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Topic 5 Conservation of Linear Momentum I. Introduction. a. One of the most important principles in physics is the law of conservation of momentum, which says that the total momentum of a system and its surroundings does not change. Whenever the momentum of a system changes, we can account for the change by the appearance or disappearance of momentum somewhere else. b. II. In this chapter, we introduce the ideas of impulse and linear momentum, and show how integrating Newton’s second law produces an important theorem known as the impulse–momentum theorem. We will also determine if the momentum of a system remains constant, and how to exploit constant momentum to solve problems involving collisions between objects. In addition, we examine a new reference frame, known as the center-of-mass reference frame, and explore situations in which a system has a continuously changing mass. Conservation of Linear Momentum. a. When Newton devised his second law, he considered the product of mass and velocity as a measure of an object’s “quantity of motion.” Today, we call the product of a particle’s mass and velocity linear momentum b. Linear momentum is a vector quantity, it is the product of a vector (velocity) and a scalar (mass). Its magnitude is and it has the same direction as The units of momentum are units of mass times speed, so the SI units of momentum are kg•m/s. c. . Momentum may be thought of as a measurement of the effort needed to bring a particle to rest. i. Thus, the net force acting on a particle equals the time rate of change of the particle’s momentum. ii. d. The total momentum of a system of particles is the vector sum of the momenta of the individual particles. i. 1 Topic 5 Conservation of Linear Momentum ii. But according to Newton’s second law for a system of particles, equals the net external force acting on the system. iii. e. When the sum of the external forces acting on a system of particles remains zero, the rate of change of the total momentum remains zero and the total momentum of the system remains constant. i. ii. This result is known as the law of conservation of momentum. iii. iv. This law is one of the most important in physics. It is more widely applicable than the law of conservation of mechanical energy because internal forces exerted by one particle in a system on another are often not conservative. The nonconservative internal forces can change the total mechanical energy of the system, though they effect no change of the system’s total momentum. If the total momentum of a system remains constant, then the velocity of the center of mass of the system remains constant. The law of conservation of momentum is a vector relation, so it is valid component by component. 2 Topic 5 Conservation of Linear Momentum f. Finding Velocities Using Momentum Conservation (Equation 8-5) i. PICTURE Determine that the net external force (or ) on the system is negligible for some interval of time. If the net force is determined not to be negligible, do not proceed. ii. SOLVE 1. Draw a sketch showing the system both before and after the time interval. Include coordinate axes and label the initial and final velocity vectors. 2. Equate the initial momentum to the final momentum. That is, write out the equation . 3. Substitute the given information into the step-2 equation(s) and solve for the quantity of interest. iii. CHECK Make sure you include any minus signs that accompany velocity components because they influence your final answer. g. Problems. i. Example 8 – 1: A Space Repair. During repair of the Hubble Space Telescope, an astronaut replaces a damaged solar panel during a spacewalk. Pushing the detached panel away into space, she is propelled in the opposite direction. The astronaut’s mass is 60 kg and the panel’s mass is 80 kg. Both the astronaut and the panel initially are at rest relative to the telescope. The astronaut then gives the panel a shove. After the shove it is moving at 0.30 m/s relative to the telescope. What is her subsequent velocity relative to the telescope? (During this operation the astronaut is tethered to the ship; for our calculations assume that the tether remains slack.) ii. Practice Problem 8 – 1. Find the final kinetic energy of the astronaut–panel system. 3 Topic 5 Conservation of Linear Momentum iii. Example 8 – 2: A Runaway Railroad Car. A runaway 14,000-kg railroad car is rolling horizontally at 4.00 m/s toward a switchyard. As it passes by a grain elevator, 2000 kg of grain suddenly drops into the car. How long does it take the car to cover the 500-m distance from the elevator to the switchyard? Assume that the grain falls straight down and that slowing due to rolling friction or air drag is negligible. iv. Practice Problem 8 – 2. Suppose that there is a small vertical chute in the bottom of the car so that the grain leaks out at 10 kg/s. Now how long does it take the car to cover the 500 m? v. Example 8 – 3: A Skateboard Workout. A 40.0-kg skateboarder on a 3.00kg board is training with two 5.00kg weights. Beginning from rest, she throws the weights horizontally, one at a time, from her board. The speed of each weight is 7.00 m/s relative to her after it is thrown. Assume the board rolls without friction. (a) How fast is she moving in the opposite direction after throwing the first weight? (b) After throwing the second weight? 4 Topic 5 Conservation of Linear Momentum vi. Practice Problem 8 – 3. How fast is the skateboarder moving if, starting from rest, she throws both weights simultaneously? The weights move at speed 7.00 m/s relative to her after they are thrown. Does she gain more speed by throwing them simultaneously or sequentially? vii. Example 8 – 4: Radioactive Decay. A thorium-227 nucleus (mass 227 u) at rest decays into a radium-223 nucleus (mass 223 u) by emitting an alpha particle (mass 4.00 u). The kinetic energy of the α particle is measured to be 6.00 MeV. What is the kinetic energy of the recoiling radium nucleus? III. Kinetic Energy of a System. a. If the net external force on a system of particles remains zero, then the total momentum of a system must remain constant; however, the total mechanical energy of the system can change. b. c. Kinetic Energy of System of Particles. 5 Topic 5 Conservation of Linear Momentum d. Theorem Proof. e. If the external force is zero, the center-of-mass velocity remains constant and the kinetic energy associated with bulk motion of the system (½ Mv2cm) does not change. Only the relative kinetic energy Krel can change in an isolated system. f. Practice Problem 8 – 4. Air-track glider A is moving at 1.0 m/s in the +x direction along a frictionless horizontal air track. An identical glider, glider B, is parked on the track ahead of glider A. The mass of each glider is 1.0 kg, and the system consists of the two gliders. (a) What is the velocity of the center of mass, and what is the velocity of each glider, relative to the center of mass? (b) What is the kinetic energy of each glider relative to the center of mass? (c) What is the total kinetic energy relative to the center of mass? (d) The gliders collide and stick together. What then is the total kinetic energy relative to the center of mass? 6 Topic 5 Conservation of Linear Momentum IV. Collisions. a. During the brief time of collision, any external forces on the two objects are usually much, much weaker than the forces of interaction between the two objects. i. Thus, the colliding objects can usually be treated as an isolated system for the duration of the collision. ii. During the collision the only significant forces are the internal interaction forces, which are equal and opposite. iii. As a result, momentum is conserved. That is, the total momentum of the system the instant before the collision is equal to the total momentum the instant following the collision. iv. In addition, the collision time is usually so short that the displacements of the colliding objects during the collision can be neglected. b. When the total kinetic energy of the two-object system is the same after the collision as before, the collision is called an elastic collision. Otherwise, it is called an inelastic collision. An extreme case is the perfectly inelastic collision, during which all of the kinetic energy relative to the center of mass is converted to thermal or internal energy of the system, and the two objects share a common velocity (often because they stick together) at the end of the collision. c. Impulse and Average Force. i. When two objects collide, they usually exert very large forces on each other for a very brief time. ii. Such forces are sometimes called impulsive forces. Such forces are sometimes called impulsive forces. iii. The impulse vector. of a force during time interval Δt = tf − ti is a iv. The impulse is a measure of both the strength and the duration of the collision force. The x component of the impulse of the force is the area under its Fx-versus-t curve, and the S.I. units of impulse are newton seconds (N • s). v. The net force acting on a particle is related to the rate of change of momentum of the particle by Newton’s second law: . 7 Topic 5 Conservation of Linear Momentum vi. Impulse-Momentum Theorem for a Particle: vii. Also, the net impulse on a system due to the external forces acting on the system equals the change in the total momentum of the system. viii. Impulse-Momentum Theorem for a Particle: ix. Average Force/Impulse and Average Force: d. Estimating the Average Force. i. PICTURE To estimate the average force we first estimate the impulse of the force . The impulse of the force equals the net impulse (assuming any other forces are negligible). The net impulse is equal to the change in momentum, and the change in momentum equals the product of the mass m and the change of velocity . An estimate of the change in velocity can be gotten from estimates of both the collision time Δt and the displacement . ii. SOLVE 1. Calculate (or estimate) the impulse and the time Δt. This estimate assumes that during the collision, the collision force on the object is very large compared to all other forces on the object. This procedure works only if the displacement during the collision can be determined. 2. Draw a sketch showing the before and after position of the object. Add coordinate axes and label the pre– and postcollision velocities displacement and . In addition, label the during the collision. 8 Topic 5 Conservation of Linear Momentum 3. Calculate the change in momentum of the object during a collision. The impulse on the object equals its change in momentum . 4. Use kinematics to estimate the collision time. This means using both and to obtain , and then solving for Δt. 5. Use to calculate the average force (Equation 8-13). iii. CHECK Average force is a vector. Your result for average force should have the same direction as the change in velocity vector. e. Problems. i. Example 8 – 5: Karate Collisions. With an expert karate blow, you shatter a concrete block. Consider your hand to have a mass 0.70 kg, to be moving 5.0 m/s as it strikes the block, and to stop 6.0 mm beyond the point of contact. (a) What impulse does the block exert on your hand? (b) What is the approximate collision time and the average force the block exerts on your hand? ii. Example 8 – 6: A Crumpled Car. A car equipped with an 80-kg crash-test dummy drives into a massive concrete wall at 25 m/s. Estimate the displacement of the dummy during the crash. iii. Example 8 – 7: A Crash Test. For the car crash described in Example 8-6, estimate the average force that the seat belt exerts on the dummy during the crash. 9 Topic 5 Conservation of Linear Momentum iv. Example 8 – 8: Hitting a Golf Ball. You strike a golf ball with a driving iron. What are reasonable estimates for the magnitudes of the (a) impulse , (b) collision time Δt, and (c) average force ? A typical golf ball has a mass m = 45 g and a radius r = 2.0 cm. For a typical drive, the range R is roughly 190m (about 210 yd). Assume the ball leaves the ground at an angle θ0 = 13° above the horizontal. f. Collisions in which the colliding objects move along the same straight line, say along the x axis, both before, during, and after the collision are called onedimensional collisions. i. Perfectly inelastic collisions In perfectly inelastic collisions, the objects have the same velocity after the collision, often because they are stuck together. 1. It is sometimes useful to express the kinetic energy, K, of a particle in terms of its momentum, p. 10 Topic 5 Conservation of Linear Momentum 2. We can apply this to a perfectly inelastic collision where one object is initially at rest. 3. After colliding, the objects move together as a single mass m1 + m2 with velocity . Momentum is conserved, so the final momentum equals Psys. 4. Example 8 – 9: A Catch in Space. An astronaut of mass 60 kg is on a space walk to repair a communications satellite when he realizes he needs to consult the repair manual. You happen to have it with you, so you throw it to him with speed 4.0 m/s relative to the spacecraft. He is at rest relative to the spacecraft before catching the 3.0-kg book. Find (a) his velocity just after he catches the book, (b) the initial and final kinetic energies of the book–astronaut system, and (c) the impulse exerted by the book on the astronaut. 5. Example 8 – 10: A Ballistic Pendulum. In a feat of public marksmanship, you fire a bullet into a hanging wood block, which is a device known as a ballistic pendulum. The block, with bullet embedded, swings upward. Noting the height reached at the top of the swing, you immediately inform the crowd of the bullet’s speed. How fast was the bullet traveling? 11 Topic 5 Conservation of Linear Momentum 6. Example 8 – 11: A Collision with an Empty Box. You repeat your feat of Example 8-10, this time with an empty box as the target. The bullet strikes the box and passes through it completely. A laser ranging device indicates that the bullet emerged with half its initial velocity. Hearing this, you correctly report how high the target must have swung. How high did it swing? 7. Example 8 – 12: Collisions with Putty. Mary has two small balls of equal mass, a ball of plumber’s putty and a one of Silly Putty. She throws the ball of plumber’s putty at a block suspended by strings. The ball strikes the block with a “thonk” and falls to the floor. The block subsequently swings to a maximum height h. If she had thrown the ball of Silly Putty (instead of the plumber’s putty) at the same speed, would the block subsequently have risen to a height greater than h? Silly Putty, unlike plumber’s putty, is elastic and would bounce back from the block. ii. Elastic collisions In elastic collisions, the kinetic energy of the system is the same before and after the collision. Elastic collisions are an ideal that is sometimes approached but never realized in the macroscopic world. 1. Only for elastic collisions is the kinetic energy the same after the collision as before the collision. 12 Topic 5 Conservation of Linear Momentum 2. Example 8 – 13: Elastic Collision of Two Blocks. A 4.0-kg block moving to the right at 6.0 m/s undergoes an elastic head-on collision with a 2.0-kg block moving to the right at 3.0 m/s. Find their final velocities. 3. Example 8 – 14: Elastic Collision of a Neutron and a Nucleus. A neutron of mass mn and speed undergoes a head-on elastic collision with a carbon nucleus of mass mc initially at rest. (a) What are the final velocities of both particles? (b) What fraction ƒ of its initial kinetic energy does the neutron lose? 13 Topic 5 Conservation of Linear Momentum 4. Practice Problem 8 – 5. Consider an elastic head-on collision between a moving object (object 1) and a second moving object of equal mass (object 2). Use Equation 8-19 and Equation 8-23 to show that the two objects exchange velocities. That is, show that the final velocity of object 2 equals the initial velocity of object 1, and vice versa. 5. The final velocity of the incoming particle and that of the originally stationary particle are related to the initial velocity of the incoming particle. 6. The coefficient of restitution Most collisions lie somewhere between the extreme cases of elastic, in which the relative velocities are reversed, and perfectly inelastic, in which there is no relative velocity after the collision. The coefficient of restitution e is a measure of the elasticity of a collision. It is defined as the ratio of the speed of recession to the speed of approach. 7. Practice Problem 8 – 6. From the picture of the golf club hitting the golf ball, estimate the coefficient of restitution of the golf ball–golf club interaction. 14 Topic 5 Conservation of Linear Momentum g. Collisions in Two and Three Dimensions. i. For one-dimensional collisions, the directions of the initial and final velocity vectors can be specified by a + or a −. For two– or threedimensional collisions, this is not the case. During such a collision, momentum is conserved in each of the x, y, and z directions. 1. Inelastic collisions For collisions in two or three dimensions, the total initial momentum is the sum of the initial momentum vectors of each object involved in the collision. 2. Example 8 – 15: A Car-Truck Collision. You are at the wheel of a 1200-kg car traveling east through an intersection when a 3000-kg truck traveling north through the intersection crashes into your car. Your car and the truck remain stuck together after impact. The driver of the truck claims you were at fault because you were speeding. You look for evidence to disprove this claim. First, there are no skid marks, indicating that neither you nor the truck driver saw the accident coming and braked hard; second, the posted speed limit for the road on which you were driving is 80 km/h; third, the speedometer of the truck was smashed on impact, leaving the needle stuck at 50 km/h; and fourth, the wreck initially skidded from the impact zone at an angle of 59° north of east. Does this evidence support or undermine the claim that you were speeding? 15 Topic 5 Conservation of Linear Momentum 3. Example 8 – 16: A Glancing Collision. An object with mass m1 and with an initial speed of 20 m/s undergoes an off-center collision with a second object of mass m2. The second object is initially at rest. After the collision the first object is moving at 15 m/s at an angle of 25° with the direction of the initial velocity of the first object. In what direction is the second object moving? ii. Elastic collisions Elastic collisions in two and three dimensions are more complicated than those we have already covered. 1. An off-center collision is often referred to as a glancing collision (as opposed to a head-on collision). 2. The distance b between the centers measured perpendicular to the direction of is called the impact parameter. 3. After the collision, object 1 moves off with velocity , making an angle θ1 with the direction of its initial velocity, and object 2 moves with velocity , making an angle θ2 with . The final velocities depend on the impact parameter and on the type of force exerted by one object on the other. 4. Linear momentum is conserved. 5. We can apply the law of conservation of momentum in component form to give us two of the needed relations among these quantities. 16 Topic 5 Conservation of Linear Momentum 6. Because the collision is elastic, we can use the conservation of kinetic energy to find a third relation. 7. We need an additional equation to be able to solve for the unknowns. The fourth relation depends on the impact parameter b and on the type of interacting force exerted by the colliding objects on each other. In practice, the fourth relation is often found experimentally, by measuring the angle of deflection or the angle of recoil. Such a measurement can then give us information about the type of interacting force between the bodies. 8. Now let us consider the interesting special case of the glancing elastic collision of two objects of equal mass when one is initially at rest. 9. Practice Problem 8 – 7. In a friendly game of pool, the cue ball moving at speed glances off the initially stationary 2 ball. The collision is elastic, the 2 ball is at rest prior to the collision, and the cue ball is deflected 30° from its pre-collision path. What is the speed of the 2 ball following the collision? (The 2 ball and the cue ball have the same mass.) 17 Topic 5 Conservation of Linear Momentum V. Collisions in the Center of Mass Reference Frame. a. If the net external force on a system remains zero, the velocity of the center of mass remains constant in any inertial reference frame. i. It is often convenient to do calculations in an alternate reference frame that moves with the center of mass. ii. Relative to the original reference frame, called the laboratory reference frame, this reference frame moves with a constant velocity relative to the laboratory frame. iii. A reference frame that moves at the same velocity as the center of mass is called the center-of-mass reference frame. iv. If a particle has velocity relative to the laboratory reference frame, then its velocity relative to the center-of-mass reference frame is . v. Because the total momentum of a system equals the total mass times the velocity of the center of mass, the total momentum is also zero in the center-of-mass frame. Thus, the center-of-mass reference frame is also a zero-momentum reference frame. b. The mathematics of collisions is greatly simplified when considered from the center-of-mass reference frame. c. After a perfectly inelastic collision, the objects remain at rest. However, for an elastic head-on collision the direction of each velocity vector is reversed without changing its magnitude. d. Consider a simple two-particle system in a reference frame in which one particle is moving with a velocity and a second particle is moving with a different velocity. 18 Topic 5 Conservation of Linear Momentum e. Example 8 – 17: The Elastic Collision of Two Blocks. Find the final velocities for the elastic head-on collision in Example 8-13 (in which a 4.0-kg block moving right at 6.0 m/s collides elastically with a 2.0-kg block moving right at 3.0 m/s) by transforming their velocities to the center-of-mass reference frame. f. Practice Problem 8 – 8. Verify that the total momentum of the system both before the collision and after the collision is zero in the center-of-mass reference frame. VI. AP Problems. a. Find the rebound speed of a 0.5-kg ball falling straight down that hits the floor moving at 5 m/s, if the average normal force exerted by the floor on the ball was 205 N for 0.02 s. b. A 5-kg mass moving at 10 m/s in the +x direction is acted upon by a force acting in the –x direction with magnitude given as a function of time by the graph below. Determine the speed and the direction of the mass after the force has stopped acting. F (x103 N) 7 0 0 2 10 t (x10-3 s) 19 Topic 5 Conservation of Linear Momentum c. Consider a one-dimensional example in which a 4-kg mass moving in the +x direction at 10 m/s collides head-on with a 2-kg mass moving at 6 m/s in the –x direction. What are the final velocities for the elastic collision? d. Consider a one-dimensional example in which a 4-kg mass moving in the +x direction at 10 m/s collides head-on with a 2-kg mass moving at 6 m/s in the –x direction. What are the final velocities for the inelastic collision and what fraction of the original kinetic energy is lost? e. Consider 4-kg mass moving in the -y direction collides and sticks to a 2-kg mass moving at 6 m/s in the +x direction. Find the final velocity components. f. A 6-kg spherical mass is currently moving horizontally at 20 m/s at a height of 45 m. A small explosion breaks the mass into a 2-kg and 4-kg piece, with the 4-kg piece falling straight down with 0 initial speed. How far does the 2-kg piece travel? 20