Goals For Chapter 7 Chapter 7 - University Of Central Florida

Goals for Chapter 7Chapter 7Impulse and MomentumWhen the bat strikes the ball, the magnitude of the force exertedon the ball rises to a maximum value and then returns to zero To study impulse and momentum. To understand conservation of momentum. To study momentum changes duringcollisions. To understand center of mass and how forcesact on the c.o.m. To apply momentum to rocket propulsion.DEFINITION OF IMPULSEThe impulse of a force is the product of the averageforce and the time interval during which the force acts: J F tThe collision time between a batand a ball is very short, often lessthan a millisecond, but the forcecan be quite large.Impulse is a vector quantity and has thesame direction as the average force.The time interval duringwhich the force acts is t,and the magnitude of theaverage force is F.Momentum transfer (collision) ‘timescales’SI unit:newton seconds (N s)Force and Impulse Collisions typically involveinteractions that happen quickly.Fsame areaF During this brief time, the forcesinvolved can be quite largeti t t big, F smallvfttf J F tvit tVfFinitialfinalThe balls are in contact for a very short time.titf t small, F big

DEFINITION OF LINEAR MOMENTUMLinear MomentumThe linear momentum of an object is the product of theobject’s mass times its velocity: p mvSI unit:kilogram meter/second (kg m/s)Momentum of a particle is defined as the product ofits mass and velocity p mv Momentum components– px m vx and py m vy– Applies to two-dimensional motion as wellMomentum (magnitude) is related to kinetic energyK Relation between Impulse and Momentum (Newton 2nd) F ma vf voa tWhen a net force acts on an object, the impulse of thisforce is equal to the change in the momentum of the objectfinal momentum mv f mv o F tfsuch that (even ifis not constant),impulse is given by Fav J Fav ( t f t i ) Fav tor: pFav t oinitial momentum F t mv f mv o J p F tImpulse and average force We can use the notion of impulse to define “averageforce”, which is a useful concept.Define average forceIMPULSE-MOMENTUM THEOREMimpulse F t mv mv 1p2mv 2 22mImpulse is a vector quantity;SI Unit: N s or kg m / simpulse change in momentum!Example: A Well Hit BallA baseball (m 0.14 kg) has an initial velocity of vo - 38m/s as it approaches a bat. The bat applies an average forcethat is much larger than the weight of the ball, and the balldeparts from the bat with a final velocity of vf 58 m/s .(a) Determine the impulse applied to the ball by the bat.(b) Assuming that the time of contact is t 1.6 10-3 s,find the average force exerted on the ball by the bat.

Example 2: A RainstormConceptual Example: Hailstones versus RaindropsRain comes down with a velocity of -15 m/s and hits the roof of acar. The mass of rain per second that strikes the roof of the car is0.060 kg/s. Assuming that rain comes to rest upon striking thecar, find the average force exerted by the rain on the roof.Instead of rain, suppose hail is falling. Unlike rain, hail usuallybounces off the roof of the car. F t mv mv fIf hail fell instead of rain, would the force be smaller than,equal to, or greater than that calculated previously ? J F t mv f mv o mv f mv oF toNeglecting the weight of the raindrops,the net force on a raindrop is simply theforce on the raindrop due to the roof. m F vo t F t mv f mv o F 0.060 kg s 15 m s 0.90 N (force on the raindrop)Fon-the-roof -0.90 F(Newton’s third law)Impulse applied to auto collisions The most important factor is the collision time orthe time it takes the person to come to a rest– This will reduce the chance of dying in a car crash Ways to increase the time– Seat belts– Air bags p m ( v f v i ) Fav t tThe air bag increases the time of the collision andabsorbs some of the energy from the body Conservation of Linear Momentum F t mv f mv oOBJECT 1 W F t m v 1121f1 m1 v o1For a raindrop, the change in velocity is from (downward) to zero.For a hailstone, the change is from (downward) to (upward).Thus hailstones have a largerConservation of Linear MomentumWORK-ENERGY THEOREM CONSERVATION OF ENERGYApply the impulse-momentum theoremto the midair collisionbetween two objects . THEOREM ?IMPULSE-MOMENTUMInternal forces – Forces that objectswithin the system exert on each other.External forces – Forces exerted one.g. Weight Wobjects by agents external to the system.Conservation of Linear Momentum W F t m v W F t m v W2 F21 t m 2 v f 2 m 2 v o 2112221212 W W F 1OBJECT 2 F t12f1 m1 v o1f2 m2 vo2Consider system:both objects involved F21 t m1 v f 1 m 2 v f 2 m1 v o1 m 2 v o 2 F12 F21 PfThe internal forces cancel out. Po

Principle of Conservation of Linear Momentum W1 W2 t Pf Po sum of average external forces t Pf PoIf the sum of the external forces is zero, then 0 Pf PoDefinition of Total Momentum for a System of Particles Pf PoFor a system of particles the totalmomentum P is the vector sum ofthe individual particle momenta: P p A p B p C .N N P pi m iv ii 1i 1 m A v A m B v B m C v C .components of total momentumPx p A , x p B,x .Py p A , y p B.y . P (p A p B p C .) Fnet tCONSERVATION OF LINEAR MOMENTUMThe total linear momentum of an isolated system is constant(conserved). An isolated system is one for which the sum of theaverage external forces acting on the system is zero. The momentum of each object will change The total momentum of the system remainsconstant if there are no external forcesExample: Assembling a Freight TrainExample: Recoil of a rifleA freight train is being assembled in a switching yard. Car 1 has a massof m1 65 103 kg and moves at a velocity of v01 0.80 m/s. Car 2,with a mass of m2 92 103 kg and a velocity of v02 1.3 m/s,overtakes car 1 and couples to it. Neglecting friction, find the commonvelocity vf of the cars after they become coupled.A marksman holds a 3.00 kg rifleloosely, allowing it to recoil freelywhen fired, and fires a bullet ofmass 5.00 g horizontally with aspeed vB 300 m/s. What is therecoil speed of the rifle ?PT P R P BBefore AfterPT 0PR PB 0mB v B m R v R 0vR mB0.005 vB 300 0.50 m / s.mR3.00Concept Test: Exploding ProjectileExample: Momentum ConservationA model rocket travels as a projectile in a parabolic path after itsfirst stage burns out. At the top of its trajectory, where its velocitypoints horizontally to the right, a small explosion separates it intotwo sections with equal masses. One section falls straight down,with no horizontal motion. What is the direction of the other partjust after the explosion ?A box with mass m 6.0 kg slides with speed v 4.0 m/s across africtionless floor in the positive direction of an x axis. It suddenlyexplodes into two pieces. One piece, with mass m1 2.0 kg, movesin the positive x-direction with speed v1 8.0 m/s.A. Up and to the leftbeforeB. Straight upC. Up and to the rightafterWhat is the velocity of the second piece, with mass m2 4.0 kg ?

Example: Conservation of Linear Momentum - Ice SkatersStarting from rest, two skaters pushoff against each other on ice wherefriction is negligible. One is a 54-kgwoman and one is a 88-kg man. Thewoman moves away with a speed of 2.5 m/s. Find the recoil velocityof the man.fo P Pm1v f 1 m 2 v f 2 0vf 2 m1v f 1m2 54 kg 2.5 m 88 kgvf 2 ms 1.5sIn Collissions Total Momentum is ConservedIn collisions, we assume that external forces eithersum to zero, or are small enough to be ignored.Hence, momentum is conserved in all collisions.– A collision may be the result of physical contact betweentwo objects– “Contact” may also arise from the electrostatic interactionsof the electrons in the surface atoms of the bodies Mathematically (for two objects):m1 v1i m 2 v 2i m1 v1f m 2 v 2 f––––Momentum is conserved for the system of objectsThe system includes all the objects interacting with each otherAssumes only internal forces are acting during the collisionCan be generalized to any number of objectsElastic CollisionsA boy stands at one end of a floating raft that is stationaryrelative to the shore. He then walks to the opposite end,towards the shore. Does the raft move (assume no friction)?1. No, it will not move at all2. Yes, it will move away from the shore 3. Yes, it will move towards the shoreNote: Since momentum is conserved in the boy-raft system and neither wasmoving at first, the raft must move in the direction opposite to theboy’s.Types of Collisions Momentum is conserved in any collision Elastic collisions– both momentum and kinetic energyare conserved Inelastic collisions– Kinetic energy is not conserved Some of the kinetic energy is convertedinto other types of energy such as heat,sound, work to permanently deform anobject– Perfectly inelastic collisions occur whenthe objects stick together Not all of the KE is necessarily lostMost collisions fall between elasticand perfectly inelastic collisionsInelastic and Elastic Collisions Elastic means that kinetic energy is conserved as wellas momentum. This gives us more constraints– We can solve more complicated problems!!– Billiards (2-D collision)– The colliding objectshave separate motionsafter the collision aswell as before.Initial First: simpler 1-D problemConcept Test: Conservation of MomentumFinalA completely inelastic collisionAn elastic collision

Applying the Principle of Conservation of Momentum1. Decide which objects are included in the system.2. Relative to the system, identify the internal andexternal forces.3. Verify that the system is isolated.4. Set the final momentum of the system equal to itsinitial momentum.Problem Solving for One-Dimensional Collisions Set up a coordinate axis and define the velocities with respectto this axis– It is convenient to make your axis coincide with one of theinitial velocities In your sketch, draw all the velocity vectors with labelsincluding all the given information Draw “before” and “after” sketches Label each object– include the direction of velocity– keep track of subscriptsRemember that momentum is a vector.Problem Solving for One-Dimensional CollisionsSketch for perfectly inelastic colllision The objects stick together Include all the velocity directions The “after” collision combines the masses Write the expressions for the momentumof each object before and after the collision– Remember to include the appropriate signs Write an expression for the total momentum before and afterthe collision --- momentum of the system is conserved If the collision is inelastic, solve the momentum equation forthe unknown --- Remember, KE is not conserved If the collision is elastic, you can use the KE equation to solvefor two unknownsPerfectly Inelastic Collisions What amount of KE lostduring collision?11m1v12i m 2 v 22i221 (1000 kg )(50 m s) 2 1.25 10 6 J2KE before 1KE after (m1 m 2 ) v f221 ( 2500 kg)(20 m s) 2 0.50 106 J2 KElost 0.75 10 6 Jlost in heat and sound Perfectly Inelastic Collisions Suppose, for example, v2i 0.Conservation of momentum becomesm1v1i m 2 v 2i (m1 m 2 ) v fm1v1i 0 (m1 m 2 ) v fE.g., if m1 1000 kg, m 2 1500 kg :(1000kg )(50 m s) 0 (2500kg) v f ,vf 5 10 4 kg m s 20 m s.2.5 103 kgExample: A Ballistic PendulumThe mass of the block of wood is 2.50-kgand the mass of the bullet is 0.0100-kg.The block swings to a maximum height of0.650 m above the initial position.Find the initial speed of the bullet.Apply conservation of momentum to the collision:m1v f 1 m 2 v f 2 m1v o1 m 2 v o 2 m1 m 2 v fv o1 m1v o1 m1 m 2 v fm1