1 / 39

Chapter 6.1 Review

Chapter 6.1 Review. 1. What is needed to cause a change in momentum?. An unbalanced force is needed to cause a change in momentum. 2. The speed of a car is tripled. By how much is the momentum changed? By how much is the kinetic energy changed?.

lillianthompson
Download Presentation

Chapter 6.1 Review

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 6.1 Review

  2. 1. What is needed to cause a change in momentum?

  3. An unbalanced force is needed to cause a change in momentum.

  4. 2. The speed of a car is tripled. By how much is the momentum changed? By how much is the kinetic energy changed?

  5. The momentum is tripled, mv.The kinetic energy is multiplied by a factor of nine, ½ mv2.

  6. 3. A ball is moving up with a speed of 8 m/s. A few seconds later, it is falling at 8 m/s. Is the momentum of the ball the same at these two points? Is the kinetic energy the same?

  7. The momentum is NOT the same. Momentum is a vector and the ball is going in two different directions.However, the kinetic energy IS the same, kinetic energy is NOT a vector.

  8. 4. A dog with a mass of 55 kg is running north with a velocity of 15 m/s. What is his momentum?

  9. p = mvp = 55 kg x 15 m/sp = 825 kgm/s

  10. 5. Toonces, a 22 kg cat is driving an 1180 kg car at a velocity of 25 m/s east. What is the total momentum of the Toonces and the car?

  11. p = mvp = (22 + 1180 kg) x 25 m/sp = 30050 kgm/s east

  12. 6. What is Toonces’ momentum in problem 5?

  13. p = mvp = (22 kg) x 25 m/sp = 550 kgm/s east

  14. 7. What is the car’s momentum in problem 5?

  15. p = mvp = (1180 kg) x 25 m/sp = 29500 kgm/s east

  16. 8. A 12 kg bird is flying 15 m/s to the north. A few moments later, the bird is flying 15 m/s to the south. Is the change in the bird’s momentum zero? If not, what is the change in the bird’s momentum?

  17. NO. The change in velocity is not zero. Since the bird is moving in the opposite direction, the velocity change is 15 + 15 m/s.p = mvp = 12 kg x 30 m/sp = 360 kgm/s south

  18. 9. Josh Hamilton hits a 38 m/s pitch. The 0.15 kg baseball leaves his bat in the opposite direction at 41 m/s. What is the change in momentum of the baseball?

  19. p = mvp = 0.15 kg x (38 + 41 m/s)p = 0.15 kg x (79 m/s)p = 11.85 kgm/s

  20. 10. How are impulse and momentum related?

  21. Impulse is equal to the change in momentum.Ft = Dmv

  22. 11. A 0.4 kg football is kicked off a tee. It leaves the foot of the kicker at 21 m/s. What impulse was delivered to the football by the kicker’s foot?

  23. Impulse is equal to change in momentum. Impulse = Dmv Impulse = 0.4 x 21 Impulse = 8.4 Ns

  24. 12. An arrow of mass 0.023 kg is moving 35 m/s. If a target slows the arrow to a stop in 0.34 seconds, what force did the target apply to the arrow?

  25. Ft = DmvF x 0.34 s = 0.023 kg x 35 m/sF x 0.34 = 0.805F = 2.37 N

  26. 13. Bengie Molina catches a 0.15 kg baseball pitched at 40 m/s. If he slows the ball to a stop in 0.15 seconds, what force did he apply?

  27. Ft = DmvF x 0.15 s = 0.15 kg x 40 m/sF x 0.15 = 6F = 40 N

  28. 14. A 50 kg ice skater throws a 15 kg box east at 10 m/s. What is the velocity of the skater?

  29. Dmv = Dmv50 x v = 15 x 1050v = 150v = 3 m/s

  30. 15. A soccer ball with a mass of 0.48 kg is moving at 20 m/s. If the ball collides with a player’s head (mass = 5 kg) and the ball rebounds in the opposite direction at 20 m/s, what is the velocity of the player’s head?

  31. mv + mv = mv + mv0.48 x 20 + 5 x 0 =0.48 x -20 + 5 x v9.6 = -9.6 + 5v19.2 = 5v v = 3.84 m/s

  32. 16. A bullet is fired at 1000 m/s to the west. The mass of the bullet is 5 grams. What is the recoil momentum of the gun?

  33. p0 = pfp = mvp = 0.005 x 1000p = 5 kgm/s

  34. 17. If the gun in question 16 has a mass of 1.1 kg, what is the gun’s recoil velocity?

  35. p = mv5 kgm/s = 1.1vv = 4.55 m/s

  36. 18. A 40 kg roller skater pushes a 60 kg roller skater. If the 60 kg skater rolls away at a velocity of 5 m/s north, what is the velocity of the 40 kg skater?

  37. Dmv = Dmv40 x v = 60 x 540v = 300v = 7.5 m/s

More Related