# Chapter 8 Conservation of Linear Momentum. Conservation of Linear Momentum

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1 Chapter 8 Conservation of Linear Momentum Physics 201 October 22, 2009 Conservation of Linear Momentum Definition of linear momentum, p p = m v Linear momentum is a vector. Units of linear momentum are kg-m/s. Can write Newton s second law in terms of momentum: d p dt = d(m v ) dt = m d v dt = m a d p dt = F net 1

2 Momentum of a system of particles The total momentum of a system of particles is the vector sum of the momenta of the individual particles: " " " P sys = mi vi = p i i i From Newton s second law, we obtain F ext = F netext = i i d P sys dt Conservation of Momentum Law of conservation of momentum: If the sum of the external forces on a system is zero, the total momentum of the system does not change. If F ext = 0 i then P sys = m i vi = M v """"" CM = const i Momentum is always conserved (even if forces are nonconservative). 2

3 Collisions m 2 before m 2 after momentum before collision = momentum after collision Always - But only if F external = 0 Explosion - I M before v 1 m 2 v 2 after Example: = M/3 m 2 = 2M/3 After explosion, which block has larger momentum? (left, right, same) 3

4 Explosion - I M before v 1 m 2 v 2 after Example: = M/3 m 2 = 2M/3 After explosion, which block has larger momentum? (left, right, same) Each has the same momentum Explosion - I M before v 1 m 2 v 2 after Example: = M/3 m 2 = 2M/3 After explosion, which block has larger momentum? (left, right, same) Each has the same momentum Which block has larger velocity? 4

5 Explosion - I M before v 1 m 2 v 2 after Example: = M/3 m 2 = 2M/3 After explosion, which block has larger momentum? (left, right, same) Each has the same momentum Which block has larger velocity? mv is the same for each block, so smaller mass has larger velocity Explosion - I M before v 1 m 2 v 2 after Example: = M/3 m 2 = 2M/3 After explosion, which block has larger momentum? (left, right, same) Each has the same momentum Which block has larger velocity? mv is the same for each block, so smaller mass has larger velocity Is kinetic energy conserved? 5

6 Explosion - I M before v 1 m 2 v 2 after Example: = M/3 m 2 = 2M/3 After explosion, which block has larger momentum? (left, right, same) Each has the same momentum Which block has larger velocity? mv is the same for each block, so smaller mass has larger velocity Is kinetic energy conserved? NO K was 0 before, it is greater after the explosion. Explosion - I M before v 1 m 2 v 2 after Example: = M/3 m 2 = 2M/3 After explosion, which block has larger momentum? (left, right, same) Each has the same momentum Which block has larger velocity? mv is the same for each block, so smaller mass has larger velocity Is kinetic energy conserved? (green=yes, red=no) NO 6

7 This is like a microscopic explosion. Momentum and Impulse p m v For single object. F ave t " I F = m a " If F = 0, then momentum conserved (p = 0) For system of objects p sys = i p i Internal forces " forces between objects in system External forces " any other forces # p sys = F ext #t Thus, if F ext = 0, then # p sys = 0 i.e. total momentum is conserved definition of impulse = m d v dt = d p # p = Ft dt 7

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10 Momentum and Impulse p m v For single object. F ave t " I F = m a " If F = 0, then momentum conserved (p = 0) For system of objects p sys = i p i Internal forces " forces between objects in system External forces " any other forces # p sys = F ext #t Thus, if F ext = 0, then # p sys = 0 i.e. total momentum is conserved definition of impulse = m d v dt = d p # p = Ft dt 10

11 F ave t " I F = m a definition of impulse = m d v dt = d p # p = Ft dt Let s estimate the average force during the collision Club speed: 50 m/s Assume that impulse is given after 5 cm F --> whiteboard ave = I t = 1 t t f " t i F dt Some Terminology Elastic Collisions: collisions that conserve kinetic energy Inelastic Collisions: collisions that do not conserve kinetic energy * Completely Inelastic Collisons: objects stick together n.b. ALL CONSERVE MOMENTUM If external forces = 0 11

12 Elastic Collision in 1-Dimension Initial Final Energy conserved (for elastic collision only) Linear momentum is conserved v 1i + m 2 v 2i = v 1 f + m 2 v 2 f 1 2 v 2 1i m 2v 2 2i = 1 2 v 2 1 f m 2 2v 2 f Elastic Collision Conservation of Momentum v 1i + m 2 v 2i = v 1 f + m 2 v 2 f (v 1i v 1 f ) = m 2 (v 2 f v 2i ) Conservation of Kinetic Energy 1 2 v 2 1i m 2v 2 2i = 1 2 v 2 1 f m 2 2v 2 f (v 2 1i v f ) = m 2 (v 2 f v 2 2i ) (v 1i v 1 f )(v 1i + v 1 f ) = m 2 (v 2 f v 2i )(v 2 f + v 2i ) Combining the above two equations v 1i + v 1 f = v 2i + v 2 f v 1i v 2i = (v 1 f v 2 f ) Magnitude of relative velocity is conserved. 12

13 Is this an elastic collision? For elastic collision only: v 1i v 2i = (v 1 f v 2 f ) 13

14 What is the speed of the golf ball, in case of an elastic collision Club speed: 50 m/s Mass of clubhead: 0.5kg Mass of golfball: 0.05kg Two unknowns: speed of club and speed of golfball after impact Problem solving strategy: - Momentum conservation - Energy conservation (or use the derived equation for relative velocities) --> whiteboard Is this an elastic collision? For elastic collision only: v 1i v 2i = (v 1 f v 2 f ) Yes, the relative speeds are approximately the same before and after collision 14

15 Result: v 1 f = m 2 + m 2 v 1i v 2 f = 2 + m 2 v 1i Special cases: 1) Golf shot: m1>>m2 Club speed almost unchanged Ball speed almost 2 x club speed 2) Neutron scatters on heavy nucleus: m1<<m2 neutron scatters back with almost same speed speed of nucleus almost unchanged Completely inelastic collision Two objects stick together and move with the center of mass: If p Ai =0: P Ai + P Bi = P Af + P Bf = P CM P Bi = P CM m B v Bi = ( m A + m B )v f 15

16 Ballistic Pendulum What is the initial velocity v li of the projectile? Known quantities:, m 2, h Two stage process: 1. m collides with M, inelastically. Both M and m then move together with a velocity V f (before having risen significantly). 2. Both ( + m 2 ) rise a height h, conserving energy E. (no non-conservative forces acting after collision) Ballistic Pendulum Stage 1: Momentum is conserved Energy is not conserved in x-direction: Stage 2 (after the collision): Energy is conserved K + U conserved : Substituting for V gives: v 1i = ( + m 2 )V f 1 ( 2 + m 2 )V 2 f = ( + m 2 )gh V f = 2gh v 1i = 1+ m \$ 2 " # % & 2gh 16

17 Ballistic Pendulum How much energy is dissipated? W thermal = U K i = ( + m 2 )gh 1 2 m 2 1v 1i = ( + m 2 )gh 1 2 m " 1 # \$ = ( + m 2 )gh m 2 + m 2 % & ' 2 2gh Fraction of kinetic energy lost is.. If m2<<m1: not much energy is lost Of m2>>m1:almost all energy is lost W thermal K i = + m 2 Inelastic collision Coefficient of restitution e=1/2 17

18 Coefficient of restitution Perfectly elastic collision: v 1i v 2i = (v 1 f v 2 f ) The coefficient of restitution is a measure of the inelasticity: e = v 1 f v 2 f v 1i v 2i Elastic collision: e=1 Perfectly inelastic collision: e=0 Collisions or Explosions in Two Dimensions y x before P total,before after " " = Ptotal,after P total,x and P total,y independently conserved *P total,x,before = P total,x,after *P total,y,before = P total,y,after 18

19 Explosions M before after A B Which of these is possible? A B both Explosions M before after A B Which of these is possible? A (p appears conserved) B (p not conserved in y direction) both 19

20 Explosions M before after A B Which of these is possible? A B neither Explosions M before after A B Which of these is possible? A (p not conserved in y direction) B neither 20

21 (Inelastic) Car truck collision Knowns: m1, m2, v1, v2, M cm =m1+m2 Unknowns: final velocity vector (in x and y) Equation(s): Momentum conservation (in x and y) Strategy: write out conservation of momentum equation. ( 2 unknowns, 2 equations, piece of cake) Elastic collision in 2 dimensions Assume we know all initial conditions, mass and momentum. 4 Unknown quantities: " " v 1 f,v2 f Equations: Momentum conservation: 2 (x and y) Energy conservation: 1 Need one more piece of information to solve the problem: often a measurement. What is unspecified above is the impact parameter (and the precise nature of the interaction) 21

22 2008 by W.H. Freeman and Company 22

23 Shooting Pool... Assuming Collision is elastic (KE is conserved) No spin is imparted Balls have the same mass One ball starts out at rest Shooting Pool Elastic collision means conservation of kinetic energy 1 2 m v 2 1 1i = p 2 1i = p 2 1 f + p 2 2 f 2 2 2m 2 if = m 2 : Conservation of momentum: if P 2i = 0 p 2 1i = p f + p 2 f P 1i + P 2i = P 1 f + P 2 f P 1i = P 1 f + P 2 f P 1 and p 2 Form a right angle 23

24 Shooting Pool... Tip: If you shoot a ball spotted on the dot, you have a good chance of scratching 24

25 Collisions in the CM frame v cm = m v " "" m2 v 2 + m 2 The transformation to the cm frame is not necessary, but it is often convenient to switch to the CM frame Subtract v cm from all velocities 25

26 Rocket equation The mass is changing Thrust is generated by impulse of exhaust of mass with velocity v: vdm 26

27 The Saturn V The Saturn V rocket: 111 m tall 10m diameter 3000 tons at start Thrust: 34 MN Rocket equation M dm Mass change Thrust dm dt Weight: Rocket equation: F = M g dm = R = const dt M (t) = M 0 Rt u ex = d(m u ex ) dt = dp dt = "F thrust M dv dt = R u ex " Mg dv dt = R u ex M 0 " Rt " Mg Integration yields: # M v = u ex ln 0 & \$ % M 0 " Rt ' ( " gt 27

28 Variable mass Newton s second law for continuously variable mass: F net,ext + dm v rel = M dv dt dt v rel = u v Where is the velocity of impacting material relative to object with mass M at a given time. 28

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