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

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

3. Why is a bullet that is thrown not as dangerous as a bullet that is fired from a rifle?

4. Think about this: Which does more damage in striking a tree, an F-150 (Ford Truck) 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? Which requires a greater stopping force? Why?

5. Sooooo There is a relationship between Mass Velocity Force

6. Momentum is inertia “mass” in motion Momentum combines the first and second law of 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.

7. the product of mass and velocity of an object momentum = mass x velocity ρ = mv Where p is momentum in kg  m/s m is mass in kg v is velocity in m/s SI units are kilogram x meters per second (kg  m/s)

8. Vector (direction matching that of the velocity) Is momentum scalar or vector?

9. impetus was the quality of an object that was moving independent of an observed force. Impetus comes from the Latin in- + petere to go to, seek -- from Greek petesthai to fly, piptein to fall, pteron wing. Also, push and pull derive from the Latin pellere.

10. Example 1 An ostrich with a mass of 146kg is running to the right with a velocity of 17m/s. Find the momentum of the ostrich. (146 kg)(17 m/s) = 2482 kg * m/s

11. Example 2 What velocity would a 5.5g bullet have, if its momentum was the same as the ostrich in the previous problem? v = p/m 5.5g = 0.0055 kg v = 2482 kg * m/s/ 0.0055 kg v = 451,273 m/s

12. 3. A car has a momentum of 20 000 kg m/s. What would be the car's new momentum if... A. its velocity were doubled. B. its velocity were tripled. C. its mass were doubled (by adding more passengers and a greater load) D. both its velocity were doubled and its mass were doubled.

13. 3. A car has a momentum of 20 000 kg m/s. What would be the car's new momentum if... A. its velocity were doubled. doubled B. its velocity were tripled. tripled C. its mass were doubled (by adding more passengers and a greater load) doubled D. both its velocity were doubled and its mass were doubled. quadrupled

14. How can you change momentum of an object? Change the velocity. (Mass could change, but then you are changing the object) 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.

16. The quantity of force applied during a time interval is called Impulse Impulse is a change in momentum Impulse = FΔt Where F = Force and t = time What unit is impulse measured in? Ns (SI Unit) FYI: The symbol of Impulse is an I or a J. We typically just write out Impulse

17. In your head….. Calculate the impulse when an average force of 10N is exerted upon a cart for 2.5 seconds. 25N * s

18. As impulse increases what happens to momentum? It increases What happens to momentum if the impulse decreases? It decreases

19. Impulse = change in momentum aka the Impulse–Momentum Theorem aka the Impulse–Momentum Theorem FΔt = m(V f -V i ) Ns = kgm/s How can these be equal? This is and FYI FΔt = mΔv Write this by the formula!

20. Impulse = change in momentum FΔt = m(V f -V i ) So what does this mean? In simple terms, a __________ acting for a long time can produce the same change in momentum as a large force acting for a __________. Small force Short time FΔt = mΔv Write this by the formula!

21. Increasing Momentum by Increasing Velocity Applies to an object. Therefore mass is usually constant. If you increase momentum you get a greater change in velocity. Why would you want to increase momentum? List some examples NOTE: the time refers to how long the force is acting on the object

22. Example 4 A hockey puck has a mass of 0.12 kg and is at rest. A hockey player makes a shot, exerting a constant force of 30.0 N on the puck for 0.1 s. With what speed does it head toward the goal?

23. Example 4 m =.12 kg v i = 0m/s t = 0.1 s F = 30N F∆t = m∆v Solve for ∆v (v f – v i ) ∆v = F∆t/m ∆v = (30 N)(0.1sec)/0. 12kg ∆v = 25 m/s

24. Example 5 A hockey puck has a mass of 0.12 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?

25. Example 5 F∆t = m∆v Solve for ∆v (v f – v i ) ∆v = F∆t/m ∆v = (30 N)(0.16sec)/0. 12kg ∆v = 40 m/s

26. Example 6 A hockey puck has a mass of 0.12 kg and is at rest. A hockey player makes a shot, exerting a constant force of 36.0 N on the puck for 0.1 s. With what speed does it head toward the goal?

27. Example 6 F∆t = m∆v Solve for ∆v (v f – v i ) ∆v = F∆t/m ∆v = (36 N)(0.1 sec)/0. 12kg ∆v = 30 m/s

28. Example 7 A hockey puck has a mass of 0.12 kg and is at rest. A hockey player makes a shot, exerting a constant force of 36.0 N on the puck for 0.16 s. With what speed does it head toward the goal?

29. Example 7 F∆t = m∆v Solve for ∆v (v f – v i ) ∆v = F∆t/m ∆v = (36 N)(0.16 sec)/0. 12kg ∆v = 48 m/s

30. The player should use more force and follow through!!!! Apply the greatest force for as long as possible

31. We just examined ways to Increase Momentum by causing an increase in velocity Can you think of an example where mass would change and therefore increase momentum? Infiniti

32. Decreasing Velocity You are driving at 50 mph and lose control of your car. You can hit a wall or a haystack. Which do you choose? Why? How is the momentum different? It isn’t Why ? Your momentum will be decreased by same impulse with either choice since momentum = impulse.

33. Decreasing Velocity So what is different? Remember impulse is Force x time Hitting the haystack increases the time, thus decreasing the Force The change in momentum is the same regardless! Actually affecting Force or Time

35. Example 8 A 1400kg car is travelling eastward at a velocity of 15m/s, when it veers off the road and collides with a pole and is brought to rest in 0.30s. How much force is exerted on the car during the collision?

36. Example 8 F∆t = m∆v Solve for F F = (m∆v)/ ∆t F = [(1400 kg)(0-15 m/s)]/0.3 s F = -70000 N Why is F negative? It is a stopping force

37. Example 9 A 1400kg car is travelling eastward at a velocity of 15m/s, when it veers off the road and instead of colliding with a pole it collides with a barrier of sand and is brought to rest in 0.70s. How much force is exerted on the car during the collision?

38. Example 9 F∆t = m∆v Solve for F F = (m∆v)/ ∆t F = [(1400 kg)(0-15 m/s)]/0.7 s F = -30000 N How does this force compare to previous example? Smaller force: hopefully less damage to the car!

39. Can you think of other examples where you would want to decrease force?

40. Effect of Collision Time Upon the Force….or why a boxer “rides” the punch Spreading impulse out over a longer time means that the force will be less; either way, the change in momentum of the boxing glove, fist, and arm will be the same.

41. What about vertical situations? Think of an object falling: its vi is 0 m/s, it accelerates and reaches a vf max just before impact. We are considering the time frame of the stopping force.

42. 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.

44. MOMENTUM PART II and Multiple Objects This section we will observe more than one object, and how they interact.

45. Law of Conservation of Momentum The momentum of any closed, isolated system does not change. individual parts of the system 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.

46. 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.

47. 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. p 1 + p 2 = p’ 1 + p’ 2

48. p = momentum before collision (kg m/s) p’ = momentum after collision (kg m/s) m 1 = mass of object 1 (kg) v 1 = velocity of object 1 before the collision (m/s) m 2 = mass of object 2 (kg) v 2 = velocity of object 2 before the collision (m/s) v 1 ’ = velocity of object 1 after the collision (m/s) v 2 ’ = velocity of object 2 after the collision (m/s) p 1 + p 2 = p’ 1 + p’ 2 Can be further extended to :

49. Solve for v 2 ’ m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ [m 1 v 1 + m 2 v 2 - m 1 v 1 ’ ] /m 2 = v 2 ’

50. There are 2 main types of collisions Inelastic collisions: two objects stick or join after the collision. They each have the same velocities after the event. Elastic collision: two objects "bounce" apart when they collide. They each have different velocities after the event. They may or may not go in the same direction Explosions: special type. One object splits into multiple objects after explosion. Momentum before is zero. Sum of momentum after is zero.

51. The animation below portrays the inelastic collision between a 1000-kg car and a 3000-kg truck. The before- and after-collision velocities and momentum are shown in the data tables.

52. The animation below portrays the elastic collision between a 3000-kg truck and a 1000-kg car. The before- and after-collision velocities and momentum are shown in the data tables.

53. When objects collide and “STICK” Two objects collide and continue jointly in same direction This is called Inelastic Train hits Car - Train crashes into Car - Train Car Accident Shocking video – YouTubeTrain hits Car - Train crashes into Car - Train Car Accident Shocking video – YouTube

54. Example A

55. Example A: A grandma is roller skating at 6 m/s and collides with a little boy who is stationary. If grandma has a mass of 80 kg and the boy has a mass of 40 kg, what is their velocity after impact? m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ Modify formula to show they stick and that you are solving for final velocity

56. Example A Grandma is m 1 Child is m 2 m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v 3 (m 1 v 1 + m 2 v 2 )/(m 1 + m 2 ) = v 3 (80 kg)(6 m/s) + (40 kg)(0 m/s)/(80 kg + 40 kg) = v 3 v 3 = 4 m/s A grandma is roller skating at 6 m/s and collides with a little boy who is stationary. If grandma has a mass of 80 kg and the boy has a mass of 40 kg, what is their velocity after impact?

57. When objects go different directions or “BOUNCE” This is called an elastic collision Exercise Ball Fails Compilation! – YouTube Determine the formula: Indicate Opposing Directions In these ex I am assuming v 1 starts right, v 2 starts left) Then Rearrange Formula to Solve for v 2 ’

58. EX B 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 velocity of the blue stone after the collision. Determine the formula: Red in m 1, blue is m 2 Indicate Directions Then Rearrange Formula to Solve for v 2 ’

59. m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ [m 1 v 1 + m 2 v 2 – m 2 v 2 ’]/m 1 = v 1 ’ (17 kg x -5 m/s) + 0 – (17 kg x 1.25 m/s) /20 kg = -5.31 m/s

60. Explosions One object explodes (seperates) into two objects moving in opposite direction Canon Recoil Recoil Music Video - YouTube

61. Ex C A 63.0kg astronaut is on a spacewalk when his tether to the shuttle breaks. He is able to throw a 10.0kg oxygen tank away from the shuttle with a velocity of 12.0m/s. Assuming he started from rest, what is his velocity? Determine the formula: Astronaut is m 1, tank is m 2 Then Rearrange Formula to Solve for v 1 ’

62. Ex C m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ 0 = m 1 v 1 ’ + m 2 v 2 ’ – m 2 v 2 ’/m 1 = v 1 ’ – (10 kg x 12 m/s)/63 kg = -1.90 m/s

64. More Practice

65. Sample Problem A A 76kg man is standing at rest in a 45kg boat. When he gets out of the boat, he steps out with a velocity of 2.5m/s to the right (onto the dock). What is the velocity of the boat?

66. m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ [m 1 v 1 + m 2 v 2 - m 1 v 1 ’]/m 2 = v 2 ’ 0 + 0 – (76 kg x 2.5 m/s)/45 kg = -4.22 m/s

67. Sample B 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?

68. Sample B (5kg)(2.4m/s) + (6kg)(4.6m/s) = (6kg)(4.1m/s) + (5kg)(? m/s) (? m/s) = (39.6 kg * m/s – 24.6 kg * m/s)/(5kg) 3m/s