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Momentum and Its Conservation - Chapter Outline Lesson 1: The Impulse-Momentum Change Theorem Lesson 2: The Law of Momentum Conservation
Lesson 1: The Impulse- Momentum Change Theorem Momentum Momentum and Impulse Connection Real-World Applications
Momentum Momentum can be defined as "mass in motion." It is a property of a moving body. The amount of momentum that an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving. Momentum is equal to the product of the body's mass and velocity. Momentum = mass velocity p = m v m: mass in kg v: velocity in m/s p: momentum in kgm/s
Momentum is a vector quantity The direction of the momentum vector is the same as the direction of the velocity of the ball. Which is the same as the direction that an object is moving.
Momentum is directly proportional to mass and momentum is directly proportional to velocity Consider a 0.5-kg physics cart loaded with one 0.5-kg brick and moving with a speed of 2.0 m/s. Its momentum is 2.0 kgm/s. If the cart was instead loaded with three 0.5-kg bricks, then the total mass of the loaded cart would be 4.0 kgm/s. A doubling of the mass results in a doubling of the momentum. Similarly, if the 2.0-kg cart had a velocity of 4.0 m/s (instead of 2.0 m/s), then the cart would have a momentum of 8.0 kgm/s (instead of 4.0 kgm/s). A doubling in velocity results in a doubling of the momentum. Mass and velocity have equal importance in momentum
Example Express your understanding of the concept and mathematics of momentum by answering the following questions Determine the momentum of a... a.60-kg halfback moving eastward at 9 m/s. b.1000-kg car moving northward at 20 m/s. c.40-kg freshman moving southward at 2 m/s.
A car possesses units of momentum. What would be the car's new momentum if... a.its velocity was doubled. b.its velocity was tripled. c.its mass was doubled (by adding more passengers and a greater load) d.both its velocity and mass was doubled. Example
Four billiard balls, each of mass.5 kg, all are traveling in the same direction on a billiard table, with speeds 2 m/s, 4 m/s, 8 m/s and 10 m/s. What is the linear momentum of this system? If all four balls collide, what is the total momentum after the collision? p total = p 1 + p 2 + p 3 + p 4 p total = (0.5 kg)(2 m/s + 4 m/s + 8 m/s + 10 m/s) p total = 12 kg∙m/s 12 kg∙m/s
A change in Momentum takes force and time p = m∙v In order to change momentum, one must change velocity, which means one must produce acceleration. To produce acceleration, one must apply force. ∆p = m∙∆v ∆p = m∙a∙∆t ∆p = F∙∆tA change in momentum takes force and time The equation states that a net external force, F, applied to an object for a certain time interval, ∆t, will cause a change in the object’s momentum. This change in the object’s momentum equals to the product of F and the time interval.
Impulse = Change in Momentum Impulse-Momentum Theorem The quantity F∆t is known as impulse, represented by letter J. J = F∙∆t F: force in Newton t: time in second J: impulse in N∙s Impulse is a vector quantity, its direction is the same as the net force F.
The impulse experienced by the object is the cause of the change in momentum. The impulse experience by the object equals the change in momentum of the object. There are two way to determine impulse: –F·t–F·t –m·∆v There are two way to determine change in momentum: –F·t–F·t –m·∆v Impulse = Change in momentum
Example If the halfback experienced a force of 800 N for 0.9 seconds, then we could say that the impulse was _______________ This impulse would cause a momentum change of ____________ In a collision, the impulse experienced by an object is always equal to the momentum change. 720 Ns 720 kgm/s Note: the unit Ns = kgm/s
A 0.50-kg cart (#1) is pulled with a 1.0-N force for 1 second; another 0.50 kg cart (#2) is pulled with a 2.0 N- force for 0.50 seconds. a.Which cart (#1 or #2) has the greatest acceleration? Explain. b.Which cart (#1 or #2) has the greatest impulse? Explain. c.Which cart (#1 or #2) has the greatest change in momentum? Explain. Example
Two cars of equal mass are traveling down Lake Avenue with equal velocities. They both come to a stop over different lengths of time. The ticker tape patterns for each car are shown on the diagram below.ticker tape patterns a.Which car (A or B) experiences the greatest acceleration? Explain. b.Which car (A or B) experiences the greatest change in momentum? Explain. c.Which car (A or B) experiences the greatest impulse? Explain. Example
1.In the diagram, a 60.-kilogram roller-skater exerts a 10.-newton force on a 30.-kilogram roller-skater for 0.20 second. What is the magnitude of the impulse applied to the 30.-kilogram rollerskater? 2.A 0.10-kilogram model rocket’s engine is designed to deliver an impulse of 6.0 newton-seconds. If the rocket engine burns for 0.75 second, what average force does it produce? 3.A 0.15-kilogram baseball moving at 20. meters per second is stopped by a catcher in second. What is the average force stopping the ball? 4.A 2.0-kilogram laboratory cart is sliding across a horizontal frictionless surface at a constant velocity of 4.0 meters per second east. What will be the cart’s velocity after a 6.0-newton westward force acts on it for 2.0 seconds? 5.A 1,000-kilogram car traveling due east at 15 meters per second is hit from behind and receives a forward impulse of 8,000 newton- seconds. Determine the magnitude of the car’s change in momentum due to this impulse. Practices
6.The instant before a batter hits a 0.14-kilogram baseball, the velocity of the ball is 45 meters per second west. The instant after the batter hits the ball, the ball’s velocity is 35 meters per second east. The bat and ball are in contact for 1.0 × 10 −2 second. Calculate the magnitude of the average force the bat exerts on the ball while they are in contact. 7.A 2.0-kilogram laboratory cart is sliding across a horizontal frictionless surface at a constant velocity of 4.0 meters per second east. What will be the cart’s velocity after a 6.0- newton westward force acts on it for 2.0 seconds? 8.A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the magnitude of the force exerted on the car during the collision.
Real-World Applications The effect of collision time upon the amount of force an object experiencesThe effect of collision time The effect of rebounding upon the velocity change and hence the amount of force an object experiences.The effect of rebounding
The relationship between Collision Time and Force of Impact The greater the time over which the collision occurs, the smaller the force acting upon the object. To minimize the effect of the force on an object involved in a collision, the time must be increased. To maximize the effect of the force on an object involved in a collision, the time must be decreased. m∙∆v = F∙∆t
Examples of Reducing force by increase time Air bags are used in automobiles because they are able to minimize the affect of the force on an object involved in a collision. Air bags accomplish this by extending the time required to stop the momentum of the driver and passenger. Padded dashboards also reduces force by increase time. A boxer rides the punch in order to extend the time of impact of the glove with their head. Nylon ropes are used in the sport of rock-climbing because of its ability to stretch. The rock climber can appreciate minimizing the effect of the force through the use of a longer time of impact.
The Effect of Rebounding Bouncing off each other is known as rebounding. Rebounding involves a change in the direction of an object; rebounding situations are characterized by –a large velocity change –a large momentum change. –a large impulse F·∆t = m·∆v
The importance of rebounding is critical to the outcome of automobile accidents. Automobiles are made with crumple zones. Crumple zones minimize the affect of the force in an automobile collision in two ways. –By crumpling, the car is less likely to rebound upon impact, thus minimizing the momentum change and the impulse. –The crumpling of the car lengthens the time over which the car's momentum is changed; by increasing the time of the collision, the force of the collision is greatly reduced.
Increase velocity by increasing time: F·∆t = m·∆v In racket and bat sports, hitters are often encouraged to follow-through when striking a ball. In this situation, both the force applied (as hard as you can) and the mass (the mass of the ball) are constant. By following through, the hitter increases the time, the result is increasing the ball’s velocity.
examples A constant force can act on an object for different lengths of time. As the length of time increases, 1.the impulse imparted to the object a)decreases b)increases c)remains the same 2.The momentum of the object a)decreases b)increases c)remains the same
Example A cannonball shot from a long-barrel cannon travels faster than one shot from a short-barrel cannon because the cannonball receives a greater a.force. b.impulse. c.both A and B d.neither A nor B A cannonball shot from a cannon receive the same force regardless of the length of its barrel. A long-barrel will take longer time (t) for the cannonball to travel. Since J = F∆t = m∆v, the longer the time, the bigger the impulse, the faster it will travel.
Example A student drops two eggs of equal mass simultaneously from the same height. Egg A lands on the tile floor and breaks. Egg B lands intact, without bouncing, on a foam pad lying on the floor. Compared to the magnitude of the impulse on egg A as it lands, the magnitude of the impulse on egg B as it lands is _________________ (less, greater, the same) as egg A.
Stopping times and stopping distances depend on the impulse-momentum theorem Two identical trucks travels with the same velocity. One is empty with mass m, and the other is loaded with mass 2m. 1.Why is loaded truck’s stopping time twice as much as the empty truck’s when acted on by the same force? 2.How do the stopping distances compare? F·∆t = m·∆v Loaded is 2x unloaded 2m m Loaded is 2x unloaded
practice 1.A 2250 car traveling to the west slows down uniformly from 20.0 m/s to 5.00 m/s. a.How long does it take the car to decelerate if the force on the car is 8450 N to the east? b.How far does the car travel during the deceleration? 2.If the maximum coefficient of kinetic friction between a car and a road is 0.50, what is the minimum stopping distance for car moving at 29 m/s? (about 65 mi/hr)
Lesson 2: The Law of Momentum Conservation 1.The Law of Action-Reaction (Revisited)The Law of Action-Reaction (Revisited) 2.Momentum Conservation PrincipleMomentum Conservation Principle 3.Isolated SystemsIsolated Systems 4.Momentum Conservation in Collisions –Using Equations as a "Recipe" for Algebraic Problem-SolvingUsing Equations as a "Recipe" for Algebraic Problem-Solving –Using Equations as a Guide to ThinkingUsing Equations as a Guide to Thinking 5.Momentum Conservation in ExplosionsMomentum Conservation in Explosions 6.Elastic and Inelastic collisions
The Law of Action-Reaction Revisited In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction in accord with Newton’s 3 rd Law. While the forces are equal in magnitude and opposite in direction, the accelerations of the objects are not necessarily equal in magnitude. According to Newton's second law of motion, Bigger mass has smaller acceleration, smaller mass has bigger accelerationNewton's second law of motion
m 1 (v 1 ’ – v 1 ) = -m 2 (v 2 ’ – v 2 ) m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ p(before) = p(after)
Momentum Conservation Principle The law of momentum conservation can be stated as follows: –For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.isolated system
Isolated Systems A system is a collection of two or more objects. An isolated system is a system which is free from the influence of a net external force which alters the momentum of the system. In reality, momentum can be approximate to be conserved for every situations due to friction.
Collision Description Isolated System? Yes or No If No, the external force is 1. Two cars collide on a gravel roadway on which frictional forces are large. 2. Hans Full is doing the annual vacuuming. Hans is pushing the Hoover vacuum cleaner across the living room carpet. 3. Two air track gliders collide on a friction- free air track. Nofriction No The friction between the carpet and the floor and the applied force exerted by Hans are both external forces. yes
Consider a fullback plunges across the goal line and collides in midair with the linebacker in a football game. The linebacker and fullback hold each other and travel together after the collision. Before the collision, the fullback possesses a momentum of 100 kg∙m/s, East and the linebacker possesses a momentum of 120 kg∙m/s, West. The total momentum of the system before the collision is _____________________. Therefore, the total momentum of the system after the collision must also be __________________ 20 kg∙m/s, West 20 kg∙m/s, West. Vector diagram for the situation: Vector diagram for conservation of momentum
consider a medicine ball is thrown to a clown who is at rest upon the ice; the clown catches the medicine ball and glides together with the ball across the ice. The momentum of the medicine ball is 80 kg∙m/s before the collision. The momentum of the clown is 0 kg∙m/s before the collision. The total momentum of the system before the collision is ______________ Therefore, the total momentum of the system after the collision must also be ________________. The clown and the medicine ball move together as a single unit after the collision with a combined momentum of 80 kg∙m/s. Momentum is conserved in the collision. 80 kg∙m/s. Vector diagram for the situation: Another vector diagram for conservation of momentum
Using Equations for Algebraic Problem-Solving Total system momentum is conserved for collisions between objects in an isolated system.Total system momentum is conserved for collisions between objects in an isolated system p before = p after m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’
Example 1 A 2.0-kilogram ball traveling north at 5.0 meters per second collides head-on with a 1.0 kilogram ball traveling south at 8.0 meters per second. What is the magnitude of the total momentum of the two balls after collision? m 1 v 1 + m 2 v 2 = p after (2.0 kg)(5.0 m/s) + (1.0 kg)(-8.0 m/s) =p after 2 kg∙m/s, north = p after m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ m 1 = 2.0 kg; v 1 = 5.0 m/s m 2 = 1.0 kg; v 2 = -8.0 m/s p after = ?
Example 2 A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motion at 15 m/s. Assuming that momentum is conserved during the collision, determine the velocity of the truck immediately after the collision kg∙m/s = (3000 kg)∙v kg∙m/s v = 5 m/s m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ It is a collision problem, momentum is conserved. m 1 = 3000 kg; v 1 = 10 m/s m 2 = 1000 kg; v 2 = 0 v 1 ’ = ? v 2 ’ = 15 m/s (3000 kg)∙(10 m/s) + 0 = (3000 kg)∙v + (1000 kg)∙(15 m/s)
Practices 1.A 15-kg medicine ball is thrown at a velocity of 20 km/hr to a 60-kg person who is at rest on ice. The person catches the ball and subsequently slides with the ball across the ice. Determine the velocity of the person and the ball after the collision. 2.A kg baseball moving at a speed of 45.0 m/s crosses the plate and strikes the kg catcher's mitt (originally at rest). The catcher's mitt immediately recoils backwards (at the same speed as the ball) before the catcher applies an external force to stop its momentum. Determine the post-collision velocity of the mitt and ball. 3.A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2.5 m/s to the right, what is the final velocity of the boat?
Types of collisions Inelastic collisionelastic collision Inelastic collisionelastic collision
Elastic and inelastic collisions The two types of collisions are elastic and inelastic collisions. To simplify matters, we will consider any collisions in which the two colliding objects stick together and move with the same post-collision speed to be an extreme example of an inelastic collision. In an elastic collision, the two objects do not stick together. they will bounce off each other.
Inelastic collision m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ Since the two objects stick together after the collision, v 1 ’ = v 2 ’ = v m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v’ The two objects stick together and have the same speed after collision.
Example Cart A (50. kg) approaches cart B (100. kg initially at rest) with an initial velocity of 30. m/s. After the collision, cart A locks together with cart B. both travels with what velocity? Before collisionafter collision B A B A m A = 50 kg, v A = 30. m/s m B = 100. kg, v B = 0 v’ = 10. m/s m A v A + m B v B = m A v A ’ + m B v B ’ (50.kg)(30.m/s) + (100.kg)(0) =(50.kg +100.kg)v’ mass increases by 3 times (50 kg to 150 kg), speed decrease by 3 times (30 m/s to 10 m/s) v A ’ = v B ’ = v’ = ?
Example A railroad diesel engine coasting at 5.0 km/h runs into a stationary flatcar. The diesel’s mass is 8,000. kg and the flatcar’s mass is 2,000. kg. Assuming the cars couple together, how fast are they moving after the collision? Before collisionafter collision v’ = 4.0 km/h Flat car diesel Flat car diesel mass increases by 1.25 times (8000 kg to kg), speed decrease by 1.25 times (50 m/s to 40 m/s)
Example (pay attention to directions) Cart A (50. kg) moving with an initial velocity of 30. m/s approaches cart B (100. kg moving with initial velocity of 20. m/s towards cart A. The two carts lock together and move as one. Calculate the magnitude and the direction of the final velocity. Before collisionafter collision B A B A v’ = 3.3 m/s to the left
Example A block of mass M initially at rest on a frictionless horizontal surface is struck by a bullet of mass m moving with horizontal velocity v. What is the velocity of the bullet-block system after the bullet embeds itself in the block? mv + M(0) mv + 0 = (m + M)v’ mv / (M+m) = v’ Before collisionafter collision (m + M)v’
Example A woman with horizontal velocity v 1 jumps off a dock into a stationary boat. After landing in the boat, the woman and the boat move with velocity v 2. Compared to velocity v 1, velocity v 2 has 1.the same magnitude and the same direction 2.the same magnitude and opposite direction 3.smaller magnitude and the same direction 4.larger magnitude and the same direction
Example A 1850 kg luxury sedan stopped at a traffic light is struck from the rear by a compact car with a mass of 975 kg. the two cars become entangle as a result of the collision. If the compact car was moving at a velocity of 220 m/s to the north before the collision, what is the velocity of the entangled mass after the collision? 7.59 m/s North
Explosions Momentum before the explosion is zero. so the momentum after the explosion is also zero. m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ 0 = m 1 v 1 ’ + m 2 v 2 ’
Consider a homemade cannon. In the exploding cannon, total system momentum is conserved. The system consists of two objects - a cannon and a tennis ball. Before the explosion, the total momentum of the system is zero since the cannon and the tennis ball located inside of it are both at rest. After the explosion, the total momentum of the system must still be zero. If the ball acquires 50 units of forward momentum, then the cannon acquires 50 units of backwards momentum. The vector sum of the individual momentum of the two objects is 0. Total system momentum is conserved. p(before) = 0 p(after) = 0 p(cannon) + p(ball) = 0
Example A 2.0-kilogram toy cannon is at rest on a frictionless surface. A remote triggering device causes a kilogram projectile to be fired from the cannon. Which equation describes this system after the cannon is fired? 1.mass of cannon + mass of projectile = 0 2.speed of cannon + speed of projectile = 0 3.momentum of cannon + momentum of projectile = 0 4.velocity of cannon + velocity of projectile = 0
Example A 2-kilogram rifle initially at rest fires a kilogram bullet. As the bullet leaves the rifle with a velocity of 500 meters per second, what is the momentum of the rifle- bullet system?
Example A 56.2-gram tennis ball is loaded into a 1.27-kg homemade cannon. The cannon is at rest when it is ignited. Immediately after the impulse of the explosion, a photogate timer measures the cannon to recoil backwards a distance of 6.1 cm in seconds. Determine the post-explosion speed of the cannon and of the tennis ball. Given: Cannon:m = 1.27 kg; d = 6.1 cm; t = s Ball:m = 56.2 g = kg The strategy for solving for the speed of the cannon is to recognize that the cannon travels 6.1 cm at a constant speed in the seconds. v cannon = d / t = (6.1 cm) / ( s) = 280 cm/s (rounded The strategy for solving for the post-explosion speed of the tennis ball involves using momentum conservation principles. m ball v ball = - m cannon v cannon v ball = 63.3 m/s
Example A 60. kg man standing on a stationary 40. kg boat throws a.20 kg baseball with a velocity of 50. m/s. With what speed does the boat move after the man throws the ball? Assume no friction between the water and the boat. v’(boat) = -0.1 m/s (in the opposite direction of the baseball)
Example In the diagram, a 100.-kilogram clown is fired from a 500.-kilogram cannon. If the clown's speed is 15 meters per second after the firing, then what is the recoil speed (v) of the cannon? 3.0 m/s
practices 1.A 1200-kilogram car moving at 12 meters per second collides with a 2300-kilogram car that is waiting at rest at a traffic light. After the collision, the cars lock together and slide. Eventually, the combined cars are brought to rest by a force of kinetic friction as the rubber tires slide across the dry, level, asphalt road surface. Calculate the speed of the locked-together cars immediately after the collision. 2.On a snow-covered road, a car with a mass of 1.1 × 10 3 kilograms collides head-on with a van having a mass of 2.5 × 10 3 kilograms traveling at 8.0 meters per second. As a result of the collision, the vehicles lock together and immediately come to rest. Calculate the speed of the car immediately before the collision. [Neglect friction.] 3.A 1,000-kilogram car traveling due east at 15 meters per second is hit from behind and receives a forward impulse of 7,000 newton-seconds. Determine the magnitude of the car’s change in momentum due to this impulse.