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Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Chapter.

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Presentation on theme: "Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Chapter."— Presentation transcript:

1 Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Chapter 8 Momentum, Impulse, and Collisions

2 Copyright © 2012 Pearson Education Inc. Goals for Chapter 8 To learn the meaning of the momentum of a particle and how an impulse causes it to change To learn how to use the principle of conservation of momentum To learn how to solve problems involving collisions

3 Copyright © 2012 Pearson Education Inc. Goals for Chapter 8 To learn the definition of the center of mass of a system and what determines how it moves To analyze situations, such as rocket propulsion, in which the mass of a moving body changes

4 Copyright © 2012 Pearson Education Inc. Introduction In many situations, such as a bullet hitting a carrot, we cannot use Newton’s second law to solve problems because we know very little about the complicated forces involved. In this chapter, we shall introduce momentum and impulse, and the conservation of momentum, to solve such problems.

5 Copyright © 2012 Pearson Education Inc. Momentum and Newton’s second law The momentum of a particle is the product of its mass and its velocity: Newton’s second law can be written in terms of momentum as What is the momentum of a 1000kg car going 25 m/s west?

6 Copyright © 2012 Pearson Education Inc. Momentum and Newton’s second law The momentum of a particle is the product of its mass and its velocity: Newton’s second law can be written in terms of momentum as What is the momentum of a 1000kg car going 25 m/s west? p = mv = (1000 kg)(25 m/s) = 25,000 kgm/s west

7 Copyright © 2012 Pearson Education Inc. Impulse and momentum Impulse of a force is product of force & time interval during which it acts. Impulse is a vector! On a graph of  F x versus time, impulse equals area under curve.

8 Copyright © 2012 Pearson Education Inc. Impulse and momentum Impulse-momentum theorem: Impulse = Change in momentum J of particle during time interval equals net force acting on particle during interval J =  p = p final – p initial(note all are VECTORS!) J = Net Force x time = (  F) x (  t) so…  p = (  F) x (  t)  F =  p/(  t)

9 Copyright © 2012 Pearson Education Inc. Compare momentum and kinetic energy Changes in momentum depend on time over which net force acts But… Changes in kinetic energy depend on the distance over which net force acts.

10 Copyright © 2012 Pearson Education Inc. Ice boats again (example 8.1) Two iceboats race on a frictionless lake; one with mass m & one with mass 2m. Wind exerts same force on both. Both boats start from rest, both travel same distance to finish line. Questions! Which crosses finish line with more KE? Which crosses with more Momentum?

11 Copyright © 2012 Pearson Education Inc. Ice boats again (example 8.1) Two iceboats race on a frictionless lake; one with mass m and one with mass 2m. The wind exerts the same force on both. Both boats start from rest, and both travel the same distance to the finish line? Which crosses the finish line with more KE? In terms of work done by wind? W =  KE = Force x distance = same; So ½ m 1 v 1 2 =  KE 1 = W =  KE 2 = ½ m 2 v 2 2

12 Copyright © 2012 Pearson Education Inc. Ice boats again (example 8.1) Two iceboats race on a frictionless lake; one with mass m and one with mass 2m. The wind exerts the same force on both. Both boats start from rest, and both travel the same distance to the finish line? Which crosses the finish line with more KE? In terms of work done by wind? W =  KE = Force x distance = same; So ½ m 1 v 1 2 =  KE 1 = W =  KE 2 = ½ m 2 v 2 2 Since m2> m1, v2 < v1 so more massive boat loses

13 Copyright © 2012 Pearson Education Inc. Ice boats again (example 8.1) Two iceboats race on a frictionless lake; one with mass m and one with mass 2m. The wind exerts the same force on both. Both boats start from rest, and both travel the same distance to the finish line? Which crosses the finish line with more p? In terms of momentum? Force on the boats is the same for each, but TIME that force acts is different. The second boat accelerates slower, and takes a longer time. Since t2> t1, p2 > p1 so second boat has more momentum

14 Copyright © 2012 Pearson Education Inc. Ice boats again (example 8.1) Check with equations! Force of wind = same; distance = same Work done on boats is the same, gain in KE same. ½ m1v1 2 =  KE1 = W =  KE2 = ½ m2v2 2 And m1v1 = p1; m2v2 = p2 ½ m1v1 2 = ½ (m1v1 ) v1 = ½ p1 v1 & same for ½ p2 v2 ½ p1 v1 = ½ p2 v2 Since V1 > V2, P2 must be greater than P1! p2 > p1 √

15 Copyright © 2012 Pearson Education Inc. A ball hits a wall (Example 8.2) A 0.40 kg ball moves at 30 m/s to the left, then rebounds at 20 m/s to the right from a wall. What is the impulse of net force during the collision, and if it is in contact for 0.01 s, what is the average force acting from the wall on the ball?

16 Copyright © 2012 Pearson Education Inc. Kicking a soccer ball – Example 8.3 Soccer ball 0.40 kg moving left at 20 m/s, then kicked up & to the right at 30 m/s at 45 degrees. If collision time is 0.01 seconds, what is impulse?

17 Copyright © 2012 Pearson Education Inc. An isolated system The total momentum of a system of particles is the vector sum of the momenta of the individual particles. No external forces act on the isolated system consisting of the two astronauts shown below, so the total momentum of this system is conserved.

18 Copyright © 2012 Pearson Education Inc. Conservation of momentum External forces (the normal force and gravity) act on the skaters, but their vector sum is zero. Therefore the total momentum of the skaters is conserved. Conservation of momentum: If the vector sum of the external forces on a system is zero, the total momentum of the system is constant.

19 Copyright © 2012 Pearson Education Inc. Remember that momentum is a vector! When applying conservation of momentum, remember that momentum is a vector quantity! Use vector addition to add momenta in COMPONENTS!

20 Copyright © 2012 Pearson Education Inc. Recoil of a rifle – 1D conservation of momentum A rifle fires a bullet, causing the rifle to recoil.

21 Copyright © 2012 Pearson Education Inc. Objects colliding along a straight line (example 8.5) Two gliders collide on an frictionless track. What are changes in velocity and momenta?

22 Copyright © 2012 Pearson Education Inc. A two-dimensional collision Two robots collide and go off at different angles. You must break momenta into x & y components and deal with each direction separately

23 Copyright © 2012 Pearson Education Inc. Elastic collisions In an elastic collision, the total momentum of the system in a direction is the same after the collision as before if no external forces act in that direction: P ix = P fx m a v ai +m b v bi =m a v af +m b v bf But wait – there’s more!

24 Copyright © 2012 Pearson Education Inc. Elastic collisions In an elastic collision, the total kinetic energy of the system is the same after the collision as before. KE i = KE f ½ m a v ai 2 + ½ m b v bi 2 = ½ m a v af 2 + ½ m b v bf 2 =

25 Copyright © 2012 Pearson Education Inc. Elastic collisions In an elastic collision, Difference in velocities initially = (-) difference in velocities finally (v ai - v bi ) = - (v af – v bf )

26 Copyright © 2012 Pearson Education Inc. Elastic collisions In an elastic collision, Difference in velocities initially = (-) difference in velocities finally (v ai - v bi ) = - (v af – v bf ) Solve generally for final velocities: v af = (m a -m b )/(m a +m b )v ai + 2m b /(m a +m b )v bi v bf = (m b -m a )/(m a +m b )v bi + 2m a /(m a +m b )v ai

27 Copyright © 2012 Pearson Education Inc. Elastic collisions

28 Copyright © 2012 Pearson Education Inc. Elastic collisions Behavior of colliding objects is greatly affected by relative masses.

29 Copyright © 2012 Pearson Education Inc. Elastic collisions

30 Copyright © 2012 Pearson Education Inc. Elastic collisions Behavior of colliding objects is greatly affected by relative masses.

31 Copyright © 2012 Pearson Education Inc. Elastic collisions

32 Copyright © 2012 Pearson Education Inc. Elastic collisions Behavior of colliding objects is greatly affected by relative masses. Stationary Bowling Ball!

33 Copyright © 2012 Pearson Education Inc. Inelastic collisions In any collision in which the external forces can be neglected, the total momentum is conserved. A collision in which the bodies stick together is called a completely inelastic collision In an inelastic collision, the total kinetic energy after the collision is less than before the collision.

34 Copyright © 2012 Pearson Education Inc. Some inelastic collisions Cars are intended to have inelastic collisions so the car absorbs as much energy as possible. Example 8.7

35 Copyright © 2012 Pearson Education Inc. The ballistic pendulum Ballistic pendulums are used to measure bullet speeds

36 Copyright © 2012 Pearson Education Inc. A 2-dimenstional automobile collision ( Example 8.9) Two cars traveling at right angles collide.

37 Copyright © 2012 Pearson Education Inc. An elastic straight-line collision

38 Copyright © 2012 Pearson Education Inc. Neutron collisions in a nuclear reactor

39 Copyright © 2012 Pearson Education Inc. A two-dimensional elastic collision (page 258)

40 Copyright © 2012 Pearson Education Inc. Center of mass of symmetrical objects It is easy to find the center of mass of a homogeneous symmetric object

41 Copyright © 2012 Pearson Education Inc. Motion of the center of mass The total momentum of a system is equal to the total mass times the velocity of the center of mass. The center of mass of the wrench at the right moves as though all the mass were concentrated there.

42 Copyright © 2012 Pearson Education Inc. Tug-of-war on the ice Start 20 meters away; each 10 m from coffee! Pull light rope; when James moved 6 m, how far is Ramon?

43 Copyright © 2012 Pearson Education Inc. External forces and center-of-mass motion When a body or collection of particles is acted upon by external forces, the center of mass moves as though all the mass were concentrated there.

44 Copyright © 2012 Pearson Education Inc. Rocket propulsion Conservation of momentum holds for rockets, too! F = dp/dt = d(mv)/dt = m(dv/dt) + v(dm/dt) As a rocket burns fuel, its mass decreases (dm< 0!)

45 Copyright © 2012 Pearson Education Inc. Rocket propulsion Initial Values mv = initial x-momentum of rocket v exh = exhaust velocity from rocket motor (This will be constant!) (v - v exh ) = relative velocity of exhaust gases m = initial mass of rocket v = initial velocity of rocket in x direction

46 Copyright © 2012 Pearson Education Inc. Rocket propulsion Changing values dm = decrease in mass of rocket from fuel dv = increase in velocity of rocket in x-direction dt= time interval over which dm and dt change

47 Copyright © 2012 Pearson Education Inc. Rocket propulsion Final values m + dm = final mass of rocket less fuel ejected v + dv = increase in velocity of rocket (m + dm) (v + dv) = final x-momentum of rocket (+x direction) [- dm] (v - v exh ) = final x-momentum of fuel (+x direction)

48 Copyright © 2012 Pearson Education Inc. Rocket propulsion Final values mv = [(m + dm) (v + dv)] + [- dm] (v - v exh ) initial momentum final momentum of the lighter rocket final momentum of the ejected mass of gas

49 Copyright © 2012 Pearson Education Inc. Rocket propulsion Final values mv = [(m + dm) (v + dv)] + [- dm] (v - v exh ) mv = mv + mdv + vdm + dmdv –dmv +dmv exh

50 Copyright © 2012 Pearson Education Inc. Rocket propulsion Final values mv = [(m + dm) (v + dv)] + [- dm] (v - v exh ) mv = mv + mdv + vdm + dmdv –dmv +dmv exh

51 Copyright © 2012 Pearson Education Inc. Rocket propulsion Final values mv = mv + mdv + vdm + dmdv –dmv +dmv exh

52 Copyright © 2012 Pearson Education Inc. Rocket propulsion Final values mv = mv + mdv + vdm + dmdv –dmv +dmv exh 0= mdv + dmdv + dmv exh

53 Copyright © 2012 Pearson Education Inc. Rocket propulsion Final values mv = mv + mdv + vdm + dmdv –dmv +dmv exh 0= mdv + dmdv + dmv exh Neglect the assumed small term: m(dv) = -dmdv – (dm)v exh

54 Copyright © 2012 Pearson Education Inc. Rocket propulsion Final values m(dv) = – (dm)v exh Gain in momentum of original rocket is related to rate of mass loss and exhaust velocity of gas!

55 Copyright © 2012 Pearson Education Inc. Rocket propulsion Final values m(dv) = – (dm)v exh Gain in momentum of original rocket is related to rate of mass loss and exhaust velocity of gas! Differentiate both sides with respect to time!

56 Copyright © 2012 Pearson Education Inc. Rocket propulsion Final values m(dv) = – (dm)v exh m(dv/dt) = – [(dm)/dt] v exh

57 Copyright © 2012 Pearson Education Inc. Rocket propulsion Final values m(dv) = – (dm)v exh m(dv/dt) = – [(dm)/dt] v exh ma = Force (Thrust!) = – [(dm)/dt] (v exh )

58 Copyright © 2012 Pearson Education Inc. Rocket propulsion Final values m(dv) = – (dm)v exh m(dv/dt) = – [(dm)/dt] v exh ma = Force (Thrust!) = – [(dm)/dt] (v exh ) rate of change of mass Velocity of gas exhausted

59 Copyright © 2012 Pearson Education Inc. Rocket propulsion Final values Thrust = – [(dm)/dt] (v exh ) Example: v exhaust = 1600 m/s Mass loss rate = 50 grams/second Thrust =?

60 Copyright © 2012 Pearson Education Inc. Rocket propulsion Final values Thrust = – [(dm)/dt] (v exh ) Example Problem 8.62 v exhaust = 1600 m/s Mass loss rate = 50 grams/second Thrust = m/s (-0.05 kg/1 sec) = +80N

61 Copyright © 2012 Pearson Education Inc. Rocket propulsion Gain in speed? m(dv) = – (dm)v exh dv = – [(dm)/m] v exh integrate both sides

62 Copyright © 2012 Pearson Education Inc. Rocket propulsion Gain in speed? m(dv) = – (dm)v exh dv = – [(dm)/m] v exh (integrate both sides) v f - v i = (v exh ) ln(m 0 /m) m 0 = initial mass m 0 > m, so ln > 1 “mass ratio” Velocity of gas exhausted gain in velocity

63 Copyright © 2012 Pearson Education Inc. Rocket propulsion Gain in speed? v f - v i = (v exh ) ln(m 0 /m) m 0 = initial mass m 0 > m, so ln > 1 “mass ratio” Velocity of gas exhausted gain in velocity Faster exhaust, and greater difference in mass, means a greater increase in speed!

64 Copyright © 2012 Pearson Education Inc. Rocket propulsion Gain in speed? v f - v i = (v exh ) ln(m 0 /m) Example: Rocket ejects gas at relative 2000 m/s. What fraction of initial mass is not fuel if final speed is 3000 m/s?

65 Copyright © 2012 Pearson Education Inc. Rocket propulsion Gain in speed? v f - v i = (v exh ) ln(m 0 /m) Example: Rocket ejects gas at relative 2000 m/s. What fraction of initial mass is not fuel if final speed is 3000 m/s? ln(m o /m) = (3000/2000) = 1.5 so m/m o = e -1.5 =.223

66 Copyright © 2012 Pearson Education Inc. Rocket propulsion Gain in speed? v f - v i = (v exh ) ln(m 0 /m) m 0 = initial mass m 0 > m, so ln > 1 “mass ratio” Velocity of gas exhausted gain in velocity STAGE rockets to throw away mass as they use up fuel, so that m 0 /m is even higher!


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