quiz993_sol

quiz993_sol - Massachusetts Institute of Technology Physics...

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Unformatted text preview: Massachusetts Institute of Technology - Physics Department Exam #3 SOLUTIONS Physics - 8.01 Fall 1999 Problem 1 35 points 8 pts a) The diagrams: Nx Ny N N2 =m2g T2 mg 2 9 pts 9 pts 9 pts T1 T2 m2 2 T1 g m1g The contact force N on the pulley must have components Nx = T2 and Ny = m2 g + T1 2 for the pulley to stay in place. g b) T2 = m2a = m2 2 a c) R(T1 ; T2 ) = I I = 1 m2 R2 = m24R2 =R 22 g g a R(T1 ; m2 2 ) = I R = 2IR = m28gR ) T1 = m82 g + m22 g = 5 m2g 8 d) m1 g ; T1 = m1a = m1 g 2 g m1 2 = T1 = 5 m2 g 8 5 ) m1 = 4 m2 1 Problem 2 30 points 6 pts a) Since m2 m1 , the bullet will leave with the same relative speed with which it came in. V1 = V1 b) The speed of the bullet after the collision is now V1 + V2 relative to m2 . Therefore, the speed of m1 is V1 = V1 + 2V2 11 c) Momentum conservation gives the speed of both masses after the collision, V = mm+Vm2 . 1 0 8 pts 0 8 pts 0 This kinetic energy brings the system to a height given by 1 2 2 (m12+2m2 )V = (m1 + m2 )gh 1 m1 V1 = g L(1 ; cos max) 2 (m1 +m2 )2 m1+m2 p2gL(1 ; cos max) Notice for max = 0, V1 = 0 as it should! ) V1 = m1 + d) If max = 90 , then cos max = 0 and V1 = m1m1m2 p2gL, which is possible. For example, m2 if L = 1 m and m1 102 , then V1 450 m=sec. 0 8 pts � 2 Problem 3 35 points 7 pts a) p = jrp F j = ;bM g sin ~ b) Ip = Ic + M b2 = 1 M R2 + M b2 2 P c) p = Ip 1 ;bM g sin = ( 2 M R2 + M b2 ) ) + 1 Rbg+b2 sin = 0 2 2 d) Under the small angle approximation, sin 7 pts e) There must be a force at P or the CM would accelerate straight downwards. 7 pts 7 pts 7 pts , and the equation of motion is given frequency given by by + 1 Rb2g+b2 = 0 which is simple harmonic with an angular r r2 1 R2 +b2 bg . The period is given by T = 2 = 2 2 ! = 1 R2 +b2 ! bg . 2 3 ...
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