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Conservation of Momentum Contents: New concept Example Whiteboards:

Conservation of Momentum Contents: New concept Example Whiteboards:. 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). B. A.

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Conservation of Momentum Contents: New concept Example Whiteboards:

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  1. Conservation of Momentum • Contents: • New concept • Example • Whiteboards:

  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

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