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Chapter 9 Linear Momentum and Collisions Momentum Linear momentum of object: mass×velocity p mv Magnitude: p=mv, Direction: same as velocity Quantity of motion, ability to acting force Newton’s second law of motion (original edition) The changing rate of momentum of an object is equal to the net force applied on it. dv dp d mv m ma F dt dt dt 2 Impulse Integral form of Newton’s second law tf dp F p p f p i Fdt ① ti dt Change in momentum is the integral of the net force over the time, the integral is called impulse J tf ti Fdt J Ft if F is constant Eq.① is called impulse-momentum theorem momentum ~ impulse energy ~ work 1) any force 3) J 2) vector F? 3 Projectile problem Example1: Object m is fired with v at angle 45°to the horizontal, no air friction. Determine: y a) Initial momentum b) Final momentum v c) Impulse during the motion 45° x Solution: a) p m v 2 mv i 2 mv j i 2 2 b) 2 2 pf mv i mv j 2 2 same magnitude different direction c) J p f pi 2 m v j J mg t 4 Conical pendulum Example2: Known: m, L, , v, T. Determine the impulse of tension and gravity respectively during one-half period. Solution: FT mg / cos L J G mgT / 2 J F mgT / 2 cos FT m FT is not a constant force! G So: p 2m v JF J p 2 G O 2 Direction of these vectors 5 Collisions and impulsive force Concept of momentum and impulse is useful when dealing with collision problems Key points: 1) Time interval is usually very short 2) Interaction of collision, or impulsive force is changing fast and may become very large The effect is shown by its impulse 3) Impulses of other forces can be ignored 6 Bend your knees Example3: A 70-kg person jumps from h=3.0m and lands on firm ground. a) Calculate the impulse experienced; b) Estimate the average force by the ground if he is stifflegged (s=1.0cm); c) average force with bent legs (s=50cm). Solution: a) Landing speed v 2 g h 7 .7 m / s Impulse J p 0 m v 5 4 0 N s x 1 2 5 b) Fnet s 0 mv F n e t m g h / s 2 .1 1 0 N 2 F n et F g r m g F g r 2 .1 1 0 5 N 6 9 0 N 2 .1 1 0 5 N c) F n e t m g h / s 4 .2 1 0 N 3 impulsive force F g r 4 .2 1 0 3 N 6 9 0 N 4 .9 1 0 3 N 7 Washing a car Example4: Water is jetting at rate of R=1.5kg/s with v=20m/s, and is stopped by a car (no splashing back). What is the force exerted on the car? Solution: In a short time interval dt, How much water hits on the car: dm=Rdt Change in momentum: dp=vdm=Rvdt Collision: ignore gravity, only force by car Fdt dp F R v 30 N What if the water splashes back? 8 Falling rope Example5: One end of a hanged rope just touch the table, then it is released. Prove: Normal force acting on table is always 3 times of weight of that part already on the table. Solution: Linear density λ, falling height h N mg F Mass of part already on table m = λ h Falling speed v 2 gh In time interval dt, dm = λ · vdt Impulse F·dt = dm·v = λ ·v2dt F= λ ·v2 = λ ·2gh =2mg N m g F 3m g dm v m mg F 9 Momentum of a system Consider a system of n interacting particles P Total momentum of the system pi Change in total momentum dP Fi f ij dt Internal forces are always in pairs f ij 0 dP Fi dt Only external forces can change the total momentum! Fi F1 f1i f i1 m1 f1n fn1 mn mi fni fin Fn 10 Conservation of momentum If the net external force is 0, then dP dt F i 0 o r P c o n s ta n t This is the law of conservation of momentum: When the net external forces on a system is zero, the total momentum remains constant. Or: The total momentum of an isolated system of bodies remains constant. 1) isolated 2) system 3) ignoring forces 11 Rifle recoiling Example6: Calculate the recoil speed of a 5kg rifle that shoots a 10g bullet at v=700m/s. Solution: Total momentum is conserved m RvR m Bv 0 mB v vR mR 1.4 m / s The rifle moves back and applies a recoil force 12 Billiard ball collision Example7: A billiard ball with v in the +x direction strikes an identical resting ball. Directions after collision are shown in figure, what are the speeds? Solution: conservation of momentum mv mv B mvY component form: x : mv mv B cos 45 mvY cos 45 y: 0 mv B sin 45 mvY sin 45 2 v B vY v 2 45 45 y o x kinetic energy is also conserved 13 Conservation in component form If F 0 , but F x 0 → x component of momentum is conserved Example8: Bullet (20g) hits into hanging ball (980g) with v=400m/s. Determine the speed after collision. Solution: No net force in horizontal Horizontal component Px is conserved m v sin 3 0 o M m v mv sin 30 o v 4 m /s M m 30 T v m M (M+m)g 14 Homework A gun carrier M moves on a frictionless incline, its speed reduces from v to 0 after shooting a canon-ball m in the horizontal direction. Is the total momentum of system (M and m) conserved in this process, and why? Find out the speed of canon-ball. v m v’ 15 Distance traveled Example9: Two objects start from rest, what is the distance traveled by m2 when m1 reaches the ground? No friction. Solution: conservation of momentum in horizontal m 1 v1 x m 2 v 2 x 0 b m 1 x1 m 2 x 2 0 m1 where x1 x 2 a b m1 S 2 x2 a b m1 m 2 So distance traveled by m2 m2 o x a 16 Challenging question Question: Someone sits on the top of an ice half sphere on the ground, and then slips down. At what angle will he leave the sphere? No friction. Thinking: Leave the half sphere: N = 0 m gcos N m v 2 / R m top M ? Relative speed & inertial frame Conservation of momentum and mechanical energy Velocity transformation 17 Elastic collisions If total kinetic energy is conserved in a collision 1 1 1 1 2 2 2 m1 v1 m 2 v 2 m1 v1 m 2 v 2 2 2 2 2 2 It is called an elastic collision Ideal model in macro world head-on collision Conservation of momentum (1-dimension) m 1 v 1 m 2 v 2 m 1 v 1 m 2 v 2 v1 v 2 v1 v 2 Relative speed: equal but opposite 18 Final speeds m 1 v 1 m 2 v 2 m 1 v 1 m 2 v 2 1 1 1 1 m1 v12 m 2 v 22 m1 v1 2 m 2 v 2 2 2 2 2 2 v1 m1 m2 2 m2 v1 v2 m1 m2 m1 m2 m2 m1 2 m1 v2 v2 v1 m1 m2 m1 m2 Special cases: 1) Equal masses v 1 v 2 , v 2 v 1 2) Target m2 at rest v 1 v 1 , v 2 0 m1 m2 m1 m 2 v 1 v 1 , v 2 2 v 1 19 Baseball batting Example10: A baseball with speed 40m/s is hit by a bat with speed 30m/s in the opposite direction. Determine the speed of baseball if the collision is elastic and mbat >> mball . Solution: final speed of ball v ball m ball m bat 2 m bat v ball v bat m bat m ball m bat m ball v b a ll 2 v b a t 1 0 0 m / s Or in the frame of bat: The ball moves in 70m/s, and rebounds in the same speed Transform to the frame of ground, it’s 100m/s. 20 Slingshot effect Example11: Spacecraft Voyager II approaches the Jupiter, it rounds the planet and departs in the opposite direction. What is the speed after this slingshot encounter? ( vS =10.4km/s, vJ =-9.6km/s ) Solution: Like an elastic collision v S v J v S v J mJ m S v J v J speed after slingshot: v S v S 2 v J 2 9 .6 k m /s 21 Inelastic collisions If kinetic energy is not conserved in a collision It is called an inelastic collision Restitution coefficient e=1: e<1: e=0: v2 v1 e v1 v2 elastic collision inelastic collision completely inelastic collision How energy transform? heat and sound … 22 Completely inelastic collision Two objects stick together in collision: v’1 = v’2 If target m2 is initially at rest Kinetic energy loss Ek m2 Ek 0 m1 m2 Depends on the mass ratio of two objects m1 m1 m2 : m2 : Almost no energy loss Hammer hit on nail … Maximum of energy loss Throw egg against rock … 23 Hit by bullet Example12: m1=0.49kg and m2= 0.50kg are related by a massless spring (k=100N/m). The system is initially at rest on frictionless horizontal plane, then a bullet m3=0.01kg hits into m1 with v =100m/s. Determine the maximum stretch. v m1 m2 Solution: Completely inelastic collision between m1 and m3 m 3 v m 1 m 3 v v 2 m /s Conservation of momentum and mechanical energy m 1 m 3 v m 1 m 2 m 3 v 1 1 1 2 2 2 m1 m 3 v m1 m 2 m 3 v kx 2 2 2 x 0 .1 m 24 Collisions in 2-dimensions Discuss an elastic collision in 2 dimensions Target initially at rest Conservation of momentum y o x m 1 v m 1 v1 c o s 1 m 2 v 2 c o s 2 1 2 0 m 1 v 1 s in 1 m 2 v 2 s in 2 Conservation of kinetic energy 1 1 1 2 2 m1 v m1v1 m 2 v 22 2 2 2 It can be solved if one of variables is known 25 Equal-mass case If it occurs between two objects of equal mass m1 m 2 Angle between final velocities is always 90° 1 2 90 This is shown to be true in many experiments You can try to prove it (Problem 57, Page 229) y o x 1 2 26 Center of mass Real body (instead of particle): general motion One point represents motion of object (system) As a particle with same mass and same net force It is called center of mass (CM) c. General motion c c = Translational motion of CM + Other motion about CM c. c (rotation, vibration…) Different from center of gravity (CG) 27 CM of several particles Definition of CM: by weighted average xC mx i M 1-dimension i or rC mr i i M more-dimensions Example13: Determine the CM of stones, where the mass of black stone is twice as much as white stone. y Solution: 25 1 (1 2 2 ) 2 ( 2 3 5 ) xC 1 3 2 3 9 1 ( 3 5 8 ) 2 ( 2 2 3 ) 10 yC 3 1 3 2 3 O x 28 CM of continuous object Continuous object: summations become integrals xC xdm or rdm rC M M Example14: CM of a thin rod with varying density. 0 x Solution: Choose a segment dx o dm dx 0 xdx The total mass CM: 1 xC M M dm 1 xdm M L 0 L 0 x dx L x 0 xdx 0 L 2 / 2 2 0 x dx L 3 2 29 CM of more objects Example15: CM of uniform thin right triangle. Solution: Choose a segment dm of the triangle ax dm bdx a xC a 0 xdm a 0 M Similarly we have y abx bx dx a ab / 2 b dx 2 a /3 O x a x yC b / 3 Comparing with Example14 What about other objects? 30 CM and translational motion rC mr i i M M rC mr i i M vC mv i i P Total momentum of a system is equal to the product of the total mass M and velocity of CM. dP MaC dt F i The sum of all the forces acting on the system is equal to the total mass M times the acceleration of CM. Or: The center of mass of a system with total mass M moves like a single particle of mass M acted by the same net external force. 31 Man on boat Example16: A man (m) moves from one side of a boat (M, L) to another side, determine the distance traveled by the boat. Initially at rest, no resistance. Solution: No net external force, CM stays at rest l l ml M ms M ( s ) 2 2 xC mM mM ml s mM y o s x 32 Two stage rocket Example17: A rocket is separated into two parts of equal mass at its highest point. Part A falls vertically to Earth from rest, where does part B lands? Solution: CM still moves along the curve It lands at x=2d So part B will land at x=3d If part A had been given a kick up or down? y A d B ? x 33 *Rocket propulsion System of variable mass & ignore external forces Gas dm is jetting with relative speed vr Speed of rocket changes from v to v+dv m v m dm v dv v vr dm v m vr dm dm m d v v r d m dv v r m v m dm m dv vr v v r ln 0 0 m0 m m Tsiolkovsky equation 34