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Momentum and Impulse 8.01 W06D2 Associated Reading Assignment: Young and Freedman: 8.1-8.5 Announcements: No Math Review Night this Week Next Reading Assignment W06D3: Young and Freedman: 8.1-8.5 Today’s Reading Questions (a) Explain the difference between the concepts of impulse and work. (b) Explain why the total force on a system of particles is only due to the sum of external forces. (a) Under what conditions does the center of mass of a system of particles move with constant velocity? Momentum and Impulse Obeys a conservation law Simplifies complicated motions Describes collisions Basis of rocket propulsion & space travel Newton’s Definition: Quantity of Motion DEFINITION II Newton’s Principia The quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjointly. The motion of the whole is the sum of the motions of all the parts; and therefore in a body double in quantity, with equal velocity, the motion is double; with twice the velocity, it is quadruple. Momentum and Impulse: Single Particle p mv • Momentum SI units [kg m s ] [N s] • Change in momentum -1 p mv tf • Impulse I Fdt ti • SI units [N s] Integral Version of Newton’s Second Law For an object with fixed mass r r r dv dp r F ma m dt dt Then impulse produces a change in momentum tf r r r dp r r r r I F dt dt d p p p(tf ) p(ti ) dt t t t tf tf i i i “The change of motion is proportional to the motive force impresses, and is made in the direction of the right line in which that force is impressed” Concept Question: Pushing Identical Carts Identical constant forces push two identical objects A and B continuously from a starting line to a finish line. If A is initially at rest and B is initially moving to the right, 1. Object A has the larger change in momentum. 2. Object B has the larger change in momentum. 3. Both objects have the same change in momentum 4. Not enough information is given to decide. Concept Question: Pushing Identical Carts Answer Answer 1: Both objects have the same mass, are pushed the same distance, by the same constant force, so they have the same acceleration. Since object A started from rest, an object B has an initial non-zero speed, object A needs a larger time interval to reach the finish than the corresponding time interval for object B Therefore the impulse on object A is larger than the corresponding impulse on object B. Hence object A has a larger change in momentum. Momentum, Kinetic Energy, and Work: Single Particle Kinetic energy and momentum for a single particle are related by 1 2 p2 K mv 2 2m Change in kinetic energy and work tf 2 2 p r r p 1 2 1 2 f W F d r K mv f mvi i 2 2 2m 2m t i Concept Question: Pushing Carts Consider two carts, of masses m and 2m, at rest on an air track. If you push one cart for 3 seconds and then the other for the same length of time, exerting equal force on each, the kinetic energy of the light cart is 1) larger than 2) equal to 3) smaller than the kinetic energy of the heavy car. Concept Question: Pushing Carts Answer 1. The kinetic energy of an object can be written as 1 2 p2 K mv 2 2m Because the impulse is the same for the two carts, the change in momentum is the same. Both start from rest so they both have the same final momentum. Since the mass of the lighter cart is smaller than the mass of the heavier cart, the kinetic energy of the light cart is larger than the kinetic energy of the heavy cart. Concept Question: Stopping Distances Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same momentum, and you exert the same force to stop each. How do the distances needed to stop them compare? 1. It takes a shorter distance to stop the ping-pong ball. 2. Both take the same distance. 3. It takes a longer distance to stop the ping-pong ball. Concept Question: Stopping Distances Answer 3. The kinetic energy of an object can be written as K = p2/2m. Because the ping pong ball and the bowling ball have the same momentum, the kinetic energy of the less massive ping pong ball is greater than the kinetic energy of the more massive bowling ball. You must do work on an object to change its kinetic energy. If you exert a constant force, then the work done is the product of the force with the displacement of the point of application of the force. Since the work done on an object is equal to the change in kinetic energy, the ping pong ball has a greater change in kinetic energy in order to bring it to a stop, so the you need a longer distance to stop the ping pong ball. Demo: Jumping Off the Floor with a Non-Constant Force Demo Jumping: Non-Constant Force • Plot of total external force vs. time for Andy jumping off the floor. Weight of Andy is 911 N. Demo Jumping: Impulse • Shaded area represents impulse of total force acting on Andy as he jumps off the floor F total (t ) N(t ) Fgrav t f 1.23 s I[ti , t f ] ti 0.11s F total (t ) dt 199 N s Demo Jumping: Height When Andy leaves the ground, the impulse is I y [0.11 s,1.23 s] 199 N s So the y-component of his velocity is vy, f I y [ti ,t f ] / m (199 N s)(9.80 m s-2 )/(911 N) 2.14 m s-1 Andy jumped (1 / 2)mvy,2 f mgh h vy,2 f / 2g 23.3 cm System of Particles: Center of Mass Position and Velocity of Center of Mass Mass for collection of discrete bodies (system): i N msys mi i1 Momentum of system: i N r r psys mi v i i1 Position of center of mass r 1 i N r R cm mi ri msys i1 Velocity of center of mass r 1 Vcm msys i N r psys i1 msys r mi v i r r psys msys Vcm Table Problem: Center of Mass of Rod and Particle A slender uniform rod of length d and mass m rests along the x-axis on a frictionless, horizontal table. A particle of equal mass m is moving along the x-axis at a speed v0. At t = 0, the particle strikes the end of the rod and sticks to it. Find a vector expression for the position of the center of mass of the system at t = 0. System of Particles: Internal and External Forces, Center of Mass Motion System of Particles: Newton’s Second and Third Laws The momentum of a system remains constant unless the system is acted on by an external force in which case the acceleration of center of mass satisfies r Fext r dpsys dt msys r dVcm dt r msys A cm Force on a System of N Particles is the External Force The force on a system of particles is the external force because the internal force is zero r r r r F Fext Fint Fext Internal Force on a System of N Particles is Zero • The internal force on the ith particle is sum of the interaction forces with all the other particles jN Fint,i Fi , j j 1 i j • The internal force is the sum of the internal force on each particle i1 r r Fint Fint,i j N r Fi, j i1 j 1 i j • Newton’s Third Law: internal forces cancel in pairs Fi , j Fj ,i • So the internal force is zero r r Fint 0 External Force and Momentum Change The momentum of a system of N particles is defined as the sum of the individual momenta of the particles i N r r r psys pi msys Vcm i1 Force changes the momentum of the system r r dpsys r r dpi F Fi dt i1 i1 dt i N i N Force equals external force r r F Fext Newton’s Second and Third Laws for a system of particles: The external force is equal to the change inr momentum of the system r Fext r dpsys dt d(msys Vcm ) dt r msys Acm External Forces and Constancy of Momentum Vector The external force may be zero in one direction but not others The component of the system momentum is constant in the direction that the external force is zero The component of system momentum is not constant in a direction in which external force is not zero Table Problem: Center of Mass of Rod and Particle Post- Collision A slender uniform rod of length d and mass m rests along the x-axis on a frictionless, horizontal table. A particle of equal mass m is moving along the xaxis at a speed v0. At t = 0, the particle strikes the end of the rod and sticks to it. Find a vector expression for the position of the center of mass of the system for t > 0. Demo : Center of Mass trajectory B78 http://tsgphysics.mit.edu/front/index.php?page= demo.php?letnum=B%2078&show=0 Odd-shaped objects with their centers of mass marked are thrown. The centers of mass travel in a smooth parabola. The objects consist of: a squash racket, a 16” diameter disk weighted at one point on its outer rim, and two balls connected with a rod. This demonstration is shown with UV light. CM moves as though all external forces on the system act on the CM so the jumper’s cm follows a parabolic trajectory of a point moving in a uniform gravitational field Table Problem: Exploding Projectile Center of Mass Motion An instrument-carrying projectile of mass m1 accidentally explodes at the top of its trajectory. The horizontal distance between launch point and the explosion is x0. The projectile breaks into two pieces which fly apart horizontally. The larger piece, m3, has three times the mass of the smaller piece, m2. To the surprise of the scientist in charge, the smaller piece returns to earth at the launching station. a) How far has the center of mass of the system traveled from the launch when the pieces hit the ground? b) How far from the launch point has the larger piece graveled when it first hits the ground? Concept Question: Pushing a Baseball Bat 1 2 3 The greatest instantaneous acceleration of the center of mass will be produced by pushing with a force F at 1. Position 1 2. Position 2 3. Position 3 4. All the same Concept Question: Pushing a Baseball Bat 1 2 3 Answer 4. The external force is equal to the total mass times the instantaneous acceleration of the center-of-mass. It doesn’t matter where the external force acts with regards to the center-of-mass acceleration. Table Problem: Landing Plane and Sandbag A light plane of mass 1000 kg makes an emergency landing on a short runway. With its engine off, it lands on the runway at a speed of 40 ms-1. A hook on the plane snags a cable attached to a sandbag of mass 120 kg and drags the sandbag along. If the coefficient of friction between the sandbag and the runway is μ = 0.4, and if the plane’s brakes give an additional retarding force of magnitude 1400 N, how far does the plane go before it comes to a stop? Strategy: Momentum of a System 1. Choose system 2. Identify initial and final states 3. Identify any external forces in order to determine whether any component of the momentum of the system is constant or not i) If there is a non-zero total external force: r dpsys r total Fext dt ii) If the total external force is zero then momentum is constant p sys,0 p sys,f Modeling: Instantaneous Interactions • Decide whether or not an interaction is instantaneous. • External impulse changes the momentum of the system. I[t , t tcol ] t tcol Fext dt (Fext )ave tcol psys t • If the collision time is approximately zero, tcol 0 then the change in momentum is approximately zero. p system 0