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homework6-solutions - ME 104 Homework 6 Solutions Prof Karl...

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ME 104: Homework 6 Solutions Prof. Karl Hedrick: University of California, Berkeley

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Chapter 14, Solution 1 There are no horizontal external forces acting during the impacts. The baggage carrier is to coast between impacts. ( a ) 15-kg suitcase tossed on carrier first: Let 1 v be the common velocity of suitcase A and the carrier after the first impact and 2 v be the common velocity of the two suitcases and the carrier of the second impact. Conservation of momentum: 1 (15)(3) (40) v 1 1.125 m/s v 20-kg suitcase tossed next: 2 (20)(2) (40)(1.125) 60 v 2 1.417 m/s v ( b ) 20-kg suitcase tossed on carrier first: Let 1 v be the common velocity of suitcase B and the carrier after the first impact and 2 v be the common velocity of all after the second impact. Conservation of momentum: 1 (20)(2) (45) v 1 0.8889 m/s   v 15-kg suitcase tossed next: 2 (15)(3) (45)(0.8889) 60 v 2 1.417 m/s   v
Chapter 14, Solution 5 The masses are m for the bullet and A m and B m for the blocks. ( a ) The bullet passes through block A and embeds in block B . Momentum is conserved. 0 0 Initial momentum: (0) (0) A B mv m m mv Final momentum: B A A B B mv m v m v 0 Equating, B A A B B mv mv m v m v 0 (6)(5) (4.95)(9) 0.0500 lb 1500 9 A A B B B m v m v m v v 0.800 oz m ( b ) The bullet passes through block A . Momentum is conserved. 0 0 Initial momentum: (0) A mv m mv 1 Final momentum: A A mv m v Equating, 0 1 A A mv mv m v 0 1 (0.0500)(1500) (6)(5) 900 ft/s 0.0500 A A mv m v v m 1 900 ft/s v

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Chapter 14, Solution 019 Let t be the time elapsed since the first collision. No external forces in the xy plane act on the system consisting of cars A , B , and C during the impacts with one another. The mass center of the system moves at the velocity it had before the collision.
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homework6-solutions - ME 104 Homework 6 Solutions Prof Karl...

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