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CONSERVATION OF LINEAR MOMENTUM Chapter 20 Objectives Know that linear momentum is conserved when no outside forces act on the system Know that linear momentum can be added to a system by forces, or by adding mass Know the connection between Newton's laws and conservation of linear momentum Be able to do calculations involving linear momentum Rat 1 Rat 1 Rat 1 Take 3 minutes to answer the following (true or false) 1. Mass can be a state quantity. 2. Velocity can be a state quantity. 3. Linear momentum can be a state quantity. 4. Linear momentum is a vector quantity. 5. Linear momentum is always conserved. Linear Momentum Newton developed the concept of linear momentum. Linear momentum, p, is defined as the product of the mass, m, and velocity, v p=mv p mv The momentum vector is an entirely different vector than the velocity vector. Care should be taken in comparing one to the other. It is safe to say that the momentum vector is in the same direction as the velocity vector as mentioned earlier. One can also say that the momentum vector is directly proportional to the velocity vector, i.e., the momentum vector doubles if the velocity vector doubles. But momentum also depends on the mass. So changing the mass of an object will also change the momentum vector. Therefore to change momentum one must change the mass or velocity or both. Regardless of what changes, the momentum vector is always in the same direction as the velocity vector. p mv As long as there are no external forces acting on a system of particles, collisions between the particles will exhibit conservation of linear momentum. This means that the vector sum of the momenta before collision is equal to the vector sum of the momenta of the particles afterwards. p mv This is known as the conservation of linear momentum. It is an extremely important concept in physics. One important area that makes use of this conservation principle is collisions. This is what you are going to explore today. p mv The collision you will study will involve two objects of equal mass colliding in a horizontal plane and then undergoing projectile motion after the collision. Since the horizontal component of velocity remains constant for a projectile in free fall, the horizontal part of the projectile motion can be used to represent the horizontal component of the momentum after collision. p mv Simple Examples of Head-On Collisions (Energy and Momentum are Both Conserved) Collision between two objects of the same mass. One mass is at rest. Collision between two objects. One at rest initially has twice the mass. Collision between two objects. One not at rest initially has twice the mass. p mv Simple Examples of Head-On Collisions (Totally Inelastic Collision, only Momentum Conserved) Collision between two objects of the same mass. One mass is at rest. Collision between two objects. One at rest initially has twice the mass. Collision between two objects. One not at rest initially has twice the mass. p mv Example of Non-Head-On Collisions (Energy and Momentum are Both Conserved) Collision between two objects of the same mass. One mass is at rest. If you vector add the total momentum after collision, you get the total momentum before collision. p mv Velocity Components in Projectile Motion (In the absence of air resistance.) Note that the horizontal component of the velocity remains the same if air resistance can be ignored. Here is another example which you can at home You will roll a ball down the curved ramp. This represents the velocity as the ball left the table because the horizontal velocity of a projectile remains constant in the absence of air resistance. p mv Linear Momentum Force is the Rate of Change of Momentum Mass is a state quantity. Velocity is also a state quantity, but because it also has direction, it is also a vector quantity. Because mass is a scalar and velocity is a vector, then momentum is also a vector. The algebraic combination of state quantities yields a state quantity, thus momentum is a state quantity (closed systems only). Momentum Momentum is a property of moving matter. Momentum describes the tendency of objects to keep going in the same direction with the same speed. Changes in momentum result from forces or create forces. Momentum The momentum of a ball depends on its mass and velocity. Ball B has more momentum than ball A. Momentum The momentum of a ball depends on its mass and velocity. Ball B has more momentum than ball A. Momentum and Inertia Inertia is another property of mass that resists changes in velocity; however, inertia depends only on mass. Inertia is a scalar quantity. Momentum is a property of moving mass that resists changes in a moving object’s velocity. Momentum is a vector quantity. Momentum Ball A is 1 kg moving 1m/sec, ball B is 1kg at 3 m/sec. A 1 N force is applied to deflect the motion of each ball. What happens? Does the force deflect both balls equally? Ball B deflects much less than ball A when the same force is applied because ball B had a greater initial momentum. Equivalent Momenta Car: m = 1800 kg; v = 80 m /s p = 1.44 ·105 kg · m /s Bus: m = 9000 kg; v = 16 m /s p = 1.44 ·105 kg · m /s Train: m = 3.6·104 kg; v = 4 m /s p = 1.44 ·105 kg · m /s Kinetic Energy and Momentum Kinetic energy and momentum are different quantities, even though both depend on mass and speed. Kinetic energy is a scalar quantity. Momentum is a vector, so it always depends on direction. Two balls with the same mass and speed have the same kinetic energy but opposite momentum. Calculating Momentum The momentum of a moving object is its mass multiplied by its velocity. That means momentum increases with both mass and velocity. Momentum (kg m/sec) Mass (kg) p=mv Velocity (m/sec) Comparing momentum A car is traveling at a velocity of 13.5 m/sec (30 mph) north on a straight road. The mass of the car is 1,300 kg. A motorcycle passes the car at a speed of 30 m/sec (67 mph). The motorcycle (with rider) has a mass of 350 kg. Calculate and compare the momentum of the car and motorcycle. 1. 2. 3. You are asked for momentum. You are given masses and velocities. Use: p = m v 4. 5. Solve for car: p = (1,300 kg) (13.5 m/s) = 17,550 kg m/s Solve for cycle: p = (350 kg) (30 m/s) = 10,500 kg m/s The car has more momentum even though it is going much slower. Conservation of Momentum The law of conservation of momentum states when a system of interacting objects is not influenced by outside forces (like friction), the total momentum of the system cannot change. If you throw a rock forward from a skateboard, you will move backward in response. Conservation of Momentum Collisions in One Dimension A collision occurs when two or more objects hit each other. During a collision, momentum is transferred from one object to another. Collisions can be elastic or inelastic. Collisions Elastic collisions Two 0.165 kg billiard balls roll toward each other and collide head-on. Initially, the 5-ball has a velocity of 0.5 m/s. The 10-ball has an initial velocity of -0.7 m/s. The collision is elastic and the 10-ball rebounds with a velocity of 0.4 m/s, reversing its direction. What is the velocity of the 5-ball after the collision? Elastic collisions 1. You are asked for 10-ball’s velocity after collision. 2. You are given mass, initial velocities, 5-ball’s final velocity. 3. Diagram the motion, use m1v1 + m2v2 = m1v3 + m2v4 4. Solve for V3 : (0.165 kg)(0.5 m/s) + (0.165 kg) (-0.7 kg)= (0.165 kg) v3 + (0.165 kg) (0.4 m/s) 5. V3 = -0.6 m/s Inelastic collisions A train car moving to the right at 10 m/s collides with a parked train car. They stick together and roll along the track. If the moving car has a mass of 8,000 kg and the parked car has a mass of 2,000 kg, what is their combined velocity after the collision? 1. You are asked for the final velocity. 2. You are given masses, and initial velocity of moving train car. Inelastic collisions 3. Diagram the problem, use m1v1 + m2v2 = (m1v1 +m2v2) v3 4. Solve for v3= (8,000 kg)(10 m/s) + (2,000 kg)(0 m/s) (8,000 + 2,000 kg) v3= 8 m/s The train cars moving together to right at 8 m/s. Collisions in 2 and 3 Dimensions Most real-life collisions do not occur in one dimension. In a two or three-dimensional collision, objects move at angles to each other before or after they collide. In order to analyze two-dimensional collisions you need to look at each dimension separately. Momentum is conserved separately in the x and y directions. Collisions in 2 and 3 Dimensions Force is the Rate of Change of Momentum Investigation Key Question: How are force and momentum related? Force is the Rate of Change of Momentum Momentum changes when a net force is applied. The inverse is also true: If momentum changes, forces are created. If momentum changes quickly, large forces are involved. Force and Momentum Change The relationship between force and motion follows directly from Newton's second law. Force (N) Change in momentum (kg m/sec) F = D p/D t Change in time (sec) Calculating force Starting at rest, an 1,800 kg rocket takes off, ejecting 100 kg of fuel per second out of its nozzle at a speed of 2,500 m/sec. Calculate the force on the rocket from the change in momentum of the fuel. 1. You are asked for force exerted on rocket. 2. You are given rate of fuel ejection and speed of rocket 3. Use F = Δp/Δt = (Δm*v) Δt = (Δm/Δt)*v 4. Thus: F = (Δm/Δt)*v = (100 kg/s)*(-2,500 m/s) = -250,000 N The fuel exerts and equal and opposite force on rocket of +250,000 N. Impulse The product of a force and the time the force acts is called the impulse. Impulse is a way to measure a change in momentum because it is not always possible to calculate force and time individually since collisions happen so fast. Force and Momentum Change To find the impulse, you rearrange the momentum form of the second law. Impulse (N•sec) FD t=Dp Change in momentum (kg•m/sec) Impulse can be expressed in kg•m/sec (momentum units) or in N•sec. Jet Engines Nearly all modern airplanes use jet propulsion to fly. Jet engines and rockets work because of conservation of linear momentum. A rocket engine uses the same principles as a jet, except that in space, there is no oxygen. Most rockets have to carry so much oxygen and fuel that the payload of people or satellites is usually less than 5 percent of the total mass of the rocket at launch. Momentum and Energy Two objects with masses m1 and m2 have equal kinetic energy. How do the magnitudes of their momenta compare? (A) p1 < p2 (B) p1 = p2 (C) p1 > p2 (D) Not enough information is given April 13, 2015 Linear Momentum and Newton’s Laws Newton’s 1st law When viewed in an inertial reference frame*, an object either remains at rest or moves at a constant velocity, unless acted upon by an external force Therefore, if there is no force acting on a body, its momentum stays constant. *Recall Coriolis and centrifugal forces Newton’s 2nd law The second law states that the net force on an object is equal to the rate of change (that is, the derivative) of its linear momentum p in an inertial reference frame F = dp/dt = d(mv)/dt For constant-mass systems, mass can be taken outside the differentiation operator F = mdv/dt = ma Linear Momentum and Newton’s Laws Newton’s 3rd law The linear momentum of one body changes the same as the other body, but in opposite directions, so the linear momentum of the universe is not affected. Forces always exist by the interaction of bodies; the force on one body is equal and opposite to the force on the other body. Newton’s third law States that all forces exist in pairs: if one object A exerts a force FA on a second object B, then B simultaneously exerts a force FB on A, and the two forces are equal and opposite: FA = −FB. The third law means that all forces are interactions between different bodies, and thus that there is no such thing as a unidirectional force or a force that acts on only one body. This law is sometimes referred to as the action-reaction law. The action and the reaction are simultaneous, It does not matter which is called the action and which is called reaction; Both forces are part of a single interaction, and neither force exists without the other. Newton stated the third law within a world-view Assumed instantaneous action at a distance between material particles. However, he was prepared for philosophical criticism of this action at a distance, and it was in this context that he stated the famous phrase "I feign no hypotheses". In modern physics, action at a distance has been completely eliminated, except for effects involving quantum entanglement. However in modern engineering in all practical applications involving the motion of vehicles and satellites, the concept of action at a distance is used extensively. Revisiting the UAE Before applying the Universal Accounting Equation (UAE): Define a system. Determine what quantity will be counted. Define time interval for counting. Best for Energy, Mass, … Care should be taken for Charge, Momentum, … Designating the System When solving linear momentum problems, the system boundary must be defined. This boundary is at the discretion of the engineer. There is no requirement that the system contain all bodies involved in the process. Thus, a system can have unbalanced forces which will change the linear momentum of the system. Revisiting the UAE The UAE equation FINAL - INITIAL = IN - OUT + GEN - CONS Accum = Net Input + Net Generation State Path Terms of the UAE FINAL AMOUNT: …in system at the END of time period INITIAL AMOUNT: …in system at the START of time period INPUT: …PASSING through boundary INTO system during time period OUTPUT: …PASSING through boundary OUT of system during time period GENERATION: …PRODUCED during time period within boundary CONSUMPTION: …DESTROYED during time period within boundary Linear Momentum Is Conserved Linear momentum when is a conserved quantity: 0 0 FINAL - INITIAL = IN - OUT + GEN - CONS 0 Accum = Net Input + Net Generation Steps to Applying the UAE To use the UAE, define the system, then compute: 1. INITIAL linear momentum for the mass in the system, 2. FINAL linear momentum for the mass in the system, 3. Linear momentum INPUT into the system, 4. Linear momentum OUTPUT from the system. Linear Momentum Change There are two main ways that the linear momentum of a system can change: mass changes velocity changes a combination of both: external forces Otherwise, the accumulation of linear momentum is zero Changing Linear Momentum by External Forces -F F Initial State mvi Positive Direction Time Passes F Final State mvf -F Changing Linear Momentum by Unbalanced Forces Momentum flow INTO or OUT OF a system can result from FORCES. Mathematically, SI Units of Force = kg·m/s2 SI Units of Momentum = kg·m/s Thus, Units of Force = Units of Momentum/s p F Analogy: Mass Flowing into Tank Assume mass flow rates are constant Dt m in out m in m Dmsystem m final minitial min mout mnet Dmsystem m in m out m net Dt dmsys m net dt out m Analogy: Momentum Flowing from Force dmsys dt dpsys dt d ( mv) dt m net (From previous slide) (By analogy) p net Fnet F a m m dv net dt Fnet m a (For constant mass in system) Newt on's 2nd Law!!! Individual Exercise #1: A 1.000-ton dragster has a jet engine that provides a thrust of 3000. lbf. The dragster starts from a dead stop at the start line and crosses the finish line 9 seconds later. Neglect mass loss. (a) What is its final velocity? (b) How long was the track? Pairs Exercise #1 A 0.100-kg hockey puck is stationary on the ice. Then it is hit with a hockey stick and 0.020 s later, the puck is traveling at 100. km/h. (a) What is the average force (N) on the puck? (b) What is the average force (N) on the hockey stick? Linear Momentum Change by Mass Transfer Suppose you define a system that contains a single object of mass m1 moving at a velocity of v1 in the positive x-direction As time passes, two more objects enter the system with masses m2 and m3 and velocities v2 and v3 respectively, also in the x-direction. Time m2v2 m1v1 Passes m3v3 m1v1 Time Passes system boundary m2v2 m1v1 m3v3 Linear Momentum Change by Mass Transfer Initial linear momentum pi = m1v1 Final linear momentum pf = (m1v1 + m2v2 + m3v3) Note:for this case we can write these as scalars if we use the magnitude of the x-component of the vectors. Then, accumulation = pf - pi is not zero. Pairs Exercise #2 A 1.0-ton (including ammo and pilot) military helicopter is flying at 40 mph. It has a machine gun that fires 60 bullets per second in the forward direction. Each 0.5-lbm bullet exits the gun at 1800 mph. After a 2-s burst of fire, what is the helicopter speed? Pairs Exercise #3 For the previous exercise, if the final system had been defined to include the fired bullets, would the answer be different? Helicopter w/ ammo mi = 1 ton vi = 40 mph Initial Helicopter w/ less ammo mf = ? vf = ? Final bullets Systems Without Net Linear Momentum Input No unbalanced forces & no mass transfer into or out of the system. UAE simplifies to: ACCUMULATION=0 FINAL AMOUNT - INITIAL AMOUNT = 0 FINAL AMOUNT = INITIAL AMOUNT Therefore, there is no change in linear momentum of the system. Individual Exercise #2 A 5.0-g bullet is fired horizontally at 300 m/s and passes through a 500-g block of wood initially at rest on a frictionless surface. The bullet emerges with a speed of 100 m/s. What is the final speed of the block? Pairs Exercise #4: In Example 20.5, what is the change in linear momentum of the white ball? The black ball? What is the total kinetic energy of the two balls before the impact and after the impact? Is energy conserved in this impact? ary, so Equation 20-24 may be used. (20-24) Pairs Exercise #5.1 The following two streams of steam are blended together. Stream A: 1 kg, 1 m/s Stream B: 1 kg, 10 m/s a. What is the mass of the blended stream? b. Assuming no friction, what is the final velocity of the blended stream? c. What is the kinetic energy before, and after, the blending process? d. What is the efficiency, i.e., what fraction of the original kinetic energy is retained in the blended stream? Conventional Jet Ejector 2 1 3 Throat Diffuser Pairs Exercise #5.2 The following two streams of steam are blended together. Stream A: 1 kg, 1 m/s Stream B: 1 kg, 2 m/s a. What is the mass of the blended stream? b. Assuming no friction, what is the final velocity of the blended stream? c. What is the kinetic energy before, and after, the blending process? d. What is the efficiency, i.e., what fraction of the original kinetic energy is retained in the blended stream? Advanced Jet Ejector 1 1 1 2.8 1 1.5 Numbers are Mach numbers. QUESTION 1: For years, space travel was believed to be impossible because there was nothing that rockets could push off of in space in order to provide the propulsion necessary to accelerate. This inability of a rocket to provide propulsion is because ... a) space is void of air so the rockets have nothing to push off of. b) gravity is absent in space. c) space is void of air and so there is no air resistance in space. d) nonsense! Rockets do accelerate in space and have been able to do so for a long time. Question 2 While driving down the road, a firefly strikes the windshield of a bus and makes a quite obvious mess in front of the face of the driver. This is a clear case of Newton's third law of motion: the firefly hits the bus and the bus hits the firefly. Which of the two forces is greater: (a) the force on the firefly (b) the force on the bus (c) none of the above QUESTION 3: In the top picture (below), Joe is pulling upon a rope that is attached to a wall. In the bottom picture, Joe is pulling upon a rope that is attached to an elephant. In each case, the force scale reads 500 Newton. Joe is pulling ... a) with more force when the rope is attached to the wall. b) with more force when the rope is attached to the elephant. c) the same force in each case