* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

# Download Chapter 6

Introduction to quantum mechanics wikipedia , lookup

Eigenstate thermalization hypothesis wikipedia , lookup

Analytical mechanics wikipedia , lookup

Center of mass wikipedia , lookup

Monte Carlo methods for electron transport wikipedia , lookup

Atomic theory wikipedia , lookup

Wave packet wikipedia , lookup

Four-vector wikipedia , lookup

Centripetal force wikipedia , lookup

Hamiltonian mechanics wikipedia , lookup

Routhian mechanics wikipedia , lookup

Work (physics) wikipedia , lookup

Classical mechanics wikipedia , lookup

Rigid body dynamics wikipedia , lookup

Special relativity wikipedia , lookup

Matter wave wikipedia , lookup

Renormalization group wikipedia , lookup

Specific impulse wikipedia , lookup

Relativistic quantum mechanics wikipedia , lookup

Old quantum theory wikipedia , lookup

Equations of motion wikipedia , lookup

Symmetry in quantum mechanics wikipedia , lookup

Tensor operator wikipedia , lookup

Laplace–Runge–Lenz vector wikipedia , lookup

Uncertainty principle wikipedia , lookup

Classical central-force problem wikipedia , lookup

Quantum vacuum thruster wikipedia , lookup

Accretion disk wikipedia , lookup

Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup

Photon polarization wikipedia , lookup

Angular momentum wikipedia , lookup

Relativistic mechanics wikipedia , lookup

Angular momentum operator wikipedia , lookup

Newton's laws of motion wikipedia , lookup

Chapter 6 — Momentum 75 Resource CD. They are organized by textbook chapter, and each animation comes within a shell that provides information on how to use the animation, exploration activities, and a short quiz. Answers to the Conceptual Questions 1. A physical quantity is conserved when its numerical value does not change. 2. If the system is “closed” so that nothing can enter or leave, the numerical value of its mass does not change. 3. Toby is correct because it is the change in momentum that matters not the value it has at any particular time. 4. Lee is correct because it is the change in momentum that matters not the value it has at any particular time. 5. Supertankers have large inertia. 6. The child has greater momentum. Momentum is mass times velocity so no matter how large the mass, if the speed is zero the momentum is zero. 7. The net force on an object is equal to the change in its momentum divided by the time required to make the change. Or the change in momentum is equal to the impulse. 8. The momentum of every object is conserved unless acted upon by an unbalanced force. 9. An impulse (F∆t) is required to change your momentum. The padding and the air pockets lengthen the time for your leg to stop thus decreasing the size of the force experienced by the leg. 10. Padded dashboards lengthen the time for the body to stop and therefore reduce the forces involved. 11. The impulses are the same because the change in momentum is the same in both cases. 12. In both cases the change in momentum is the same. Therefore, the impulses are also the same. But in the deep powder, the interaction takes a longer time and the forces are therefore smaller. 13. Because the changes in momentum are the same, the impulses are equal. 14. The carpet lengthens the time for the egg to stop thus decreasing the sizes of the forces the egg experiences. The carpet also spreads the forces out over a larger fraction of the egg's surface. 15. The initial momentum is 8 kilogram-meters per second down. It changes to zero, so the impulse is 8 kilogram-meters per second directed up. 16. The initial momentum is 8 kilogram-meters per second down. The final momentum is directed upward so the change must be greater than 8 kilogram-meters per second. 17. Momentum is a vector. Both flowerpots have the same initial momentum. The change in momentum is greater for the pot that bounces and so it must experience the greater impulse. Jeff’s head must apply the greater force to the pot and therefore, by Newton’s third law, he must feel the greater force. 18. Momentum is a vector. Both balls have the same initial momentum. The change in momentum is greater for the ball that bounces and so it must experience the greater impulse. 19. The large padded gloves used in amateur fights lengthen the interaction time between the fist and the face, decreasing the forces involved. 20. The change in momentum and therefore the impulse is the same in both cases. In the second case the time of impact is reduced, which increases the average force. 21. At each interaction there are equal and opposite momentum changes. This includes the interactions involved in throwing and catching the ball as well as the interactions of the feet and the Earth. The total momentum of the Earth-ball-people system remains constant at all times. 22. The ball experiences three interactions; one with your hand, one with the floor and one with the Earth (gravity). The first two interactions produce impulses that cause large, rapid momentum changes for the ball (both speed and direction changes). The steady impulse of gravity causes the ball to experience 76 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. Chapter 6 — Momentum a continual slow momentum change in the downward direction. However, if the Earth-ball-you-floor system is considered, all of these impulses are equal and opposite and the total system momentum remains constant during the motion. 3 N × 5 s = 15 N⋅s and 4 N × 4 s = 16 N⋅s. The second produces the larger impulse and therefore the larger change in momentum. 3 N × 3 s = 9 N⋅s and 4 N × 2 s = 8 N⋅s. Therefore, the first case produces the larger impulse. Conservation of linear momentum requires the bullet to go one way and the rifle the other. You can reduce the recoil by reducing the momentum given to the bullet or by increasing the mass of the rifle, but you can never eliminate the recoil. So-called recoilless rifles are really tubes for firing rockets. The recoil momentum is taken up by the gases escaping from the rocket engine. Only external forces can change a system’s momentum. With or without the plate, the exhaust gases exert an external force on the rocket that should cause it to accelerate. With the plate, the exhaust gases also exert an additional (backward) force on the plate. This yields a nearly zero net external force on the rocket-plate system. Assuming that the air leaves the board moving toward the sides, the momentum transferred by the fan to the air molecules is canceled by that due to the air molecules colliding with the screen. If the molecules have some recoil, there would be a small net forward force. Momentum is only conserved for systems that have no external forces acting on them. The momentum of the Earth-ball system is conserved but not the momentum of just the ball. The Earth’s large mass allows it to acquire an equal and opposite momentum to the ball without a measurable velocity. Momentum is only conserved for systems that have no external forces acting on them. There is an external force for systems a and b, which changes their momenta. All forces in system c are internal so its momentum is conserved. Momentum is a vector. The initial momentum of the system is zero. The only way that the momentum of the two stuck together can be zero is if they are at rest. The initial momentum of the system is to the left, which means that after the collision the pair will be moving to the left. The initial momentum of the system is to the left, which means that after the collision the pair will be moving to the left. The initial momentum of the system is to the right, which means that after the collision the pair will be moving to the right. The magnitude of the total momentum stays constant at 250 kilogram-meters per second because the forces between the teacher and the board are internal to the system. The total momentum remains zero because the forces between your friend and the board are internal to the system. Therefore, the board must have a momentum of 150 kilogram-meters per second to the left. The momentum remains zero because the forces are internal. Because the initial momentum is zero, the gliders must have equal and opposite momenta. Thus, glider A with twice the mass will have half the velocity of glider B, namely 1 meter per second. As they jump toward the shore, they give the rowboat momentum away from the dock. As the astronaut pushes off the wall, the Shuttle experiences an impulse in the other direction, increasing its momentum in the direction of the impulse. The total momentum is conserved because the forces of the explosion are internal. The total momentum of the fragments changes in the same way as if the rocket never exploded. Conservation of linear momentum says that it is impossible for the astronaut to move anywhere unless something is thrown in the opposite direction. There are a variety of solutions. They all share the requirement that Al create a change in the momentum of his body, moving it toward the spaceship. Some students suggest pushing off the bag of 77 Chapter 6 — Momentum 45. 46. 47. 48. gold. A more creative solution is to throw coins away from the spaceship, one at a time. The magnitude of the momentum of the northbound object is greater than the magnitude of the momentum of the eastbound object. Therefore the final momentum must be more north than east, which is path B. The magnitude of the momentum of the northbound object is greater than the magnitude of the momentum of the eastbound object. Therefore the final momentum must be more north than east, which is path B. The magnitude of the momentum of the eastbound object is greater than the magnitude of the momentum of the northbound object. Therefore the final momentum must be more east than north, which is path D. The magnitude of the momentum of the eastbound object is equal to the magnitude of the momentum of the northbound object. Therefore the final momentum must be northeast, which is path C. Answers to the Exercises 1. p = m v = (1200 kg ) ( 30 m s ) = 36,000 kg ⋅ m s 2. pDE = m v = (120 kg ) ( 6 m s ) = 720 kg ⋅ m s pRB = m v = (100 kg ) ( 8 m s ) = 800 kg ⋅ m s Therefore, the running back has the larger momentum. (10 g ) ( 900 m s ) = 62 m s m v 3. vbb = bul bul = 145 g mbb 4. v you = m18 v18 = m you ( 24,000 kg ) (1 mph ) 80 kg = 300 mph 5. F = (1500 kg ) ( 30 m s ) = 5625 N mv ∆p = = 8s ∆t ∆t 6. F = ( 60 kg ) ( 44.7 m s ) = 89.4 N mv ∆p = = 30 s ∆t ∆t F F 89.4 N 89.4 N = = = = 0.15 2 W mg ( 60 kg ) (10 m/s ) 600 N 7. F ∆t = ∆ ( m v ) = 0 − (1400 kg ) ( 25 m s ) = − 35,000 N ⋅ s 8. F ∆t = ∆ ( m v ) = 0 − ( 0.5 kg ) ( 6 m s ) = − 3 N ⋅ s If the ball had bounced, it would have experienced a greater change in momentum, which requires a greater impulse. 9. F ∆t = ∆ ( m v ) = 0 − (1500 kg ) ( 30 m s ) = − 45,000 N ⋅ s F = 10. F = ∆ (m v ) − 45,000 N ⋅ s = = − 5625 N 8s ∆t ( 0.145 kg ) ( 50 m s ) = 181 N mv ∆p = = 0.04 s ∆t ∆t 78 Chapter 6 — Momentum 11. F = ( 0.6 kg ) (8 m s ) − ( 0.6 kg ) ( −8 m s ) = 240 N (upward) ∆p = 0.04 s ∆t 12. F = − ( 0.2 kg ) (8 m s ) − ( 0.2 kg ) (12 m s ) ∆p = = − 100 N; magnitude would be 100 N 0.04 s ∆t 13. F = 14. vgun ∆p m∆v ( 0.01 kg )( 900 m/s − 0 ) = = = 900 N 0.001 s ∆t ∆t ⎛ 0.01 kg ⎞ m = vbullet bullet = ( 800 m s ) ⎜ ⎟ = 2m s mgun ⎝ 4 kg ⎠ 15. v father = mson vson = m father ( 40 kg ) ( 3 m s ) 80 kg = 1.5 m s 17. ⎛ ⎞ 50 kg m1 = (6 m s) ⎜ ⎟ = 3.75 m s m2 ⎝ 50 kg + 30 kg ⎠ pi = p f = m1 v1 + m2 v2 = ( 3 kg ) ( 4 m s ) + ( 4 kg ) ( −3 m s ) = 0 18. pi = p f = m1 v1 + m2 v2 = ( 4 kg ) ( 4 m s ) + ( 5 kg ) ( −2 m s ) = + 6 kg ⋅ m s 19. pi = p f = m1 v1 + m2 v2 = (1200 kg ) (14 m s ) + ( 2000 kg ) ( 25 m s ) 16. v2 = v1 = 66,800 kg ⋅ m s 20. v f = pf m1 + m2 = ( right ) ( north ) 66,800 kg ⋅ m s = 20.9 m s 1200 kg + 2000 kg 21. v final = m1 v1 + m2 v2 m v − m1 v1 = 1 1 = 0 m1 + m2 m1 + m2 22. v final = m1 v1 + m2 v2 mv + 0 v 10 m s = = = = 2.5 m s m1 + m2 m + 3m 4 4 Answers to the Problems in Problem Solving 1. 2. 3. 4. I = p f − pi = m ( v f − vi ) = ( 2000 kg )(10 m/s − 20 m/s ) = −20,000 kg ⋅ m/s (+ is forward) I = p f − pi = m ( v f − vi ) = ( 6000 kg )( 30 m/s − 25 m/s ) = +30,000 kg ⋅ m/s (+ is forward) I = p f − pi = m ( v f − vi ) = ( 0.145 kg )( −60 m/s − 40 m/s ) = −14.5 kg ⋅ m/s (+ is forward) I = p f − pi = m ( v f − vi ) = ( 5 kg ) (1 m/s − ( −3 m/s ) ) = +20 kg ⋅ m/s (+ is upward) 5. ∆t = 6. ∆t = ∆ (m v ) (1600 kg ) ( 20 m s ) = 40 s = F 800 N ∆ (m v) F = (1200 kg ) ( 30 m s ) 2000 N = 18 s 79 Chapter 6 — Momentum ∆ (m v ) ( 70 kg ) ( 24 m s ) = 8400 N = ∆t 0.2 s 7. F = F F 8400 N = = = 12.2 W mg ( 70 kg ) ( 9.8 m s2 ) 8. ∆ (m v) F = ∆t 12. F = ∆ (m v) ∆t ∆ (m v) ∆t 14. vboy = 16. = 2800 N = = ( 0.145 kg ) ( 50 m s ) − ( 0.145 kg ) ( − 40 m s ) 0.05 s ( 5 kg ) (1 m s ) − ( 5 kg ) ( − 3 m s ) 0.1 s = 261 N = 200 N ( 0.2 kg ) ( 3 m s ) = 0.15 m s mb vb = 4 kg mc 13. vc = 15. v = 0.6 s ∆ ( mv ) 50 m s mv v = = = = 0.51 s 2 10 m g 10 g F max 98 m s 10. ∆t ≥ F = ( 70 kg ) ( 24 m s ) ∆ (m v) 30 m s mv v = = = = 3.06 s 9.8 m s2 Fmax mg g 9. ∆t ≥ 11. = mb vb ( 0.5 kg ) ( 25 m s ) = 0.313 m s = mboy 40 kg (80 kg ) ( 5 m s ) = 4.76 m s p mv + 0 = = 80 kg + 4 kg M M pb = ptot − pw = ( 60 kg ) ( 5 m s ) − ( 50 kg ) ( 3 m s ) = 150 kg ⋅ m s vb = pb 150 kg ⋅ m s = = 15 m s mb 10 kg ( + 2 m s ) m − (3 m s ) m = − 1 m s ptot − p2 = ( + is right ) m m ( 4 m s ) m − (3 m s ) m = + 1 m s p − pr 18. vb = tot = ( + is right ) m m The red ball cannot pass the blue ball. 19. ptotal = m1 v1 + m2 v2 = ( 2 kg ) ( 4 m s ) + ( 4 kg ) ( − 2 m s ) = 0 17. v1 = v4 = 20. ( + is right ) ptotal = m1 v1 + m2 v2 = ( 2 kg ) ( 6 m s ) + ( 4 kg ) ( − 4 m s ) = − 4 kg ⋅ m s v4 = 21. 0 − ( 2 kg ) ( − 2 m s ) ptotal − p2 = = +1m s m4 4 kg pf = − 4 kg ⋅ m s − ( 2 kg ) ( − 1 m s ) ptotal − p2 = = − 0.5 m s m4 4 kg p12 + p12 = ( + is right ) ⎣⎡ (1200 kg ) ( 25 m s ) ⎤⎦ + ⎣⎡ (1600 kg ) ( 35 m s ) ⎤⎦ = 63,500 kg ⋅ m s 2 2 80 Chapter 6 — Momentum vf = 22. pf mtot = 63,500 kg ⋅ m s = 22.7 m s 2800 kg ⎡⎣ (1200 kg ) ( 25 m s ) ⎤⎦ + ⎡⎣ (1500 kg ) ( 20 m s ) ⎤⎦ = 42, 400 kg ⋅ m s southwest pf = 2 pc2 + pt2 = 2 23. To end up traveling directly northeast after the collision, the two cars must have had the same initial momenta. Therefore, (1000 kg ) ( 30 m s ) = 21.4 m s m v ve = n n = 1400 kg me 24. ⎡⎣ ( 50 kg ) ( 6 m s ) ⎤⎦ + ⎡⎣ (100 kg ) ( 4 m s ) ⎤⎦ ∆p 500 kg ⋅ m s = = 250 N F = ∆t 2s pi = p12 + p22 = 2 2 = 500 kg ⋅ m s