Elastic Collisions Definition Examples Work and Energy Definition of work Examples. Physics 201: Lecture 10, Pg 1

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1 Phyic 131: Lecture Today Agenda Elatic Colliion Definition i i Example Work and Energy Definition of work Example Phyic 201: Lecture 10, Pg 1

2 Elatic Colliion During an inelatic colliion of two object, ome of the mechanical energy i diipated inide the object a thermal energy. A colliion in which mechanical energy i conerved i called a perfectly elatic colliion. Colliion between two very hard object, uch a two billiard ball or two teel ball, come cloe to being perfectly elatic. Phyic 201: Lecture 10, Pg 2

3 Colliion Two kind of colliion Elatic-KE i conerved Inelatic-KE i not conerved Momentum i almot alway conerved during a colliion i (external force are generally mall compared to colliion force) Phyic 201: Lecture 14, Pg 3

4 Inelatic Colliion A box ma = 20 kg liding acro a frictionle floor with contant velocity v = 5 m/ collide with another box ma = 50 kg which h i at ret. If the two boxe tick together, what will be their peed right after the colliion? v = 5 m/ Phyic 201: Lecture 14, Pg 4

5 Initial Inelatic Colliion v = 5 m/ Box 1 Box 2 Final ma = 20 kg ma = 50 kg v =? Box 1 Box 2 20 kg 50 kg object 1 object 2 object 1 object 2 mv i + mv i = mv f + mv f (20kg)(5 m/) + (50kg) (0) = (20kg + 50kg) v f Phyic 201: Lecture 14, Pg 5

6 Inelatic Colliion (20kg)(5 m/) = (20kg + 50kg) v f 100 kg-m/ = (70 kg) v f v f v f 100kg - m/ 70 kg 1.43 m/ v = 5 m/ Box 1 Box 2 Phyic 201: Lecture 14, Pg 6

7 Inelatic Colliion Momentum i conerved! P initial = m 1 v 1i = (20 kg)(5 m/) = 100 kg m/ P final =(m 1+ m 2 2) v f = (70 kg)( 1.43 m/) = kg m/ You can alway prove thi to yourelf afterward! Kinetic Energy i not conerved! K i = ½m 1 (v 1i ) 2 = ½(20 kg)(5 m/) 2 = 250 J K F = ½ (m 1 + m 2 )(v f ) 2 = ½(70 kg)(1.43 m/) 2 = J Energy i lot in heat, ound, deformation etc. when object tick together! Phyic 201: Lecture 14, Pg 7

8 Elatic Colliion During an elatic colliion i both momentum and mechanical energy are conerved: m 1v1f m2v2f m1 v1i m2v2i (1) 1 2 m 1 v m2v 2 m1 v 2 m2 (2) 2 v 1f 2f 1i 2i Combining i equation (1) and (2) we get that t peed of approach equal peed of receion: v v v 2f 1f 2i v 1i Together with equation (1) you can olve for jut about any 1-d elatic colliion i Phyic 201: Lecture 10, Pg 8

9 Elatic Colliion: Quick Example A 200 g ball move to the right at 2.0 m/. It ha a headon, perfectly elatic colliion with a 100 g ball that i moving toward it at 3.0 m/. What are the final velocitie of both ball? Phyic 201: Lecture 10, Pg 9

10 m v m v m v m v 1 1f 2 2f 1 1i 2 2i m m 0.2kg 2 0.1kg m 3 1v1f m2v2f m v m v 0. 1kg m 1 1f 2 2f v v v v 2 f 1f 2i 1i m m v 3 v 3 2 2f 1f v 5 m v 2 f 1f Phyic 201: Lecture 10, Pg 10

11 m v m v 0. 1kg m 1 1f 2 2f v 5 m v 2f 1f 5 m v1 f 0. kg m v m 1 1 1f 2 0.2kgv kg m 1f v1 f 0. 1kg 0.2kgv kgv kg m 1f 0.1 1f kg m m m 0.3kgv1 f 0. 4 v 1. 3 m 1f kg m v 3. 7 m 2f Phyic 201: Lecture 10, Pg 11

12 The Baic Energy Model W > 0: The environment doe work on the ytem and the ytem energy increae. W < 0: The ytem doe work on the environment and the ytem energy decreae. Phyic 201: Lecture 10, Pg 12

13 Work by a Contant Force in Same Direction a Motion To do work you need two thing AForce Motion I lift my book with force F through a ditance of Work = F = Force (Diplacement) Only work with a contant t force Only true when force i in ame direction a diplacement Phyic 201: Lecture 10, Pg 13

14 F = 25 N Example = 35 m I puh a box with a contant force F = 25 N a ditance of 35 m. How much work do I do? F = 25 N = 35 m Work = F = 25 N 35 m = 875 N*m = 875 J Phyic 201: Lecture 10, Pg 14

15 Give it a try I puh a box with a contant force F = 25 N a ditance of 35 m. Thi time the force i at an angle of 30. How much work do I do? (a) 234 J (b) 435 J (c) 760 J (d) (e) 875 J 1000 J F = 25 N 30 = 35 m Phyic 201: Lecture 10, Pg 15

16 F = 25 N Clicker Quetion 2: 30 = 35 m F Co 30 =component of force in the direction of diplacement. F Co 30 = 25 N Co 30 = N Work = F Co =2165N m = J More general definition of work!! Work = (F Co ) ) F i magnitude of Force; i diplacement i angle between the two Phyic 201: Lecture 10, Pg 16

17 Work example = 20 m mg F N An object of ma 35 kg goe 20 m. How much work doe the normal force do? F N = mg but force ha no component in direction of diplacement! Work = F N Co =0 Co 90 = 0 Either figure it out with equation, or conceptually! Phyic 201: Lecture 10, Pg 17

18 Intereting thing about work You have to move the object to DO work!! I hold up a dumbbell, but it doen t move I do no work!! Definition of work! If force i perpendicular to motion you do no work! You can have poitive or negative work Work i a calar (eaier for u!) Phyic 201: Lecture 10, Pg 18

19 Poitive v. negative work If component of force point in ame direction a diplacement, work i poitive If component of force point in oppoite direction a diplacement, work i negative Good example: Friction (often doe negative work) Phyic 201: Lecture 10, Pg 19

20 Work and Kinetic Energy Conider a force acting on a particle which move along the -axi. The force component F caue the particle to peed up or low down, tranferring energy to or from the particle. The force doe work on the particle: The unit of work are N m, where 1 N m = 1 kg m 2 / 2 = 1 J. Phyic 201: Lecture 10, Pg 20

21 Work total Work total i the work on an object by all force acting. W W tot tot Work total i the work done by the net force Phyic 201: Lecture 10, Pg 21

22 Give it a try: A block i pulled by a force of 400 N a in the picture. However a frictional force of 200 N oppoe the motion. If the block move a ditance of 20 m, what i the total work done? (a) W S = 1770 J (b) W S = J (c) W S = J (d) W S = 7517 J (e) W S = 3517 J 20 Phyic 201: Lecture 10, Pg 22

23 Give it a try: F P 20 F K Work total i the um of all the work done. We have two force doing work, friction and F P So W friction = (F co ) = (200 N co 180 )20 m =-4000 J And W Pull = (F co ) = (400N co 20 )20 m = 7517J W tot = W friction + W Pull = J J = 3517 J Phyic 201: Lecture 10, Pg 23

24 Kinetic Energy: Motion Apply contant force along x-axi to a point particle m. W = F x x = ma x x recall: axx x ( vx vx0 ) = ½ m (v f2 v 02 ) 2 Work change ½ m v 2 Define Kinetic Energy K = ½ m v 2 W = K For Point Particle Phyic 201: Lecture 10, Pg 24

25 The Work-Energy Theorem W 2 tot = ½mv f2 - ½mv 2 i Put poitive work into an object, it peed increae! Put negative work into an object, it peed decreae! Be careful, we are intereted in total work done!!! We mut look at all force acting, not jut one! Phyic 201: Lecture 10, Pg 25

26 Give it a try A block i pulled by a force of 400 N to the left a in the picture, 20 m. However a frictional force of 200 N oppoe the motion. If the block tart from ret and ha ma 100 kg what i it final velocity? (a) (b) (c) (d) (e) v F = 3.22 m/ v F = 4.27 m/ v F =535m/ 5.35 v F = 6.78 m/ v F = 8.38 m/ 20 Phyic 201: Lecture 10, Pg 26

27 Clicker Quetion 8: F P 20 K We found total work done in our lat example: W tot = 3517 J W tot = ½mv f2 -½mv i 2 (v i =0) Block tart at ret W t = 2 t =mv 2 tot ½mv f lead to (2W tot ) f (2W tot ) /m =v f 2 v f = 8.38 m/ v F v 2(W m tot Phyic 201: Lecture 10, Pg 27 )