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# 20 (10) - hernandez(ejh742 oldhomework 20 Turner(58120 This...

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hernandez (ejh742) – oldhomework 20 – Turner – (58120) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Given: A uniform flexible chain whose mass is 5 . 4 kg and length is 7 m. A table whose top is frictionless. Initially you are holding the chain at rest and one-half of the length of the chain is hung over the edge of the table. When you let loose of the chain it falls downward. The acceleration of gravity is 9 . 8 m / s 2 . a 7 m 4 . 27 m Radius of table is negligible compared to the length of chain Mass of chain is 5 . 4 kg . Find the acceleration a of the chain when the length of the chain hanging vertically is 4 . 27 m . Correct answer: 5 . 978 m / s 2 . Explanation: Note: The initial condition does not enter into the consideration for the acceleration. Let : g = 9 . 8 m / s 2 , L = 7 m , = 4 . 27 m , and m = 5 . 4 kg . The linear density of the chain is λ = m L = 5 . 4 kg 7 m = 0 . 771429 kg / m . F = ℓ λ g cm The free body diagram in the vertical di- rection gives summationdisplay F y = ℓ λ g = L λ a . Therefore a = L g (1) = 4 . 27 m 7 m (9 . 8 m / s 2 ) = 5 . 978 m / s 2 . 002 (part 2 of 2) 10.0 points Find the magnitude of the velocity of the of the chain when 4 . 27 m of the chain is hanging vertically. Correct answer: 2 . 89414 m / s. Explanation: The change in kinetic energy is Δ K = 1 2 m v 2 = 1 2 λ L v 2 . (2) Let i = L 2 and f = . Using the table top as the origin of the y -coordinate and down as the positive y di- rection y cm = m on table parenleftBig 0 parenrightBig + m hanging parenleftbigg 2 parenrightbigg m on table + m hanging

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hernandez (ejh742) – oldhomework 20 – Turner – (58120) 2 y cm i = parenleftbigg L - L 2 parenrightbigg λ 0 + L 2 λ parenleftbigg L 4 parenrightbigg λ L y cm f = ( L - ) λ 0 + ℓ λ parenleftbigg 2 parenrightbigg λ L The vertical center of mass difference Δ y cm is Δ y cm = y cm f - y cm i = λ ℓ 2 - λ L 2 L 4 λ L = 1 8 L [4 2 - L 2 ] . (3) The change in potential energy is Δ U = λ L g Δ y cm = 1 8 λ g [4 2 - L 2 ] . (4) From conservation of energy Δ K = Δ U , Eq. 2 and Eq. 4, we have 1 2 λ L v 2 = 1 8 λ g (4 2 - L 2 ) v 2 = g 4 L [4 2 - L 2 ] . (5) Therefore v = radicalbigg g 4 L [4 2 - L 2 ] (6) = radicalBigg (9 . 8 m / s 2 ) 4 (7 m) [4 (4 . 27 m) 2 - (7 m) 2 ] = 2 . 89414 m / s . Alternative: You can use the kinematic expression and remember that the center of mass accelerates at g [not Eq. 1], since the acceleration of gravity is not a function of mass. v 2 = 2 a Δ y cm = 2 g 1 8 L [4 2 - L 2 ] = g 4 L [4 2 - L 2 ] , (7) where Δ y cm is obtained from Eq. 3. Note: Equation 7 is identical to Eq. 5. 003 10.0 points Two particles of masses m and 3 m are moving toward each other along the x -axis with the same speed v . They undergo a head- on elastic collision and rebound along the x - axis. m v 3 m v Determine the final speed of the heavier object.
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20 (10) - hernandez(ejh742 oldhomework 20 Turner(58120 This...

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