Chapter 7 Linear Momentum and Collisions

1 Chapter 7 Linear Momentum and The Important Linear MomentumThelinear momentumof a particle with massmmoving with velocityvis defined asp=mv( ) Linear Momentum is avector. When giving the Linear Momentum of a particle youmustspecify itsmagnitudeanddirection. We can see from the definition that its units must bekg ms. Oddly enough, this combination of SI units does not have a commonly used named sowe leave it askg ms!The Momentum of a particle is related to the net force on that particle in a simple way;since the mass of a particle remains constant, if we take the time derivative of a particle smomentum we finddpdt=mdvdt=ma=Fnetso thatFnet=dpdt( ) Impulse, Average ForceWhen a particle moves freely then interacts with another system for a (brief) period andthen moves freely again, it has a definite change in Momentum ;we define this change as theimpulse Iof the interaction forces:I=pf pi= pImpulse is a vector and has the same units as we integrate Eq.

2 We can show:I= tftiFdt= p155156 Chapter 7. Linear Momentum AND COLLISIONSWe can now define theaverage forcewhich acts on a particle during a time interval t. It is:F= p t=I tThe value of the average force depends on the time interval Conservation of Linear MomentumLinear Momentum is a useful quantity for cases where we have afew particles (objects)which interact witheach otherbut not with the rest of the world. Such a system is calledanisolated often have reason to study systems where a few particles interact with each other verybriefly, with forces that are strong compared to the other forces in the world that they mayexperience. In those situations, and for that brief period of time, we can treat the particlesas if can show that when two particles interactonlywith each other ( they are isolated)then their total Momentum remains constant:p1i+p2i=p1f+p2f( )or, in terms of the masses and velocities,m1v1i+m2v2i=m1v1f+m2v2f( )Or, abbreviatingp1+p2=P(total Momentum ), this is:Pi= is important to understand that Eq.

3 Is avectorequation; it tells us that the totalxcomponent of the Momentum is conserved,andthe totalycomponent of the momentumis CollisionsWhen we talk about acollisionin physics (between two particles, say) we mean that twoparticles are moving freely through space until they get close to one another; then, for ashort period of time they exert strong forces on each other until they move apart and areagain moving such an event, the two particles have well-defined momentap1iandp2ibefore thecollision event andp1fandp2fafterwards. But the sum of the momenta before and afterthe collision is conserved, as written in Eq. thetotal momentumis conserved for a system of isolated colliding particles, themechanical energymay or may not be conserved. If the mechanical energy (usually meaningthe total kinetic energy) is the same before and after a collision, we say that the collision iselastic.

4 Otherwise we say the collision two objects collide, stick together, and move off as a combined mass, we call this aperfectly inelasticcollision. One can show that in such a collision more kineticenergy islost than if the objects were to bounce off one another and moveoff THE IMPORTANT STUFF157 When two particles undergo anelasticcollision then we also know that12m1v21i+12m2v22i=12m1v21f+ the special case of a one-dimensional elastic collision between massesm1andm2wecan relate the final velocities to the initial velocities. The result isv1f=(m1 m2m1+m2)v1i+(2m2m1+m2)v2i( )v2f=(2m1m1+m2)v1i+(m2 m1m1+m2)v2i( )This result can be useful in solving a problem where such a collision occurs, but it isnotafundamental equation. So don t memorize The Center of MassFor a system of particles (that is, lots of em) there is a special point in space known as thecenter of masswhich is of great importance in describing the overall motion of the point is a weighted average of the positions of all the mass the particles in the system have massesm1,m2.

5 MN, with total massN imi=m1+m2+ +mN Mand respective positionsr1,r2, .. ,rN, then the center of massrCMis:rCM=1MN imiri( )which means that thex,yandzcoordinates of the center of mass arexCM=1MN imixiyCM=1MN imiyizCM=1MN imizi( )For an extended object ( a continuous distribution of mass) the definition ofrCMisgiven by anintegralover the mass elements of the object:rCM=1M rdm( )which means that thex,yandzcoordinates of the center of mass are now:xCM=1M x dmyCM=1M y dmzCM=1M z dm( )158 Chapter 7. Linear Momentum AND COLLISIONSWhen the particles of a system are in motion then in general their center of mass is alsoin motion. The velocity of the center of mass is a similar weighted average of the individualvelocities:vCM=drCMdt=1MN imivi( )In general the center of mass will accelerate; its acceleration is given byaCM=dvCMdt=1MN imiai( )IfPis the total Momentum of the system andMis the total mass of the system, thenthe motion of the center of mass is related toPby:vCM=PMandaCM= The Motion of a System of ParticlesA system ofmanyparticles (or an extended object) in general has a motion forwhich thedescription is very complicated, but it is possible to make asimple statement about themotion of its center of mass.

6 Each of the particles in the system may feel forces from theother particles in the system, but it may also experience a net force from the (external)environment; we will denote this force byFext. We find that when we add up all theexternalforces acting on all the particles in a system, it gives the acceleration of thecenter of massaccording to:N iFext, i=MaCM=dPdt( )Here,Mis the total mass of the system;Fext, iis the external force acting on words, we can express this result in the following way: Fora system of particles, thecenter of mass moves as if it were asingleparticle of massMmoving under the influence ofthe sum of the external Worked Linear Momentum1. kgparticle has a velocity of( )ms. Find itsxandycomponentsof Momentum and the magnitude of its total the definition of Momentum and the given values ofmandvwe have:p=mv= ( kg)( )ms= ( )kg WORKED kg10 m/s10 m/sxyFigure :Ball bounces off wall in Example the particle has Momentum componentspx= + msandpy= magnitude of its Momentum isp= p2x+p2y= ( )2+ ( 12.)

7 2kg ms= Impulse, Average Force2. A child bounces a superball on the sidewalk. The Linear impulse delivered bythe sidewalk N sduring the1800sof contact. What is the magnitude of theaverage force exerted on the ball by the magnitude of the change in Momentum of (impulse delivered to) the ball is| p|=|I|= N s. (Thedirectionof the impulse is upward, since the initial Momentum of theball was downward and the final Momentum is upward.)Since the time over which the force was acting was t=1800s = 10 3sthen from the definition of average force we get:|F|=|I| t= N 10 3s= 103N3. kgsteel ball strikes a wall with a speed of10msat an angle of60 withthe surface. It bounces off with the same speed and angle, as shown in Fig. the ball is in contact with the wall s, what is the average force exertedon the wall by the ball?

8 160 Chapter 7. Linear Momentum AND COLLISIONS750 m/smmmmm = 35 gxFigure :Simplifiedpicture of a machine gun spewing out bullets. An external force is necessary to holdthe gun in place!The average force is defined asF= p/ t, so first find the change in Momentum ofthe ball. Since the ball has the same speed before and after bouncing from the wall, it isclear that itsxvelocity (see the coordinate system in Fig. ) stays the same and so thexmomentum stays the same. But theymomentumdoeschange. The initialyvelocity isviy= (10ms) sin 60 = the finalyvelocity isvf y= +(10ms) sin 60 = + the change inymomentum is py=mvf y mviy=m(vf y viy) = ( kg)( ( )) = 52kg msThe averageyforce on the ball isFy= py t=Iy t=(52kg ms)( s)= 102 NSinceFhas noxcomponent, the average force has magnitude 102N and points in theydirection (away from the wall).

9 4. A machine gun gbullets at a speed If the gun can fire200 bullets/min, what is the average force the shooter must exert to keep the gunfrom moving?Whoa! Lots of things happening here. Let s draw a diagram andtry to sort things a picture is given in Fig. gun interacts with the bullets; it exerts a brief, strongforce on each of the bulletswhich in turn exerts an equal and opposite force on the gun s force changes thebullet s Momentum fromzero(as they are initially at rest) to the final value ofpf=mv= ( kg)(750ms) = WORKED EXAMPLES161so this is also thechangein Momentum for each , since 200 bullets are fired every minute (60 s), we shouldcount the interaction timeas the time to fireonebullet, t=60 s200= sbecause every s, a firing occurs again, and theaverageforce that we compute will bevalid for a length of time for which many bullets are fired.

10 So the average force of the gunon the bullets isFx= px t= s= NFrom Newton s Third Law, there must an average backwards forceof the bullets on thegunof magnitude N. If there were no other forces acting on the gun, it would acceleratebackward! To keep the gun in place, the shooter (or the gun s mechanical support) mustexert a force of N in the forward can also work with the numbers as follows: In one minute, 200 bullets were fired, andatotalmomentum ofP= (200)( ms) = 103 kg mswas imparted to them. So during this time period (60 seconds!) the average force on thewhole set of bullets wasFx= P t= 103kg s= before, this is also the average backwards force of thebullets on the gunand the forcerequired to keep the gun in Collisions5. gbullet is stopped in a block of wood (m= kg).