# Sol-Ch09 - CHAPTER 9 E xercises 1 p = 700 kg m/s(a v = p/m...

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Exercises 1. p = 700 kg m/s. (a) v = p/m = 3.5 × 10 4 m/s ; (b) 0.47 m/s 2. (a) p = 3 × 10 5 kg m/s, so v = 250 m/s; (b) K = 4.5 × 10 4 J, so v = 86.6 m/s 3. (a) K = p 2 /2m, thus K B /K R = m R / m B = 3000; (b) p = (2mK) 1/2 , so p B /p R = (m B /m R ) 1/2 = 0.0183. 4. p x ; 0 = m(20) – m(15cos 45°) + m v 3 x’ so v 3x = –9.39 m/s p y : 0 = m(15cos 45°) + mv 3 y ; so v 3 y = –10.6 m/s v 3 = 14.2 m/s at 48.5° S of W 5. 60i = 5(2i j   ) +5v, so v = 10i +  j   m/s 6. p x : 23.96 = 6cos 52° – 6 +v 3 x’ , so v 3 x = 26.35 m/s p y : –18.05 = 6sin 53° + v 3 y’ , so v 3 y = –22.84 m/s Thus v 3 is 34.9 m/s at 40.9° S of E 7. p x : 0 = –3v 1 cos 60° + 2v 2 sin 25°; p y : –18 = –3v 1 sin 60° – 2v 2 cos 25° Solve to find v 1 = 3.1 m/s, v 2 = 5.5 m/s. 8. K f = 0.75K i = 6.75 J = 1/2 (0.5 + m)v 2 . p: 3 = (0.5 + m) v. Substitute for v into K f to find v = 4.5 m/s, then m = 167 g 9. 4.5 = 2.7v, so v = 1.85 m/s 10. p: 2u 1 = 5 V. Also K i – K f = u 1 2 – 2.5v 2 = 60 J. Use V = 0.4u 1 in the second equation to find u 1 = 10 m/s 11. p x : 0 = 3cos 37° – 4cos 53° + p 3 x’ , so p 3 x = 0.01 × 10 –24 kg m/s p y : 0 = 3sin 37° – 4sin 53° + p 3 y’ , so p 3 y = 1.39 × 10 –24 kg m/s 12. (a) (1.2) (–i + 3 j ) + (1.8)(3i + 4 j   ) = (1.2) (2i +1.5 j ) + 1.8v B . Thus v B = 1 + 5 j m/s; (b) K i = 28.5 J, K f = 27.15 J, thus ΔK = –1.35 J CHAPTER 9

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(b) 84v – 9, so v = 0.107 m/s; (c) 0.107 m/s 14. p x : u 1 = v 1 cos 30° + 2v 2 cos 45°; where v 2 = 10 m/s. p y : 0 = v 1 sin 30° – 2v 2 sin 45°, so v 1 = 2.83v 2 = 28.3 m/s Substitute v 1 into p x to find U 1 = 38.6 m/s. 15. (a) v = 2m/s, ΔK/K = –2/3 : (b) v = 4 m/s, ΔK/K = –1/3. 16. p : u = (1 + m) v. Also, (0.4) (1/2 u 2 ) = 1/2(1+m)v 2 Solve to find m = 1.5 kg 17. (a) F = – μ (m 1 + m 2 )g = (m 1 + m 2 )a, thus a = μ g = –5.88 m/s 2 . Using Δ x = 4 m n v 2 = 2aΔx, we find the speed just after the collision is v = 6.86 m/s. (b) p : 1400 u = 2400(6.86), thus u = 11.8 m/s. 18. (1500)(20) = (2500)v, so v = 12 m/s just after the collision. F : – μ (m 1 + m 2 )a, so a = –4.9 m/s 2 . Use v 2 = 2aΔx to find Δ x = 14.7 m 19. (a) p = 825i + 720 j   = 200v, thus v = 4.13i + 3.60 j m/s. (b) ΔK = 3000 – 5270 = –2970 J 20. p = 64 × 10 3 = (1.5 × 10 6 )v, so v = 4.27 cm/s 21. K α = 1/2mv α 2 = 6.72 × 10 –13 J, thus v α = 1.42 × 10 4 m/s. (a) (222) v R = 4v α , v R = 2.56 × 10 5 m/s (b) K R = 1.21 × 10 –14 J 22. (a) (5 × 10 8 )(10 4 ) = (6.0 × 10 24 )v, so v = 8.3 × 10 –13 m/s; (b) ΔK = K i = 2.5 × 10 16 J = 6.0 megatons 23. (a) p x : –164 × 10 7 + 17.4 × 10 7 = 5.8 × 10 7 v x , so v x = –25.3 km/h; p y : 47.9 × 10 7 = 5.8 × 10 7 v y , so v y = 8.26 km/h v = 26.6 km/h (7.39 m/s) at 72° W of N (b) K i = 1/2 (4.1 × 10 7 ) (11.1) 2 + 1/2(1.7 × 10 7 ) (8.33) 2 = 3.12 × 10 9
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Sol-Ch09 - CHAPTER 9 E xercises 1 p = 700 kg m/s(a v = p/m...

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