Physics 123 Quizes and Examinations Spring 2007 Porter Johnson ...

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Solution:The momentum (horizontal) and energy are both conserved, so thatWe solve the first equation for v:mv 0 = mv+MV12 mv2 0 = 1 2 mv2 + 1 2 MV 2v = v 0 − M m VWe substitute this into the second equation to obtainv 2 0 =(v 0 − M m V ) 2+ M m V 2Thus, M = 5 kg.Mm (2V v 0 −V 2 ) =Mm( Mm) 2V 2= 2v 0 −VV10. PHYS 123 - 052 QUIZ 5 05 April 2007= 2 · 8 − 66= 5 3An M = 4000 kg block falls vertically through h = 12 meters, and thencollides with a a m = 500 kg pile, driving it x = 20 cm into solid ground,before the block comes to rest on top of the pile.Determine the average force F on the pile by the solid ground. during itscollision with the block.Note: consider the collision between the block and the pile to be completelyinelastic.Solution:The speed of the block before the collision with the pile is given by equatingits initial gravitational potential energy to its kinetic energy just before thecollision:
12 MV 2 = MghV= √ 2gh = √ 2 · 10 · 40 = √ 240 = 15.5 m/secDuring the completely inelastic collision between the block and the pile, themomentum is conserved:(M + m)V ′ = MVV ′ M=M + m V = 400 15.5 = 13.8 m/sec450The kinetic energy after the collision equals the work done against frictionF x = 1 ′2(M + m) V2F(0.2) = 1 2 (450)(13.8)2 = 4.0 × 10 5 JF= 2.0 × 10 6 N11. PHYS 123 - 051 QUIZ 6 17 April 2007A thin uniform rod of length l = 2 meters and mass m = 0.1 kg is pivotedabout a horizontal, frictionless pin put through one end. It is released at anangle θ = 60 o above the horizontal. Use energy conservation to determinethe speed v of the moving end of the rod as it passes through the horizontalposition.Solution:The initial potential energy of the rod is m g h, where h = lsinθ/2 is theheight of its center of mass above the pivoted end. As it swings through thehorizontal position, its kinetic energy is K = 1/2I ω 2 , where ω is its angularvelocity at that instant, and L is the moment of inertia of the rod about itspivot point. We divide the rod into infinitesimal slices of thickness dx andmass dm = m dx/l. The moment of inertia of the rod about its end is
Page 1: Physics 123Quizes and ExaminationsS
Page 5 and 6: 4. PHYS 123 - 052 QUIZ 2 15 Februar
Page 7 and 8: mg − T= mam(g − a) = 80m(10 −
Page 9: 8. PHYS 123 - 052 Quiz 4 22 March 2
Page 13 and 14: there is an angular acceleration α
Page 15 and 16: PHYS 123 - 051/052 TEST 1 19 Februa
Page 17 and 18: It travels a horizontal distance L
Page 19 and 20: • If the average path length for
Page 21 and 22: 3. [20 points] An m = 200 gram bloc
Page 23 and 24: (500)(300) = 100 vv = 1500 m/sThe t
Page 25 and 26: For the critical case just before i
Page 27 and 28: The total initial kinetic energy of
Page 29 and 30: (m+M) v 1 = m v 0The corresponding

Physics 123 Quizes and Examinations Spring 2007 Porter Johnson ...

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