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Momentum & Collisions Chapter 6 What is this chapter about? • To this point, we have learned about the motion of single objects. • This chapter is about collisions, the interaction of a moving object with other objects. Important Concepts We will learn about… • Momentum – Its transfer – Its conservation • Impulse – Relates force to changes in momentum • Types of Collisions – Elastic – Inelastic 6.1 Momentum and Impulse rieberer.com/photogallery/photo77123 Linear Momentum • Momentum describes motion • A vector quantity defined as the product of an object’s mass and velocity • p = mv • Units: kg·m/s • Momentum increases with increasing mass and/or increasing velocity Why is momentum important? It allows us to discuss the interaction of two objects… Particularly collisions collisions between moving objects collision of a moving object with a stationary object An ostrich with mass of 145 kg escapes its pen with a velocity of +17m/s (~ 38 mi/h). Find the momentum of the bolting bird. p = mv = 145 kg x 17 m/s = 2500 kg∙m/s Impulse • Changing momentum requires force and time • Impulse is the product of an average external force and the time during which it acts on an object • Impulse causes a change in momentum • Impulse·momentum theorem F∆t = ∆p F∆t = ∆p = mvf – mvi Sometimes symbolized as J or I. Impulse • The quantity F∆t is the impulse of the force F acting for the time interval ∆t • Is related to Newton’s second law • F = ma • F = m(∆v/∆t) F∆t = m∆v • F∆t = ∆p • Shows that the longer a force is applied, or the greater the force… • the greater the change in momentum, e.g. follow through in batting a baseball or in tennis. Plane & Duck Collison • Estimate the average impact force between an airliner traveling at 600 mi/hr (268 m/s) and a 1 pound (0.454 kg) duck whose collison time is 0.0011 sec. Plane & Duck Collison • Estimate the average impact force between an airliner traveling at 600 mi/hr (268 m/s) and a 1 pound (0.454 kg) duck whose collison time is 0.0011 sec. Ft p mv f mvi mv f vi 0.454kg268m / s 0 F t 0.0011s F 1.1105 N 12tons Stopping Times and Distances Impulse-momentum theorem can be used to solve these problems A force of 8410 N is applied to a car (2240 kg), which slows uniformly from 20.0 m/s to 5.00 m/s. How long does it take for the car to slow down? FΔt = Δp What distance is required to slow down? vf2= vi2 + 2ax Stopping Time and Distance A force of 8410 N is applied to a car (2240 kg), which slows uniformly from 20.0 m/s to 5.00 m/s. How long does it take for the car to slow down? Ft p p (mv f mvi ) t 4.00s F F How far did the car travel while slowing down? vavg x t x vavg t What was the acceleration of the car? v a t 6.2 Conservation of Momentum Objectives 1. Describe the interaction between two objects in terms of the change in momentum of each object 2. Compare the total momentum of two objects before and after they interact 3. State the law of conservation of momentum 4. Predict the final velocities of objects after collisions, given the initial velocities Momentum is Conserved • • • • Law of Conservation of Momentum Total initial momentum = total final momentum pi = pf mvi = mvf Total momentum of all objects interacting with one another remains constant • Momentum is conserved in collisions and when objects push away from one another m1v 1i m2 v 2i m1v 1f m2 v 2f Practice Problems 6D An 85.0 kg fisherman jumps from a dock into a 135.0 kg rowboat at rest. If the velocity of the fisherman is 4.30 m/s, what is the final velocity of the fisherman and the boat? What do you know? m1 = 85.0 kg; v1 = 4.30 m/s; v2 = 0 m/s; m2 = 135 kg What principle is involved in this problem? Conservation of momentum; ∑pi = ∑pf What is the unknown variable? Velocity of the boat How would you set up this problem? p1i + p2i = p1f + p2f m1v1i + m2v2i = m1v1f + m2v2f but…the two objects join, so m1v1i + m2v2i = (m1+ m2)vf What is your answer? 1.66 m/s 6.3 Elastic and Inelastic Collisions Objectives 1. Identify different types of collisions 2. Determine the decrease in kinetic energy during perfectly inelastic collisions 3. Compare conservation of momentum and conservation of kinetic energy in perfectly inelastic collisions 4. Find the final velocity of an object in perfectly inelastic and elastic collisions Types of Collisions Collision Momentum Kinetic Energy Perfectly inelastic Conserved Not Conserved Elastic Conserved Conserved Perfectly Inelastic Collisions • When two objects collide, they stick together. • If the objects stick together, then the consolidated mass has a single velocity • So, if m1v1i + m2v2i = m1v1f + m2v2f • Then m1v1i + m2v2i = (m1+ m2)vf Practice 6E A 2500 kg car traveling at 20.0 m/s runs into a 5000 kg truck which is stopped at a traffic light. If the two vehicles stick together and move together after a collision, what is the final velocity of the two-vehicle mass? m1v1i + m2v2i = (m1+ m2)vf 6.7 m/s KE is not conserved during inelastic collisions 6F #1 a. A 0.25 kg arrow with velocity 12 m/s strikes and pierces a 6.8 kg target What is the final velocity of the combined mass? m1v1i + m2v2i = (m1+ m2)vf (0.25kg)(12m/s) + (6.8kg)(0 m/s) = (0.25 + 6.8kg)(vf) vf = 0.43 m/s b. What is the decrease in kinetic energy during the collision? ΔKE = KEf-KEi [KE = ½mv2] ΔKE = KE combined mass – (KE arrow + KE target) ΔKE = 17 J