# Chapter 7 Linear Momentum Definition Of Linear Momentum M

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Chapter 7 Linear MomentumDefinition of linear momentum An object with mass m and moving with velocity v has linear momentum p p mvMany Particles If many particles m1, m2, etc. are moving with velocities v1 , v 2 , etc., the total linear momentumof the system is the vector sum of the individual momenta, p pi mi v iiiLinear momentum is a vector quantity. We have to use our familiar rules for vector additionwhen dealing with momentum.Impulse-Momentum TheoremStarting with Newton’s second law F maand using the definition of acceleration va twe have v F m tIf the mass is constant, we can pull it inside the operator (mv ) F t p F t p F tThis states that the change in linear momentum is caused by the impulse. The quantity

F tis called the impulse. For situations where the force is not constant, we use the average force, impulse Fav tRestatement of Newton’s second lawThere is a more general form of Newton’s second law p F limt 0 tThe net force is the rate of change of momentum.The applications of the impulse-momentum theorem are unlimited. In an automobile we havecrumple zones, air bags, and bumpers. What do seat belts do?Here is another example:

The idea is to lengthen the time t during which the force acts, so that the force is diminishedwhile changing the momentum a defined amount.Problem 10. What average force is necessary to bring a 50.0-kg sled from rest to a speed of 3.0m/s in a period of 20.0 s. Assume frictionless ice.Solution: A sketch of the situation would beNfmgWe would use the impulse-momentum theorem p F tTo use a vector equation we need to take components px Fx tThe change in momentum is px p fx pix mv fx mvix 0 (50 kg )(3.0 m/s) 150kg m/s px Fx t Fav tFav px 150 kg m/s 7.5 kg m/s 2 7.5 N t20 sWhat is the meaning of the negative sign?Conservation of Linear MomentumConsider the collision between two pucks. When they collide, they exert forces on each other, F12 and F21By Newton’s third law, F12 F21

The total force acting on both pucks is then F F12 F21 0By the impulse-momentum theorem p F t 0Which leads to pi p fIn a system composed of more than two objects, interactions between objects inside the systemdo not change the total momentum of the system – they just transfer some momentum from onepart of the system to another. Only external interactions can change the total momentum of thesystem. The total momentum of a system is the vector sum of the momenta of each object in thesystem External interactions can change the total momentum of a system. Internal interactions do not change the total momentum of a system.The Law of Conservation of Linear MomentumIf the net external force acting on a system is zero, then the momentum of the system isconserved.If Fext 0, pi p f

Linear momentum is always conserved for an isolated system. Of course, we deal withcomponentspix p fxandpiy p fyI like to say that momentum is conserved in collisions and explosions.Note: The total momentum of the system is conserved. The momentum of an individual particlecan change.Problem 18. A rifle has a mass of 4.5 kg and it fires a bullet of mass 10.0 g at a muzzle speed of820 m/s. What is the recoil speed of the rifle as the bullet leaves the gun barrel?Solution: Draw a sketch of the initial and final situations.vrInitialvbFinalSince only internal forces act, linear momentum is conserved. pi p fThe motion is in the x-direction,pix p fxInitially, everything is at rest and pix 0. The final momentum is the vector sum of the momentaof the bullet and the rifle. From the diagram,p fx mbvb mr vrUsing conservation of momentumpix p fx0 mb vb mr vrvr mb vb (0.010 kg )(820 m/s) 1.82 m/smr4.5 kgThe heavier the rifle, the smaller the recoil speed.

Center of MassWe have seen that the momentum of an isolated system is conserved even though parts of thesystem may interact with other parts; internal interactions transfer momentum between parts ofthe system but do not change the total momentum of the system. We can define point called thecenter of mass (CM) that serves as an average location of the system.What if a system is not isolated, but has external interactions? Again imagine all of themass of the system concentrated into a single point particle located at the CM. The motion ofthis fictitious point particle is determined by Newton’s second law, where the net force is thesum of all the external forces acting on any part of the system. In the case of a complex systemcomposed of many parts interacting with each other, the motion of the CM is considerablysimpler than the motion of an arbitrary particle of the system.For the two particles pictured above, the CM isxCM m1 x1 m2 x2 m1 x1 m2 x2 m1 m2MNotice that the CM is closer to the more massive object.For many particles the definition is generalized to rCM m ri iM

The more useful component form isxCM m xi iMyCM m yizCM iM m zi iMProblem 33. Find the x-coordinate of the CM of the composite object shown in the figure. Thesphere, cylinder, and rectangular solid all have uniform composition. Their masses anddimensions are: sphere: 200 g, diameter 10 cm; cylinder: 450 g, length 17 cm, radius 5.0cm; rectangular solid: 325 g, length in x-direction 16 cm, height 10 cm, depth 12 cm.Solution: Because the different pieces have uniform composition, the center of mass of eachpiece is located at the geometrical center of that piece. (See the note at the top of page 240.) Forthe sphere, ms 200 g, xs 5 cm. For the cylinder, mc 450 g, xc 10 cm 8.5 cm 18.5 cm.For the rectangular solid, mr 325 g, xr 10 cm 17 cm 8 cm 35 cm. The center of mass isxCM m xi iMm x mc xc mr xr s sms mc mr(200 g)(5 cm) (450 g)(18.5 cm) (325 g)(35 cm)(200 g) (450 g) (325 g) 21.2 cm Motion of the Center of MassHow is the velocity of the CM related to the velocities of the various parts of the system ofparticles? During a short time interval t, each particle changes position by ri . The CM changes rCM m riiMDivide each side by the time interval t r m i rCM i t tM

In general, the velocity is defined as rv tUsing the definition of velocity v CM m viiM Mv CM mi v iThe right hand side is the total linear momentum of the system of particles. We have the veryuseful relation p total Mv CMFor a complicated system consisting of many particles moving in different directions, the totallinear momentum of the system can be found from the total mass of the system and the velocityof the center of mass.We have shown that for an isolated system, the total linear momentum of the system isconserved. In such system, the equation above implies that the CM must move with constantvelocity regardless of the motions of the individual particles. On the other hand, what if thesystem is not isolated? If a net force acts on a system, the CM does not move with constantvelocity. Instead, it moves as if all the mass were concentrated there into a fictitious pointparticle with all the external forces acting on that point. The motion of the CM obeys thefollowing statement of Newton’s second law: Fext MaCMA complicated system is reduced to treating the system as a point particle located at its center ofmass reacting only to external forces!Collisions in One DimensionIn general, conservation of momentum is not enough to predict what will happen after acollision. Here is a collision between to objects with the same mass. Many other outcomes arepossible besides the two given,

Some vocabulary A collision in which the total kinetic energy is the same before and after is called elastic. When the final kinetic energy is less than the initial kinetic energy, the collision is said tobe inelastic. Collisions between two macroscopic objects (for example, billiard balls) aregenerally inelastic to some degree, but sometimes the change in kinetic energy is so smallthat we treat them as elastic. When a collision results in two objects sticking together, the collision is perfectlyinelastic. The decrease of kinetic energy is a perfectly inelastic collision is as large aspossible (consistent with the conservation of momentum).Problem-Solving Strategy for Collisions Involving Two Objects (page 245)1. Draw before and after diagrams of the collision2. Collect and organize information on the masses and velocities of the two objects beforeand after the collision. Express the velocities in component form (with correct algebraicsigns).3. Set the sum of the momenta of the two before and after the collision equal to the sum ofthe momenta after the collision. Write one equation for each direction:m1v1ix m2v2ix m1v1 fx m2v2 fxm1v1iy m2v2iy m1v1 fy m2v2 fy4. If the collision is known to be perfectly inelastic, set the final velocities equal:v1 fx v2 fx and v1 fy v2 fy

5. If the collision is known to be perfectly elastic, then set the final kinetic energy equal tothe initial kinetic energy:12m1v1i 12 m2v2i 12 m1v1 f 12 m2v2 f22226. Solve for the unknown quantities.Example: A one-dimensional elastic tion of momentum givesm1v1i m2v2i m1v1 f m2v2 fElastic collision means that kinetic energy is conserved12m1v1i 12 m2v2i 12 m1v1 f 12 m2v2 f2222Cancelling the ½ and grouping like masses on the same side of the equation,m1 (v1i v1 f ) m2 (v2 f v2i )2222Doing the same with the linear momentum equationm1 (v1i v1 f ) m2 (v2 f v2i )Divide the two equationsm1 (v1i v1 f )22m1 (v1i v1 f )m2 (v2 f v2i )2 2m2 (v2 f v2i )v1i v1 f v2 f v2iThis can be rewritten asv1i v2i v2 f v1 fThe equation states that m1 approaches m2 with the same speed as m2 moves away from m1 afterthe collision. Important result. Solve the above for v2f

v2 f v1 f v1i v2iand substitute into the momentum equationm1v1i m2v2i m1v1 f m2v2 f m1v1 f m2 (v1 f v1i v2i ) m1v1 f m2v1 f m2v1i m2v2i(m1 m2 )v1i 2m2v2i (m1 m2 )v1 f m m2 2m2 v1i v2iv1 f 1 m1 m2 m1 m2 Using very similar reasoning, we can find an expression for the final speed of m2 2m1 m m1 v1i 2 v2iv2 f m1 m2 m1 m2 Be careful when you use these equations. A mass moving to the left would have a negativespeed.If m2 is initially at rest, v2i 0 and we have m m2 v1iv1 f 1m m12 2m1 v1iv2 f m1 m2 After the collision, m2 moves to the right.If m1 m2, m1 continues to move to the right.If m1 m2, m1 stops and m2 to move to the right with speed v1i.If m1 m2, m1 bounces back to the left.Example: Two dimensional elastic collision between two identical masses. One mass is initiallyat rest.m1v1im2v1fm1m2v2f

Elastic collisions conserve kinetic energy12m1v1i 12 m1v1 f 12 m2v2 f222Cancelling the ½ and the masses, since m1 m2,v1i v1 f v2 f222The conservation of linear momentum condition is m1v1i m1v1 f m2 v 2 fAgain, we can cancel the masses v1i v1 f v 2 fThe two boxed equations can be interpreted by the picture v1 f v2 f v1iThe velocities of the two masses after collision are perpendicular to each other.

Linear momentum is always conserved for an isolated system. Of course, we deal with components ix fx iy fy p p and p p I like to say that momentum is conserved in collisions and explosions. Note: The total momentum of the system is conserved. The momentum of an individual particle can change. Problem 18.

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