MOMENTUM AND IMPULSE Chapter 7 Linear Momentum, p Momentum is a measure of how hard it is to stop or turn a moving object. It is moving inertia. p mv (single particle) ptotal pn (system of particles) Units of momentum: kg(m/s) or (N)(S) Since velocity, v, is a vector, momentum, p, is a vector. p is in the same direction as v Linear Momentum, p p mv Which car has more momentum? A or B A B The faster car, A. Linear Momentum, p p mv Which car has more momentum? A or B A B The more massive vehicle, B. Newton’s 2nd Law mv m(v v0 ) m v m v0 F m a t t t p F t The rate of change of momentum of a body is equal to the net force applied to it. Example - Washing a car: momentum change and force. Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 m/s and is aimed at the side of a car, which stops it (that is we ignore any splashing back). What is the force exerted by the water on the car? In each second, 1.5 kg of water v 20m / s leaves hose and has v=20 m/s. pi mvi 1.5(20) 30kg m / s p f mvf 0 px p f pi 0 30 Fx t t 1 30N By Newton’s 3rd law, force exerted by water on the car is +30 N Example - Washing a car: momentum change and force. What if the water splashes back from the car? Would the force on the car be more or less? pi pf In each second, 1.5 kg of water leaves hose and has v=20 m/s. pi mvi 30kg m / s If the water splashes back p f mvf 0 Change in momentum would be greater and so the force should be greater. The car exerts a force on the water not only to stop it, but an extra force to give it momentum in the opposite direction vf vi Newtons 2nd Law v Fnet p mv t t Slope ~ Fnet Fnet Impulse, J Contact begins Contact ends Fnet t J p area = J=p Impulse-Momentum Theorem p Fnet t J Fnet t p Impulse is the product of a net external force and time which results in a change in momentum Units are N s or kg m/s Impulsive forces are generally of high magnitude and short duration. Impulse-Momentum Theorem J Fnet t p “riding the punch” Impulse is the area under an F-t graph. 5 F (N) 4 3 2 1 0 0 2 4 6 t (sec) 8 10 DO NOW – Force to stop a car: momentum change, force and impulse. A 2200 kg vehicle traveling at 26 m/s can be stopped in 21 s by gently applying the brakes. It can be stopped in 3.8 s if the driver slams on the brakes, or in 0.22s if it hits a concrete wall. What impulse is exerted on the vehicle in eachof these stops? What net force is exerted in each case? v 26m / s For all three J x Fx t px p fx pix J x m(v fx vix ) 2200(0 26) 57,200 kg m / s J x DO NOW – Force to stop a car: momentum change, force and impulse. What net force is exerted on the vehicle in each of these stops? v 26m / s STOP BY: Gentle brake Slam brake J x Fnetxt Jx Fx t Fnet J 57,200 2734 N t 21 0.13 Gs Fnet J 57,200 15,053 N t 3.8 0.70 Gs Concrete wall Fnet J 57,200 260 ,000 N t 0.22 12 Gs Problem: This force acts on a 1.2kg object moving at 120.0m/s. The direction of force is aligned with velocity. What is the new velocity of the object? vf = 328 m/s 3000 Fnet (N) 2000 1000 0 0 0.2 0.4 t (s) 0.6 0.8 1 Problem: This force acts on a 1.2kg object moving at 120.0m/s. The direction of force is aligned with velocity. What is the new velocity of the object? 0 0.2 0.4 t (s) 0.6 0.8 1 0 F (N) -1000 -2000 vf = -88.3 m/s -3000 Impulse Impulse EXAMPLE: A 100 g ball is dropped from a height of h = 2.00 m above the floor. It rebounds vertically to a height of h'= 1.50 m after colliding with the floor. (a) Find the momentum of the ball immediately before it collides with the floor and immediately after it rebounds, (b) Determine the average force exerted by the floor on the ball. Assume that the time interval of the collision is 0.01 sec. pA=0 2m pD=0 1.5m pC pB p=? Fnet=? Example A 100 g ball is dropped from a height of h = 2.00 m above the floor. It rebounds vertically to a height of h'= 1.50 m after colliding with the floor. (a) Find the momentum of the ball immediately before it collides with the floor and immediately after it rebounds, (b) Determine the average force exerted by the floor on the ball. Assume that the time interval of the collision is 0.01 seconds. E A EB U GA K B 2 m ghA 1 m vB 2 vB 2 gho 2 * 9.8 * 2 6.26 m / s EC ED K C U gD vC 2 gh 2 * 9.8 *1.5 5.4 m / s p mv pbefore 0.100(6.26) 0.626 kg * m / s pafter 0.100(5.4) 0.54 kg * m / s Ft mv m(v vo ) F (0.01) 0.100(5.4 (6.26)) F 116.6 N Problem: A 150-g baseball moving at 40 m/s 15o below the horizontal is struck by a bat. It leaves the bat at 55 m/s 35o above the horizontal. What is the impulse exerted by the bat on the ball? If the collision took 2.3 ms, what was the average force of the bat on the ball? vf = 55 35o 15o J Fnet t p vi = 40 J x px m vfx m vix J y p y m vfy m viy 0.15(45 (38.6)) 12.54 N ( S ) I I x2 I 2y 12.9 N s, 14.20 0.15(31.5 10.4) 3.17 N ( s) Fnet 5624N , 14.2o Impulse-Momentum Theorem Work-Energy Theorem I Fnet t p Wnet Fnet d KE Impulse on a graph: area under the F-t curve VECTOR Work on a graph: area under the F-x curve SCALAR Problem: a tennis player receives a shot with the ball (0.6 kg) travelling horizontally at 50.0 m/s and returns the shot with the ball travelling horizontally at 40.0m/s in the opposite direction. A) what is the impulse delivered to the ball by the racket? B) what work does the racquet do on the ball? I x p x m vf m vi 0.6(40 50) 54 Ns vi=50 vf=40 Wnet KE 12 m v2f 12 m vi2 0.3(40 50 ) 270 J 2 2 The figure gives the momentum vs time for a particle moving along an axis. A force directed along the axis acts on the particle. a) Rank the four indicated regions according to the 1 > 3 > 2 =4 magnitude of the force, greatest first. b) In which region is the particle slowing? 3 p 2 1 3 4 t Conservation of Momentum 1 F12 m1v’1 1 m2v2 m1v1 1 2 F21 2 F ext 2 0 m2v’2 momentum before = momentum after m1v1 m2v2 m1v1 'm2v2 ' as long as NO EXTERNAL FORCE ACTS on the SYSTEM Newtons 2nd Law p F t p Fnet t BALL 1 BALL 2 1 m1v’1 m1v1 F12 m2v2 1 2 2 F21 1 2 m2v’2 Newtons 2nd Law p1 m1v1 'm1v1 F12 t p2 m2v2 'm2v2 F21t F12 t m1v1 'm1v1 m2v2 'm2v2 m1v1 m2v2 m1v1 'm2v'2 p1 p2 p1 ' p2 ' Conservation of Momentum Newtons 3rd Law Conservation of Momentum can be extended to include any number of interacting bodies F F psystem system system 0 psystem 0 t Total momentum of system (vector sum of momenta of all objects) If there are NO external forces LAW OF CONSERVATION OF MOMENTUM – The total momentum of an isolated system of bodies remains constant LAW OF CONSERVATION OF MOMENTUM The total momentum of an isolated system of bodies remains constant. psystem 0 A system is a set of objects that interacts with each other. An isolated system in one in which the only forces present are those between the objects of the system and those will be zero because of Newtons 3rd law. ( Fisolated system 0 ) Momentum in space Example. Railroad cars collide: momentum conserved. A 10,000 kg railroad car traveling at a speed of 24 m/s strikes an identical car at rest. If the cars lock together as a result of the collision, what is their common speed afterward? v1 = 24 pbefore m1v1 10000(24) 240,000kg (m / s) pafter 2m1v' 20000v' pbefore pbefore v' 12 m / s v2= 0 v’ DO NOW Rifle recoil. Calculate the recoil velocity of a 5.0 kg rifle that shoots a 0.050 kg bullet at a speed of 120 m/s. before shooting after shooting v’R v’B pbefore 0 pafter mR v'R mB v'B 5v'R 0.05(120) 5v'R 6 pbefore pbefore 0 5v'R 6 v'R 1.2 m / s Types of Collisions In all collisions where ΣFext = 0, momentum is conserved Elastic Collisions No deformation occurs. Kinetic energy is also conserved. Inelastic Collisions: Deformation occurs. Kinetic energy is lost. Explosions Reverse of perfectly inelastic collision, kinetic energy is gained. Perfectly Inelastic Collisions Objects stick together, kinetic energy is lost. K 1 2 2 mv Elastic Collisions Inelastic Collisions: http://www.science-animations.com/support-files/collisions.swf Example. Railroad cars. A railroad car of mass 3000 kg, moving at 20 m/s eastward, strikes head-on a railroad car of mass 1000 kg that is moving at 20 m/s westward. The railroad cars stick together after the impact. a) What is the magnitude and direction of the velocity of the combined trains after the collision? b) What is the impulse exerted on the smaller train by the larger train? On the larger train by the smaller? c) Prove that the collision is inelastic by kinetic energy analysis. v1 = 20 v2 = 20 v’ pbefore m1v1 m2 v2 3000(20) 1000(20) 40000N ( s) pafter (m1 m2 )v' (3000 1000)v' 4000v' pbefore pafter 40000 4000v' v' 10.0 m / s Example. Railroad cars. A railroad car of mass 3000 kg, moving at 20 m/s eastward, strikes head-on a railroad car of mass 1000 kg that is moving at 20 m/s westward. The railroad cars stick together after the impact. a) What is the magnitude and direction of the velocity of the combined trains after the collision? b) What is the impulse exerted on the smaller train by the larger train? On the larger train by the smaller? c) Prove that the collision is inelastic by kinetic energy analysis. v1 = 20 v2 = 20 J1 J2 v’ v' 10.0 m / s Impulse on smaller train is equal to its change in momentum J 2 p2 m2 v'm2 v2 1000[10 (20)] 3000Ns J1 J 2 3000Ns Newton’s 3rd Law Example. Railroad cars, inelastic collision. v1 = 20 v2 = 20 v’ v' 10.0 m / s K before K1 K 2 1 2 m1v12 1 2 m2v22 800,000J K after K1 ' K 2 ' 1 2 (m1 m2 )v'2 200,000J K is reduced so collision is inelastic Example. Old cannons were built on wheeled carts, both to facilitate moving the cannon and to allow the cannon to recoil when fired. When a 150 kg cannon and cart recoils at 1.5 m/s, at what velocity would a 10 kg cannonball leave the cannon? v’c = 1.5 v’B = ? pbefore 0 pafter mC v'C mB v' B (150)(1.5) 10(v' B ) pbefore pbefore 0 225 10v' B v' B 22.5 m / s KEbefore 0 KEafter 1 2 1 m v '2 m v ' 2 C C 2 B B 2700J Example. Pool or billiards. A billiard ball of mass 0.5 kg moving with a velocity of 3 m/s collides head on in an elastic collision with a second ball of equal mass at rest (v2 = 0). What are the speeds of the 2 balls after the collision? v1 = 3 v2= 0 pbefore m1v1 m2v2 0.5(3) 0 1.5 N ( s) pafter m1v'1 m2v'2 0.5v'1 0.5v'2 pbefore pbefore 1.5 0.5(v'1 v'2 ) 3 v'1 v'2 Example. Pool or billiards. v1 = 3 from conservation of momentum: 3 v'1 v'2 v2= 0 v'1 0 v' 2 3 Since collision is elastic, kinetic energy is also conserved: K before K1 K 2 2 m v 2 m v 2.25J K after K1 ' K 2 ' 1 2 m v' 1 2 m v' 0.25(v'12 v'22 ) K before K after 2.25 0.25(v'12 v'22 ) 9 v'12 v'22 (3 v'2 ) 2 v'22 1 2 1 1 2 1 1 1 2 2 2 2 2 2 Truck Collision In a head-on collision: Which truck will experience the greatest force? Which truck will experience the greatest impulse? Which truck will experience the greatest change in momentum? Which truck will experience the greatest change in velocity? Which truck will experience the greatest acceleration? Which truck would you rather be in during the collision? Truck Collision same same same In a head-on collision: Which truck would you rather be in during the collision? Example. A nuclear collision. A proton of mass 1.0 u (unified atomic mass units) traveling with a speed of 20,000m/s has an elastic, head-on collision with a Helium (He) nucleus (mHe = 4.00 u) at initially rest. What are the velocities of the proton and Helium nucleus after the collision? vP = 20,000 vHe= 0 pbefore mP vP mHe vHe 1(20000) 0 20,000 pafter mP v'P mHe v'He 1v'P 4v'He pbefore pbefore 20,000 1v'P 4v'He Example. A nuclear collision. vP = 20,000 vHe= 0 v'P 12,000 v'He 8000 from conservation of 20,000 1v'P 4v'He momentum: Since collision is elastic, kinetic energy is also conserved: K before K after K before 8 2 x10 2 m v 2 m v 2 x10 2 2 1 1 2 m v' 2 m v' 0.5v' P 2v' He K after 2 2 0.5v' P 2v' He 2 2 0.5(20000 4v' He ) 2v'He 1 2 P P 2 P P 1 2 He He 2 He He 8 Example. Propulsion in space: explosion. Rocket F p fuel t J Ft Ion Thruster 5000N Rocket thrust: LARGE force over short time pion F 0.092N t J Ft Ion Thruster: small force over LONG time Example. Propulsion in space: explosion. An astronaut at rest in space fires a thruster pistol that expels 35 g of hot gas at 875 m/s. The combined mass of astronaut and pistol is 84 kg. How fast and in what direction is the astronaut moving after firing the pistol? pbefore 0 v’A = ? v’G = 875 pafter m Av' A mG v'G 84v' A 0.035(875) pbefore pbefore 0 84v' A 30.63 v' A 0.36 m / s K before 0 K after 1 2 mAv'2A 1 2 mG v'G2 13,404J Conservation of Momentum can also be applied in 2 or 3 dimensions For 2-dimensional collisions • Set coordinate system up with x-direction • • • the same as one of the initial velocities Label and resolve velocity vectors into x and y components in a sketch Resolve momentum vectors into x and y components when working the problem Use conservation of momentum independently for x and y dimensions. 2 Dimensional Problem: A pool player hits the 14- ball in the xdirection at 0.80 m/s. The 14-ball knocks strikes the 8-ball, initially at rest, which moves at a speed of 0.30 m/s at an angle of 35o angle below the x-axis. Determine the angle of deflection of the 14-ball. v’1 v1=0.8 q1 q2=350 q1= 17.20 v’1= 0.58 m/s v’2=0.3 A 2.5 g bullet is fired into a 215 g wooden block of a ballistic pendulum. After the collision, the block and embedded bullet rise to a maximum height of 1.20m. Find the speed of the bullet. pbefore pafter pb pP pb P mb vb 0 (mb mP )v' (mb mP )v' vb mb 0.2175v' vb 87v' 0.0025 EB ET 2 1 m v ' mB gh B 2 v' 2 gh 4.85 vb 87v' 422m / s The diagram depicts the beforeand after-collision speeds of a car that undergoes a head-oncollision with a wall. In Case A, the car bounces off the wall. In Case B, the car crumples up and sticks to the wall. a. In which case is the change in velocity the greatest? b. In which case is the change in momentum the greatest? c. In which case is the impulse the greatest? d. In which case is the force that acts upon the car the greatest (assume same contact times)?