Be on time Switch off mobile phones. Put away laptops. Being present = Participating actively

Transcription

1 A couple of house rules Be on time Switch off mobile phones Put away laptops Being present = Participating actively

2 Collisions: amazing physics

3 Het basisvak Toegepaste Natuurwetenschappen Applied Natural Sciences Leo Pel e mail: 3nab0@tue.nl

4 Chapter 8 Momentum, Impulse, and Collisions PowerPoint Lectures for University Physics, Thirteenth Edition Hugh D. Young and Roger A. Freedman Lectures by Wayne Anderson Copyright 2012 Pearson Education Inc.

5 LEARNING GOALS The meaning of the momentum of a particle, and how the impulse of the net force acting on a particle causes its momentum to change. The conditions under which the total momentum of a system of articles is constant (conserved). How to solve problems in which two bodies collide with each other. The important distinction among elastic, inelastic, and completely inelastic collisions. The definition of the center of mass of a system, and what determines how the center of mass moves.

6 Summary Newtons laws F m a F AB F BA In some situations easier: work r 2 W F dr r 1

7 Summary Work and kinetic energy W K Potential energy conservative force W U energy-conservation W niet not U K Law of energy conservation

8 (part 2) Momentum Impulse Type of collisions Note: the wording in Dutch is quite confusing: Impuls (Dutch) = Momentum (English) Stoot (Dutch) = Impulse (English)

9 Momentum F ma dv m dt Newton s 2 nd law m=constant F dm ( v) dt dp dt p mv momentum Measure for the amount of motion or force = time difference in motion (Work is change in kinetic energy)

10 dm ( v) F dt Hence we see: or: dp dt F p/ t p p F t 2 1 The difference in momentum equals the force times the time over which the force acts. Ft impulse (dutch stoot )

11 Example: What is the force of a jet of water that you spray on a car. The water has a velocity of 20 m / s, and a flux of 1.5 kg / s. Answer: In 1s: m = 1.5 kg with 20 m / s, hence p = 30 kg m / s= the momentum of the water. The water comes to a standstill? Therefore final momentum = 0. Thus F = p/t = 30 kgm/s 2 = 30 N

12 Force during collision Tennisbal has a momentum -mv 1, and is hit back: momentum +mv 2, (e.g.. m=0.1kg, v 1 =20m/s, v 2 =30m/s, t = 10 ms (i.e. the contact time). What is the force of the racket on ball? Momentum change: m(v 1 +v 2 )=Ft. Contact time is: t = 10 ms, So F = 0.5 kn.

13 Force during collision: bend your knees Example: person 70 kg jump of 7.7 m/s (so ~ 1 meter) Momentum= 7.7x70 kgm/s F=p/t Straight legs: F= N Bend legs: F=2695 N 70 kg So force difference factor of 100 Break in 2 ms or 0.2 s.

14 Example: A Karate Collision With an expert karate blow, you shatter a concrete block. Consider that you hand has a mass of 0.70 kg, is initially moving downward at 5.0 m/s, and stops 6.0 mm beyond the point of contact. (a) What impulse does the block exert on your hand? (b) What is the approximate collision time and average force that the block exerts on your hand? 1 v5.0 m/s( yˆ ) y vavt vt 2 I pmv (0.70 kg)(5.0 m/s yˆ) 3.5 N s yˆ 2y 2(0.006 m) t s v 5.0 m/s I (3.5 N s yˆ ) Fav = 1,500 N yˆ t ( s)

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16 Momentum m=constant F dm ( v) dt dp dt p mv momentum Measure for the amount of motion or Newton s 2 nd law F force = time difference in motion dp 0 0 p dt Momentum is conserved constant

17 Experiment

18 Conservation of momentum F dm ( v) dt dp dt There is no net external force ISOLATED SYTEM Newton s 3 rd law pb F AB F AB F BA dt F BA ( p pb) 0 A dt p A Conservation of momentum is a universal law

19 Isolated system of 2 objects: same mass p=0 p left =p right mv left =mv right

20 Example Before shot p =0 After shot p =0 p R = p B m gun V r = m bullit v bullit Recoil of gun

21 Example Recoil of gun Propulsion of rocket.

22 Collisions Before B A During D After C ISOLATED SYSTEM MOMENTUM CONSERVED

23 Examples of collisions

24 Classification of collisions Momentum If no external forces act Conservation of total momentum Kinetic energy Elastic collisions Kinetic energy is conserved Inelastic collisions Kinetic energy is not conserved Completely Inelastic collision Objects stick together after collision

25 Inelastic collisions H 2 H 2 Nonconservative forces Mechanical energy is transformed into internal energy (e.g. heat, deformation etc)

26 Super inelastic car collision of 193 km/h = 120 mph with concrete block

27 Car collision Ford focus : 1200 kg, length 2.5 m and speed 193 km/h=54 m/s Kinetic energy ~ J What was the force on this car in this complete inelastic collision? 1. Not enough info ,000 N ,000 N 4. 1,400,000 N.

28 F p t p is easy p= x 54 = kg m/s PROBLEM is to estimate t However we know car length 2.5 m Speed 54 m/s, so average speed during crash estimate : 54/2 m/s F= N = 1.4 MN How many people? ~ 1750 of 80 kg Hence rough estimate t=1.25/27=46 ms

29 Balistic pendulum : measure speed bullit bullit BEFORE Stays in the wood Completely inelastic collision AFTER

30 Balistic pendulum Momentum conservation BEFORE Energy conservation AFTER So in total:

31 What about the energy loss? b b before v m K b w b b w b w after v m m v m m m K Before After, with approximation m b <<m w before w b after K m m K Very large loss Heat / damage speed of block

32 Elastic collisions Conservative forces No mechanical energy lost

33 1D collision in general Particle collision: A A B v A v B Momentum conservation mv A A mv B B Pmw A A mw B B 0 unknown 2 unknown Elastic so energy conservation mv mv K mw mw A 2 B 2 A 2 B A B A B 1 vgl 2 equations 2 unknowns solvable + 1 vgl 0 unknown 2 unknown

34 Newton s Cradles

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36 Quiz What will happen after collision? 1. 1 ball will go to the left 3 times as fast 2. 2 balls will go to to left 1.5 times as fast 3. 3 balls will go to the left 4. 2 balls to left and one to the right

37 momentum conservation v P = 3m x v = 3m x v = 2m x 1.5 v = 1m x 3v = 2m x 2v 1m x 1v many solutions when only apply momentum conservation

38 Momentum and energy conservation P = 3m x v K = ½ 3mv 2 1) 1 ball will go to the left 3 times as fast = 1m x 3v ½ 1m 9v 2 v 2) balls will go to to left 1.5 times as fast = 2m x 1.5 v ½ 2m 2.25 v 2 3) balls will go to the left = 3m x v = ½ 3mv 2 4) balls to left and one to the right = 2m x 2v 1m x 1v ½ 2m 4v 2 + ½ m v 2

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40 Quiz collision Same mass m V 1 V 2 =0 What will happen after collision? 1. Ball 1 will bounce back with same speed 2. Ball 1 stops and ball 2 Is moving with speed V 1 3. Both ball will move at same speed

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42 Before 1 2 V 1 V 2 =0 momentum energy After mv 1 mv 2 m0 mw mw v m w 1 w 2 ( v w mw 2 ) w 2 mw 2 2 v w w 0 w ( w v1) w v 1 w 2 W 2 =0 : situation t=0 W 2 =V 1 and W 1 =0

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44 Relative velocities for elastic collisions before m v M V Large ball mv MV P mw MW mv MV K mw MW after w W Large ball

45 Derivation mv MV P mw MW mv MV K mw MW mw v MV W ( ) ( ) mw ( v)( wv) MV ( W)( VW) mw ( v) MV ( W) w vv W 2V wv2v w W Large bal Final speed = initial speed + 2x speed bal

46 Space sling shot Gravitational potential energy Launce on 1997, arival july 2004 Saturnus: m from sun Aarde: m from sun GmM U R U GM m r r Z ( ) ( ) A S MJ / kg Energy needed for Saturnus U GM MJ kg m A( ) / 6 RA 610 Energy needed to leave the earth

47 Space sling shot m v M V Jupiter w W Jupiter w v 2V = x 9.6 = 29.6 km/s

48 Space sling shot

49 Collisions in general 2D/3D Can also be in 2D: take all components

50 CERN : particle physics From collisions find new particles

51 How to make billard boring

52 Examination 25 jan 2017

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54 How about large objects (week 1): still valid? Each object is considered to be a point mass mass M point mass M book: particle mass M

55 Center of Mass The center of mass of a system is the point where the system can be balanced in a uniform gravitational field.

56 For two objects: The center of mass is closer to the more massive object. Note that we can also apply this relation to the velocities and accelerations of the objects and their center of mass.

57 Center of mass Center of mass is defined as (massamiddelpunt): x CM mx m mx m More general center of mass: C M O

58 Center of mass The center of mass need not be within the object:

59 Center of mass To determine the center of mass we can again split the calculation up in x and y x = (1.03 m a m b ) / (m a + m b ) y = (0.10 m a m b ) / (m a + m b )

60 Center of mass: Fosbury flop 1968 Summer Olympics

61 Centre of mass and momentum system of particles i F ext Centre of mass r CM 1 M Total mass m r i i Mr CM mr i i Mm i

62 Centre of mass and momentum system of particles i F ext So momentum is equal to the total mass times the velocity of the centre of mass

63 Centre of mass and momentum system of particles i F ext So only the center of mass is influenced by external force

64 Center of mass What is this telling us about center of mass? 1. The CM moves as a single particle in which the total net external force is: where F= the total external force F, M= the total mass a= the acceleration of the CM. 2. The total momentum of a system of particles is: M v CM p tot where M= the total mass v= the velocity of the CM p= total momentum

65 Fire works

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67 Center of mass motion

68 Center of mass collision: same mass

69 Before 1 2 V 1 V 2 =0 V cm mv1 m0 2m 1 2 v 1 After 2 1 V 1 =0 V 2 =V 1 V cm m0 mv 2m v 1

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71 If no external force : which part starts to move?

72 Better known as canoe problem As the girl start to walk the canoe will move in the opposite direction. However the CM will stay at the same position,

73 Better known as canoe problem A 45.0-kg woman stands up in a 60.0-kg canoe 5.00 m long. She walks from a point 1.00 m from one end to a point 1.00 m from the other end. If you ignore resistance to motion of the canoe in the water, how far does the canoe move during this process?

74 No external force Additional relation (she walks 1.5 meter from CM canoe)

75 Youtube: anti gravity double cone

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77 Summary Conservation of momentum is a universal law It always holds when no net force is acting In contrast to the conservation of mechanical energy, which depends on the type of collision Center of Mass When a body or a collection of particles is acted on by external forces, the center of mass moves just as though all the mass were concentrated at that point

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