homework4-solutions - ME 104: Homework 4 Solutions Prof....

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ME 104: Homework 4 Solutions Prof. Karl J. Hedrick: University of California, Berkeley Chapter 13, Solution 160 (a) First collision (between A and B). The total momentum is conserved. mv A mvB v0 mv A vA mvB vB (1) Relative velocities. vA vB e v0 e vB vB vA vA (2) v0 (1 e) 2 v0 (1 e) 2 Solving Equations (1) and (2) simultaneously, vA vB (b) Second collision (between B and C). The total momentum is conserved. mvB mvC mvB mvC Using the result from (a) for vB v0 (1 e) 2 0 vB vC (3) Relative velocities. ( vB 0)e vC vB Substituting again for vB from (a) v0 (1 e) ( e) 2 vC vB (4) Solving equations (3) and (4) simultaneously, vC 1 v0 (1 e) 2 2 v0 (1 e) (e) 2 vC vB v0 (1 e)2 4 v0 (1 e2 ) 4 PROBLEM 13.160 (Continued) (c) For n spheres n balls n 1th collision, we note from the answer to Part (b) with n 3 vn v3 vC v0 (1 e)2 4 1) 1) or Thus, for n balls v3 v0 (1 e)(3 2(3 vn v0 (1 e)( n 2( n 1) 1) (d) For n 6, e 0.95, 6 1) from the answer to Part (c) with n vB v0 (1 0.95)(6 2(6 1) v0 (1.95)5 (2)5 vB 0.881 v0 Chapter 13, Solution 165 Before vA (v A ) n (v A )t vB (vB )t 0 6 m/s (6)(cos 40 ) 6(sin 40°) ( vB ) n 4 m/s After 4.596 m/s 3.857 m/s t-direction. Total momentum conserved. m A (v A )t mB (vB )t 2.314 m/s m A (vB )t 0.6 (v A )t mB (vB )t (1 kg)(vB )t ( vB ) t (0.6 kg)( 3.857 m/s) 0 (0.6 kg)(v A )t (1) Ball A alone. Momentum conserved. mA v A t mA v A t 3.857 (v A ) t (v A )t 3.857 m/s (2) Replacing (v A )t in (2) in Eq. (1) 2.314 (0.6)( 3.857) (vB )t 2.314 ( vB ) t 0 2.314 (vB )t n-direction Relative velocities. [(v A ) n (vB ) n ]e 6.877 ( vB ) n ( vB ) n ( vB ) n (v A ) n (v A ) n (v A ) n [(4.596) ( 4)](0.8) (3) PROBLEM 13.165 (Continued) Total momentum conserved. m A (v A ) n mB (vB ) n m A (v A ) n (1 kg)(vB ) n mB (vB ) n (0.6 kg)(v A ) n (0.6 kg)(4.596 m/s) (1 kg)( 4 m/s) 1.2424 (vB ) n 0.6 (v A ) n (4) Solving Eq. (4) and (3) simultaneously, (v A ) n (vB ) n 5.075 m/s 1.802 m/s Velocity of A. tan |(v A )t | |(v A ) n | 3.857 5.075 37.2 vA (3.857) 6.37 m/s vA 6.37 m/s 1.802 m/s 77.2 40 2 40 77.2 (5.075) 2 Velocity of B. vB Chapter 13, Solution 166 (v A ) n (v A )t ( vB ) n (vB )t (3 m/s) cos 20° ( 3 m/s) sin 20° ( 3 m/s) cos 20° 2.819 m/s 1.0261 m/s 2.819 m/s mA mB (3 m/s) sin 20°=1.0261 m/s t-direction Momentum of A is conserved. m A (v A )t (v A )t m A (v A )t 1.0261 (v A )t 1.0261 m/s Momentum of B is conserved. mB (vB )t ( vB )t mB (vB )t 1.0261 (vB )t 1.0261 m/s n-direction Total momentum is conserved. m A (v A ) n mA mB 2.819 2.819 (v A ) n (v A ) n ( vB ) n ( vB ) n mB (vB ) n m A (v A ) n mB ( v B ) n PROBLEM 13.166 (Continued) Relative velocities (coefficient of restitution). e1 [(v A ) n (vB ) n ]e ( vB ) n (v B ) n 5.638 5.638 (v A ) n (v A ) n [2.819 ( 2.819)](1) ( vB ) n ( vB ) n (v A ) n 2(v A ) n (v A ) n (v A ) n 2.819 m/s ( vB ) n 2.819 m/s vA vB 3.00 m/s 3.00 m/s 40 40 Chapter 13, Solution 170 Momentum in t direction is conserved. mv sin 30 (15)(sin 30 ) vt mvt vt 7.5 m/s Coefficient of restitution in n-direction. (v cos 30 )e (15)(cos 30 )(0.9) vn vn vn 11.69 m/s Writing v in terms of x and y components (v x ) 0 (v x ) 0 (v y ) 0 (v y ) 0 vn cos 30 vn sin 30 vt sin 30 6.374 m/s vt cos 30 (11.69)(cos 30 ) (7.5)(sin 30 ) (11.69)(sin 30 ) (7.5)(cos 30 ) 12.340 m/s Motion of a projectile. (origin at 0) y y0 (v y ) 0 t ( gt 2 ) 2 t2 2 y 1.2 (12.340 m/s)t (9.81 m/s 2 ) PROBLEM 13.170 (Continued) Time to reach Point B ( yB 0) 0 1.2 12.340t B 9.81 2 tB 2 tB 2.610 s x x xB xB d x0 (v x ) 0 t 0 6.374t (6.374)(t B ) (6.374 m/s)(2.610 s) 16.63 m xB 1.2 cot 60° 15.94 m d 15.94 m Chapter 12, Solution 102 Using Eq. (12.39), 1 rA 1 rB B GM h2 GM h2 A C cos A and But so that Adding, cos C cos 180 , B. B. A cos 1 rA 1 rB 1 r0 1 r1 2GM h2 Chapter 12, Solution 103 For earth, R 3960 mi 20.909 106 ft GM rA rB gR 2 (32.2)(20.909 106 ) 2 4296 mi 14.077 1015 ft 3 /s 2 3960 40.3 4000.3 mi 3960 336 21.1216 106 ft 22.6829 ft Elliptic trajectory. Using Eq. (12.39), But Adding, 1 rA 1 rA B GM h2 A C cos A and 1 rB A GM h2 cos C cos B B. 180 , so that cos 1 rB h rA rB rA rB 2GM h2 2GMrA rB rA rB (2)(14.077 1015 )(21.1216 106 )(22.6829 106 ) 43.8045 106 554.911 109 ft 2 /s (a) Speed v0 at A. v0 vA h rA 554.911 109 21.1216 106 v0 26.3 103 ft/s (vB )1 h rA 554.911 109 22.6829 106 24.464 103 ft/s For a circular orbit through Point B, (vB )circ GM rB 14.077 1015 22.6829 106 24.912 103 ft/s (b) Increase in speed at Point B. vB (vB )circ 448 ft/s (vB )1 vB 448 ft/s Chapter 12, Solution 105 First note Rearth rA rB 3960 mi 20.9088 106 ft 485.76 109 ft 202 106 mi 1066.56 109 ft 92 106 mi From the solution to Problem 12.102, we have for any elliptic orbit about the sun 1 r1 1 r2 2GM sun h2 (a) For the elliptic orbit AB, we have r1 rA , r2 rB , h hA rA v A Also, GM sun G[(332.8 103 ) M earth ] 2 gRearth (332.8 103 ) using Eq. (12.30). Then 1 rA 1 rB vA 2 2 gRearth (332.8 103 ) (rA v A ) 2 Rearth 665.6 g 103 1 1 rA r r A B 1/ 2 or 1/ 2 3960 mi 665.6 103 32.2 ft/s 2 1 202 106 mi 1066.56 109 ft 485.761 109 ft 52, 431 ft/s or vA 52.4 103 ft/s (b) From Part (a), we have 2GM sun (rA v A )2 1 rA 1 rB Then, for any other elliptic orbit about the sun, we have 1 r1 1 r2 (rA v A ) 2 1 rA 1 rB h2 For the elliptic transfer orbit AB , we have r1 rA , r2 rB , h htr rA (v A ) tr PROBLEM 12.105 (Continued) Then 1 rA 1 rB (rA v A ) 2 1 rA 1 rB [rA (v A ) tr ]2 vA 1 rA 1 rA 1 rB 1 rB 1/ 2 1 vA 1 1/ 2 or (v A ) tr rA rB rA rB 1/ 2 (52, 431 ft/s) 51,113 ft/s 1 1 202 92 202 85.5 Now Then htr ( hA ) tr ( hB ) tr : rA (v A ) tr rB (vB ) tr (vB ) tr 202 106 mi 51,113 ft/s 120,758 ft/s 85.5 106 mi For the elliptic orbit A B , we have r1 rA , r2 rB , h 1 rA 1 rB rB vB Then 1 rA 1 rB vB (rA v A ) 2 (rB vB )2 r vA A rB 1 rA 1 rA 1 rB 1 rB 1 202 106 1 164.5 106 1 92 106 1 85.5 106 1/ 2 1/ 2 or 202 106 mi (52, 431 ft/s) 85.5 106 mi 116,862 ft/s Finally, or or and or or (v A ) tr vA vA vA (51,113 52, 431) ft/s | v A | 1318 ft/s vB (vB ) tr vB vB (116,862 120, 758) ft/s 3896 ft/s | vB | 3900 ft/s Chapter 12, Solution 117 First we note R rA rB 3960 mi 20.9088 106 ft 4310 mi 4035 mi 6 (3960 350) mi 22.7568 10 ft (3960 75) mi For the circular orbit, we have vcirc gR 2 rA 6 [Eq. (12.44)] 1/ 2 32.2 ft/s 2 20.9088 10 ft 22.7568 106 ft 24,871 ft/s Now (v A ) AB vcirc vA (24,871 500) ft/s 24,371 ft/s For the elliptic descent trajectory, we have 1 r GM h2 C cos [Eq. (12.39)] Noting that Point A is at the apogee of this trajectory, we have at A, 180 : 1 rA C GM h2 GM h2 C 1 rA or at B, B 180 AOB : 1 rB C GM h2 1 rA GM h2 1 cos 1 cos 1 rB C cos 1 rB 1 rB B or B GM h2 GM h2 Then B or Now and cos B GM h2 GM h2 1 rA h ( hA ) AB rA (v A ) AB GM gR 2 [Eq. (12.30)] PROBLEM 12.117 (Continued) gR 2 GM h2 From above, Then rA (vcirc )2 rA (vcirc ) 2 [rA (v A ) AB ]2 1 rB 1 rA vcirc (v A ) AB 2 [Eq. (12.44)] vcirc 1 rA (v A ) AB 2 rA rB 2 so that cos vcirc (v A ) AB vcirc (v A ) AB 2 2 B 1 rA vcirc ( v A ) AB 1 rA 2 1 4310 mi 4035 mi 24,871 ft/s 43,371 ft/s 2 24,871 ft/s 24,371 ft/s 1 0.64411 or Finally, or B 49.901 AOB 180 49.901 AOB 130.1 Chapter 12, Solution 133 For Earth, R GM 3960 mi gR 2 20.909 106 ft, g 32.2 ft/s 2 14.077 1015 ft 3 /s 2 (32.2)(20.909 106 ) 2 4000 mi 4130 mi (a) For the elliptic orbit, rA rB 3960 40 3960 170 21.12 106 ft 21.8064 106 ft a b 1 (rA rB ) 21.5032 106 ft 2 rA rB 21.4605 106 ft GM h2 GM 2 rB A Using Eq. 12.39, and But 1 rA 1 rB C cos A C cos cos B B A 180 , so that cos B Adding, 1 rA 1 rB h rA rB rA rB GMb 2 a 2 ab h 2a b2 2GM h2 or Periodic time. 2 ab a GMb 2 2 a3/ 2 GM 5280.6 s 1.4668 h 2 (21.5032 106 )3/ 2 14.077 1015 The time to travel from A to B is one half the periodic time AB 0.7334 h AB 44.0 min (b) For the circular orbit, a circ b rB 21.4065 106 ft 2 (21.8064 106 )3/ 2 14.077 1015 5393 s 89.9 min 2 a3/ 2 GM 1.498 h circ circ ...
View Full Document

Ask a homework question - tutors are online