Physics131_L15WI10

Physics131_L15WI10 - Physics 131 - Mechanics Le cture15 Fe...

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Physics 131 - Mechanics Lecture 15 February 17, 2010 Homeyra Sadaghiani
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October 24, 2010 Physics 131 - Lecture 15 Announcements HW#7 is due at 9 a.m. on Wed. February 24 th Tutorial Homework is due today.
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Week Date L# Lecture topic Text reading HW 5 1-Feb 10 Equilibrium, using the 2nd Law 6.1-3 (12 P)   3-Feb 11 2nd Law Application: Friction, Drag 6.4-6 (12 P) HW#4   5-Feb 12 Newton’s 3rd Law: Interaction 7.1-5 (19 P) 6 8-Feb 13 Uniform circular motion 8.1-7 (18 P)   10-Feb 14 Momentum, Impulse & conservation of momentum 9.1-3 (13 P) HW#5   12-Feb F1 Conservation of momentum Tutorial 7 15-Feb H2 Holiday (Presidents' Day)   17-Feb 15 Inelastic collisions   19-Feb 16 Energy 8 24-Feb 18 Work & kinetic energy 11.1-3 (10 P) HW#3 26-Feb E2 EXAM 2- Chapters 5-9 October 24, 2010 Physics 131 - Lecture 15 Lecture Schedule We are here!
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A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time interval and gives the cart a final speed. To reach the same speed using a force that is half as big, the force must be exerted for a time interval that is F air track cart 10090002_1 A. Four times as long. B. Twice as long. C. The same length. D. Half as long. E. A quarter as long. October 24, 2010 Physics 131 - Lecture 15 Clicker Question
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Two carts—A and B—on frictionless air tracks are initially at rest. Cart A is twice as massive as cart B. Now you exert the same constant force on both carts for 1 second. One second later, the momentum of cart A is: F air track cart A 10090002v1_4 A. Twice the momentum of cart B B. The same as the momentum of cart B C. Half the momentum of cart B D. Not enough information to determine F air track cart B October 24, 2010 Physics 131 - Lecture 15 Clicker Question
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Two identical carts, A and B, initially are moving on frictionless air tracks. The initial speed of cart A is twice as that of cart B. You then exert the same constant force on the two carts over 1 second. One second later, the change in momentum of cart A is: 10090002v2_5 F air track cart B F air track cart A A. Non-zero and twice the change in momentum of cart B B. Non-zero and the same as the change in momentum of cart B C. Zero. D. Non-zero and half the change in momentum of cart B E. Not enough information to determine October 24, 2010 Physics 131 - Lecture 15 Clicker Question
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October 24, 2010 Physics 131 - Lecture 15 Example: Hitting a Baseball (1) A 150 g baseball is thrown at a speed of 20 m/s. It is hit straight back to the pitcher at a speed of 40 m/s. The interaction force is as shown here. What is the maximum force F max that the bat exerts on the ball? a) What is the average force F av that the bat exerts on the ball?
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October 24, 2010 Physics 131 - Lecture 15 Solution: Hitting a Baseball (2) Use the impulse approximation: Neglect all other forces on ball during the brief duration of the collision.
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This note was uploaded on 10/24/2010 for the course PHY PHYSICS 13 taught by Professor Mylander during the Spring '09 term at Cal Poly Pomona.

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Physics131_L15WI10 - Physics 131 - Mechanics Le cture15 Fe...

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