Copyright © 2010 Pearson Education, Inc. Lecture Outline Chapter 9 Physics, 4 th Edition James S. Walker

Copyright © 2010 Pearson Education, Inc. Chapter 9 Linear Momentum and Collisions

Copyright © 2010 Pearson Education, Inc. Units of Chapter 9 Linear Momentum Momentum and Newton’s Second Law Impulse Conservation of Linear Momentum Inelastic Collisions Elastic Collisions

Copyright © 2010 Pearson Education, Inc. Center of Mass Systems with Changing Mass: Rocket Propulsion Units of Chapter 9

Copyright © 2010 Pearson Education, Inc. 9-1 Linear Momentum Momentum is a vector; its direction is the same as the direction of the velocity. Linear momentum is defined as the product of the mass m and velocity V of an object.

Copyright © 2010 Pearson Education, Inc. 9-1 Linear Momentum Change in momentum: (a) mv (b) 2 mv Change in momentum A beanbag bear and a rubber ball with the same mass m and the same downward speed v, hit the floor. (a) The beanbag bear comes to rest on hitting the floor. Its change in momentum is mv upward. (b). The rubber ball bounces upward with a speed. Its change in momentum is 2mv upward.

Copyright © 2010 Pearson Education, Inc. 9-1 Linear Momentum At a city park, a person throws some bread into a duck pond. Two 4.0-kg ducks and a 9.0-kg goose paddle rapidly toward the bread, as shown in our sketch. If the ducks swim at 1.10 m/s, and the goose swims with a speed of 1.30 m/s., find the magnitude and direction of the total momentum of the three birds. Use x and y unit vectors to express the momentum of each bird in vector form: P d1 = m a v d x = (4.0kg)(1.10 m/s)x = 4.40 kg.m/s)x P d2 = - m a v d y = - (4.0kg)(1.10 m/s)y= 4.40 kg.m/s)y P g = m g v g y = (9.0kg)(1.30 m/s)y = 11.7 kg.m/s)y P total = P d1 + P d2 + P g P total = (4.40 kg.m/s)x + [ kg.m/s]y = (4.40 kg.m/s)x + (7.30 kg.m/s) y Calculate the magnitude of the total momentum: P total = (P 2 total,x + P 2 total,y ) 1/2 = ((4.40 kg.m/s) 2 + (7.30 kg.m/s) 2 ) 1/2 P total = 8.52 kg.m/s Calculate the direction of the total momentum: Θ = tan -1 ((7.30 kg.m/s) / (4.40 kg.m.s)) = 58.9 O

Copyright © 2010 Pearson Education, Inc. 9-2 Momentum and Newton’s Second Law Newton’s second law, as we wrote it before: is only valid for objects that have constant mass. Here is a more general form, also useful when the mass is changing:

Copyright © 2010 Pearson Education, Inc. 9-3 Impulse Impulse is a vector, in the same direction as the average force. The pitcher delivers a fastball, the batter takes a swing, and with a crack of the bat the ball that was approaching home plate at 95.0 mi/h is now heading toward the pitcher at 115 mi/h. We say the bat has delivered an impulse, I to the ball.

Copyright © 2010 Pearson Education, Inc. 9-3 Impulse We can rewrite as So we see that The impulse is equal to the change in momentum.

Copyright © 2010 Pearson Education, Inc. 9-3 Impulse Therefore, the same change in momentum may be produced by a large force acting for a short time, or by a smaller force acting for a longer time. Hitting a baseball A batter hits a ball, sending it back toward the pitcher’s mound. The impulse delivered to the ball by the bat changes the ball’s momentum from –p i x to p f x

Copyright © 2010 Pearson Education, Inc. 9-4 Conservation of Linear Momentum The net force acting on an object is the rate of change of its momentum: If the net force is zero, the momentum does not change:

Copyright © 2010 Pearson Education, Inc. 9-4 Conservation of Linear Momentum Internal Versus External Forces: Internal forces act between objects within the system. As with all forces, they occur in action-reaction pairs. As all pairs act between objects in the system, the internal forces always sum to zero: Therefore, the net force acting on a system is the sum of the external forces acting on it.

Copyright © 2010 Pearson Education, Inc. 9-4 Conservation of Linear Momentum Furthermore, internal forces cannot change the momentum of a system. However, the momenta of components of the system may change.

Copyright © 2010 Pearson Education, Inc. 9-4 Conservation of Linear Momentum An example of internal forces moving components of a system: Two groups of canoeists meet in the middle of a lake. After a brief visit, a person in canoe 2 with a force of 46 N to separate the canoes. If the mass of canoe 1 and its occupants is 130-kg, and the mass of canoe 2 and its occupants is 250-kg, find the momentum of each canoe after 1.20 s of pushing. -Newton’s 2 nd law : a 2,x = ∑F 2,x / m 2 = 46 N/250 kg = 0.18 m/s 2 -a 1,x = ∑F 1,x / m 1 = - 46 N/130 kg = m/s 2 -V 2,x = a 2,x t = (0.18 m/s 2 ) (1.20 s) = 0.22 m/s -V 1,x = a 1,x t = ( m/s 2 ) (1.20 s) = m/s -p 1,x = m 1 v 1, x = (130 kg) ( m/s) = - 55 kg.m / s -p 2,x = m 2 v 2, x = (250 kg) (0.22 m/s) = 55 kg.m / s -Note that the sum of the momenta of the two canoes is zero. The canoes start at rest with zero momentum, there is zero net external force acting on the system, hence the final momentum must also be zero. The final velocities do not add to zero; it is momentum (mv) that is conserved, not velocity (v). Choose the positive direction to point from canoe 1 to canoe 2. With this choice, the force exerted on canoes 2 is F 2 = (46 N) x and the force exerted on canoe 1 is F 1 = (-46 N) x.

Copyright © 2010 Pearson Education, Inc. 9-5 Inelastic Collisions Collision: two objects striking one another Time of collision is short enough that external forces may be ignored Inelastic collision: momentum is conserved but kinetic energy is not Completely inelastic collision: objects stick together afterwards

Copyright © 2010 Pearson Education, Inc. 9-5 Inelastic Collisions A completely inelastic collision: Railroad cars collide and stick together A moving train car collides with an identical car that is stationary. After the collision, the cars stick together and move with the same speed.

Copyright © 2010 Pearson Education, Inc. 9-5 Inelastic Collisions Solving for the final momentum in terms of the initial momenta and masses:

Copyright © 2010 Pearson Education, Inc. 9-5 Inelastic Collisions On a touchdown attempt, a 95.0-kg running back runs toward the end zone at 3.75 m/s. A 111-kg linebacker moving at 4.10 m/s meets the runner in a head-on collision. If the two players stick together, (a) what is their velocity immediately after the collision? (b) What are the initial and final kinetic energy of the system? V f = (95.0 kg) (3.75 m/s) x + (111 kg)(-4.10 m/s) x 95.0 kg kg V f = ( m/s) x (b)Calculate the initial kinetic energy of the two players K i = 1/2m 1 v ½ m 2 v 2 2 = ½(95.0 kg)(3.75 m/s) 2 + ½(111-kg)(-4.10 m/s) 2 K i = J Calculate the final kinetic energy of the two players noting that they both move with the same velocity after the collision. K f = ½(m 1 + m 2) v f 2 = ½(95-kg kg)(-0.48 m/s) 2 = 23.7 J K f = 23.7 J

Copyright © 2010 Pearson Education, Inc. 9-5 Inelastic Collisions Ballistic pendulum: the height h can be found using conservation of mechanical energy after the object is embedded in the block. Our sketch shows the physical setup of a ballistic pendulum. Initially, only the object of m is moving, and it moves in the positive x direction with a speed v o. Immediately after the collision, the bob and object move together with a new speed, v f, which is determined by momentum conservation. Finally, the pendulum continues to swing to the right until its speed decreases to zero and it comes to rest at the height h.

Copyright © 2010 Pearson Education, Inc. 9-5 Inelastic Collisions In a ballistic pendulum, an object of mass m is fired with an initial speed v o at the bob of a pendulum. The bob has a mass M, and is suspended by a rod of negligible mass. After the collision, the object and the bob stick together and swing through an arc, eventually gaining a height h. Find the height h in terms of m, M, v o and g. Set the momentum just before the bob-object collision equal to the momentum just after the collision. mv o = (M + m)v f v f = (m / (M + m) ) v o Calculate the kinetic energy just after the collision K f = ½ (M + m) v f 2 = ½ (M + m) ((m / (M + m) 2 ) v o 2 K f = ½ m v o 2 (m ) (M + m) K f = (M + m) gh = ½ m v o 2 (m ) (M + m) h = (m / M + m) 2 (v o 2 /2g)

Copyright © 2010 Pearson Education, Inc. 9-5 Inelastic Collisions in Two Dimensions For collisions in two dimensions, where we must conserve the momentum component by component. To do this, we set up a coordinate system and resolve the initial momentum into x and y components as the initial momentum. That is, p x,i = p x, f p y, i = p y,f

Copyright © 2010 Pearson Education, Inc. 9-5 Inelastic Collisions in Two Dimensions A car with a mass of 950 kg and a speed of 16 m/s approaches an intersection as shown below. A 1300-kg minivan traveling at 21 m/s is heading for the same intersection. The car and the minivan collide and stick together. Find the speed and direction of the wrecked vehicles just after the collision, assuming external forces can be ignored. In our sketch, we align the x and y axes with the crossing streets. With this choice, V 1 (the car’s velocity) is in the positive x direction and V 2 (the minivan's velocity) is in the positive y direction. The problem indicates that m 1 = 950 kg and m 2 = 1300 kg. After the collision, the two vehicles move together ( as a unit) with a speed V f in a direction θ with respect to the positive x axis. Set the initial x component of momentum equal to the final x component of momentum: m 1 v 1 = (m 1 + m 2 )v f cos θ eq. 1 Set for the y component of momentum m 2 v 2 = (m 1 + m 2 )v f sin θ (eq. 2) Divide eq 2. by eq. 1  m 2 v 2 = (m 1 + m 2 )v f sin θ = tan θ m 1 v 1 (m 1 + m 2 )v f cos θ θ = tan -1 (m 2 v 2 / m 1 v 1 ) = tan -1 ((1300kg)(21 m/s) / (950 kg)(16 m/s)) θ = tan -1 (1.8) = 61 O The final speed v f = m 1 v 1 / (m 1 + m 2 ) cos θ = (950 kg)(16 m/s) / (950 kg kg) cos 61 o V f = 14 m /s

Copyright © 2010 Pearson Education, Inc. 9-6 Elastic Collisions In elastic collisions, both kinetic energy and momentum are conserved. That is, p i = p f and K i = K f One-dimensional elastic collision: An elastic collisions between two carts In the case pictured, V 1,f is to the right (positive), which means that m 1 is greater than m 2. In fact, we have chosen m 1 = 2m 2 for this pilot, therefore, v 1,f = V o /3 and V 2,f = 4V o /3. if m1 were less than m 2, cart 1 would bounce back toward the left, meaning that V 1,f would be negative.

Copyright © 2010 Pearson Education, Inc. 9-6 Elastic Collisions We have two equations (conservation of momentum and conservation of kinetic energy) and two unknowns (the final speeds). Solving for the final speeds:

Copyright © 2010 Pearson Education, Inc. 9-6 Elastic Collisions in Two Dimensions Two-dimensional collisions can only be solved if some of the final information is known, such as the final velocity of one object:

Copyright © 2010 Pearson Education, Inc. 9-6 Elastic Collisions in Two Dimensions Two astronauts on opposite ends of a spaceship are comparing launches. One has an apple, the other has an orange. They decide to trade. Astronauts-1 tosses the kg apple toward astronaut 2 with a speed of 1.11 m/s. The kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.21-m/s. Unfortunately, the fruits collide, sending the orange off with a speed of 1.16 m/s at an angle of with respect to its original direction of motion. Find the final speed and direction of the apple, assuming an elastic collision. Give the apple’s direction relative to its original direction of motion. 1.Calculate the initial kinetic energy of the system: K i = 1/2m 1 v 1,i 2 + 1/2m 2 v 2,1 2 K i = ½(0.130 kg)(1.11 m/s) 2 + ½(0.160 kg)(1.21 m/s) 2 K i = J. 2. Calculate the final Kinetic energy of the system in terms of V 1,f : K i = 1/2m 1 v 1f 2 + 1/2m 2 v 2,f 2 K f = ½(0.130 kg)(V 1,f m/s) 2 + ½(0.160 kg)(1.16 m/s) 2 = ½(0.130 kg)(V 1,f m/s) J V 1,f = (2(0.197 J J)) / 0.130) ½ = 1.17 m/s 3. Set the final y component of momentum equal to zero to determine the angle θ 0 = m 1 V 1,f sin θ - m2V 2,f sin 42.0 O sin θ = m 2 V 2,f sin 42.0 O / m 1 V 1,f = (0.160 kg) (1.16 m/s) sin 42.0 O / (0.130 kg) (1.17 m/s) θ = sin -1 (0.817) = 54.8 O

Copyright © 2010 Pearson Education, Inc. 9-7 Center of Mass The center of mass of a system is the point where the system can be balanced in a uniform gravitational field.

Copyright © 2010 Pearson Education, Inc. 9-7 Center of Mass For two objects: The center of mass is closer to the more massive object.

Copyright © 2010 Pearson Education, Inc. 9-7 Center of Mass The center of mass need not be within the object: Locating the center of mass In an object of continuous, uniform mass distribution, the center of mass is located at the geometric center of the object. In some cases, this means that the center of mass is not located within the object.

Copyright © 2010 Pearson Education, Inc. 9-7 Center of Mass Motion of the center of mass:

Copyright © 2010 Pearson Education, Inc. 9-7 Center of Mass The total mass multiplied by the acceleration of the center of mass is equal to the net external force: The center of mass accelerates just as though it were a point particle of mass M acted on by

Copyright © 2010 Pearson Education, Inc. 9-8 Systems with Changing Mass: Rocket Propulsion If a mass of fuel Δ m is ejected from a rocket with speed v, the change in momentum of the rocket is: The force, or thrust, is SI unit : newton, N

Copyright © 2010 Pearson Education, Inc. Summary of Chapter 9 Linear momentum: Momentum is a vector Newton’s second law: Impulse: Impulse is a vector The impulse is equal to the change in momentum If the time is short, the force can be quite large

Copyright © 2010 Pearson Education, Inc. Summary of Chapter 9 Momentum is conserved if the net external force is zero Internal forces within a system always sum to zero In collision, assume external forces can be ignored Inelastic collision: kinetic energy is not conserved Completely inelastic collision: the objects stick together afterward

Copyright © 2010 Pearson Education, Inc. Summary of Chapter 9 A one-dimensional collision takes place along a line In two dimensions, conservation of momentum is applied separately to each Elastic collision: kinetic energy is conserved Center of mass:

Copyright © 2010 Pearson Education, Inc. Summary of Chapter 9 Center of mass:

Copyright © 2010 Pearson Education, Inc. Summary of Chapter 9 Motion of center of mass: Rocket propulsion: