 2 (4.0 102 m) 2 1.6 103 m 2 0 BAcos 0.20 T (1.6 103 m 2 ) cos 0 3.2 104 Wb f BAcos 0.50 T (1.6 103 m 2 ) cos 0 8.0 104 Wb f 0 8.0 104 Wb – 3.2 104 Wb 4.8 104 Wb N t 4.8 104 Wb 0.30 s –1.6 103 V Not for Reproduction b) operating full Vback – Ir 120 V – 15.0 A (3.00 ) 75.0 V 3. a) Vback – Ir 120 V – 9.0 A (5.0 ) 75 V b) no back emf 35 Physics 12 SNAP ANSWERS AND SOLUTIONS c) initial Vback 0 Lesson 15—EMF Continued Vback – Ir 0 120 V – I (5.0 ) 120 V I 5.0 24 A 4. PRACTICE EXERCISES ANSWERS AND SOLUTIONS 1. B lv 2. Blv b) Back emf has decreased c) B 3. , which tells us that the t induced emf varies directly with the rate of change in the magnetic flux. The back emf is the emf induced by the motor. Therefore, if the rate of rotation is halved, the back emf is halved. 0.150 T 0.220 m 1.25 m/s 0.0413 V IR R 0.0413 V 2.25 1.83 102 A counter-clockwise 4. a) initial Vback 0 Vback – Ir Vback r I 120 V 0 r 10.0 A 120 V 10.0 A 12.0 a) IR 5.6 10 –2 A 1.5 8.4 10 –2 V Again B lv B lv 8.4 10 2 V (15 10 2 m)(0.95 m/s) 0.59 T b) operating full speed Vback – Ir 120 V – 3.0 A (12.0 ) 84 V CASTLE ROCK RESEARCH B lv I 45.0 V We can use N 6. lv 0.075 V (0.28 m)(0.80 m/s) 0.33 T i) Vback – Ir 120 V – 3.6 A (6.0 ) 98 V ii) Vback – Ir 120 V – 8.4 A (6.0 ) 70 V 5. 0.75 T 0.35 m 1.5 m/s 0.39 V a) Greater current—more heat (Remember: P I 2 r ) b) counter-clockwise 36 Copyright Protected ANSWERS AND SOLUTIONS 5. B lv W t W Pt 0.181 W 1.0 102 s P 0.950 T 0.300 m 1.50 m/s 0.428 V IR I 1.8 103 J R 0.428 V 3.25 0.132 A Fm BIl 0.950T 0.132 A 0.300 m 3.75 10 –2 N 6. B lv 4.70 10 2.79 10 7. –6 3 10. IR 5.2 10 –2 A 2.9 0.151 V Blv v 0.151 V (0.65 T)(0.50 m) 0.46 m/s T 6.25 m 95.0 m/s V 11. a) 0.600 T 0.300 m 3.00 m/s 0.540 V 2 b) clockwise 12. a) 0.75 T 1.2 m 2.5 m/s 2.25 V IR ( B lv) N 1.1 T 0.18 m 2.7 m/s 5 2.67 V IR I R 2.67 V 3.5 0.76 A I 2.25 V 0.45 A 5.0 9. 1.30 T 0.625 m 1.80 m/s 1.463 V IR R 1.463 V 1.50 0.975 A B lv R B lv I R (0.540 V) 2 2.25 0.130 W W P t W Pt 0.130W 15.0 s 1.94 J 8. B l B lv P b) clockwise B lv 0.95 T 1.0 m 3.0 m/s 2.85 V P 2 R (2.85 V) 2 45.0 0.181 W Not for Reproduction 37 Physics 12 SNAP ANSWERS AND SOLUTIONS I s Vp I p Vs 0.267 A 1.20 102 V Ip 20 V I p 4.4 10 –2 A Lesson 16—Transformers (Induction Coil) PRACTICE EXERCISES ANSWERS AND SOLUTIONS 1. 2. 3. Is Np I p Ns 0.95 A 1.0 103 0.25 A Ns Ns 2.6 102 a) Pp I pVp Pp Ip Vp 5.0 101 W 1.20 10 2 V 0.42 A 7. Pp I pVp 1.0 A 1.20 102 V The power in the secondary coil is also 1.2 10 2 W (same in both cells) 8. Pp I pVp Pp Ip Vp 1.0 102 W 9.0 V 11.1 A Is Np I p Ns Np 0.65 A 11.1 A 1.5 103 N p 88 Np Vs N s 1.20 102 V 1.50 102 Vs 25 Vs 20 V Vs I s Rs V Is s Rs 20 V 75 0.27 A CASTLE ROCK RESEARCH 6. 1.2 102 W Ps I sVs P Is s Vs 2.4 103 W 1.20 102 V 2.0 101 A 3.0A 1.20 102 V Power delivered to the secondary coil is also 3.6 10 2 W (same in both cells) Pp I pVp Pp Ip Vp 2.4 103 W 3.6 103 V 0.67 A Vp Pp I pVp 3.6 102 W Is Np I p Ns 5.00 A 1.20 103 Ip 1.50 02 I p 0.625 A b) 4. 5. 38 Copyright Protected ANSWERS AND SOLUTIONS 5. Practice Test ANSWERS AND SOLUTIONS 1. 2. Gm1m2 1 or Fg 2 (r is doubled, because 2 r r the final distance from Earth’s centre is 2r) 1 Fg 4 2.0 N Fg 4 0.50 N Fg kq1q2 r2 2 9 Nm 6 9.00 10 (2.00 10 C) 2 C (2.00 10 6 C) (2.00 102 m) 2 90.0 N Fe GM r2 1 g 2 r This is an inverse square relationship. An inverse square relationship is graphed as: g The force between the pith balls is 90.0 N. 6. F2 F1 3. The equation Ep mgh cannot be used in the case of an object separated from Earth’s surface by a large distance because the magnitude of g is not constant—g decreases as the distance from the earth increases. 4. Ep kq1q2 r2 2 9 Nm 6 2 9.00 10 (3.00 10 C) 2 C (0.333 m) 2 0.730 N Gm1m2 r 2 11 N m (5.98 1024 kg)(1.25 103 kg) 6.67 10 2 kg 4.00 106 m + 6.38 106 m –4.80 1010 J Fnet F12 F2 2 (0.730 N) 2 (0.730 N) 2 1.03 N The net electric force on sphere B is 1.03 N. Not for Reproduction 39 Physics 12 SNAP ANSWERS AND SOLUTIONS 7. Fnet m 2.02 102 N 5.00 104 kg 4.05 105 m/s 2 kq1q2 1 or Fe 2 r2 r 1 r decreases by 3 2 r changes by 0.111 Fe changes by 9 times New force = 9.0F a Fe The initial acceleration of sphere A was 4.05 105 m/s 2 . The new force between the two point charges is 9.0F. 8. 10. Fnet ma a Fnet In this case the net force is the electric force. Fnet Fe kq1q2 r2 1 r2 This expression tells that when the distance 1 increases 3 times, the force becomes . 9 1 if the force decreases to , then according to 9 Fnet the expression, a Fnet The negatively charged rod will induce the electrons in the electroscope to move away from the rod, i.e., move from the head to the leaves. Therefore, the head of the electroscope will be positive. 11. 1 the acceleration is also . 9 New acceleration is 1.00 1010 m/s2 19 1.11109 m/s 2 1 2 (6.65 10 27 kg)v (3.20 10 2 v 2 The initial acceleration of sphere A was 1.11 109 m/s 2 . 9. Ek Ep 0 1 2 1 mv mv0 2 qV 0 2 2 1 2 1 mv m(0) 2 qV 2 2 19 2(3.20 10 C)( 2.00 10 V) 19 3 C)( 2.00 10 V) 3 (6.65 10 27 kg) v 4.39 10 m/s 5 The maximum speed of the alpha particle is 4.39 105 m/s . Fnet ma a Fnet 12. In this case the net force is the electric force. kq q Fnet Fe 12 2 r 2 9 Nm 6 2 9.00 10 (3.00 10 C) 2 C (2.00 102 m) 2 2.03 102 N The electron will accelerate in the direction of the electric force which as away from the negative plate (opposite direction to the electric field). 13. Electric potential is the potential electric energy per unit of charge. CASTLE ROCK RESEARCH 40 Copyright Protected ANSWERS AND SOLUTIONS kqq r2 According to this equation, as the distance between the charges decreases, the electric force increases. If the electric force increases, the electric potential energy increases. 14. Fe 19. Np I s Ns I p I 1 s 10 I p I p 10 I s This shows that the current in the primary coil is 10 times greater than the current in the secondary coil. 15. The direction of an electric field is defined in terms of the direction of the electric force on a positive test charge. 20. A “step-up” transformer is used to increase (hence step up) the voltage. But, if the transformer increases the voltage, the current must decrease. Remember the law of conservation of energy. The power in each coil must be the same. Therefore, a step-up transformer is used to increase voltage and decrease current. 16. The electric field is away from a positive object toward a negative object. Therefore, the sketch should show that the field is directed down toward the negative plate. 21. A transformer makes use of the principle of electromagnetic induction. This means that there must be a change in the magnetic field inside the secondary coil. In order to change this magnetic field, a switch is placed in the primary circuit. kq1 r 1 V r 17. V 22. Lenz’s law is an example of conservation of energy. The magnitude of the electric potential (V) as a function of the distance (r) from a point charge is an inverse relationship. This can be shown as a curve approaching the x-axis as it moves away from the y-axis. 23. Use the right-hand rule. The opposing magnetic force must be in the opposite direction to its motion (i.e., to the left). Place your right palm to face the left along the conductor, and direct your fingers into the page (in the direction of the magnetic field). Your thumb will now point along the conductor from end B toward end A. This is the direction of the induced electric current. Blv and I R From these mathematical relationships, if the velocity decreases, the current also decreases. Therefore, the current within rod AB will be directed toward end A, and it will decrease as the rod passes through the magnetic field. 18. Lenz’s law states that the induced field will oppose the original change of flux of the magnet. In other words, the induced magnetic field inside the solenoid must oppose the motion of the magnet. That means that the right side of the solenoid is the induced magnetic north. Use the right-hand rule to determine the direction of the current within the circuit. Point your thumb of your right hand to the right and curl your fingers in the direction of the electric current. For this to happen, the current flows from A to B through G. Not for Reproduction 24. Note that the alpha particle is moving parallel to the magnetic field. Therefore, there is no magnetic force acting upon the alpha particle. Fm qvBsin 41 Physics 12 SNAP ANSWERS AND SOLUTIONS 33. Apply Lenz’s law here. Use the right-hand rule for conventional current. In this scenario, as the north pole of the magnet moves down into the loop, the magnetic field within the loop will point upward out of the loop. Point your right thumb in this direction and curl your fingers to identify the direction of the current. Your fingers should be pointing in a counterclockwise direction. However, when the magnet is pulled out of the loop, the motion of the magnet, the direction of the magnetic field, and the direction of the current all change direction. Therefore, the direction of the current as the magnet is pulled out of the loop is in the clockwise direction. Therefore, as viewed from above the loop, the direction of the induced current in the loop would first be counterclockwise and then clockwise. 25. Use the right-hand rule. Point your thumb east. Point your fingers north. Your palm should be directed up. The magnetic force on the proton is directed upward. 26. Lenz’s law states that the induced current flows in such a way that the magnetic field inside the coil opposes the magnetic field of the magnet. Using the right-hand rule, curl the fingers on your right hand in a counterclockwise direction. Your right thumb will thus point upward. The magnetic field inside of the loop is directed upward. 27. Use the left-hand rule (electrons). Point your thumb north. When your left-hand fingers curl above your thumb, they should be pointing west. Therefore, the magnetic field above the conductor will be pointed west. 34. J.J. Thomson determined the charge-to-mass ratio of the electron by balancing the electric and magnetic forces. 28. Blv If the period (time) of the rotation decreases, the loop will be rotating at a higher speed. From the above equation, as the speed increases, so does the emf. Therefore, if the period of rotation was cut in half, the induced voltage would increase as well. 35. A cathode ray is a beam of electrons. 36. Statements i) and ii) are true. Cathode rays are electrons, and charged particles can be deflected by electric and magnetic fields. Therefore, cathode rays can be deflected by both electric and magnetic fields. 29. Generators convert mechanical energy into electric energy. 30. Motors convert electric energy into mechanical energy. Statement iii) is incorrect. Only electromagnetic radiation travels at 3.00 108 m/s . Electromagnetic radiation does not consist of charged particles as cathode rays do. 31. Both transformers and generators operate on the principle of electromagnetic induction. 32. 37. The greater the deflection, the smaller the radius. Therefore, in order to increase the deflection, determine which variables in the following equation will result in a smaller radius. Fm Fc mv 2 qvB R mv R qB The radius depends on the mass, speed, charge, and magnetic field strength. In order to decrease the radius, increase the charge of the particle, increase the strength of the magnetic field, or decrease either the mass or speed of the particle. Fm qvB Fm v and Fm does not depend on the mass. if the speed is halved, you will halve the magnetic force. Fm = F 2 CASTLE ROCK RESEARCH 42 Copyright Protected ANSWERS AND SOLUTIONS 38. If electrons pass through superimposed electric and magnetic fields without deflection, that means the electric and magnetic forces acting on the electrons must be equal but in opposite directions. Fe Fm q | E | qvB |E| v B Therefore, the speed of the electrons can be determined from the strengths of the superimposed fields. 5. MOMENTUM 6. p mv 0.673 kg 3.0 m/s north 2.0 kg m/s north Lesson 1—Momentum PRACTICE EXERCISES ANSWERS AND SOLUTIONS 1. 2. 3. p mv 4.0 kg 12.0 m/s east 48 kg m/s east 7. p mv p v m 25.0 kg m/s west 5.0 kg 5.0 m/s west v v0 t v 0 2 –9.81 m/s 0.25 s v –2.45 m/s a p mv 5.0 kg –2.45 m/s –12 kg m/s =12 kg m/s down p mv p m v 36.0 kg m/s south 8.0 m/s south 4.5 kg 8. Impulse Fnet t 17.0 N east 2.5 102 s 4.3 101 N s east p mv p m v 29 kg m/s east 2.0 m/s east 14.5 kg Impulse Fnet t Impulse t Fnet 7.00 N s west 11.2 N west 0.625 s 10. Consider north direction as positive Fnet t mv Fnet 2.60 s 26.3 kg –21.0 m/s Fnet –212 N or 212 N south Fg mg 14.5 kg 9.81 m/s 2 1.4 102 N Not for Reproduction d t 2.6 m west 1.1 s 2.36 m/s west v p mv 7.0 kg 2.36 m/s west 17 kg m/s west 9. 4. Fg mg Fg m g 6.6 N 9.81 m/s 2 0.673 kg 43 Physics 12 SNAP ANSWERS AND SOLUTIONS 11. Magnitude of impulse Fnet t mv 31.6 N t 15.0 kg 10.0 m/s t 4.75 s 17. Consider up as positive, Fnet ma F a net m 1.5 105 N 9.5 103 kg 15.8 m/s 2 v v0 a t v 0 2 15.8 m/s 15 s v 2.4 102 m/s 12. Fnet t mv Therefore, mv 25.0 N 7.20 101 s =18.0 N s north 13. Impulse mv 5.00 kg 15.0 m/s east 75.0 kg m/s east 14. or Fnet t mv 1.5 10 N 15 s 9.5 103 kg v 5 d t 26.3 m west 3.2 s 8.22 m/s west v v0 vav 2 v 0 8.22 m/s west 2 v 16.4 m/s west p mv 11.0 kg 16.4 m/s west 181 kg m/s west vav v 2.4 102 m/s 18. 19. The magnitude of impulse Fnet t mv 225 N t 1.0 103 kg 5.0 m/s 2.0 m/s t 13 s 15. The magnitude of momentum p mv p v m 6.0 kg m/s 3.0 kg 2.0 m / s v 2 v0 2 2ad 20. 16. Fnet t mv 95 N north 1.65 s 15 kg v Lesson 2—Collisions PRACTICE EXERCISES ANSWERS AND SOLUTIONS 2 1. p1 mv1 1.0 kg –2.0 m/s –2.0 kg m/s p2 mv2 1.0 kg 1.6 m/s 1.6 m/s p p2 – p1 1.6 kg m/s 2.0 kg m/s 3.6 kg m/s upward CASTLE ROCK RESEARCH v 1.0 101 m/s north 2 9.81 m/s 2 d d 0.20 m The object has fallen 0.20 m when its momentum is 6.0 kg m/s down. 2.0 m/s Fnet t mv Fnet 0.75 s 5.4 kg 3.0 m / s east Fnet 22 N east Consider right as positive and left as negative m1 30.0 kg m2 20.0 kg v1 1.00 m/s v2 –5.00 m/s p1 30.0 kg m/s p2 –100 kg m/s psys(before) p1 p2 –70.0 kg m/s 44 Copyright Protected ANSWERS AND SOLUTIONS 3. m1 30.0 kg v1 ? p1 ? m1 925 kg v1 18.0 m/s p1 1.67 104 kg m/s m2 20.0 kg v2 –1.25 m/s p2 25.0 kg m/s m2 ? v2 0 m/s p2 0 kg m/s pbefore p1 p2 1.67 104 kg m/s psys(after) psys(before) 70.0 kg m/s after collision psys(after) p1 p2 70.0 kg m/s p1 70.0 kg m/s (25.0 kg m/s) p1 45.0 kg m/s m? v 6.50 m/s p pafter =1.67 104 kg m/s p1 m1v1 p v1 1 m1 45.0 kg m/s 30.0 kg 1.50 m/s p v 1.67 104 kg m/s 6.50 kg 2.57 103 kg m2 2.57 103 kg 925 kg 1.65 103 kg m The velocity of the 30.0 kg ball after the collision is 1.50 m/s left 2. 4. Consider east as positive m1 4.50 103 kg v1 5.0 m/s p1 2.25 104 kg m/s m2 6.50 103 kg v2 0 m/s p2 0 kg m/s m1 0.050 kg v1 ? p1 ? pbefore p1 p2 2.25 104 kg m/s m m1 m2 7.05 kg v 5.00 m/s p pafter 35.25 kg m/s pbefore pafter 35.25 kg m/s p1 pbefore 35.25 kg m/s p v1 1 m1 35.25 kg m/s 0.050 kg 705 m/s m m1 m2 1.10 104 kg v ? p pafter =2.25 104 kg m/s p mv p v m 2.25 104 kg m/s 1.10 104 kg 2.1 m/s east Not for Reproduction m2 7.00 kg v2 0 m/s p2 0 kg m/s 45 Physics 12 SNAP ANSWERS AND SOLUTIONS pafter po +ps =0 ps 4.4 kg m/s 5. ps m 4.4 kg m/s 76 kg 0.058 m/s right vs Consider right as positive and left as negative m1 40.0 g v1 9.00 m/s p1 3.60 10 2 g m/s 7. m2 55.0 g v2 6.00 m/s p2 3.30 102 g m/s m 1.13 103 kg v0 p0 pbefore p1 p2 3.00 101 g m/s pbefore 0 m m1 m2 95.0 kg v ? p pafter pbefore =3.00 101 g m/s Consider east as positive mP 25 kg mL 1.1 103 kg vP 325 m/s vL ? pL ? pP 8.13 103 kg m/s p m 3.00 101 g m/s 95.0 g 0.316 m / s right v pafter pbefore 0 pafter pL +pP 0 pL 8.13 103 kg m/s 6. pL mL 8.13 103 kg m/s 1.1 103 kg –7.4 m/s 7.4 m/s west vL m 76 kg v0 p0 pbefore 0 8. m? v0 p0 Consider left as negative mo 0.20 kg ms 76 kg vo 22 m/s vs ? po 4.4 kg m/s ps ? pbefore 0 pafter pbefore =0 CASTLE ROCK RESEARCH 46 Copyright Protected ANSWERS AND SOLUTIONS mC 1.02 103 kg vC 25 m/s pC 2.55 104 kg m/s Consider right as positive mG 4.5 102 kg mV ? vG 1.4 103 m/s vV 45 m/s pV ? pG 6.3 105 kg m/s pbefore pT pC 1.28 105 kg m/s pafter pbefore =0 pafter pV +pG 0 pV 6.3 105 kg m/s pV vV 6.3 105 kg m/s 45 m/s 1.4 104 kg mV m 1.12 104 kg v ? p pafter 1.28 105 kg m/s p m 1.28 105 kg m/s 1.12 104 kg 11 m/s north v 9. m 7.0 kg v0 p0 pbefore 0 11. a) Consider right as positive m1 225 g v1 30.0 cm/s p1 6.75 103 g cm/s Consider right as positive mB 5.0 kg mA 2.0 kg vB ? vA 10.0 m/s pB ? pA 20.0 kg m/s m2 125 g v2 10.0 cm/s p2 1.25 103 g cm/s pafter pbefore 0 pafter pA +pB 0 pB 20.0 kg m/s pbefore p1 p2 8.00 103 g cm/s pB mB 20.0 kg m/s 5.0 kg 4.0 m/s left vB m1 225 g v1 ? p1 ? m2 125 g v2 24 cm/s p2 3.00 103 g cm/s psys(after) psys(before) 8.00 103 g cm/s 10. p1 5.00 103 g cm/s p v1 1 m 5.00 103 g cm/s 225 g 22.2 cm/s right Consider north as positive mT 1.02 104 kg vT 15 m/s pT 1.53 105 kg m/s Not for Reproduction 47 Physics 12 SNAP ANSWERS AND SOLUTIONS b) Kinetic energy calculations, unlike momentum, require converting units to standard units. Also, consider speed. Mechanical (kinetic) energy lost in collision 1.075 102 J – 9.1445 103 J 1.61103 J collision is inelastic. 12. a) m1 = 0.225 kg v1 = 0.300 m/s m2 = 0.125 kg v2 = 0.100 m/s Consider right as positive m1 10.0 g m2 30.0 g v1 20.0 cm/s v2 0 p1 200 g cm/s p2 0 m1 = 0.225 kg v1 = 0.222 m/s m2 = 0.125 kg v2 = 0.240 m/s pbefore p1 p2 200 g cm/s Kinetic energy of object 1 before 1 Ek1 m1v12 2 1 (0.225 kg)(0.300 m/s) 2 2 1.0125 10 2 J m1 10.0 g v1 –6.0 cm/s p1 –60 g cm/s psys(after) psys(before) Kinetic energy of object 2, before 1 Ek2 m2 v2 2 2 1 (0.125 kg)(0.100 m/s) 2 2 6.25 104 J m2 30 g v2 ? p2 ? 200 g cm/s p2 200 g cm/s (60 g cm/s) 260 g cm/s p2 m 260 g cm/s 30.0 g 8.67 cm/s right v2 Kinetic energy of object 1, after 1 2 m1 v1 Ek1 2 1 (0.225 kg)(0.222 m/s) 2 2 5.5445 103 J b) Kinetic energy calculations, unlike momentum, require converting units to standard units. Kinetic energy of object 2, after 1 2 m2 v2 Ek2 2 1 (0.125 kg)(0.240 m/s) 2 2 3.60 103 J m1 = 0.0100 kg v1 = 0.200 m/s m2 = 0.0300 kg v2 = 0 m1 = 0.0100 kg = –0.060 m/s m2 = 0.0300 kg v2 = 0.0867 m/s Total kinetic energy before collision 1.0125 102 J 6.25 104 J 1.075 102 J Total kinetic energy after collision 5.5445 103 J 3.60 103 J 9.1445 103 J CASTLE ROCK RESEARCH 48 Copyright Protected ANSWERS AND SOLUTIONS Kinetic energy of object 1 before 1 Ek1 m1v12 2 1 (0.0100 kg)(0.200 m/s) 2 2 2.00 104 J Lesson 3—Two-Dimensional Interactions PRACTICE EXERCISES ANSWERS AND SOLUTIONS Kinetic energy of object 1 after 1 2 m1 v1 Ek1 2 1 (0.0100 kg)( 0.060 m/s) 2 2 1.8 105 J 1. m1 2.0 103 kg v1 35 km/h north p1 7.00 104 kg km/h north Kinetic energy of object 2 before 1 Ek2 m2 v2 2 2 1 (0.0300 kg)(0) 2 2 0 m2 1.4 103 kg v2 37.0 m/s west p2 28.0 kg m/s west Kinetic energy of object 2 after 1 2 m2 v2 Ek2 2 1 (0.0300 kg)(0.0867 m/s) 2 2 1.13 104 J m m1 m2 3.4 103 kg v ? p? Total kinetic energy before collision 2.00 104 J 0 2.00 104 J First find out the momentum of the system before collision ( psys(before) ) Total kinetic energy after collision is 1.8 105 J 1.13 104 J 1.31104 J Mechanical (kinetic) energy lost in collision 2.00 104 J –1.31104 J 6.9 05 J collision is inelastic. c) The magnitude of the resultant momentum is pR Most of this energy was converted to thermal energy—some to sound, etc. p12 p2 2 (7.00 104 kg km/h) 2 (5.18 104 kg km/h)2 8.71 104 kg km/h Now, find the direction p tan 2 p1 5.18 104 kg km/h 7.00 104 kg km/h 0.750 36.5 Not for Reproduction 49 Physics 12 SNAP ANSWERS AND SOLUTIONS Therefore psys(before) pR 8.71 104 kg km/h 36.5 W of N p m 33.6 kg m/s 56 W of N 14.2 kg 24 m/s 65 W of N v Again p psys(after) psys(before) pR p m 8.71 04 kg km/h 36.5 W of N 3.4 103 kg 26 km/h 37 W of N v 3. m1 4.0 104 kg v1 8.0 m/s west p1 3.2 105 kg m/s west 2. m2 3.0 104 kg v2 5.0 m/s south p2 1.5 105 kg m/s south mA 6.2 kg vA 3.0 m/s north pA 18.6 kg m/s north mB 8.0 kg vB 3.5 k m/s west pB 28.0 kg m/s west after collision 1+2 m m1 m2 7.0 10 4 kg v ? p? m mA mB 14.2 kg v ? p? pR p12 p2 2 (3.2 105 kg m/s) 2 (1.5 105 kg m/s) 2 3.53 105 kg m/s pR pA pB 2 p2 p1 1.5 105 kg m/s 3.2 105 kg m/s 0.469 25 tan 2 (18.6 kg m/s)2 (28.0 kg m/s)2 33.6 kg m/s pB pA 28.0 kg m/s 18.6 kg m/s 1.51 56.4 56 tan p pR 3.53 105 kg m/s 26 S of W p m 3.53 105 kg m/s 26 S of W 7.0 104 kg 1 5.0 10 m/s 25 S of W v p pR 33.6 kg·m/s 56 W of N CASTLE ROCK RESEARCH 50 Copyright Protected ANSWERS AND SOLUTIONS 4. a) m1 50.0 kg v1 ? p1 ? m2 60.0 kg v2 0 p2 0 p2 x p2 cos 378 kg m/s cos 322 298 kg m/s p2 y p2 sin 378 kg m/s sin 322 – 233 kg m/s m1 50.0 kg v1 6.00 m/s 50.0 N of E p1 300 kg m/s 50.0 N of E Do the vector addition for the components of the momentum px p1x p2 x =193 kg m/s 298 kg m/s 491 kg m/s m2 60.0 kg v2 6.30 m/s 38.0 S of E p2 378 kg m/s 38.0 S of E py p1y p2 y =230 kg m/s –233 kg km/s –3.00 kg m/s Find horizontal and vertical components of 300 kg·m/s 50.0° N of E. Now add p x and p y using Pythagoras theorem, to find out the magnitude of the resultant momentum. pR ( px ) 2 ( p y ) 2 (491 kg m/s) 2 (3.00 kg m/s) 2 491 kg m/s tan To solve this problem consider east and north direction as positive. p1x p1x cos 300 kg m/s cos 50.0 193 kg m/s px 3.00 kg m/s 491 kg m/s 0.35 0 p1y p1 sin 300 kg m/s sin 50.0 230 kg m/s psys(after) pR 491 kg m/s east psys(after) psys(before) p1 p1 m1v1 p v1 m 491 kg m/s east 50.0 kg 9.82 m/s east Find horizontal and vertical components of 378 kg·m/s 38.0° S of E. 38.0° S of E = 322° heading counter clockwise from positive x-axis Not for Reproduction py 51 Physics 12 SNAP ANSWERS AND SOLUTIONS Mechanical (kinetic) energy lost in collision is 2.41 × 103 J – 2.10 × 103 J = 3.10 × 102 J, collision is inelastic. Kinetic energy was not conserved. b) m1 50.0 kg v1 9.82 m/s m2 60.0 kg v2 0 c) 5. m1 50.0 kg v1 6.00 m/s a) m2 60.0 kg v2 6.30 m/s m1 15.0 kg v1 7.0 m/s east p1 105 kg m/s east NOTE: Do not include direction with the magnitude of velocity because direction is not necessary in calculating kinetic energy. Kinetic energy of object 1 before 1 Ek1 m1v12 2 1 (50.0 kg)(9.82 m/s) 2 2 2.41 103 J m2 10.0 kg v2 0 p2 0 m1 15.0 kg v1 4.2 m/s 20.0 S of E p1 63 kg m/s 20.0 S of E Kinetic energy of object 1 after 1 2 m1 v1 Ek1 2 1 (50.0 kg)(6.00 m/s) 2 2 9.00 10 2 J m2 10.0 kg v2 ? p2 ? pafter pbefore pbefore 105 kg m/s east pafter 105 kg m/s east Kinetic energy of object 2 before 1 Ek2 m2 v2 2 2 1 (60.0 kg)(0)2 2 0 Find horizontal and vertical components of 63 kg·m/s 20.0° S of E. 20.0° S of E = 340° heading counter clockwise from positive x-axis Kinetic energy of object 2 after 1 2 m2 v2 Ek2 2 1 (60.0 kg)(6.30 m/s) 2 2 1.20 103 J p1x p1x cos 63 kg m/s cos 340 59.2 kg m/s Total kinetic energy before collision = 2.41 × 103 J + 0 = 2.41 × 103 J p1y p1 sin 63 kg m/s sin 340 –21.5 kg m/s Total kinetic energy after collision = 9.0 × 102 J + 1.20 × 103 J = 2.10 × 103 J CASTLE ROCK RESEARCH Most of the loss in mechanical energy was converted to thermal energy—some to sound, etc. 52 Copyright Protected ANSWERS AND SOLUTIONS Kinetic energy of object 1, before 1 Ek1 m1v12 2 1 (15.0 kg)(7.0 m/s) 2 2 3.7 102 J After collision, the 10.0 kg object must have a horizontal component of 105 kg·m/s – 59.2 kg·m/s = 45.8 kg·m/s east After collision, the 10.0 kg object must have a vertical component of Kinetic energy of object 1, after 2 1 m1 v1 Ek1 2 1 (15.0 kg)(4.2 m/s) 2 2 1.3 102 J 0 – (–21.5 kg·m/s) = 21.5 kg·m/s north Now, add 45.8 kg·m/s east and 21.5 kg·m/s north using Pythagoras theorem. Kinetic energy of object 2, before 1 Ek2 m2 v2 2 2 1 (10.0 kg)(0) 2 2 0 pR ( px ) 2 ( p y ) 2 (45.8 kg m/s) 2 (21.5 kg m/s) 2 50.6 kg m/s Kinetic energy of object 2, after 1 2 m2 v2 Ek2 2 1 (10.0 kg)(5.1 m/s) 2 2 1.30 102 J Magnitude of velocity p v2 2 m 50.6 kg m/s 10.0 kg 5.1 m/s Total kinetic energy before collision 3.7 102 J 0 3.7 102 J Now find out the direction using py tan px 21.5 kg m/s 45.8 kg m/s 25 N of E v2 5.1 m/s 25 N of E Total kinetic energy after collision 1.3 102 J 1.3 102 J 2.6 102 J Mechanical (kinetic) energy lost in collision is 3.7 102 J – 2.6 102 J 1.1 102 J , collision is inelastic. Kinetic energy was not conserved. b) c) m1 = 15.0 kg v1 = 7.0 m/s m1 = 15.0 kg v1 = 4.2 m/s Not for Reproduction m2 = 10.0 kg v2 = 0 6. Most of the loss in mechanical energy was converted to thermal energy—some to sound, etc. before explosion m = 3m v =0 p =0 m2 = 10.0 kg v2 = 5.1 m/s 53 Physics 12 SNAP ANSWERS AND SOLUTIONS after explosion Now add the x- and y-components of p3 . 1 2 m1 m v1 15 m/s east p1 15m m/s east m2 m v2 10 m/s 45 S of E p2 10m m/s 45 S of E 3 m3 m v3 ? p3 ? p3(R) ( p3 x ) 2 ( p3 y ) 2 psys(after) psys(before) p1 p2 p3 0 ((22.1m) m/s) 2 ((7.07 m) m/s) 2 (23.2m) m/s Find the horizontal and vertical components of p1 . p1x 15m m/s east p1 y 0 tan p3 y p3 x (7.07m) m/s (22.1m) m/s 17.8 N of W p3 p3R (23.2m) m/s 17.8° N of W Find the horizontal and vertical components of p2 . 45.0° S of E = 315° heading counter clockwise from positive x-axis p3 m (23.2m) m/s 17.8° N of W m 23.2 m/s 17.8 N of W v3 Practice Test p2 x p2 cos 10m m/s cos 315 7.07m m/s ANSWERS AND SOLUTIONS p2 y p2 sin 1. 10m m/s sin 315 7.07m m/s p 15m m/s 7.07m m/s x p y p x Ep = mgh Since the height (h) increases, the gravitational potential energy increases. 22.1m m/s 0 –7.07m m/s –7.07m m/s p = mv Like kinetic energy, since both the mass and the velocity remain constant, the momentum remains constant. should be zero, x-component of p3 = (–22.1 m) m/s p y 1 2 mv 2 Since both the mass and the velocity remain constant, the kinetic energy remains constant. Ek 2. Momentum is conserved in all collisions and explosions. 3. Ft = mv Units of Ft = N·s Units of mv = kg·m/s units of impulse can be: N·s or kg·m/s. should be zero, y-component of p3 = (7.07 m) m/s CASTLE ROCK RESEARCH 54 Copyright Protected ANSWERS AND SOLUTIONS 4. If the object has momentum, it has mass and velocity; and if it has mass and velocity, it has kinetic energy also. 10. Find the velocity of the spacecraft. Mechanical energy is the sum of the gravitational potential energy and the kinetic energy. Therefore, if it has kinetic energy, it must also have mechanical energy. m 1.30 102 kg 2.80 103 kg =2.93 103 kg v 0 p mv 0 Note: The object does not need to have potential energy. 5. 6. An elastic collision is defined as a collision in which both the momentum and kinetic energy are conserved. m1 1.30 102 kg m2 2.80 103 kg v1 9.00 m/s v2 ? p1 1.17 103 kg·m/s p2 –1.17 103 kg m/s 1 2 mv 2 2 Ek v m Ek p2 m2 1.17 103 kg m/s 2.80 103 kg 0.418 m/s v2 Again magnitude of the momentum p mv 2 Ek p m m m 2 2 Ek 2 p m p m 2 Ek or 2mEk 7. 8. 9. relative velocity of the astronaut with respect to the spacecraft is 9.00 m/s – (– 0.418 m/s) = 9.42 m/s d v t d t v 22.0 m 9.42 m/s 2.34 s p = mv If you triple the velocity, you triple the momentum. p m 3v p 3mv p 3 p Inertia is a measure of mass only. It will take 2.34 s for the safety line to reach its full extension. If the object slides along a horizontal frictionless surface, there will be no acceleration (i.e., constant velocity). If the velocity remains constant, then the object’s momentum, potential energy, and kinetic energy all remain constant. 11. p m vf – v0 Note that the final velocity of the subatomic particle has a magnitude approximately equal to the initial velocity; but the direction is opposite to its original direction. i.e., if v0 = v, then vf = –v, or if v0 = –v, then vf = v. p m v v m 2v 2mv The change in momentum for the subatomic particle is 2mv. Impulse can be defined as the change in momentum. Therefore, when an object experiences an impulse, it is experiencing a change in its momentum. Not for Reproduction 55 Physics 12 SNAP ANSWERS AND SOLUTIONS 12. A perfect elastic collision is defined as a collision in which both kinetic energy and momentum are conserved. 13. impulse mv 0.25 kg vf – v0 0.25 kg –2.5 m/s – 3.0 m/s 1.4 N s The impulse on the ball was 1.4 N∙s. 14. Consider east as positive m1 1.1103 kg v1 25 km/h p1 2.75 104 kg·km/h m2 2.3 103 kg v2 –15 km/h p2 –3.45 104 kg km/h pbefore p1 p2 –7.0 103 kg·km/h m 3.4 103 kg v ? p –7.0 103 kg·km/h p m –7.0 103 kg km/h 3.4 103 kg –2.1 km/h or 2.1 km/h west v The velocity of the locked cars after the collision was 2.1 km/h west. CASTLE ROCK RESEARCH 56 Copyright Protected