Understanding Momentum and Conservation Laws in Physics

Chapter 7 Momentum

Conservation Laws In Ch. 6, we developed the Law of Conservation of Energy. This is a powerful law that tells us the total energy of system does not change over time In fact, any quantity that obeys conservation is important to study as we can relate one moment of time to the next because we know that the quantity is constant Another law that we are interested in is the Law of Conservation of Momentum which works in the same way The conservation of momentum is particularly useful when dealing with collisions, which will be a large topic of this chapter We can define momentum as an objects mass times velocity which has units of ?? ?/? ? = ? ?

Momentum and Force There is a direct relationship between momentum and force. In fact, Newton first developed his 2nd law in terms of momentum Originally, Newton s second law of motion was described in terms of momentum The rate of change of momentum of an object is equal to the net force applied to it ? = ? ? Momentum is related to force as velocity is related to acceleration. We can see that a change in momentum must mean a change in velocity, so we can consider momentum as the velocity version of force ? ?= ? ? ?= ??

Impulse Deconstructing this form of Newton s second law, we can define a quantity called the impulse, which is defined as total change in momentum ? = ? = ? ?

Impulse Deconstructing this form of Newton s second law, we can define a quantity called the impulse, which is defined as total change in momentum ? = ? = ? ? During a collision, the time of impact is usually very small (milliseconds) which results in very large forces acting on objects This can cause deformations on objects during the collision ? If we were to graph the force over the time of the collision, we would see a spike in a very large force during the time of the collision ? ?

Example: Impulse Against a Wall Estimate the impulse and average force delivered by a 1 ?? ball against a wall when the ball is moving at 10 ?/? when it hits the wall and comes to rest. Assume that the ball is in contact with the wall for 2 ?? Remember that the impulse can be defined as the change in momentum ? ? = ? ?0= 10 ?/? ? = ? ? = ?? ??0 ? = 10 ?? ?/? 0.002 ? ? = (1 ??)(0 10 ?/?) ? = 0 ?/? ? = 10 ?? ?/? ? = 5,000 ?

Conservation of Momentum The power of discussing momentum comes from our ability to use the Law of Conservation of Momentum, which states The total momentum of the system remains constant We can use momentum in a similar fashion to how we dealt with energy. We only need to consider the changes between initial and final states of a system The most common use of momentum is when we deal with collisions of objects We can define this with the conservation of momentum which we define as, + ?? ??+ ??= ?? + ?? ?? ?? ??+ ?? ??= ?? ??

Example: Train Collision A 10,000 ?? railroad car, A, is traveling at a speed of 24.0 ?/? strikes an identical car, B, at rest. IF the cars lock together as a result of the collision, what is their common speed afterwards? Initial State Final State ??= 24.0 ?/? ??= 0 ?/? ??= ??= ? = 10,000 ?? ? ? ? ? We can state the conservation of momentum to be, + ??? ???+ ???= ??? The cars stick together after the collision so they share a common velocity, ?? = ?? = ? ???+ 0 = 2?? ? =1 ? = 12.0 ?/? 2??

Types of Collisions When two objects collide, there are two types of collisions that can occur Elastic Collision: During this type of collision, the total kinetic energy of the system is conserved ??+ ??= ?? + ?? 1 2???? 2+1 2=1 2+1 2 2???? 2???? 2???? The conservation of kinetic energy reduces to a helpful relation between velocities = ??+ ?? ??+ ?? We always use this condition with the conservation of momentum for elastic collisions

Example: Elastic Collision A 0.060 ?? tennis ball, moving with a speed 5.50 ?/?, has a head on collision with a 0.090 ?? ball, initial moving in the same direction at a speed of 3.00 ?/?. Assuming a perfectly elastic collision, determine the speed and direction of each ball after the collision? Because momentum obeys the conservation law, we must consider the initial and final states. Initial State Final State ? ?= ? ?= ??= 5.50 ?/? ??= 3.00 ?/? The conservation of momentum is then, ????+ ????= ??? ?+ ??? ?

Example: Elastic Collision A 0.060 ?? tennis ball, moving with a speed 5.50 ?/?, has a head on collision with a 0.090 ?? ball, initial moving in the same direction at a speed of 3.00 ?/?. Assuming a perfectly elastic collision, determine the speed and direction of each ball after the collision? Notice that we have two unknowns, the final velocities. For elastic collisions, we always have two equations that we can use: Conservation of Momentum and the Kinetic Energy condition ????+ ????= ??? ?+ ??? ? = ??+ ?? ??+ ?? We use substitution to solve for the velocities ? ?= ? ?= = ?? ??+ ?? ?? ????+ ????= ???? ??+ ?? + ??? ? ????+ ???? ????+ ????= ??+ ??? ?

Example: Elastic Collision A 0.060 ?? tennis ball, moving with a speed 5.50 ?/?, has a head on collision with a 0.090 ?? ball, initial moving in the same direction at a speed of 3.00 ?/?. Assuming a perfectly elastic collision, determine the speed and direction of each ball after the collision? ????+ ???? ????+ ????= ??+ ??? ? ? ?=2????+ ???? ???? ??+ ?? ? ?=2 0.060 ?? 5.50 ?/? + 0.090 ?? 3.00 ?/? 0.060 ?? 3.00 ?/? 0.150 ?? ? ?= ? ?= = ?? ??+ ?? ? ?= 5.00 ?/? ?? = 3.00 5.50 + 5.00 = 2.50 ?/? ?? ??

Inelastic Collisions A Collision in which kinetic energy is not conserved is called an inelastic collision During an inelastic collision energy that is lost is converted into other forms of energy Often its is a change to thermal energy The energy lost manifests as the velocity of the objects after collision having a lower velocity than before the collision 1 2???? 2+1 2=1 2+1 2+ ??? ?? 2???? 2???? 2???? + ?? + ?? = ?? + ??? ?? ??

Completely Inelastic Collisions A manageable type of inelastic collision is when two objects collide and combine. Initially there are two objects moving, and after the collision we have a single massive object When this happens, the collision is said to be completely inelastic In these cases, with inelastic collisions, the kinetic energy is not conserved, but the total energy is always conserved Situations like explosions or ballistic pendulums are examples of completely inelastic collisions

Example: Ballistic Pendulum The ballistic pendulum is used to show how fast a projectile is moving. A projectile of mass 0.500??, is fired into a block of mass 2.00 ??, which is suspended like a pendulum. If the block is pushed a up a vertical distance of 0.300 ?, how fast was the arrow initially moving?

Example: Ballistic Pendulum The ballistic pendulum is used to show how fast a projectile is moving. A projectile of mass 0.500??, is fired into a block of mass 2.00 ??, which is suspended like a pendulum. If the block is pushed a up a vertical distance of 0.300 ?, how fast was the arrow initially moving?

Example: Ballistic Pendulum The ballistic pendulum is used to show how fast a projectile is moving. A projectile of mass 0.500??, is fired into a block of mass 2.00 ??, which is suspended like a pendulum. If the block is pushed a up a vertical distance of 0.300 ?, how fast was the arrow initially moving? Let s consider all phases of this problem 1. Initially, the arrow is moving a speed ?0, while the blocks initial speed is zero 2. The collision happens. At this point the arrow and block combine, and have some shared velocity ? 3. The system then moves to a maximum height where its velocity is reduced to zero There are two parts to this. We can use the conservation of momentum to discuss the collision, and the conservation of energy to discuss how high the block goes

Example: Ballistic Pendulum The ballistic pendulum is used to show how fast a projectile is moving. A projectile of mass 0.500??, is fired into a block of mass 2.00 ??, which is suspended like a pendulum. If the block is pushed a up a vertical distance of 0.300 ?, how fast was the arrow initially moving? We can use energy to find the velocity immediately after the collision ?1+ ?1= ?2+ ?2 1 2??+ ???2+ ??+ ??? 1=1 2+ ??+ ??? 2 2??+ ???? 1 2??+ ???2= ??+ ??? 2 ? = 2? 2 ? = 2.42 ?/? Now we know the velocity immediately after the collision, we can use momentum to find the velocity of the arrow before the collision

Example: Ballistic Pendulum The ballistic pendulum is used to show how fast a projectile is moving. A projectile of mass 0.500??, is fired into a block of mass 2.00 ??, which is suspended like a pendulum. If the block is pushed a up a vertical distance of 0.300 ?, how fast was the arrow initially moving? ???0+ ????= ??+ ??? ???0= ??+ ??? ??+ ??? ?? ?0= ?0= 12.1 ?/?

Collisions in Two Dimensions Conservation of momentum can be applied when a collision occurs in more than one dimension So far, we have only looked at head on collisions which occur in only one direction A common situation is when on object hits another, and they both go off and different angles ?? A B

Collisions in Two Dimensions Conservation of momentum can be applied when a collision occurs in more than one dimension So far, we have only looked at head on collisions which occur in only one direction A common situation is when on object hits another, and they both go off and different angles A B

Collisions in Two Dimensions Conservation of momentum can be applied when a collision occurs in more than one dimension So far, we have only looked at head on collisions which occur in only one direction A common situation is when on object hits another, and they both go off and different angles B ?? ?? A

Collisions in Two Dimensions When we have collisions in two dimensions, we must consider the vector components of the momentum Since momentum is a vector, it can be broken into ? and y- components We know that the total momentum of a system is conserved, and it has the nice property the components are conserved as well ??? B x- components: ??? ???+ ???= ??? + ??? y- components: ??? A ???+ ???= ??? + ??? ???

Collisions in Two Dimensions Considering this system, we note that before the collision, object A moves horizontally and collides with object B After the two have collided, they go off in different directions, which means that the momentum picks up ? components These components can be written as x- components: ??? B cos(??) ??? = ?? ??? cos(??) ??? = ?? ??? y- components: A ??? sin(??) ??? = ?? sin(??) ??? = ??

Example: Billiard Balls A billiard ball A moving with a speed ??= 3.0 ?/? in the +x direction strikes an equal-mass ball B initially at rest. The two balls are observed to move off at 45 to the x axis, ball A above the axis and ball B below the axis. What are the speeds of the two balls after the collision? ?? ?? ?? ?? ??

Example: Billiard Balls A billiard ball A moving with a speed ??= 3.0 ?/? in the +x direction strikes an equal-mass ball B initially at rest. The two balls are observed to move off at 45 to the x axis, ball A above the axis and ball B below the axis. What are the speeds of the two balls after the collision? =??? ?/? =??? ?/? = ?? = 45 ?? ?? ??= ??= ? ??= 3.0 ?/? ?? Momentum in both directions is conserved, so we have 2 set s of equations ? direction ? direction ???+ ???= ??? + ??? ???+ ???= ??? + ??? 0 = ???? + ???? ????= ???? + ???? cos(45 ) + ??? sin 45 ??? cos(45 ) sin(45 ) ????= ??? 0 = ???

Example: Billiard Balls A billiard ball A moving with a speed ??= 3.0 ?/? in the +x direction strikes an equal-mass ball B initially at rest. The two balls are observed to move off at 45 to the x axis, ball A above the axis and ball B below the axis. What are the speeds of the two balls after the collision? =??? ?/? =??? ?/? = ?? = 45 ?? ?? ??= ??= ? ??= 3.0 ?/? ?? ? direction ? direction cos(45 ) + ??? sin 45 ??? cos(45 ) sin(45 ) ????= ??? 0 = ??? cos 45 + ?? sin 45 ?? cos(45 ) sin(45 ) ??= ?? 0 = ?? sin(45 ) sin(45 ) cos 45 + ?? cos(45 ) ??= ?? = ?? ?? ?? = cos(45 ) ?? ??= 2?? 2cos(45 ) = ?? ?? = 2.1 ?/? ??

Center of Mass So far, when dealing with physical situations we have always made the assumption that our objects are point objects These objects have no volume, and all their mass is at one point

Center of Mass So far, when dealing with physical situations we have always made the assumption that our objects are point objects These objects have no volume, and all their mass is at one point Real objects are not point particles, although when an object undergoes some type of motion, there will be one point that moves in the same path that a particle move if it were subjected to the same force This point is called the center of mass (CM) which is where we say forces act upon

Center of Mass The center of mass is defined to be at the position ??? If we consider a system of only two particles, with masses ?? and ??, we can define the center of mass position as ???=????+ ???? ??+ ?? The center of mass lies on the line joining masses ?? and ?? This can be extrapolated to a many bodied system ???=????+ ????+ ????+ ??+ ??+ ??+

Example: Center of Mass Three people of roughly equal mass m sit along the x- axis at positions ??= 1.0 ?, ??= 5.0 ?, and ??= 6.0 ?, measured from the left-hand side. Find the position of the center of mass. ? 6 5 0 1 12 ? ???=????+ ????+ ???? ??+ ??+ ?? ?(1.0 ?) + ? 5.0 ? + ? 6.0 ? 4.0 ? 3 3?

Example: Center of Mass in 2 Dimensions Consider three objects, each with a mass 10 ?? that are placed along the cartesian axis. If the masses are placed at ?1= 2, 2 , ?2= 3, 2 , ?3= 3,4 , where is the center of mass? We can build this up one mass at a time to find the CoM. Since there are two dimensions, we have an ? CM and a ? CM ???=?1 2 = 2 ?1 ???=?1 2 = 2 ?1 ???=?1 2 + ?23 ?1+ ?2 = 20 + 30 20 =1 2 ???=?1 2 + ?2 2 ?1+ ?2 = 20 20 20 ?1 ?2 = 2

Example: Center of Mass in 2 Dimensions Consider three objects, each with a mass 10 ?? that are placed along the cartesian axis. If the masses are placed at ?1= 2,2 , ?2= 3 2 , ?3= 3,4 , where is the center of mass? ?3 We can build this up one mass at a time to find the CoM. Since there are two dimensions, we have an ? CM and a ? CM ???=?1 2 = 2 ?1 ???=?1 2 = 2 ?1 ???=?1 2 + ?23 ?1+ ?2 = 20 + 30 20 =1 2 ???=?1 2 + ?2 2 ?1+ ?2 = 20 20 20 ?1 ?2 = 2

Example: Center of Mass in 2 Dimensions Consider three objects, each with a mass 10 ?? that are placed along the cartesian axis. If the masses are placed at ?1= 2,2 , ?2= 3 2 , ?3= 3,4 , where is the center of mass? ?3 We can build this up one mass at a time to find the CoM. Since there are two dimensions, we have an ? CM and a ? CM = 20 + 30 + 30 30 ???=?1 2 + ?23 + ?33 ?1+ ?2+ ?3 = 1.33 ???=?1 2 + ?2 2 + ?34 ?1+ ?2+ ?3 = 20 20 + 40 30 = 0 ?1 ?2

Example: Center of Mass in 2 Dimensions Consider three objects, each with a mass 10 ?? that are placed along the cartesian axis. If the masses are placed at ?1= 2,2 , ?2= 3 2 , ?3= 3,4 , where is the center of mass? ?3 We can build this up one mass at a time to find the CoM. Since there are two dimensions, we have an ? CM and a ? CM = 20 + 30 + 30 30 ???=?1 2 + ?23 + ?33 ?1+ ?2+ ?3 = 1.33 ???=?1 2 + ?2 2 + ?34 ?1+ ?2+ ?3 = 20 20 + 40 30 = 0 ?1 ?2

Center of Mass and Translational Motion When we consider the motion of the CM for a system of particles, this motion is directly related to the net force acting on the system as a whole. Consider the center of mass of a system of particles ???=????+ ????+ ???? ??+ ??+ ?? The denominator is the total mass of the system: ? = ?1+ ?2+ ?3 ????= ????+ ????+ ???? 1 If the system is moving, then we can consider its rate as a velocity of each particle ???? ? ?? 2 ?? ?+ ?? ?? ?+ ?? ?? ? = ?? 3

Center of Mass and Translational Motion When we consider the motion of the CM for a system of particles, this motion is directly related to the net force acting on the system as a whole. The motion of each particle can then be related to the velocity of the center of mass ???? ?? ?+ ?? ?? ?+ ?? ?? ? = ?? ? ????= ????+ ????+ ???? We can consider the rate again to get the acceleration of the system 1 ???? ?? ?+ ?? ?? ?+ ?? ?? ? = ?? ? ?? 2 ????= ????+ ????+ ???? 3

Center of Mass and Translational Motion When we consider the motion of the CM for a system of particles, this motion is directly related to the net force acting on the system as a whole. Which we recognize as the force acting on the system ???= ??+ ??+ ?? We see why we can now approximate objects as point particles. A force applied to an object at the center of mass equals the sum of the forces on all the particles in the system 1 Forces applied to the center of mass creates linear motion as we have studied ?? 2 Next, we consider how force act on objects with volume, and how this can cause rotations in an object 3