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 r m m = r G G 3 2 1 y x ( ) ( ) 1 2 3 1 1 ˆ ˆ ˆ ˆ ˆ 3 3 cm r r r r i j j k = + + = + + + + G G G G k ( ) 1 ˆ ˆ ˆ 2 2 3 cm r i j = + + G k ( ) ( ) 1 2 3 1 1 ˆ ˆ ˆ ˆ ˆ 2 3 3 cm cm d v r v v v i j i j dt = = + + = + + + + G G G G G k ( ) 1 ˆ ˆ ˆ 3 2 3 cm v i j = + + G k 3 G G From eqn 7.1.3, 1 2 i i i p m v v v v = = + G G + G ˆ ˆ ˆ 3 2 p i j = + + G k 7.2 (a) From eqn. 7.2.15, 2 1 2 i i i = T m v ( ) 2 2 2 2 2 1 2 1 1 1 1 4 2 T = + + + + = (b) From Prob. 7.1, ( ) 1 ˆ ˆ ˆ 3 2 3 cm v i j = + + G k ( ) 2 2 2 2 1 1 1 3 3 2 1 2 2 2 9 cm mv = × × + + = G G G 1 3 v (c) From eqn.7.2.8, i i i L r m = × ( ) ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 L i j i j k j k i j k = + × + + × + × + + G G ˆ 2 k omentum i ( ) ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 2 L k i j i i j = − + − + = − + 7.3 b g v v v = D G G G Since m s conserved and the bullet and gun were initially at rest: 0 b g mv Mv = + G G g b v v γ = − G G , m M γ = ( ) 1 b v v γ = + D G G G 1 b v v γ = + D G 1
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1 g v v γ γ = − + D G G 7.4 Momentum is conserved: 2 blk v mv m Mv = + D D 1 2 blk v v γ = D m M γ = 2 2 2 1 1 1 2 2 2 2 2 i f v v T T mv m M γ = + D D D 2 2 1 1 1 1 2 4 mv M m 4 γ = D 3 4 4 i f i T T T γ = v 0 60 o θ v 0 / 2 7.5 At the top of the trajectory: ˆ ˆ cos60 2 v v iv i = = D D D G Momentum is conserved: 2 ˆ ˆ 2 2 2 2 v m v m im j v  = +   D D G 2 ˆ ˆ 2 v v iv j = 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 v v = + = D D 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 2 2 2 2 v v h h g g ε ε ′ = = = Similarly, the height of the second bounce, h h 2 4 h ε ε ′′ = = 2
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Total distance 2 4 2 0 2 2 1 2 n n h h h h ε ε ε = = + + + = − + Now 0 1 n n a r = = ar , 1 r < . 2 2 2 2 0 2 1 1 2 1 1 1 n n ε ε ε ε = + − + = − + = total distance 2 2 1 1 h ε ε + = For the first fall, 2 1 2 gt h = D , so 2 h g = D t For the fall from height h : 1 2 2 h h g g ε = = t Accounting for equal rise and fall times: Total time ( ) 2 0 2 2 1 2 2 1 2 n n h h g g ε ε ε = + + = − + = + Total time 2 1 1 h g ε ε + = - v 0 /2 4m v 0 m 7.7 From eqn. 7.5.5: ( ) ( ) 1 2 1 2 2 1 1 2 m m x m m x x m m ε ε + + ′ = + ± ± ± 2 ) v t ( ) ( 1 1 1 2 1 2 1 2 m m x m m x x m m ε ε + + ′ = + ± ± ± 2 v c 1 1 4 4 4 0 5 4 4 2 4 5 c v v m m v m m m v m m m  + + +   = = + D D D 2 2 v = − D 1 1 4 4 4 4 t v m m v m m v m m  + +   = + D D 2 5 15 4 4 2 5 8 v mv m v m + = = D D D
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