A4 9th Edition Solution

# A4 9th Edition Solution - 4320 hA = 4300 km A R = 6370 km...

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Unformatted text preview: 4320 hA = 4300 km A R = 6370 km 13.87 PROBLEM 13.85 km Knowing that the velocity of an experimental space probe fired from the earth has a magnitude v A 32.32Mm/h at point A, determine the velocity 32.5 of the probe as it passes through point B. vA hB = 12 640 12700 km km B vB SOLUTION rA hA rA 10.69 Mm 10.67 Mm rB R hB 4.32Mm 6.37Mm 4.3Mm + 6.37 Mm R 72.7Mm 6.37MmMm 12.64 Mm + 6.37 rB 19.07 Mm 19.01 Mm At A, vA TA 32.5 Mm/h 9028 m/s 32.32 Mm/h = 8978 m/s 1 8978 m 9028 m/s 2 GMm rA VA rA R 10.69 Mm 10.67 Mm 6 40.302 10 m 40.752 x 10 6m gR 2m rA 10.69 10.67 106 m 6.37 106 m 6370 km 9.81 m/s 2 6.37 106 m VA At B 2 2 TB rB 12 mvB ; VB 2 19.01 Mm 19.07 Mm 37.237 37.306 106 m m 10.69 10.67 106 m gR 2m rB GMm rB 19.01 19.07 106 m 2 9.81 m/s 2 6.37 106 m m VB TA VA TB 2 vB 20.939 20.874 106 m 19.01 19.07 106 m 40.302 VB ; 40.752 106 m 2 40.302 106 40.752 2 vB vB 37.237 37.306 106 m 37.237 37.306 106 12 mvB 2 20.939 20.874 20.939 20.874 106 m 106 48.01 48.64 106 m 2 /s 2 6.929 6.9742 103 m/s 24.944 25.107 Mm/h vB 25.1 24.9 Mm/h ! 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"#'&! 250 m 250 m %&)* 0.5 m/s! 120 mm Conservation of angular momentum about O Conservation of angular momentum about O mvq (0.25) = mv¢q (0.12) 0.5 m/s "#%+&! 120 mm v¢ = q "#+'&! ! mv 0.25 v! q (0.25) = mv¢q (0.12) = 2.08 (0.5) = 1.04 m/s 0.12 v¢ = q 0.25 v! = 2.08 (0.5)== .04 m/sm/s v¢q 1 1.04 J 0.12 v¢q = 1.04 m Conservation of energy T + V = T ¢ + V¢ Conservation of energy 1 2.5 2 m (v 2 + v q ) = (0 + 0.52) = 0.3125 r 2 2 1 1 T + V = T ¢ + V¢ V= k (r – rB)2 = (750) (0.25 – 0.15)2 = 3.75 2 2 T= 1 2.5 2 T vq = 1.04 (v 2 + v q ) = 20.736) + 0.52) = 0.3125 = m m/s (0 ¢ r 2 2 1 2.5 2 T ¢ = m (v ¢ 2+ vr¢ 2V== 1(vkr2 (r – .04) 2 = 1 v(¢750)1.352) – 0.15)2 = 3.75 ) ¢ + 1 r ) = (1.25 r + (0.25 q B 2 2 2 20.736 2 1 1 =2.5 V ¢ = k ( r ¢ – r B)2 = (750 N/m) (0.12 – 0.15)2 = 0.3375 N.m 2 2 vq = 1.04 m/s ¢ 20.736 T + V = T ¢ + V ¢ : 0.3125 + 3.75 = 1.25v¢r2 + 1.352 + 0.3375 1 48.804 2 2.5 2 m (v ¢ +2 vr¢ 2 ) = (v ¢r2 + 1.04) = (1.25v ¢r + 1.352) q 0.3125 + 3.75 = 1.25v¢ r + 1.352 + 0.3375 2 2 119,522 1 2 V 1.25v¢r 2 k 2.373 rB) = ¢ = = (r ¢ – (750 N/m) (0.12 – 0.15)2 = 0.3375 N.m 2 2 4.418 m/s T¢ = ¢ ! 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