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CHAPTER 9a

CHAPTER 9a. Momentum. Momentum. So What’s Momentum ?. Momentum = mass x velocity This can be abbreviated to : . momentum = mv Or, if direction is not an important factor : . . momentum = mass x speed

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CHAPTER 9a

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  1. CHAPTER9a Momentum Momentum

  2. So What’s Momentum ? • Momentum = mass x velocity • This can be abbreviated to : . momentum = mv • Or, if direction is not an important factor : . . momentum = mass x speed • So, A really slow moving truck and an extremely fast roller skate can have the same momentum.

  3. Impulse and Momentum • If momentum changes, it’s because mass or velocity change. • Most often mass doesn’t change so velocity changes and that is acceleration. • And mass x acceleration = force • Applying a force over a time interval to an object changes the momentum • Force x time interval = Impulse • Impulse = F t or Ft = m v Ft = m v

  4. MOMENTUM • An object at rest has no momentum, why? • Because anything times zero is zero • (the velocity component is zero for an object at rest) • To INCREASE MOMENTUM, apply the greatest force possible for as long as possible. • Examples : • pulling a sling shot • drawing an arrow in a bow all the way back • a long cannon for maximum range • hitting a golf ball or a baseball . (follow through is important for these !) FORCE TIME

  5. MOMENTUM • SOME VOCABULARY : • impulse : impact force X time (newton.sec) . Ft = impulse • impact : the force acting on an object (N) . usually when it hits something. • impact forces : average force of impact

  6. MOMENTUM • Decreasing Momentum • Which would it be more safe to hit in a car ? • Knowing the physics helps us understand why hitting a soft object is better than hitting a hard one. Ft mv mv Ft

  7. MOMENTUM • In each case, the momentum is decreased by the same amount or impulse (force x time) • Hitting the haystack extends the impact time (the time in which the momentum is brought to zero). • The longer impact time reduces the force of impact and decreases the deceleration. • Whenever it is desired to decrease the force of impact, extend the time of impact !

  8. DECREASING MOMENTUM • If the time of impact is increased by 100 times (say from .01 sec to 1 sec), then the force of impact is reduced by 100 times (say to something survivable). • EXAMPLES : • Padded dashboards on cars • Airbags in cars or safety nets in circuses • Moving your hand backward as you catch a fast-moving ball with your bare hand or a boxer moving with a punch. • Flexing your knees when jumping from a higher place to the ground. or elastic cords for bungee jumping • Using wrestling mats instead of hardwood floors. • Dropping a glass dish onto a carpet instead of a sidewalk.

  9. EXAMPLES OF DECREASING MOMENTUM Ft = change in momentum • Bruiser Bruno on boxing … • Increased impact time reduces force of impact • Barney Jervais on bungee Jumping … Ft = change in momentum POOF ! CRUNCH ! Ft = Δmv applies here. mv = the momentum gained before the cord begins to stretch that we wish to change. Ft = the impulse the cord supplies to reduce the momentum to zero. Because the rubber cord stretches for a long time the average force on the jumper is small.

  10. Questions : • When a dish falls, will the impulse be less if it lands on a carpet than if it lands on a hard ceramic tile floor ? • The impulse would be the same for either surface because there is the same momentum change for each. It is the forcethat is less for the impulse on the carpet because of the greater time of momentum change. There is a difference between impulse and impact. • If a boxer is able to increase the impact time by 5 times by “riding” with a punch, by how much will the force of impact be reduced? • Since the time of impact increases by 5 times, the force of impact will be reduced by 5 times.

  11. Example, Page 149 • A force of 20 N acts on a 2.0 kg mass for 10 s. Compute the impulse and the change in velocity of the mass. • Impulse: • Ft = (20N)(10s) • Ft = 200 Ns • Change in velocity: • Ft = m∆v • ∆v = Ft/m • ∆v = 200 Ns/2.0 kg = 100 m/s (in direction of F)

  12. Example, Page 149 • A car that weighs 7840 N is accelerated from rest to a velocity of 25.0 m/s east by a force of 1000 N. What was the car’s change in momentum? • ∆p = m∆v • ∆p = (7840N/9.81 m/s2) (25.0 m/s-0m/s) • ∆p = 19 9979.613 kg m/s, east • ∆p = 20 000 kg m/s east

  13. Example, Page 149 • A car that weighs 7840 N is accelerated from rest to a velocity of 25.0 m/s east by a force of 1000 N. How long did the force act? • Ft = ∆p • t = ∆p/F • t = (19 979.613 kg m/s)/1000 N • t = 19.979613 s = 20 s

  14. COLLISIONS • ELASTIC COLLISIONS • INELASTIC COLLISIONS Momentum transfer from one Object to another . Is a Newton’s cradle like the one Pictured here, an example of an elastic or inelastic collision?

  15. Collisions • Regardless of the type of collision • net p before collision = net p after collision • Important! • SIGN of velocity indicates direction!

  16. Problem Solving #1 • A 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is at rest. Find the velocity of the fish immediately after “lunch”. • net momentum before = net momentum after • (net mv)before = (net mv)after • (6 kg)(1 m/sec) + (2 kg)(0 m/sec) = (6 kg + 2 kg)(vafter) • 6 kg.m/sec = (8 kg)(vafter) • vafter = 6 kg.m/sec /8 kg • 8 kg • vafter = ¾ m/sec vafter =

  17. Problem Solving #2 • Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is swimming towards it at 2 m/sec. Find the velocity of the fish immediately after “lunch”. • net momentum before = net momentum after • (net mv)before = (net mv)after • (6 kg)(1 m/sec) + (2 kg)(-2 m/sec) = (6 kg + 2 kg)(vafter) • 6 kg.m/sec + -4 kg.m/sec = (8 kg)(vafter) • vafter = 2 kg.m/sec /8 kg • 8 kg • vafter = ¼ m/sec vafter =

  18. Problem Solving #3 & #4 • Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is swimming towards it at 3 m/sec. • (net mv)before = (net mv)after • (6 kg)(1 m/sec) + (2 kg)(-3 m/sec) = (6 kg + 2 kg)(vafter) • 6 kg.m/sec + -6 kg.m/sec = (8 kg)(vafter) • vafter = 0 m/sec • Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is swimming towards it at 4 m/sec. • (net mv)before = (net mv)after • (6 kg)(1 m/sec) + (2 kg)(-4 m/sec) = (6 kg + 2 kg)(vafter) • 6 kg.m/sec + -8 kg.m/sec = (8 kg)(vafter) • vafter = -1/4 m/sec

  19. CONSERVATION OF MOMENTUM • To accelerate an object, a force must be applied. • The force or impulse on the object must come from outside the object. • EXAMPLES : The air in a basketball, sitting in a car and pushing on the dashboard or sitting in a boat and blowing on the sail don’t create movement. • Internal forces like these are balanced and cancel each other. • If no outside force is present, no change in momentum is possible.

  20. The Law of Conservation of Momentum • In the absence of an external force, the momentum of a system remains unchanged. • This means that, when all of the forces are internal (for EXAMPLE: the nucleus of an atom undergoing . radioactive decay, . cars colliding, or . stars exploding the net momentum of the system before and after the event is the same.

  21. Example! • What is the recoil velocity of a 1.20 x 103 kg launcher if it projects a 20.0 kg mass at a velocity of 6.00 x 102 m/s? • Remember, p before = p after! • Before: (1.20 x 103 kg)(0 m/s) + (20.0 kg)(0 m/s) p = 0 kg m/s • After: (1.20 x 103 kg)(v) + (20.0 kg)(6.00 x 102 m/s) = 0 kg m/s

  22. Example, Continued! • After: (1.20 x 103 kg)(v) + (20.0 kg)(6.00 x 102 m/s) = 0 kg m/s • v = (- 1.20 x 104 kg m/s) • 1.20 x 103 kg • V = - 1.00 x 101 m/s

  23. QUESTIONS • 1. Newton’s second law states that if no net force is exerted on a system, no acceleration occurs. Does it follow that no change in momentum occurs? • If no acceleration occurs, no change occurs in velocity , so there is no change in momentum. • 2. Newton’s 3rd law states that the forces exerted on a cannon and cannonball are equal and opposite. Does it follow that the impulse exerted on the cannon and cannonball are also equal and opposite? • Since the time interval and forces are equal and opposite, the impulses (F x t) are also equal and opposite.

  24. MOMENTUMVECTORS • Momentum can be analyzed by using vectors • The momentum of a car accident is equal to the vector sum of the momentum of each car A & B before the collision. A B

  25. MOMENTUMVECTORS(Continued) • When a firecracker bursts, the vector sum of the momenta of its fragments add up to the momentum of the firecracker just before it exploded. • The same goes for subatomic elementary particles. The tracks they leave help to determine their relative mass and type.

  26. CHAPTER #9 - MOMENTUM • Finish

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