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1/3/11 Welcome Back!!!!

1/3/11 Welcome Back!!!!. Today: Review concept of work and power through lab activity. 1/4/11. Today: Review concept of work, energy, and momentum through lab activity. If you did not turn in power lab, place it in blue sorter now!

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1/3/11 Welcome Back!!!!

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  1. 1/3/11 Welcome Back!!!! Today: Review concept of work and power through lab activity.

  2. 1/4/11 Today: Review concept of work, energy, and momentum through lab activity. If you did not turn in power lab, place it in blue sorter now! Those of you who missed the test on momentum can take it in class today or tomorrow.

  3. 1/4/11 Today: Finish review of concept of work, energy, and momentum through lab activity. When completed, place it in blue sorter. Those of you who missed the test on momentum can take it in class today or tomorrow.

  4. Momentum and Collisions http://www.physicsclassroom.com/Class/momentum/momtoc.html

  5. Think about this: Which does more damage in striking a tree, an F-150 or a Mini-Cooper? • Is this always true? • What other information do you need to determine your response? • What term do you think describes this?

  6. Define Inertia • The property of any body to resist changes in its state of motion. • The measure of Inertia is: • Mass • Which of Newton’s Laws is this associated with? • First Law

  7. Momentum is inertia in motion • The linear momentum of an object of mass m moving with velocity v is defined as the product of the mass and the velocity. Momentum is represented by the symbol p.

  8. Momentum p = mv • Where • p is momentum in kgm/s • m is mass in kg • v is velocity in m/s • Is momentum scalar or vector? • Vector (direction matching that of the velocity) • SI units are kilogram x meters per second (kgm/s)

  9. p = mv • Determine the momentum of a 60-kg halfback moving eastward at 9 m/s. 540 kg m/s

  10. Example 1 • Determine the momentum of a F 150 truck moving northward at 45 mph. Assume a weight 5000 pounds. • Remember the SI unit for momentum in kg * m/s • 1 mile = 1.6 km • 1 kg = 2.2 pounds 45400 kg * m/s

  11. Example 2 • How fast (in mph) would a Mini Cooper (2500 pounds) need to be traveling to have the same momentum as the truck in example 1? 1 mile = 1.6 km • 1 kg = 2.2 pounds 90 mph you could do this in your head, just look at respective weights!

  12. Example 3 • Calculate the momentum of the Titanic, of mass 4.2 x 107 kg, moving at 14 knots (1 knot = 1.852 km/h). 3.02 x 108 kg * m/s

  13. 12/13 Almost done….. • Goal: Discuss Impulse Momentum Theory and Conservation of Momentum • Tomorrow you will be working on problem solving • Wednesday we will consider special situations • Thursday we will have a test.

  14. How can you change momentum of an object? • Change the velocity. • What term describes a change in velocity? • Acceleration • How do you change the velocity, ie cause acceleration? • Apply a Net force.

  15. As force increases, what happens to momentum? • It increases. • Will the change be instantaneous? • No. It takes time. • As time increases, what happens to momentum? • It increases.

  16. The quantity of force applied during a time interval is called Impulse • Impulse = F∆t • As impulse increases what happens to the change in momentum? • It increases • What happens to change in momentum if the impulse decreases? • It decreases

  17. Impulse-Momentum Theorem • The expression is called the impulse-momentum theorem. • Impulse = Change in Momentum • F∆t = m∆v = ∆p = mvf – mvi • What part of the equation describes the impulse? • FΔt • Is impulse a scalar or vector quantity? • Vector quantity in the direction of the force. • What units is Impulse is measured in? • Ns

  18. In other words • A net force, F, applied to an object for a certain time interval, t, willcause a change in the object’s momentum equal to the product of the force and time interval. • In simple terms, a small force acting for a long time can produce the same change in momentum as a large force acting for a short time.

  19. Increasing Momentum • Teeing Off a golf ball or Swinging at a baseball • How would you increase the momentum? • Apply the greatest force for as long as possible • In other words: FOLLOW THROUGH!!!

  20. Impulse = F∆t • Calculate the impulse when an average force of 10N is exerted upon a cart for 2.5 seconds. • 25N*s

  21. Answer example 4 and 5

  22. Example 4 • A hockey puck has a mass of 0.115 kg and is at rest. A hockey player makes a shot, exerting a constant force of 30.0 N on the puck for 0.16 s. With what speed does it head toward the goal?

  23. Example 4 • F∆t = m∆v • Solve for ∆v (vf – vi) • ∆v = F∆t/m • ∆v = (30 N)(0.16sec)/0.0115kg • ∆v = 41.7 m/s • How would the velocity of the puck change if the hockey player didn’t follow through with the shot? • Velocity would be less giving goalie more time to react and block the shot .

  24. Example 5 • If the momentum of the NASA space shuttle as it leaves the atmosphere is 3.75 x 108 kg m/s, and its mass is 75,000 kg, what is its speed? • (3.75 x 108 kg m/s) / 75,000 kg • 5000 m/s

  25. Decreasing “Momentum” • You lose control of your car. You can hit a wall or a haystack. Which do you choose? Why? • Your momentum will be decreased by same impulse with either choice since momentum = impulse. • So what is different? • Remember impulse is Force x time • Hitting the haystack increases the time, thus decreasing the Force

  26. Effect of Collision Time Upon the Force….or why a boxer “rides” the punch

  27. Example 6 • A 2200-kg sport utility vehicle traveling at 26 m/s can be stopped in 21 s by gently applying the brakes, in 5.5 s in a panic stop, or in 0.22 s if it hits a concrete wall. What is the momentum in all of these situations?What average force is exerted on the SUV in each of these stops?

  28. Example 6 2200 kg at 26m/s F∆t = m∆v • Determine momentum • mv = (2200kg)(0m/s - 26m/s) = -57200 kg*m/s • Why is momentum negative? • We are stopping. • Set momentum equal to impulse and solve for Force • At 21sec: F(21s) = -57200 kg*m/s • F = (-57200 kg*m/s)/21s = -2720 N

  29. Example 6 • At 5.5sec: F(5.5s) = -57200 kg*m/s • F = (-57200 kg*m/s)/5.5s = -10400 N • At .22sec: F(.22) = -57200 kg*m/ • F = (-57200 kg*m/s)/.22s = -260000 N

  30. Think….. • When a dish falls, will the impulse be less if it lands on a carpet than if it lands on a hard floor? • No. The impulse will be the same for either surface because the same momentum change occurs for each. Force is less on carpet because of greater time for momentum change.

  31. Momentum and Multiple Objects • In the last section, we looked at the momentum of objects one at a time. This section we will observe more than one object, and how they interact. • Take for example a game of billiards…

  32. What happens… • …when one billiard ball is knocked into another?

  33. Several Outcomes • …but all have something in common. • Either the cue ball stops completely or slows down, and, • The ball it hits goes from being still to moving at some speed.

  34. If… • If the two balls were BOTH moving, there would be even more possible outcomes, such as… • The cue ball would bounce back while the other ball went forward, • The cue ball knocks the other forward, and it keeps going forwards, too, • The cue ball and the other ball roll backwards.

  35. AND, what if… • If you imagine playing billiards with different sized balls, you get even more possible outcomes! • What if the ball you were hitting with the cue ball was a bowling ball? What would happen?

  36. Conservation of Momentum • In all of those examples, momentum is conserved. • Conservation of momentum means that in any collision, the total amount of momentum remains constant. • In other words, when two objects collide, the momentum of object A plus the momentum of object B initially is equal to the momentum of object A plus the momentum of object B at the end.

  37. Most Generally: • The total momentum of all objects interacting with one another remains constant regardless of the nature of forces between the objects.

  38. How to calculate? • Compare the total momentum of two objects before and after they interact. • The momentum of each object changes before and after an interaction, but the total momentum of the two objects together remains constant.

  39. Law of Conservation of Momentum • The law of conservation of momentum states: The momentum of any closed, isolated system does not change. • During any event, such as an explosion or a collision, individual parts of the system of objects involved may experience changes in momentum. • However, the total momentum of the system before the event must equal the total momentum of the system after the event.

  40. To solve conservation of momentum problems, use the formula: • The sum of the momenta before the collision equals the sum of the momenta after the collision. p1 + p2 = p’1 + p’2

  41. p1 + p2 = p’1 + p’2 Can be further extended to : p = momentum before collision (kg m/s) p’ = momentum after collision (kg m/s) m1 = mass of object 1 (kg) v1 = velocity of object 1 before the collision (m/s) m2 = mass of object 2 (kg) v2 = velocity of object 2 before the collision (m/s) v1’ = velocity of object 1 after the collision (m/s) v2’ = velocity of object 2 after the collision (m/s)

  42. A Two ice skaters—one guy and one girl—are standing, facing each other on a frozen pond. They push off of each other, and the guy, who has a mass of 64.2kg, travels backwards at a velocity of 5.6m/s. If the girl has a mass of 49.3 kg, what will be her velocity in the opposite direction?

  43. Insert one of these to solve for mass

  44. Example 7 • A 5.0 kg bowling ball with a velocity of 6.0 m/s strikes a 1.5 kg standing pin squarely. If the ball continues on at a velocity of 3.0 m/s what will be the velocity of the pin after the collision?

  45. Example 8 REWRITE ANSWER IN long equation set up • Before the collision: • (5kg)(6m/s) = 30 kg*m/s • (1.5kg)(0m/s) = 0 kg*m/s • Total p = 30 kg*m/s + 0 kg*m/s = 30 kg*m/s • After the collision: • Total p = 30 kg*m/s (From before collision) • (5kg)(3m/s) + (1.5kg)(? m/s) = 30 kg*m/s • (? m/s) = 30 kg*m/s – [(5kg)(3m/s)]/ (1.5kg) • 10 m/s

  46. Example 9 • A 5 kg bowling ball is rolling in the gutter towards the pins at 2.4 m/s. A second bowling ball with a mass of 6 kg is thrown in the gutter and rolls at 4.6 m/s. It eventually hits the smaller ball and the 6 kg ball slows to 4.1 m/s. What is the resulting velocity of the 5 kg ball?

  47. Example 9 • Before the collision: • (5kg)(2.4m/s) = 12 kg*m/s • (6kg)(4.6m/s) = 27.6 kg*m/s • Total p = 12 kg*m/s + 27.6 kg*m/s = 39.6 kg*m/s • After the collision: • Total p = 39.6 kg*m/s (From before collision) • (6kg)(4.1m/s) + (5kg)(? m/s) = 39.6 kg*m/s • (? m/s) = 39.6 kg*m/s – [(6kg)(4.1m/s)]/(5kg) • 3m/s

  48. Example 10 • Two people are practicing curling. The red stone is sliding on the ice towards the west at 5.0 m/s and has a mass of 17.0 kg. The blue stone has a mass of 20.0 kg and is stationary. After the collision, the red stone moves east at 1.25 m/s. Calculate the momentum and velocity of the blue stone after the collision.

  49. Example 10 • Before the collision: • (17kg)(-5m/s) = -85.0 kg*m/s • (20kg)(0m/s) = 0 kg*m/s • Total p = kg*m/s + 0 kg*m/s = -85.0 kg*m/s • After the collision: • Total p = - 85.0 kg*m/s (From before collision) • (17kg)(1.25m/s) + (20kg)(? m/s) = - 85kg*m/s • (? m/s) = -85.0 kg*m/s – [(17kg)(1.25m/s)]/ (20kg) • -5.31m/s blue stone • p = -106 kg*m/s

  50. Sample Problem 9 4 m/s

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