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Chapter 9

Chapter 9. Momentum & Its Conservation. Determining Impulse. F = ma a = D v / D t. Thus. F = m D v / D t or F D t = m D v. Impulse. The product of a force times the amount of time the force is applied. F D t. Determining Momentum. D v = v f – v i thus m D v = m v f – m v i.

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Chapter 9

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  1. Chapter 9 Momentum & Its Conservation

  2. Determining Impulse F = ma a = Dv/Dt

  3. Thus F = mDv/Dt or FDt = mDv

  4. Impulse • The product of a force times the amount of time the force is applied. • FDt

  5. Determining Momentum Dv = vf – vi thus mDv = mvf – mvi

  6. Momentum (p) • The product of mass times velocity • p = mv

  7. Change in Momentum Dp = mDv

  8. FDt = mDv • Impulse = momentum change

  9. FDt = mDv = mvf-mvi= pf-pi

  10. The Equation below is called the Impulse-Momentum TheoremFDt = pf-pi

  11. A 750 kg car is traveling east at 180 km/hr. Calculate the magnitude & direction of its momentum.

  12. A 250 kg car is traveling east at 360 km/hr. Calculate the magnitude & direction of its momentum.

  13. A 250 kg car collides with a 10.0 Mg shed & remains in contact with the shed for 0.500 s. Calculate the force of the collision & the impulse imparted onto the shed.

  14. A force of 25 N is applied to a 5.0 kg object for 5.0 seconds. Calculate: impulse, Dp & Dv:

  15. A force of 75 N is applied to a 5.0 kg object for 15.0 seconds. Calculate: impulse, Dp & Dv:

  16. A 250 kg sled is accelerated from 6.0 m/s to 18 m/s over 120 s. Calculate: a, pi, pf, Dp, & impulse

  17. A 150 g ball pitched at 40.0 m/s is batted in the opposite direction at 40.0 m/s. Calculate: Dp, & impulse

  18. A 60.0 kg man drives his car into a tree at 25 m/s. The car comes to rest in 0.20 s. Calculate: Dp & F on the man.

  19. Calculate the momentum change when a 100.0 kg block accelerates for 10.0 s down a 37o incline with a frictional coefficient of 0.25

  20. Conservation of Momentum • In a closed system, momentum is conserved • pf = pi or p1 = p2

  21. Conservation of Momentum • In collisions, momentum is conserved • (p1 + p2)b = (p1 + p2)a

  22. Book Notation of Momentum (p1 + p2)b = (p1 + p2)a (pA + pB)1 = (pA + pB)2 pA1 + pB1 = pA2 + pB2

  23. Book Notation of Momentum pA1 + pB1 = pA2 + pB2 mAvA1 + mBvB1 = mAvA2 + mBvB2

  24. Collision Momentum mAvA + mBvB = mAvA’ + mBvB’

  25. A 200. Mg freight car moving at 2.5 m/s collides with the same sized car at rest where they remain connected. Calculate vf:

  26. A 125 g hockey puck moving at 40.0 m/s is caught in a glove by a 75 kg goalie. Calculate vf of the goalie.

  27. A 35 g bullet strikes a 2.5 kg stationary block at 750 m/s. The bullet exits the block at 350 m/s.Calculate vf of the block.

  28. A 250 g ball at 4.0 m/s collides head on with a 1.0 kg ball 2.0 m/s. the 250 g ball bounced backwards at 5.0 m/s. Calculate vf of the other.

  29. A 750 g ball at 4.0 m/s collides head on with a 1.0 kg ball 5.0 m/s. The 750 g ball bounced backwards at 8.0 m/s. Calculate vf of the other.

  30. A 25 g ball at 40.0 m/s collides head on with a 2.0 kg ball 2.0 m/s. the 25 g ball bounced backwards at 50.0 m/s. Calculate vf of the other.

  31. A 250 g ball at 4.0 m/s collides head on with a 2.0 kg ball 5.0 m/s. the 250 g ball bounced backwards at 40.0 m/s. Calculate vf of the other.

  32. A 1.0 kg bat swung at 50.0 m/s strikes a 250 g ball thrown at 40.0 m/s. The bat continues at 10.0 m/s. Calculate vf of the ball.

  33. Explosion Momentum • The momentum before the explosion must = the momentum after the explosion. • The momentum before the explosion = 0

  34. Explosion Momentum • pA = pB • pB = 0 thus • pA = 0

  35. Explosion Momentum • The summation of all parts after the explosion = 0

  36. Explosion Momentum mAvA + mBvB + etc = 0

  37. Explosion Momentum with only 2 parts mAvA + mBvB = 0

  38. Explosion Momentum with only 2 parts mAvA = -mBvB

  39. A 50.0 kg gun fired a 150 g bullet at 500.0 m/s. Calculate the recoil velocity of the gun.

  40. A 500.0 Mg cannon fired a 150 kg projectile at 1500.0 m/s. Calculate the recoil velocity of the gun.

  41. A 250 g cart is connected to a 1.5 kg cart. When disconnected, a compressed spring pushes the smaller cart 4.0 m/s east. Calculate the velocity of the larger cart.

  42. A 2.0 kg block is tied to a 1.5 kg block. When untied, a compressed spring pushes the larger block 6.0 m/s east. mblock = 0.25 Calculate: vi, a, t, d for the smaller block

  43. A 5.0 kg block is tied to a 2.0 kg block. When untied, a compressed spring pushes the larger block 1.0 m/s east. mblock = 0.20 Calculate: vi, a, t, d for the smaller block

  44. Two Dimensional Collisions

  45. A 5.0 kg ball moving at 40.0 m/s collides with a stationary 2.0 kg. The 2.0 kg ball bounced at a 30o angle from the path at 50.0 m/s. Calculate vf of the other.

  46. A 2.0 kg ball is dropped from a 14.7 m high ledge collides with a stationary 10.0 kg ball hanging at a height of 9.8 m. The 2.0 kg ball bounced straight up at 4.9 m/s. Calculate vi, vf, & tair of the 10 kg ball.

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