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121ss06m2_sol - Phy 121 Spring 2006 Midterm 2 Solutions...

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Phy 121 Spring 2006 Midterm 2 Solutions: Exam A=B=C=D 1. (1) (D) R GmM / - (2) (E) ) /( h R GmM + - (3) (C) 2 1 / 2 GmM R mv - + (4) (A) 2 1/ 2 ' /( ) mv GmM R h - + (5) (E) 2 9 1 ' / /( ) 2.9 10 2 mv GmM R GmM R h J + - + + × ( 6) (C) 2 9 1 ' (1/ 1/( )) 2.9 10 2 mv GmM R R h J = + - + + × = 11 24 2 9 9 6 1 6.67 10 12000 6.0 10 12000 580 (1/ 6.4 1/(6.4 0.5)) 2.9 10 59.3 10 2 10 J J - × × × × = × × + × - + + × = × 9 2 2 59.3 10 3144 / 12000 KE v m s m × × = = = 2. (7) (B) The total momentum is conserved (8) (B) The total mechanical energy is conserved (9) (D) ( ) ' mv m M v = + (10) (A) 2 1 ( ) ' ( ) 2 m M v m M gh + = + (11) (C) 2 2 2 2 1 1 0.085 61 ( ) ( ) 0.26 2 2 0.085 2.2 9.81 m v m m M g = = + + 3. (12) (D) 2 1 2 MR (13) (A) 2 1/ 2 MR τ (14) (E) t α (15) (C) 2 2 1/ 4 M R ϖ (16) (C) 2 2 2 2 2 2 2 2 2 2 2 1 1 2.3 4.7 ( ) ( ) 818.1 4 4 2.7 0.23 rot t KE MR t MR t J MR MR τ τ α × = = = = = 1/2 × 1
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4. (17) ( D) The magnitude of the displacement is minimal at 1 and 3 s (18) (A) 0 0 2 2 2 cos( ) cos( ) a t v t T T T π π π - = - (19) (D) 2 11.1 / m s (20) (D) The total energy of the system is equal to the potential energy at 0, 2 and 4 s , where the kinetic energy is zero. (21) (B) 2 2 0 0 1/ 2 1/ 2 ( ) 2 T k x k v π = = (22) (B) 2 2 0 1 1 0.86 10 43.0 2 2 tot E mv J = = × × = ( or complicated from step (21): 2 2 2 0 0 0 2 2 1 1 1 ( ) ; 2 4 ; 4 ( ) 2 2 2 2 2 tot tot T m m m T E k v
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