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Fowles07 - Chapter 7 Dynamics of Systems of Particles 7.1...

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Chapter 7 Dynamics of Systems of Particles z 7.1 From eqn. 7.1.1, 1 cm i i i rm m = r GG 3 2 1 y x () 123 11 ˆˆ ˆˆˆ 33 cm rr r ri j j k = + + = ++++ G G k 1 ˆ 22 3 cm j =++ G k ( ) ˆ ˆˆˆˆ 2 cm cm d vrv v v i j i j dt ==+ + =+ + + + GGG G G k 1 ˆ 32 3 cm vi j + G k 3 From eqn 7.1.3, 12 ii i p mv v v v == + G G + G ˆ p ij G k 7.2 (a) From eqn. 7.2.15, 2 1 2 i = Tm v 22 222 1 21 111 4 2 T  + + + =  (b) From Prob. 7.1, 1 ˆ 3 cm j + G k 2 2 1 2 9 cm mv =×× G 1 3 v (c) From eqn.7.2.8, i Lr m () () ( ) ˆ ˆ ˆ 2 Li ji j k jk i j k =+×++×+×+ + G G ˆ 2 k omentum i ()()( ) ˆ ˆ ˆ Lki j j =− +− + − =− + − 7.3 bg vvv D Since m s conserved and the bullet and gun were initially at rest: 0 mv Mv g b vv γ , m M = 1 b D G 1 b v v = + D G 1
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1 g v v γ =− + D G G 7.4 Momentum is conserved: 2 blk v mv m Mv  =+   D D 1 2 blk vv = D m M = 22 2 11 1 2 2 2 if TT m v m M    −= +       DD D 2 2 1 1 24 mv M m 4 D 3 44 i T v 0 60 o θ v 0 / 2 7.5 At the top of the trajectory: ˆˆ cos60 2 v vi v i == D D D G Momentum is conserved: 2 2 2 vm v m im j v G 2 2 v vj D D G Direction: 1 2 tan 26.6 v v θ D D D below the horizontal. Speed: 1 2 2 2 2 1.118 2 v = D v 7.6 When a ball reaches the floor, 2 1 2 mv mgh = . As a result of the bounce, v v ε = . The height of the first bounce: 2 1 2 mgh mv = 2 2 hh gg == = Similarly, the height of the second bounce, h ′′ 2
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Total distance 24 2 0 22 12 n n hhh h ε εε =  =+ + + = −+   Now 0 1 n n a r = = ar , 1 r < . 2 2 0 21 1 11 n n = + −+ =−+ = −− total distance 2 2 1 1 h + = For the first fall, 2 1 2 gt h = D , so 2 h g = D t For the fall from height h : 1 hh gg == t Accounting for equal rise and fall times: Total time () 2 0 12 2 1 2 n n = + + =− + Total time 1 h g + = - v 0 /2 4m v 0 m 7.7 From eqn. 7.5.5: 122 1 mm xmm x x + ′ = + ±± ± 2 ) v t ( 121 2 x x ++ ′ = + ± 2 v c 44 4 0 5 2 45 c vv vmm m v m  − + +−+   + DD D 2 2 v = − D 4 4 t v v mm v       = + D D 2 51 5 442 58 v mv m v m +− D D D Both car and truck are traveling in the initial direction of the truck with speeds 2 v D and 8 v D , respectively. 3
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7.8 From eqn. 7.2.15, 222 11 2 2 111 ii i v m v m == + Tm v Meanwhile: () 2 2 22 1 2 12 1 2 cm mv mm mv v m v v m µ +  += +  +  G G G G 2 2 1 21 2 1 2 1 2 1 2 mv mv vv mmv v vv m  =+ + + +  G GG G Therefore, cm v v 7.9 From Prob. 7.8, cm v v and since QTT =− cm cm = : Qv v From eqn. 7.5.4, v v ε = 1 1 2 Qv 7.10 Conservation of momentum: ′′ 2 1 m v m 2 2 1 2 2 2 v v v + 2 2 Conservation of energy: 2 2 2 2 212 2 1 1 m mvv v m 2 =−+ + 2 1 2 1 10 2 vm v v m +− = 2 2 v ′ = + 2 22 2 1 1 1 2 11 1 2 T T v −= = = + 2 4
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2 1 1 11 2 1 2 4 1 2 m v TT m Tm mv µ == 7.11 From eqn. 7.2.14, cm cm i i i i Lr m v rm v + × G G G GG 12 2 ii i i v rm v r m v ×= ×+ × G 2 G From eqn. 7.3.2, 2 22 1 mm m m 1 1 R rr r  + =+ = =   G G Since from eqn. 7.3.1, 2 12 1 m m =− 2 2 m R r G G 2 2 i i v Rm v v µµ ×= ×+ × G G () R vv R v −= × G G Lr cm cm m v 7.12 Let m s
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Fowles07 - Chapter 7 Dynamics of Systems of Particles 7.1...

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