basicphy_tut1_soln

# basicphy_tut1_soln - Suggested solution of tutorial 1 1...

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Suggested solution of tutorial 1 1. Considering the Horizontal motion: Lcos (40) = ut Lcos (40) = 5 t Considering the vertical motion: Lsin (40) = at 2 2 Lsin (40) = 9 . 8 t 2 2 Solve for L and t (only L is needed though): Lsin (40) = 9 . 8( Lcos (40) 5 ) 2 2 L = 5 . 59 m 2. (a) (b) Consider vertical force equilibrium: 70 cos (30) = mg + Fcos (30) 70 cos (30) = 2 × 9 . 8 + Fcos (30) F = 47 N (c) For circular motion: F = mv 2 r F = Horizontal component of tension: 70 sin (30) + 47 sin (30) = 58 . 5 r = 1 . 5 sin (30) = 0 . 75 v = 4 . 68 m/s 3. (a) a = GM r 2 = 6 . 67 E - 11 × 5 . 98 E 24 (6370 E 3+1000 E 3) 2 a = 7 . 34 ms - 2 (b) Initial KE of the brick = mv 2 2 = m × 8000 2 2 = 3 . 2 E 7 m All KE then turn into Gravitational PE: GMm r = 3 . 2 E 7 m r = 1 . 25 E 7 m (c) i. G-Force from star1 = GM star 1 m comet (50 E 9) 2 = 6 . 67 E 7 N G-Force from star2 = GM star 2 m comet (50 E 9) 2 = 10 . 67 E 7 N Total Force = 6 . 67 E 7 2 + 10 . 67 E 7 2 = 1 . 26 E 8 In the direction of tan -

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## This note was uploaded on 09/06/2010 for the course BSC PHY1417 taught by Professor Prof during the Spring '08 term at HKU.

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basicphy_tut1_soln - Suggested solution of tutorial 1 1...

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