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Chapter 6: Momentum and Collisions Momentum and Impulse Linear momentum • Linear momentum p of an object of mass m moving with velocity v is the product of its mass and velocity: p mv SI unit: kilogram-meter per second (kg m/s) p ( p x , p y ) (mvx , mvy ) Chang of momentum and force v (mv ) p Fnet ma m where m and Fnet are constant. t t t v (mv ) p Fnet ma m lim t 0 lim t 0 lim t 0 with m const. t t t Momentum and Impulse Momentum conservation p 0 if Fnet 0 t The (linear) momentum of an object is conserved when Fnet = 0. Impulse • If a constant force F acts on an object, the impulse I delivered to the object over a time interval t is given by : I Ft SI unit: kilogram-meter per second (kg m/s) • When a single constant force acts on an object, Ft p mv f mvi • When the force is not constant, then Impulse-momentum theorem lim t 0 Ft lim t 0 p lim t 0 (mv f mvi ) Momentum and Impulse Impulse-momentum theorem lim t 0 Ft lim t 0 p lim t 0 (mv f mvi ) The impulse of the force acting on an object equals the change in momentum of that object as long as the time interval t is taken to be arbitrarily small. An example of impulse Average force Fav Favt p The magnitude of the impulse delivered by a force during the time interval t is equal to the area under the force vs. time graph or, equivalently, to Favt. Momentum and Impulse Examples • Example 6.1 : Teeing off A golf ball is struck with a club. The force on the ball varies from zero at contact and up to the max. value. (a) Find the impulse. I p p f pi p f 2.2 kg m/s (b) Estimate the duration of the collision and the average force. x t 9.110 2 s vav Fav p 2.4 103 N t m = 5.0x10-2 kg vi = 0, vf = 44 m/s Momentum and impulse Examples • Example 6.2 : How good are the bumpers (a) Find the impulse delivered to the car. pi 2.25 10 4 kg m/s p f 0.390 10 4 kg m/s I p p f pi 2.64 10 4 kg m/s (b) Find the average force. Fav p 1.76 105 N t t=0.150 s Momentum and Impulse Injury in automobile collisions • A force of about 90 kN compressing the tibia can cause fracture. • Head accelerations of 150g experienced for about 4 ms or 50g for 60 ms are fatal 50% of the time. • When the collision lasts for less than about 70 ms, a person will survive if the whole-body impact pressure (force per unit area) is less than 1.9x105 N/m2. Death results in 50% of cases in which the whole-body impact pressure reaches 3.4x105 N/m2. Consider a collision involving 75-kg passenger not wearing s seat belt, traveling at 27 m/s who comes to rest in 0.010 s after striking an unpadded dashboard. Fatal F mv f mvi 2.0 105 N 90 kN 150g Fav a 3.4 105 N/m 2 Fav / A av t v 2700 m/s 2 2 a 2700 m/s g 280 g 2 t 9.8 m/s Fav / A 4 105 N/m 2 Conservation of Momentum Conservation of momentum average force on 1 by 2 F21t m1v1 f m1v1i F12t m2v2 f m2v2i average force on 2 by 1 F21 F12 m1v1 f m1v1i m2v2 f m2v2i m1v1i m2 v2i m1v1 f m2v2 f Conservation of momentum When no net external force acts on a system, the total momentum of the system remains constant in time Conservation of Momentum An example • Example 6.3 : The archer m1=60 kg (man + bow) m2=0.500 kg (arrow) pi p f 0 m1v1 f m2 v2 f m2 v1 f v2 f 0.417 m/s m1 The archer is moving opposite the direction of the arrow speed of arrow v2=50.0 m/s Collisions Three types of collisions • Inelastic collision A collision in which momentum is conserved, but kinetic energy is not. • Perfectly inelastic collision A collision between two objects in which both stick together after the collision. • Elastic collision A collision in which both momentum and kinetic energy are conserved. Collisions Perfectly inelastic collisions • Consider two objects with mass m1 and m2 moving with known initial velocities v1i and v2i along a straight line. • They collide head-on and after the collision, they stick together and move with a common velocity vf. m1v1i m2 v2i (m1 m2 )v f m1v1i m2 v2i vf m1 m2 Collisions Examples of perfect inelastic collision • Example 6.4 : An SUV vs. a compact (a) Find the final speed after collision. m1v1i m2 v2i (m1 m2 )v f m1v1i m2 v2i vf 5.00 m/s m1 m2 m1=1.80x103 kg m2=9.00x102 kg v1i=15.0 m/s v2i=-15.0 m/s (b) Find the changes in velocity. v1 v f v1i 10.0 m/s v2 v f v2i 20.0 m/s (c) Find the change in kinetic energy of the system. KEi 1 1 1 m1v12i m2 v22i , KE f (m1 m2 )v 2f KE 2.70 105 J 2 2 2 Collisions Examples of perfect inelastic collision • Example 6.5 : Ballistic pendulum Find the initial speed of bullet. Right after collision At the height h KEi PEi KE f PE f 1 (m1 m2 )v 2f 0 0 (m1 m2 ) gh 2 v 2f 2 gh v f 2 gh 0.990 m/s Before collision Right after collision m1v1i m2 v2i (m1 m2 )v f m1v1i 0 (m1 m2 )v f v1i (m1 m2 )v f m1 199 m/s m1=5.00 g m2=1.00 kg h = 5.00 cm Collisions Elastic collisions • Consider two objects with mass m1 and m2 moving with known initial velocities v1i and v2i along a straight line. • They collide head-on and after the collision, they leave each other with velocities v1f and v2f . 1 1 1 1 m1v12i m2 v22i m1v12f m2 v22 f 2 2 2 2 m1 (v12i v12f ) m2 (v22 f v22i ) m1 (v1i v1 f )(v1i v1 f ) m2 (v2 f v2i )(v2 f v2i ) (1) m1v1i m2 v2i m1v1 f m2 v2 f m1 (v1i v1 f ) m2 (v2i v2 f ) (2) (1) /( 2) : v1i v1 f v2 f v2i v1i v2i v1 f v2 f Collisions An example of elastic collision • Example 6.7 : Two blocks and a spring (a) Find v2f when v1f=+3.00 m/s. m1v1i m2 v2i m1v1 f m2v2 f v2 f m1v1i m2v2i m1v1 f m2 1.74 m/s (b) Find the compression of the spring. 1 1 1 1 1 m1v12i m2 v22i m1v12f m2 v22 f kx2 2 2 2 2 2 x 0.173 m m1=1.60 kg m2=2.10 kg v1i=+4.00 m/s v2i=-2.50 m/s k=6.00x102 N/m Glancing Collisions Collisions in 2-dimension • Momentum conservation in 2-D m1v1i m2 v2i m1v1 f m2v2 f m1v1ix m2 v2ix m1v1 fx m2 v2 fx m1v1iy m2 v2iy m1v1 fy m2 v2 fy m1v1ix 0 m1v1 f cos m2v2 f cos 0 0 m1v1 f sin m2 v2 f sin Glancing Collisions An example of a collision in 2-D • Example 6.8 : A perfect inelastic collision at an intersection Find the magnitude and direction of the velocity of the wreckage. 4 p m v 3 . 75 10 kg m/s ix car car p (m p p fx car ix mvan )v f cos mcar=1.50x103 kg fx 3.75 104 kg m/s (4.00 103 kg) v f cos 4 p m v 5 . 00 10 kg m/s iy van van p (m p p fy iy car mvan )v f sin fy 5.00 104 kg m/s (4.00 103 kg) v f sin mvan=2.50x103 kg Glancing Collisions An example of a collision in 2-D (cont’d) • Example 6.8 : A perfect inelastic collision at an intersection (cont’d) Find the magnitude and direction of the velocity of the wreckage. 3.75 104 kg m/s (4.00 103 kg) v f cos 5.00 104 kg m/s (4.00 103 kg) v f sin mcar=1.50x103 kg 5.00 104 kg m/s tan 1.33 4 3.75 10 kg m/s 53.1 5.00 104 kg m/s vf 15.6 m/s 3 (4.00 10 kg) sin 53.1 mvan=2.50x103 kg Rocket Propulsion Principle (hand-waving argument) • The driving force of motion of ordinary vehicles such as cars and locomotives is friction. A car moves because a reaction to the force exerted by the tire produces a force by the road on the wheel. • What is then driving force of a rocket? When an explosion occurs in a spherical chamber with fuel gas in a rocket engine the hot gas expands and presses against all sides of the chamber uniformly. So all forces are in balance-no net force. If there is a hole as in (b), part of the hot gas escapes from the hole (nozzle), which breaks the balance of the forces. This unbalance create a net upward force. Rocket Propulsion Principle (detailed argument) • At time t, the momentum of the rocket plus the fuel is (M+m)v. • During time period t, the rocket ejects fuel of mass m whose speed ve relative to the rocket and gains the speed to v+v. From momentum conservation: (M m)v M (v v) m(v ve ) Mv ve m • The increase m in the mass of the exhaust corresponds to an equal decrease in the mass of the rocket so that m=-M. Mv ve M M : mass of rocket m : mass of fuel to be ejected in t Rocket Propulsion Principle (detailed argument) • Using calculus: Mv ve M Mi v f vi ve ln M f Thrust • is defined as the force exerted on the rocket by the ejected exhaust gases. Instantaneous thrust v M Ma M ve t t M : mass of rocket m : mass of fuel to be ejected in t Rocket Propulsion An example • Example 6.0 : Single stage to orbit (b) Find the thrust at liftoff. M M f M i M m ve t : mass of rocket 1.00x105 kg : burnout mass 1.00x104 kg : exhaust velocity 4.50x103 m/s : blast off time period 4 min 1.00 104 kg 1.00 105 kg 9.00 104 kg M 3.75 10 2 kg/s t vf Thrust Th ve M 1.69 106 N t (c) Find the initial acceleration. Ma F Th Mg a Th g 7.10 m/s 2 M