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Momentum and Collisions Linear Momentum The linear momentum of a particle or an object that can be modeled as a particle of mass m moving with a velocity v is defined to be the product of the mass and velocity: p=mv We will usually refer to this as “momentum”, omitting the “linear”. Momentum is a VECTOR. It has components. Don’t forget that. Linear Momentum The dimensions of momentum are ML/T The SI units of momentum are kg · m / s Momentum can be expressed in component form: px = m v x py = m v y pz = m v z A 3.00-kg particle has a velocity of . (a) Find its x and y components of momentum. (b) Find the magnitude and direction of its momentum. Newton and Momentum Newton called the product mv the quantity of motion of the particle Newton’s Second Law can be used to relate the momentum of a particle to the resultant force acting on it dv d mv dp F ma m dt dt dt with constant mass. Newton’s Second Law The time rate of change of the linear momentum of a particle is equal to the net force acting on the particle This is the form in which Newton presented the Second Law It is a more general form than the one we used previously This form also allows for mass changes Applications to systems of particles are particularly powerful Two Particles (1 dimension for now) dptotal dp1 dp2 dt dt dt IF there are NO external forces then the only forces acting on this system are those that are one particle producing a force on the other. Continuing dptotal dp1 dp2 dt dt dt dptotal Fexternal 0 Fint ernal 0 dt NO EXTERNAL FORCES: ptotal is constant Force on P1 is from particle 2 and is F21 Force on P2 is from particle 1 and is F12 F21=-F12 Summary The momentum of a SYSTEM of particles that are isolated from external forces remains a constant of the motion. Conservation of Linear Momentum Textbook Statement Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant The momentum of the system is conserved, not necessarily the momentum of an individual particle This also tells us that the total momentum of an isolated system equals its initial momentum Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them). A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. (a) What is the speed of the block of mass M? (b) Find the original elastic potential energy in the spring if M = 0.350 kg. An Easy One… A particle of mass m moves with momentum p. Show that the kinetic energy of the particle is K = p2/2m. (b) Express the magnitude of the particle’s momentum in terms of its kinetic energy and mass. Consider a particle roaming around that is suddenly subjected to some kind of FORCE that looks something like the last slide’s graph. NEWTON : p mv dp dv m ma F dt dt p constant if F 0 If F 0 then p will change Let’s drop the vector notation and stick to one dimension. dp F dt dp Fdt f tf i ti dp Fdt tf p f pi Fdt ti Change of Momentum Impulse NEW LAW IMPULSE = CHANGE IN MOMENTUM A 3.00-kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle (Fig. P9.9). If the ball is in contact with the wall for 0.200 s, what is the average force exerted on the ball by the wall? LETS TALK ABOUT Consider two particles: v2 v1 m1 m2 1 1 V1 2 2 V2 What goes on during the collision? N3 Force on m2 =F(12) Force on m1 =F(21) The Forces Equal and Opposite More About Impulse: F-t The Graph Impulse is a vector quantity The magnitude of the impulse is equal to the area under the force-time curve Dimensions of impulse are ML/T Impulse is not a property of the particle, but a measure of the change in momentum of the particle Impulse The impulse can also be found by using the time averaged force I = F Dt This would give the same impulse as the time-varying force does SKATEBOARD DEMO Conservation of Momentum, Archer Example The archer is standing on a frictionless surface (ice) Approaches: Newton’s Second Law – no, no information about F or a Energy approach – no, no information about work or energy Momentum – yes Archer Example, 2 Let the system be the archer with bow (particle 1) and the arrow (particle 2) There are no external forces in the x-direction, so it is isolated in terms of momentum in the x-direction Total momentum before releasing the arrow is 0 The total momentum after releasing the arrow is p1f + p2f = 0 Archer Example, final The archer will move in the opposite direction of the arrow after the release Agrees with Newton’s Third Law Because the archer is much more massive than the arrow, his acceleration and velocity will be much smaller than those of the arrow An estimated force-time curve for a baseball struck by a bat is shown in Figure P9.7. From this curve, determine (a) the impulse delivered to the ball, (b) the average force exerted on the ball, and (c) the peak force exerted on the ball. Overview: Collisions – Characteristics We use the term collision to represent an event during which two particles come close to each other and interact by means of forces The time interval during which the velocity changes from its initial to final values is assumed to be short The interaction force is assumed to be much greater than any external forces present This means the impulse approximation can be used Collisions – Example 1 Collisions may be the result of direct contact The impulsive forces may vary in time in complicated ways This force is internal to the system Momentum is conserved Collisions – Example 2 The collision need not include physical contact between the objects There are still forces between the particles This type of collision can be analyzed in the same way as those that include physical contact Types of Collisions In an elastic collision, momentum and kinetic energy are conserved Perfectly elastic collisions occur on a microscopic level In macroscopic collisions, only approximately elastic collisions actually occur In an inelastic collision, kinetic energy is not conserved although momentum is still conserved If the objects stick together after the collision, it is a perfectly inelastic collision Collisions, cont In an inelastic collision, some kinetic energy is lost, but the objects do not stick together Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types Momentum is conserved in all collisions Perfectly Inelastic Collisions Since the objects stick together, they share the same velocity after the collision m1v1i + m2v2i = (m1 + m2) vf A 10.0-g bullet is fired into a stationary block of wood (m = 5.00 kg). The relative motion of the bullet stops inside the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.600 m/s. What was the original speed of the bullet? . Velcro couplers make the carts stick together after colliding. Find the final velocity of the train of three carts. (b) What If ? Does your answer require that all the carts collide and stick together at the same time? What if they collide in a different order? Most of us know intuitively that in a head-on collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that experienced by the truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false. Newton’s third law tells us that both objects experience forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at 8.00 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 80.0 kg. Including the drivers, the total vehicle masses are 800 kg for the car and 4 000 kg for the truck. If the collision time is 0.120 s, what force does the seatbelt exert on each driver? SPRING BREAK BREAK March 20, 2006 Welcome Home!! • The pace now quickens! • 5 weeks (+ 1 session) remain in the semester • Next Exam March 31 – Details on next 2 slides • This week we return to momentum • Watch for another WA soon • Quiz on Friday - momentum APPROXIMATE SCHEDULE 20Mar Ch 9: Momentum Ch 9: Momentum 27Mar Ch 10: Rotation Ch 10: Rotation 3-Apr Ch 10: Rotation Ch 9: Momentum EXAM #3 Ch 11: Angular Mom. Ch 11: Angular Mom. 10Apr Ch 11: Angular Mom. Ch 11: Angular Mom.+ Chapt 12. Ch 12: Statics 17Apr Ch 12: Statics Ch 15: Oscillations Ch 15: Oscillations 24Apr Ch 15: Oscillations 1May FINAL EXAM No Class No Class No Class No Class Next Exam • Will Include – Potential Energy Chapter – Conservation of Momentum – Rotation – Material covered prior to exam • Includes Wednesday before exam! SORRY! Now Where Were We??? Oh yeah …. Elastic Collisions Both momentum and kinetic energy are conserved m1v1i m2 v 2 i m1v1 f m2 v 2 f 1 1 2 m1v1i m2 v 22i 2 2 1 1 2 m1v1 f m2 v 22 f 2 2 Elastic Collisions, cont Typically, there are two unknowns to solve for and so you need two equations The kinetic energy equation can be difficult to use The energy equation, along with conservation of momentum, can be used to solve for the two unknowns It can only be used with a one-dimensional, elastic collision between two objects The solution is shown on pages 262-3 in the textbook. (Lots of algebra but nothing all that difficult. We will look at a special case. One Dimension Energy is Conserved Let’s look at the case of equal masses. Second Particle initially at rest Explains the demo! Elastic Collisions, final Example of some special cases m1 = m2 – the particles exchange velocities When a very heavy particle collides head-on with a very light one initially at rest, the heavy particle continues in motion unaltered and the light particle rebounds with a speed of about twice the initial speed of the heavy particle When a very light particle collides head-on with a very heavy particle initially at rest, the light particle has its velocity reversed and the heavy particle remains approximately at rest Collision Example – Ballistic Pendulum Perfectly inelastic collision – the bullet is embedded in the block of wood Momentum equation will have two unknowns Use conservation of energy from the pendulum to find the velocity just after the collision Then you can find the speed of the bullet As shown in the figure, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? Two Dimensional Collisions Two-Dimensional Collisions The momentum is conserved in all directions Use subscripts for identifying the object indicating initial or final values the velocity components If the collision is elastic, use conservation of kinetic energy as a second equation Remember, the simpler equation can only be used for onedimensional situations Two-Dimensional Collision, example Particle 1 is moving at velocity v1i and particle 2 is at rest In the x-direction, the initial momentum is m1v1i In the y-direction, the initial momentum is 0 Two-Dimensional Collision, example cont After the collision, the momentum in the xdirection is m1v1f cos q m2v2f cos f After the collision, the momentum in the ydirection is m1v1f sin q m2v2f sin f Problem-Solving Strategies – TwoDimensional Collisions Set up a coordinate system and define your velocities with respect to that system It is usually convenient to have the x-axis coincide with one of the initial velocities In your sketch of the coordinate system, draw and label all velocity vectors and include all the given information Problem-Solving Strategies – TwoDimensional Collisions, 2 Write expressions for the x- and y-components of the momentum of each object before and after the collision Remember to include the appropriate signs for the components of the velocity vectors Write expressions for the total momentum of the system in the x-direction before and after the collision and equate the two. Repeat for the total momentum in the y-direction. Problem-Solving Strategies – TwoDimensional Collisions, 3 If the collision is inelastic, kinetic energy of the system is not conserved, and additional information is probably needed If the collision is perfectly inelastic, the final velocities of the two objects are equal. Solve the momentum equations for the unknowns. Problem-Solving Strategies – TwoDimensional Collisions, 4 If the collision is elastic, the kinetic energy of the system is conserved Equate the total kinetic energy before the collision to the total kinetic energy after the collision to obtain more information on the relationship between the velocities Two-Dimensional Collision Example Before the collision, the car has the total momentum in the xdirection and the van has the total momentum in the ydirection After the collision, both have x- and ycomponents Two automobiles of equal mass approach an intersection. One vehicle is traveling with velocity 13.0 m/s toward the east and the other is traveling north with speed v2i. Neither driver sees the other. The vehicles collide in the intersection and stick together, leaving parallel skid marks at an angle of 55.0o north of east. The speed limit for both roads is 35 mi/h and the driver of the northward-moving vehicle claims he was within the speed limit when the collision occurred. Is he telling the truth? New Topic: Center of Mass Momentum of a SYSTEM of particles. The Center of Mass There is a special point in a system or object, called the center of mass, that moves as if all of the mass of the system is concentrated at that point The system will move as if an external force were applied to a single particle of mass M located at the center of mass M is the total mass of the system Consider a SYSTEM of N particles There are N point particles. Each point “I” has a mass mi The TOTAL mass of the system is given by M where M=mi Let the FORCE on particle “I” be given by Fi, a VECTOR Quantity. Let’s see what Newton says about such a SYSTEM The MATH … Newton : Fsystem Ma some kind of average Fsystem is the TOTAL force on the system of particles asome… is some average acceleration that we need to deal with. We use a vector approach … Consider a system of only two particles external 1 F F2 on 1 m1a1 F2external F1 on 2 m2a 2 F2 on 1 F1 on 2 F external F1external F2external F external m1a1 m2a 2 F external m1a1 m2a 2 2 2 d d m1r1 m2r2 external F 2 (m1r1 m2r2 ) M 2 ( ) dt dt M where M m1 m2 This would look like F ma if we define the center of mass vector R as m1r1 m2r2 R CM M and F Ma cm We therefore define the center of mass as 1 R CM m1ri M i M mi i Center of Mass, Coordinates The coordinates of the center of mass are xCM m x i i i M yCM m y i i M i zCM m z i i i M where M is the total mass of the system Center of Mass, position The center of mass can be located by its position vector, rCM rCM m r i i i M ri is the position of the i th particle, defined by ri xi ˆi yi ˆj zi kˆ Center of Mass, Example Both masses are on the x-axis The center of mass is on the x-axis The center of mass is closer to the particle with the larger mass Center of Mass, Extended Object Think of the extended object as a system containing a large number of particles The particle distribution is small, so the mass can be considered a continuous mass distribution Center of Mass, Extended Object, Coordinates The coordinates of the center of mass of the object are xCM 1 1 x dm yCM M M 1 zCM z dm M y dm Center of Mass, Extended Object, Position The position of the center of mass can also be found by: rCM 1 r dm M The center of mass of any symmetrical object lies on an axis of symmetry and on any plane of symmetry Density r mass r unit - volume m rV dm rdv Usually, r is a constant but NOT ALWAYS! Center of Mass, Example An extended object can be considered a distribution of small mass elements, Dm=rdxdydz The center of mass is located at position rCM For a flat sheet, we define a mass per unit area … s mass s unit volum e dm sda Linear Density m m mL dm mdx Center of Mass, Rod Find the center of mass of a rod of mass M and length L The location is on the x-axis (or yCM = zCM = 0) xCM = L / 2 A golf club consists of a shaft connected to a club head. The golf club can be modeled as a uniform rod of length L and mass m1 extending radially from the surface of a sphere of radius R and mass m2. Find the location of the club’s center of mass, measured from the center of the club head. R L m2 m1 Motion of a System of Particles Assume the total mass, M, of the system remains constant We can describe the motion of the system in terms of the velocity and acceleration of the center of mass of the system We can also describe the momentum of the system and Newton’s Second Law for the system Velocity and Momentum of a System of Particles The velocity of the center of mass of a system of particles is mv vCM dr CM dt i i i M The momentum can be expressed as Mv CM m v i i i p i p tot i The total linear momentum of the system equals the total mass multiplied by the velocity of the center of mass Acceleration of the Center of Mass The acceleration of the center of mass can be found by differentiating the velocity with respect to time aCM dv CM 1 dt M m a i i i Forces In a System of Particles The acceleration can be related to a force MaCM Fi i If we sum over all the internal forces, they cancel in pairs and the net force on the system is caused only by the external forces Newton’s Second Law for a System of Particles Since the only forces are external, the net external force equals the total mass of the system multiplied by the acceleration of the center of mass: Fext = M aCM The center of mass of a system of particles of combined mass M moves like an equivalent particle of mass M would move under the influence of the net external force on the system Momentum of a System of Particles The total linear momentum of a system of particles is conserved if no net external force is acting on the system MvCM = ptot = constant when Fext = 0 Motion of the Center of Mass, Example A projectile is fired into the air and suddenly explodes With no explosion, the projectile would follow the dotted line After the explosion, the center of mass of the fragments still follows the dotted line, the same parabolic path the projectile would have followed with no explosion Rocket Propulsion The operation of a rocket depends upon the law of conservation of linear momentum as applied to a system of particles, where the system is the rocket plus its ejected fuel Rocket Propulsion, 2 The initial mass of the rocket plus all its fuel is M + Dm at time ti and velocity v The initial momentum of the system is pi = (M + Dm) v Rocket Propulsion, 3 At some time t + Dt, the rocket’s mass has been reduced to M and an amount of fuel, Dm has been ejected The rocket’s speed has increased by Dv Rocket Propulsion, 4 Because the gases are given some momentum when they are ejected out of the engine, the rocket receives a compensating momentum in the opposite direction Therefore, the rocket is accelerated as a result of the “push” from the exhaust gases In free space, the center of mass of the system (rocket plus expelled gases) moves uniformly, independent of the propulsion process Rocket Propulsion, 5 The basic equation for rocket propulsion is Mi v v v ln rocket f i e in increase M f The speed is proportional to the speed of the escape gases (ve) So, the exhaust speed should be very high The increase in rocket speed is also proportional to the natural log of the ratio Mi/Mf So, the ratio should be as high as possible, meaning the mass of the rocket should be as small as possible and it should carry as much fuel as possible Thrust The thrust on the rocket is the force exerted on it by the ejected exhaust gases dv dM Thrust = M v dt e dt The thrust increases as the exhaust speed increases The thrust increases as the rate of change of mass increases The rate of change of the mass is called the burn rate