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Physics 30

Physics 30. Unit 1 – Momentum and Impulse. To accompany Pearson Physics. Momentum. “quantity of motion” - Newton is the product of mass and velocity Is a vector, has units of. Momentum. Newton’s 2 nd Law, , can be rewritten as

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Physics 30

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  1. Physics 30 Unit 1 – Momentum and Impulse To accompany Pearson Physics

  2. Momentum • “quantity of motion” - Newton • is the product of mass and velocity • Is a vector, • has units of

  3. Momentum • Newton’s 2nd Law, , can be rewritten as as your text shows on page 450. We’ll use this later to determine a quantity called impulse • Try Practice Problem 1 page 451

  4. Momentum Practice Problem 1, page 451: • m = 65 kg + 535 kg = 600 kg Since velocity is a vector, direction must be given along with the numerical answer as shown on page 451

  5. Momentum • Momentum “varies directly as” or “is directly proportional to” bothStudy Example 9.2 page 452 • Try Practice Problem 1 page 452

  6. Momentum • Practice Problem 1, page 452

  7. Do questions 1 – 3, 5, 6, 8, 9, 11 page 453 Momentum

  8. Momentum and Impulse • Quick Lab 9.2 page 455 (discuss or do and discuss) • Earlier you saw that Newton’s 2nd Law could be rewritten as: This equation can be rewritten as:

  9. Momentum and Impulse • The change in momentum is equal to the net force times the time for the change, This quantity is called IMPULSE and is equal to the change in momentum

  10. Momentum and Impulse • Since , For a given change in momentum (or change in velocity if mass is constant), a fast change will mean a large force while a slower change will mean a smaller force

  11. Practical applications? Momentum and Impulse • cushioned soles on runners • airbags • crumple zones in cars • others? Additional examples p. 463 - 465

  12. Momentum and Impulse • Units of impulse: • Units of momentum: Since impulse = change in momentum, these units must be same. Your text shows this on page 457.

  13. Direction of Impulse??? Momentum and Impulse Same as the direction of the change of momentum If you are travelling west in a car at constant speed, and hit the brakes, Impulse is east If you are travelling west in a car at constant speed, and step on the gas, Impulse is west

  14. Try Practice Problems 1 and 2 page 458 Momentum and Impulse

  15. Momentum and Impulse • Practice Problem 1, p. 458

  16. Momentum and Impulse • Practice Problem 2, p. 458Since velocity is a vector, it is necessary to take direction into account when determining change in velocity

  17. When the net force is not constant, the impulse can be calculated by finding the area under an F vs t graph Your textbook has a number of examples on pages 459 – 461 Discuss golf ball graph page 462 Momentum and Impulse

  18. Review Example 9.4, page 462 Do Practice Problem 1, page 462 Momentum

  19. Momentum and Impulse c) A1 A2 b) Impulse = total area = A1+A2 • Practice Problem 1, page 462 a)

  20. Do questions 1, 2, 4, 7, 8, 9b, 10, page 467 Momentum and Impulse

  21. Note that in collisions friction is present as a constant force before, during, and after the collision Since collision times are relatively short, friction can be ignored in collision questions provided you are using only momentum of the objects immediately before or after the collision This will always be the case for you Momentum: 1d Collisions

  22. Lab 9.5 page 471 Momentum

  23. Momentum : 1d Collisions • Law of Conservation of Momentum: When no external net force acts on a system, the momentum of the system remains constant • Review pages 474 – 479including examples 9.5, 9.6, 9.7, and 9.8

  24. Try Practice Problem 1 on page 476, Practice Problem 2 on page 477, Practice Problem 1 on page 478, and Practice Problem 1 on page 479 Types of interactions: explosions including:???? hit and stick hit stationary and bounce back hit moving and bounce back Momentum : 1d Collisions

  25. Momentum : 1d Collisions • Practice Problem 1, page 476The negative sign on the velocity means it is opposite in direction to the astronaut, that is, towards the astronautFinal expression of answer???

  26. Momentum:1d Collisions • Practice Problem 2, page 477 Since initial and final velocities are in the same direction, there is no need to assign positive and negative values

  27. Momentum : 1d Collisions • Practice Problem 1 page 478 • 1.2 m/s [ W ] • Note the use of positive and negative on the velocities: + right, [E], up, [N] - left, [ W ], down, [S]

  28. Momentum : 1d Collisions • Practice Problem 1 page 479 • 0.62 m/s [E]

  29. Momentum : Elastic and Inelastic Collisions • Elastic collisions: Ek conserved • Since Ek is a scalar direction of v is unimportant • No true elastic collisions in the macroscopic world:

  30. Inelastic collisions: Ek not conserved Macroscopic collisions are always inelastic to some degree 100% inelastic – both objects are deformed and stick together Collisions between vehicles always involve deformation even when the vehicles bounce off each other Review examples 9.9 and 9.10 Do Practice Problem 1 on each of pages 484 and 485 Momentum : Elastic and Inelastic Collisions

  31. Practice Problem 1, page 484 Energy conserved in pendulum swing: Momentum conserved in collision of bullet and block Momentum : Elastic and Inelastic Collisions

  32. Practice Problem 1 page 485 Momentum : 1d Collisions

  33. Do Check and Reflect Questions, page 486 1, 2a, 3, 6, 8, 10 Momentum : 1d Collisions

  34. Real collisions are most often 3 dimensional, but the techniques to solve them are the same as those for 2 dimensional Because momentum is a vector, vector analysis is required: Momentum : 2d Collisions Components!

  35. Momentum : 2d Collisions • Look at the information available on the Physics 30 Data Sheets: • I recommend, that like in Physics 20, you measure all angles with respect to the positive x-axis

  36. If you do this, x and y components of any vector R will be found by: The angle between the x axis positive or negative and the vector will always be determined by: Momentum : 2d Collisions

  37. You can save work by drawing a good vector diagram and using the law of sines and law of cosines to solve even non-right angle triangles I don’t recommend this Review example 9.12 – I’ll redo it here Momentum : 2d Collisions

  38. Momentum : 2d Collisions • First change 12.0°[E of N] to 78.0° [N of E] (standard position) • Since both masses are the same we can ignore mass in the calculation – work with velocities only

  39. Momentum : 2d Collisions We now have everything needed to find !

  40. Momentum : 2d Collisions Since x is negative and y is positive the resultant is in the 2nd quadrant – therefore 12.9° [N of W ] or 167°

  41. Try Practice Problem 1, page 491 Momentum : 2d Collisions

  42. Momentum : 2d Collisions

  43. Momentum : 2d Collisions

  44. Practice Problem 1, page 491, continued Momentum : 2d Collisions Since both x and y are positive, this is a 1st quadrant angle: [15.7° N of E]

  45. Review Example 9.13, page 492 Try Practice Problem 2, page 492 Momentum : 2d Collisions

  46. Momentum : 2d Collisions • Practice Problem 2, page 492

  47. Since x and y are both negative this is in a 3rd quadrant angle 71.7° [S of W ] Momentum : 2d Collisions

  48. Review Example 9.14, page 494 Try Practice Problem 1, page 494 Momentum : 2d Collisions

  49. Practice Problem 1, page 494 Mass of missing 3rd fragment = 0.058 kg – 0.018 kg – 0.021 kg = 0.019 kg Momentum : 2d Collisions

  50. Momentum : 2d Collisions Practice Problem 2, page 494

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