solns2A - v bottom 2 + 2 gh = 8.20 m/s 3) Total momentum...

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EXAM 2A – SOLUTIONS Page 1 of 1 Part II. Word Problems [30 pts each] 1) (A) At the top of the loop, F Net = N + mg = mv 2 r . Set N = 0 to obtain minimum speed v min = gr = 7.80 m/s (B) Use conservation of total mechanical energy: 1 2 kx 2 = 1 2 mv min 2 + mg (2 r ) x = 5 mgr k = 2.87 m 2) (A) E top + W fric = E bottom mg ( d sin35 ° ) μ mg cos35 ° d = 1 2 mv bottom 2 v bottom = 2 gd (sin35 ° cos35 ° ) = 2.89 m/s (B) Energy conservation from table to floor: 1 2 mv bottom 2 + mgh = 1 2 mv floor 2 v floor =
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Unformatted text preview: v bottom 2 + 2 gh = 8.20 m/s 3) Total momentum conservation: P x before = P x after (2)(3.5 10 6 ) + 2(1.8 10 6 cos64 ) = 4 v x v x = 2.14 10 6 m/s P y before = P y after (2)(0) + 2(1.8 10 6 sin64 ) = 4 v y v y = 0.809 10 6 m/s (A) v = v x 2 + v y 2 = 2.29 10 6 m/s (B) = tan 1 ( v y / v x ) = 20.7 N mg...
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This note was uploaded on 03/03/2012 for the course PHYS 152 taught by Professor Button during the Fall '08 term at IUPUI.

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