An Introduction to Momentum (Doodle Science)

Transcription

1 Momentum

2 An Introduction to Momentum (Doodle Science)

3 Intro to Momentum part one

4 Momentum Momentum is a way of describing the inertia of an object in motion. Momentum = Mass x Velocity P = m v When direction isn t important, we can say Momentum = Mass x Speed Unit is mass unit x speed unit (exp: kg. m/s) Momentum increases with either an increase in mass or in speed. Momentum is a vector quantity

5 Momentum: FYI regarding units P = m v Momentum = Mass x Speed Momentum is measured in kg. m/s A Newton of force is the same amount of force necessary to accelerate a 1 kg object at 1 m/s 2 therefore 1 N = 1 kg m/s 2

6 Circle diagram for the Momentum formula P = mv P m = P V V = P M M V Units most commonly used: Momentum = kg x m/s Mass = kg Velocity = m/s P = Momentum M = Mass V = Velocity

7 Momentum The greater the amount of momentum, the harder it is to stop something that is moving. Since momentum requires a value for both velocity (or speed) and mass, if an object isn t in motion it has no momentum.

8 Intro to Momentum part two

9 Conservation of Momentum Law of Conservation of Momentum states: In the absence of a net external force, the momentum remains unchanged. The momentum of an object doesn t change unless its mass, velocity, or both change. Momentum, however can be transferred from one object to another. consider a cue ball transferring some of its momentum to the other pool balls during the break, the momentum that the balls gained is equal to the momentum that the cue ball lost. If no other forces act upon the balls the total momentum is conserved it isn t lost or gained.

10 Law of Conservation of Momentum

11 Conservation of Momentum continued. Collisions between objects are governed by laws of momentum and energy. when a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the objects, the momentum of all objects before the collision equals the momentum of all objects after the collision. If there are only two objects involved in the collision, then the momentum lost by one object equals the momentum gained by the other object. The total system momentum is 60 kg*cm/s. The momentum of the cart-dropped brick system is conserved. The momentum lost by the cart (40 kg*cm/s) is gained by the dropped brick.

12 Stacked Ball Drop Conservation of Momentum

13 Momentum and Isolated Systems For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2. Total system momentum is conserved if the system is isolated A system is a collection of two or more objects, and it is isolated if there isn t the presence of a net external force that alters the momentum of the system. In real life, there almost always is a net external force, but the law remains a good tool for approximating speeds after collisions

14 Elastic Collisions and Momentum An elastic collision occurs whenever two objects bounce apart when they collide. In general, elastic collisions are characterized by a large velocity change, a large momentum change, a large impulse, and a large force. The total momentum before and after the collision is conserved. The animation below portrays the elastic collision between a 1000-kg car and a 3000-kg truck. The before- and after collision velocities and momentum are shown in the data tables.

15 Conservation of Momentum in Elastic Collisions Yoga Ball Mayhem!

16 Inelastic Collisions, Energy, & Momentum An Inelastic Collision occurs whenever two objects collide and do not bounce away from each other. The objects act as one object once they collide. The total momentum before and after the collision is conserved. Inelastic Collisions have less of an impact force than elastic collisions Kinetic Energy, KE, (energy of movement) is not conserved as some of the kinetic energy that was present before the collision is transferred into heat due to friction. If a larger object strikes a smaller object, it doesn t experience much loss of KE a car striking a fly isn t effected much, but the fly the smaller item striking the larger item the car, is effected much more and slowed more

17 Inelastic Collisions and Momentum The animation below portrays the elastic collision between a 1000-kg car and a 3000-kg truck. The before- and after collision velocities and momentum are shown in the data tables.

18 Inelastic vs. Elastic Collisions involving rubber balls a demonstration of differences in impact force

19 Tom Cruise in Mission: Impossible 2 demonstrates an Inelastic Collision

20 Example 1: Inelastic Collision Momentum Before = Momentum After P of ball + P of skater = P of ball with skater (15 kg x 20 km/hr) + (60 kg x 0) = (15 kg + 60 kg) x V 300 kg km/hr = 75 kg x V 4 km/hr = V

21 Example 2: Elastic Collision Momentum before = Momentum after P Car A + P Car B = P Car A + P Car B (1000 kg x 10 m/s) + (2000 kg x -2 m/s) = (1000 kg x V car A ) + ( 2000 kg x 4m/s) kg m/s kg m/s = (1000 kg x V car A ) kg m/s 6000 kg m/s = (1000 kg x V car A ) kg m/s kg m/s = 1000 kg x V car A -2 m/s = V car A

22 Example #3: Elastic Collision A 30 kg figure skater threw a 2 kg ball to the left resulting in her moving at 3 m/s to the right, what was the velocity of the ball after it was thrown. Momentum before = Momentum after 0 = (2 kg x V ball ) + ( 30 kg x 3 m/s) 0 = (2 kg x V ball ) + 90 kg m/s 0-90 kg m/s = 2 kg x V ball - 90 kg m/s = 2 kg x V ball 90 kg m/s 2 kg = V ball -45 m/s = V ball

23 Example 4 : Inelastic Collision A 0.15-kg baseball moving at a speed of 45 m/s crosses the plate and strikes the 0.25-kg catcher's mitt (originally at rest). The catcher's mitt immediately recoils backwards (at the same speed as the ball) before the catcher applies an external force to stop its momentum. Determine the post-collision velocity of the mitt and ball. Momentum Before = Momentum After P of ball + P of mitt = P of ball and mitt (0.15 kg x 45 m/s) + (0.25 kg x 0) = (0.15 kg kg) x V 6.8 kg m/s = 0.4 kg x V 17 m/s = V

24 Impulse Physics

25 In order to stop any object with momentum, a force must be applied against it over a length of time. The change in momentum in a period of time is known as Impulse. For momentum to change there must be a force applied. Force x time = Impulse Impulse Force x Time = mass x Δ velocity or you could simplify the formula as F. T = Δ MV The equation is known as the impulse-momentum change equation In the metric system, Impulse is commonly expressed in Newton seconds, N. s (Momentum is usually given in kg. m/s)

26 IMPULSE CONTINUED If the change in momentum occurs over a short period of time (like if a car crashed into a brick wall), the momentum is concentrated and so the force of impact that is experienced will be larger. If the change occurs over a longer period of time the force experienced would be smaller. Impulse is a vector quantity

27 Impulse and an egg

28 The Impulse Formula as seen on the STAAR Formula Chart Note: a capital J is used to denote Impulse in formulas but don t confuse it with the J for Joule

29 Impulse and Change in Momentum Remember: Impulse = change in momentum Impulse = Δp Therefore if the impulse was 20 Ns then the change in momentum was 20 kg * m/s So in problems, when necessary, you can use change in momentum in place of impulse and vice versa

30 Impulse and Change in Momentum Recall from earlier notes this year that when we found the change in something (velocity, time, speed, etc.) we found it by subtracting the initial value from the final value Momentum = Final Momentum Initial Momentum Δp = P 2 P 1 Note: change in momentum is also equal to mass x change in velocity Δp = m x Δv = m (v 2 v 1 )

31 Circle diagram for the Impulse formula J = Ft J t = J F F = J t F t Units: Impulse = N x s Force = N time = s J = Impulse F = force t = time

32 Impulse & Δp Note: since J = Δp, it can replace J in the circle diagram Δp = Ft Δp t = Δp F F = Δp t F t Units: Δp = kg x m/s Force = N time = s Remember that 1 N = 1 kg x m/s 2 Δp= change in momentum F = force t = time

33 Important Unit Conversion Info: time t = Δp F t = t = t = kg xm/s N kg xm /s kg x m/s 2 kg xm /s kg x m/s s t = kg x m/s x Remember: when Δp is in kg x m/s and force is in N, that time must be in seconds! s kg x m/s t = seconds

34 Important Unit Conversion Info: Force F = Δp t F = kg xm/s s kg x m s F = s Note: when Δp is in kg x m/s and time is in seconds, that force must be in Newtons! Remember that 1 N = 1 kg x m/s 2 F = kg x m/s 2 F = Newtons

35 Important Unit Conversion Info: Change in Momentum or Impulse Δp = F x t Δp = N x s Δp = kg x m/s 2 x s Δp = kg xm/s s Δp = kg x m/s x s Remember that a N x s is the unit for Impulse Note: when force is in Newtons and time is in seconds, that change in momentum must be in kg x m/s

36 Impulse Examples (finding force when given mass, change in velocity, & time) Example 1: Jennifer, who has a mass of 50.0 kg, is riding at 35.0 m/s in her red sports car when she must suddenly slam on the brakes to avoid hitting a deer crossing the road. She strikes the air bag, that brings her body to a stop in s. What average force does the seat belt exert on her? F = (mass * velocity change)/time F = (50 * 35) / F = 3500 N

37 Impulse Examples (finding force when given mass, change in velocity, & time) Example 2: If Jennifer had not been wearing her seat belt and not had an air bag, then the windshield would have stopped her head in s. What average force would the windshield have exerted on her? F = (mass * velocity change)/time F = (50 * 35)/0.002 F = N Note that a 250-fold decrease in the time corresponds to a 250-fold increase in the force.

38 Relationship of Force and Time Force and Time are inversely related. In the examples below, note that 100 units of impulse are required to stop 100 units of momentum Many combinations of force and time could be used to produce the impulse Combinations of Force and Time Required to Produce 100 units of Impulse Force Time Impulse

39 Rebounding vs. Sticking Rebounding is a specific type of collision which involved both direction and velocity change. When the direction changes, change in velocity (Δv) is found by V f - V i In general, elastic collisions result not only in large velocity change, but also large change in momentum, large impulse, large amounts of force compared to inelastic collisions. Bouncing (elastic) collisions are more forceful than sticky (inelastic) collisions.

40 Impulse Practice (solving for impulse & for force) 1 2 #1) Impulse = Force x Time Impulse = 1200 N x 0.1 s Impulse = 120 Ns #2) Impulse = Force x Time 120 Ns = F x 0.65 s N = F 180 N F

41 Impulse Practice solving for time The impulse of a ball leaving a bat is 1.26 N * s and the force delivered by the bat is 2520 N. How long was the bat in contact with the ball? Impulse = Force x Time T = Impulse / force t = J f 1.26 Ns t = 2520 N t = seconds t = 5 x 10 4 sec

42 Momentum Review Problem 1 Total Momentum Before = Total Momentum After (3000 kg x 10 m/s) + (1000 kg x 0 m/s) = ( 3000 kg x V truck ) + (1000 kg x 15 m/s) (30000 kgm/s) + (0 kgm/s) = (3000 kg x V truck ) + (15000 kgm/s) kgm/s = (3000 kg x V truck ) + (15000 kgm/s) kgm/s = 3000 kg x V truck 5 m/s = V truck

43 Momentum Practice Problem 2 Beginning mass of loaded cart = 50. kg Mass of brick = 20. kg Initial velocity of loaded cart = 22. m/s Initial velocity of brick = 0 m/s What is the final velocity of cart and bricks? (50. kg x 22. m/s) + (20 kg x 0 m/s) = (50. kg kg) x V final 1100 kgm/s = 70. kg x V final m/s = V final 16 m/s V final

44 Momentum Practice Problem 3 A B Two ice skaters, initially at rest, push off of one another. Skater A has a velocity of 2 m/s. Skater B has a mass of 60 kg and a velocity of 3m/s. What is the mass of Skater A? Momentum Before = Momentum After 0 = (m x 2m/s) + ( 60kg x 3 m/s) 0 = (m x 2m/s) kgm/s kgm/s = m x -2m/s 90 kg = m

45 Impulse & Change in Momentum Example: Part One finding P What is the change in momentum of a 40 kg runner that travels from 5 m/s to 11 m/s? P = P final - P initial P = (40 kg x 11 m/s ) - ( 40 kg x 5 m/s) P = (440 kg m/s ) - ( 200 kg m/s) P = 240 kg m/s

46 Impulse & Change in Momentum Example: Part Two finding force What was the force applied by the runner if the change in momentum took 1.5 seconds? P = 240 kg m/s J = P J = FT FT = P FT = 240 kg m/s F (1.5 sec) = 240 kg m/s F = 240 kg m/s 1.5 sec F = 160 kg m/s 2 F = 160 N

47 Impulse & Change in Momentum Example: Part Three finding Impulse What impulse did the runner experience? J = FT J = 160 N x 1.5 sec J = 240 N sec