Chapter 6: Momentum And Collisions

PHU 205Mechanics for Life SciencesChapter 6:Momentum and CollisionsProf. Liliana Braescu &Prof. Nouredine Zettili

Required textCollege PhysicsRaymond A. Serway and Chris Vuille9th Edition, 2012BROOKS/COLE CENGAGE LearningISBN 10: 1-111-42745-3ISBN 13: 978-1-111-42745-01

Course OutlineChapter 6Momentum and Collisions6.1 Momentum and Impulse6.2 Conservation of Momentum6.3 Collisions6.4 Glancing Collisions2

Chapter 6Momentum and Collisions6.1 Momentum and Impulse6.2 Conservation of Momentum6.3 Collisions6.4 Glancing Collisions3

6.1 Momentum and Impulseu Linear momentum – the linear momentum of anobject of mass m moving at speed v is expressed!!asp mvu SI unit of momentum kg m/su Newton used the concept of momentumto!express his second law: F! ma! mΔ vΔt!!! v f vi F m Δt !!!! p f pi ΔpF 4Δt Δt

6.1 Momentum and Impulseu Force is also a measurement of change in linearmomentum over a !giventimeinterval.!!! p f pi Δp F ΔtΔtu For an object !in equilibriumFext 0 Δp 0 pi p f If the sum of all of the external forces equalszero, momentum remains constant and istherefore conserved.5

6.1 Momentum and Impulseu Impulse – In the initial seconds of a collision,there is an impulse force on the object.u This force is defined as the change in linear"!"!!!!momentum: I FΔt Δp mv mvfiu In order to change the momentum of an object,a force must be applied.u The time rate of change of momentum of anobject is equal to the !net forceacting on it!!Δp m(v f v i ) ! FnetΔtΔt6

6.1 Momentum and Impulse!!!Δp m(v f v i ) ! FnetΔtΔtu It gives an alternative statement of Newton’s! !second lawI FΔtu Impulse is a vector quantity, the direction is thesame as the direction of the force.Impulse-Momentum Theoremu The theorem states that the impulse acting onthe object is equal to the change in momentum ofthe object ! !!!!I FΔt Δp mv f mv i7

6.1 Momentum and Impulseu If the force is not constant, use the averageforce applied.Average Force in Impulseu The average force can be thought of as theconstant force that would give the same impulse tothe object in the time interval as the actual timevarying force gives in the interval:8

6.1 Momentum and ImpulseAverage Force in Impulseu The impulse imparted by a force during the timeinterval Δt is equal to the area under the force-timegraph from the beginning to the end of the timeinterval.u Or, the impulse is equal to the average forcemultiplied by the time interval,!!Fav Δt Δp9

6.1 Momentum and ImpulseImpulse Applied to Auto Collisionsu The most important factor is the collision time orthe time it takes the person to come to a rest.E.g., this will reduce the chance of dying in a car crash.u Ways to increase the time: Seat belts Air bags10

6.1 Momentum and ImpulseAcceleration That Would Kill a Humanu When placing a human under high acceleration thebrain, lungs, ribs (any bones), etc., all affect the death of thehuman.u Since different body parts have different densities,certain organs will undergo more g-force than others.u Pilots, roller coaster rides, and car crashes are sufficientexamples of g-force.u Pilots are trained to undergo accelerations of 9 g's to pullquick maneuvers during flight for less than a second.u If a force of 4 to 6 g's is held for more than a fewseconds, the results could be devastating.11

6.1 Momentum and ImpulseTypical Collision Valuesu For a 75 kg person traveling at 27 m/s andcoming to stop in 0.010 su F -2.0 x 105 Nu a 280 g: Almost certainly fatalu The 1997 car crash that killed Princess Dianawas estimated to range somewhere between 70–100 g's.u In general, high velocity doesn't kill; what kills isthe high acceleration or deceleration sustainedover a certain time interval.12

6.1 Momentum and ImpulseSurvival secret: Increase timeu Seat belt: Restrain people so it takes more timefor them to stop: about 0.15 secondsu Air Bag: It increases the time of the collision It will also absorb some of the energy from thebody It will spread out the area of contact (decreasesthe pressure, it helps prevent penetration wounds)13

Problemsu Problem 6.1An estimated force vs. time curve for a baseballstruck by a bat is shown in Figure. From thiscurve, determine:(a) the impulse delivered to the ball and(b) the average force exerted on the ball.14

Problemsu Problem 6.1!!Fav Δt ΔpSolution(a) The impulse imparted by the force during thetime interval Δt is equal to the area under theforce-time graph from the beginning to the end ofthe time interval:1.5(ms) 18000( N ) 1.5 18( N s)! !Δp Fav Δt 13.5( N s)2!! 2(b) The average force: Fav Δp / Δt(13.5 Ns)/1.5 ms 9000 (N)15

Chapter 6Momentum and Collisions6.1 Momentum and Impulse6.2 Conservation of Momentum6.3 Collisions6.4 Glancing Collisions16

6.2 Conservation of Momentumu When two isolated objects collide, the impulseexerted by m2 on m1 is!"!"!"Δt F 1 m1 v1 f m1 v1iwhile the impulse exerted by m1 on m2 is!"!"!"Δt F 2 m2 v 2 f m2 v 2i!"!"F 1 F 2u According to Newton’s third law,u Hence, combining above Eq. we prove thatmomentum is conserved during the collision:17

6.2 Conservation of MomentumMomentum is conserved during the collision:()!"!"!"!"!" !"Δt F 1 F 2 m1 v1 f m1 v1i m2 v 2 f m2 v 2i!"!"!"!"()m1 v1 f m2 v 2 f m1 v1i m2 v 2i 0" "p f pi 0""p f piThe principle of conservation of momentum states: when noexternal forces act on a system consisting of two objects thatcollide with each other, the total momentum of the systemremains constant in time.18

6.2 Conservation of Momentumu More precisely, the total momentum before thecollision will equal the total momentum after thecollision.u The momentum of eachobject will change.u The total momentum ofthe system remainsconstant.19

6.2 Conservation of MomentumForces in a collisionu The force with whichobject 1 acts on object 2 isequal and opposite to theforce with which object 2 actson object 1.u Impulses are also equaland opposite.20

6.2 Conservation of Momentum!!!!u Mathematically: m1v1i m2 v 2i m1v1 f m2 v 2 fu Momentum is conserved for the system ofobjects.u The system includes all the objects interactingwith each other.u Assumes only internal forces are acting duringthe collision.u Can be generalized to any number of objects.21

Problemsu Problem 6.2A rifle with a weight of 30 N fires a 5.0-g bulletwith a speed of 300 m/s.Questions:(a) Find the recoil speed of the rifle.(b) If a 700-N man holds the rifle firmly againsthis shoulder, find the recoil speed of man andrifle.22

Problemsu Problem 6.2Solution(a) Choosing the direction of the bullet's motionas positive,mRvR mbvb 0, which givesvR -vb(mb / mR) -300 m/s (5.0 x 10-3/3.06) -0.49 m/s.(b) The mass of the man plus rifle is 74.5 kg. Weuse the same approach as in (a), to findv -300 m/s (5.0 x 10-3/74.5) -2.0 x 10-2 m/s.23

Chapter 6Momentum and Collisions6.1 Momentum and Impulse6.2 Conservation of Momentum6.3 Collisions6.4 Glancing Collisions24

6.3 Collisionsu Inelastic collisions – aftertwo masses collide, theymove together as one unit.u Characteristics of an inelasticcollision:! ! momentum is conserved pi p f kinetic energy is not conserved K i K f!!! before collision pi m1 v1 m2 v 2!! after collision p f ( m1 m2 ) v fii25

6.3 CollisionsPerfectly Inelastic CollisionsWhen two objects stick together after thecollision, they have undergone a perfectlyinelastic collision.u Conservation of momentum becomesu !!!!m1v1i m2 v 2i m1v1 f m2 v 2 f!!!m1v1i m2 v 2i ( m1 m2 ) v f26

6.3 Collisionsu Elastic collisions – aftertwo masses collide, theyseparate again.u Characteristics of anelastic collision: momentum is conserved kinetic energy is conserved,also27

6.3 Collisionsu Both momentum and kinetic energy areconserved.u Typically have two unknowns!!!!m1v1i m2 v 2i m1v1 f m2 v 2 f1 !2 1 !2 1 !2 1 !2m1 v1i m2 v 2i m1 v1 f m2 v 2 f2222u Solve the equations simultaneously28

6.3 CollisionsElastic Collisionsu A simpler equation can beused in place of the KE equation!!!!v1i v 2i v1 f v 2 f()29

6.3 CollisionsSummary of types of Collisionsu In an elastic collision, both momentum andkinetic energy are conserved.u In an inelastic collision, momentum is conservedbut kinetic energy is not.u In a perfectly inelastic collision, momentum isconserved, kinetic energy is not, and the twoobjects stick together after the collision, so theirfinal velocities are the same.30

Problemsu Problem 6.3Three carts of masses 4.0 kg, 10 kg, and 3.0 kgmove on a frictionless horizontal track withspeeds of 5.0 m/s, 3.0 m/s, and -4.0 m/s, asshown in the figure below. The carts sticktogether after colliding.Questions:(a) Find the final velocity of the three carts.(b) Does your answer require that all cartscollide and stick together at the same time?31

Problemsu Problem 6.3Solution(a) Using conservation of momentum Pi Pf, wefind(4.0)(5.0) (10)(3.0) (3.0)(-4.0) (4.0 10 3.0)v,where v is the speed of the three mass systemafter collision.Therefore,v 2.2 m/s, or 2.2 m/s toward the right.32

Problemsu Problem 6.3Solution(b) No. For example, if the 10 kg and 3.0 kgmass were to stick together first, they would movewith a speed given by solving(13)v1 (10)(3.0) (3.0)(-4.0),So v1 1.38 m/s.Then, when this 13 kg combined mass collideswith the 4.0 kg mass, we have(17)v (4.0)(5.0) (13)(1.38), so thatv 2.2 m/s, just likepart (a).33

Problemsu Problem 6.4A 0.400-kg bead slides on a curved frictionlesswire, starting from rest at point A, where h 1.50m. At point B the bead collides elastically with a0.600-kg bead at rest.Questions:Find the distance the bead moves up the wire.34

Problemsu Problem 6.4Solution:-We shall first use conservation of energy to findthe speed of the bead just before it strikes theball. The zero level of potential energy is at thelevel of point B. We have,(1/2)mvi2 mgyi (1/2)mvf2 mgyf0 (0.400 kg)(g)(1.50 m) (1/2)(0.400)v1f2 0,giving v1f 5.42 m/s.-We now treat the collision to find the speed of theball immediately after the collision. Momentumconservation gives:35

Problemsu Problem 6.4Solution:-Momentum conservation gives(0.400 kg)(5.42 m/s) 0 (0.400 kg)v1f (0.600 kg)v2f-For an elastic collisionv1f v1i v2f v2i v1f 5.42 v2f 0-Solving simultaneously both Eqs: v2f 4.34 m/s.-Apply conservation of energy to the 0.6 kg ballafter impact to find½(0.600 kg)(4.34 m/s)2 0 0 (0.600 kg)(g)H,or H 0.96 m.36

Chapter 6Momentum and Collisions6.1 Momentum and Impulse6.2 Conservation of Momentum6.3 Collisions6.4 Glancing Collisions37

6.4 Glancing Collisionsu The “after” velocities have x and y componentsu Momentum is conserved in the x direction and inthe y direction.u Apply conservation of momentum separately toeach direction.38

Problemsu Problem 6.5A 3.00-kg steel ball strikes amassive wall at 10.0 m/s at anangle of θ 60.0 with the plane ofthe wall. It bounces off the wallwith the same speed and angle.Questions:If the ball is in contact with thewall for 0.200 s, what is theaverage force exerted by the wallon the ball?39

Problemsu Problem 6.5Solution:Hints-For x direction:compute momentum replace it in the formula!F !!p f piΔt !ΔpΔtand find the force.-Similarly for y direction average force .40

Chapter 6EndsQuiz 6 is coming . before Chapter 7For review,Study the following problems: 6.1 page 169, 6.2 page 170,6.3 page 174, 6.4 page 177, 6.6 page 180, 6.7 page 181.Solve the following problems: 15, 16 page 191; 38 page 193,46 page 194, 54 page 194.Textbook “College Physics”Raymond A. Serway and Chris Vuille9th Edition, 201259

6.1 Momentum and Impulse ! Impulse – In the initial seconds of a collision, there is an impulse force on the object. ! This force is defined as the change in linear momentum: ! In order to change the momentum of an object, a force must be applied. ! The time rate of change of momentum of