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Conservation of Momentum Contents: New concept Example Whiteboards: Brennen and the railcars

Learn about the concept of conservation of momentum and how to apply it with different examples. Understand why momentum is conserved and how to calculate the velocities of objects involved in collisions.

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Conservation of Momentum Contents: New concept Example Whiteboards: Brennen and the railcars

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  1. Conservation of Momentum • Contents: • New concept • Example • Whiteboards: • Brennen and the railcars

  2. Conservation of Momentum - Definition For a given interaction between objects, the net change in momentum is Zero. Or Momentum is conserved. (It don’t change)

  3. B A Conservation of Momentum - Why is it conserved? pA = FBt pB = FAt FB = -FA t = t pA = -pB Total p = pA+ pB = 0

  4. Conservation of Momentum - Using it. Example: 60. kg Sally going 4.5 m/s collides head on with 80. kg Bob who is going 2.3 m/s. What is the velocity of their embracing bodies?

  5. + Step 1:Draw a picture (decide +/-) Conservation of Momentum - Using it. 4.5 m/s 60 kg 2.3 m/s 80 kg v = ? (80. kg)(2.3 m/s) + (60. kg)(-4.5 m/s) = (140. kg)v Step 2: Write down the momentum of every object. (p = mv) Step 3: Solve for the unknown

  6. Conservation of Momentum - Using it. + 4.5 m/s 60 kg 2.3 m/s 80 kg v = ? (80 kg)(2.3 m/s) + (60 kg)(-4.5 m/s) = (140 kg)v 184 kgm/s + -270kgm/s = (140 kg)v -86kgm/s = (140 kg)v (-86kgm/s)/(140 kg) = -.61 m/s They are moving in Sally’s direction (to the left) at .61 m/s They live happily ever after

  7. Whiteboards

  8. + Before After 3.2 m/s 1.2 m/s v = ? 4.0 kg (Stuck together) 5.0 kg (5.0kg)(3.2m/s) + (4.0kg)(1.2m/s) = (9.0kg)v 20.8kgm/s = (9.0kg)v v = (20.8kgm/s)/ (9.0kg) = 2.3 m/s 2.3 m/s

  9. + Before After 0 m/s 0 m/s v = ? 3.6 m/s 17 kg 17 kg 13 kg 13 kg 0 = (13kg)v + (17kg)(3.6m/s) 0 = (13kg)v + 61.2kgm/s -61.2kgm/s = (13kg)v (-61.2kgm/s)/(13kg) = -4.7 m/s = v -4.7 m/s

  10. + Before After 16.0 m/s 10.0 m/s v = ? 281 g (Stuck together) 231 g (.231kg)(16m/s) + (.281kg)(-10m/s) = (.512kg)v .886 kgm/s = (.512kg) v (.886 kgm/s)/ (.512kg) = 1.73 m/s 1.73 m/s

  11. + Before After 6.20m/s 1.20 m/s v = ? 17.0 kg 17.0 kg 13.0 kg 13.0 kg (13kg+17kg)(6.2m/s) = (13kg)(-1.2m/s)+(17kg)v 186kgm/s = -15.6kgm/s+(17kg)v 201.6kgm/s = (17kg)v (201.6kgm/s)/(17kg) = 11.9 m/s = v 11.9 m/s

  12. A 3452 kg truck going 23.0 m/s collides with a 1253 kg car at rest. What is the velocity of the wreckage just after the collision? (3452kg)(23.0m/s)=(3452kg+1253kg)v 79396kgm/s = (4705kg)v (79396kgm/s)/(4705kg) = 16.9 m/s 16.9 m/s

  13. A 215 gram bullet going 542 m/s goes through a stationary 3.40 kg block of wood. After this, the block is traveling at 12.0 m/s in the same direction, (a) what is the bullet’s velocity? (. 215kg)(542 m/s) = (3.4kg)(12m/s) + (.215kg)v 116.53kgm/s = 40.8kgm/s + (.215kg)v 75.73kgm/s = (.215kg)v (75.73kgm/s)/(.215kg) = 352 m/s 352 m/s

  14. A 215 gram bullet going 542 m/s goes through a stationary 3.40 kg block of wood. After this, the block is traveling at 12 m/s in the same direction, (b) if ripping through the block took .0250 s, what force did the bullet exert on the block? use mv = Ft: (3.4 kg)(12m/s) = (F)(.025 s) F = (3.4 kg)(12m/s)/(.025 s) = 1632 N = 1630 N 1630 N

  15. A 2450. kg airplane (including ammunition) going 52.40 m/s fires a 5.120 kg shell forward at 890.0 m/s. What is the velocity of the plane afterward? (2450kg)(52.40m/s) = (2444.88 kg)v+(5.12kg)890m/s 128,380kgm/s = (2444.88 kg)v+4556.8kgm/s 123,823.2kgm/s = (2444.88 kg)v (123,823.2kgm/s)/(2444.88 kg) = 50.65 m/s 50.65 m/s

  16. A 153 gram bullet going 452 m/s goes through the first of two stationary 3.50 kg blocks of wood, and sticks in the second. After this, the first block is traveling at 6.50 m/s in the same direction, what speed are the second block and bullet going? (.153kg)(452m/s) = (3.5kg)(6.5m/s)+(.153kg+3.5kg)v 69.156kgm/s = 22.75kgm/s + (3.653kg)v 47.406kgm/s = (3.653kg)v (47.406kgm/s)/(3.653kg) = 12.7 m/s 12.7 m/s

  17. A 153 gram bullet going 452 m/s goes through the first of two stationary 3.50 kg blocks of wood, and sticks in the second. After this, the first block is traveling at 6.50 m/s in the same direction, What is the bullet’s velocity between the blocks? (.153kg)(452m/s) = (3.5kg)(6.5m/s)+(.153kg)v 69.156kgm/s = 22.75kgm/s + (.153kg)v 46.406kgm/s = (.153kg)v (46.406kgm/s)/(.153kg) = 303.31 = 303 m/s 303 m/s

  18. A 153 gram bullet going 452 m/s goes through two stationary 3.50 kg blocks of wood. After this, the first block is traveling at 6.50 m/s and the second is going 8.30 m/s in the same direction. What is the bullet’s velocity after this? (.153kg)(452m/s)= (3.5kg)(6.5m/s)+(3.5kg)(8.3m/s)+(.153kg)v 69.156kgm/s = 22.75kgm/s+29.05kgm/s+(.153kg)v 17.356kgm/s = (.153kg)v (17.356kgm/s)/(.153kg) = 113.44 = 113 m/s 113 m/s

  19. A 153 gram bullet going 452 m/s is heading toward two stationary 3.50 kg blocks of wood. What is the velocity if the bullet sticks in the first block, and the two blocks then slide and stick together? (.153kg)(452m/s)= (3.5kg+3.5kg+.153kg)v 69.156kgm/s = (7.153kg)v (69.156kgm/s)/(.153kg) = 9.668 = 9.67 m/s 9.67 m/s

  20. B A 60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track 1. What is the velocity of car A when he is in midair? 0 = (60kg)(+5.2 m/s) + (560 kg)v v = -.557 m/s -.56 m/s

  21. B A 60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track 2. What is the velocity of car B when he leaves it? (60 kg)(5.2 m/s) + 0 = (60 kg)(3.4 m/s) + (780 kg)v v = +.138 m/s +.14 m/s

  22. B A 60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track 3. What would have been the velocity of car B had he remained there, and not jumped off? (60 kg)(5.2 m/s) + 0 = (60 kg+ 780 kg)v v = +.371 m/s +.37 m/s

  23. B A 60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track 4. What would the velocity of car B have been had he jumped off the back of it to give himself a velocity of zero? (60 kg)(5.2 m/s) + 0 = (60 kg)(0 m/s)+ (780 kg)v v = +.40 m/s +.40 m/s

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