Linear Momentum
The linear momentum of a particle, or an Week 8: Chapter 9 object that can be modeled as a particle, of mass m moving with a velocity v is defined to Linear Momentum and Collisions be the product of the mass and velocity: pv m The terms momentum and linear momentum will be used interchangeably in the text
Linear Momentum, cont Newton’ Law and Momentum
Linear momentum is a vector quantity Newton’s Second Law can be used to relate Its direction is the same as the direction of the the momentum of a particle to the resultant velocity force acting on it The dimensions of momentum are ML/T ddvpdm v Famm The SI units of momentum are kg · m / s dt dt dt with constant mass Momentum can be expressed in component form:
px = m vx py = m vy pz = m vz
Conservation of Linear Momentum Conservation of Momentum, 2
Whenever two or more particles in an Conservation of momentum can be expressed mathematically in various ways isolated system interact, the total momentum p = p + p = constant of the system remains constant total 1 2 p1i + p= 2i p 1f + p 2f The momentum of the system is conserved, not In component form, the total momenta in each necessarily the momentum of an individual direction are independently conserved
particle pix = pfx piy = pfy piz = pfz This also tells us that the total momentum of an Conservation of momentum can be applied to isolated system equals its initial momentum systems with any number of particles This law is the mathematical representation of the momentum version of the isolated system model
1 Conservation of Momentum, Archer Example Archer Example, 2
The archer is standing Conceptualize on a frictionless surface The arrow is fired one way and the archer recoils in the (ice) opposite direction Approaches: Categorize Newton’s Second Law – Momentum no, no information about Let the system be the archer with bow (particle 1) and the F or a arrow (particle 2) Energy approach – no, There are no external forces in the x-direction, so it is no information about isolated in terms of momentum in the x-direction work or energy Analyze Momentum – yes Total momentum before releasing the arrow is 0
Archer Example, 3 Impulse and Momentum dp Analyze, cont. From Newton’s Second Law, F dt The total momentum after releasing the arrow is Solving for d p gives ddtpF pp0 12ff Integrating to find the change in momentum Finalize over some time interval The final velocity of the archer is negative tf pp p Fdt I Indicates he moves in a direction opposite the arrow fit i Archer has much higher mass than arrow, so velocity is much lower The integral is called the impulse, , of the force acting on an object over t I
Impulse-Momentum Theorem More About Impulse
This equation expresses the impulse- Impulse is a vector quantity The magnitude of the momentum theorem: The impulse of the impulse is equal to the area force acting on a particle equals the change under the force-time curve The force may vary with in the momentum of the particle time pI Dimensions of impulse are M L / T This is equivalent to Newton’s Second Law Impulse is not a property of the particle, but a measure of the change in momentum of the particle
2 Impulse-Momentum: Crash Impulse, Final Test Example
The impulse can also Categorize be found by using the Assume force exerted by time averaged force wall is large compared with other forces IFt Gravitational and normal This would give the forces are perpendicular same impulse as the and so do not effect the time-varying force does horizontal momentum Can apply impulse approximation
Collisions – Example 1 Collisions – Example 2
Collisions may be the The collision need not result of direct contact include physical contact The impulsive forces between the objects may vary in time in There are still forces complicated ways between the particles This force is internal to This type of collision can the system be analyzed in the same Observe the variations in way as those that include the active figure physical contact Momentum is conserved
Types of Collisions Collisions, cont
In an elastic collision, momentum and kinetic In an inelastic collision, some kinetic energy energy are conserved is lost, but the objects do not stick together Perfectly elastic collisions occur on a microscopic level In macroscopic collisions, only approximately elastic Elastic and perfectly inelastic collisions are collisions actually occur limiting cases, most actual collisions fall in Generally some energy is lost to deformation, sound, etc. between these two types In an inelastic collision, kinetic energy is not conserved, although momentum is still conserved Momentum is conserved in all collisions If the objects stick together after the collision, it is a perfectly inelastic collision
3 Clicker Question Perfectly Inelastic Collisions In a perfectly inelastic one-dimensional collision between two moving objects, what condition alone is Since the objects stick necessary so that the final kinetic energy of the system together, they share the is zero after the collision? same velocity after the collision A. It is not possible mm11vvii 2 2 mm 1 2 v f B. The objects must have momenta with the same magnitude but opposite directions. C. The objects must have the same mass. D. The objects must have the same velocity. E. The objects must have the same speed, with velocity vectors in opposite directions.
Elastic Collisions Elastic Collisions, cont
Both momentum and Typically, there are two unknowns to solve for and so you kinetic energy are need two equations The kinetic energy equation can be difficult to use conserved With some algebraic manipulation, a different equation can be used mm11vvii 2 2 v –v = v + v mmvv 1i 2i 1f 2f 11ff 2 2 This equation, along with conservation of momentum, can be 11 used to solve for the two unknowns mmvv22 2211ii 2 2 It can only be used with a one-dimensional, elastic collision between two objects 11 mmvv22 2211ff 2 2
Collision Example – Ballistic Elastic Collisions, final Pendulum
Example of some special cases Conceptualize Observe diagram m1 = m2 – the particles exchange velocities Categorize When a very heavy particle collides head-on with a very Isolated system of projectile and light one initially at rest, the heavy particle continues in block motion unaltered and the light particle rebounds with a Perfectly inelastic collision – the speed of about twice the initial speed of the heavy particle bullet is embedded in the block of wood When a very light particle collides head-on with a very heavy particle initially at rest, the light particle has its Momentum equation will have velocity reversed and the heavy particle remains two unknowns Use conservation of energy from approximately at rest the pendulum to find the velocity just after the collision Then you can find the speed of the bullet
4 Two-Dimensional Collision, Two-Dimensional Collisions example
The momentum is conserved in all directions Particle 1 is moving at Use subscripts for velocity v 1 i and particle Identifying the object 2 is at rest Indicating initial or final values In the x-direction, the The velocity components initial momentum is If the collision is elastic, use conservation of kinetic m1v1i energy as a second equation In the y-direction, the Remember, the simpler equation can only be used for one- initial momentum is 0 dimensional situations
Two-Dimensional Collision, Clicker Question example cont
What is the direction of After the collision, the motion ofv m after the momentum in the x-direction 1i 2 is m v cos m v cos collision? 1 1f 2 2f After the collision, the A. up-left momentum in the y-direction
B. up-right is m1v1f sin m2v2f sin C. down-left If the collision is elastic, apply the kinetic energy equation D. down-right This is an example of a E. Right only glancing collision
Two-Dimensional Collision Example The Center of Mass
Conceptualize There is a special point in a system or object, See picture called the center of mass, that moves as if Choose East to be the positive x-direction and all of the mass of the system is concentrated North to be the positive y- at that point direction Categorize The system will move as if an external force Ignore friction were applied to a single particle of mass M Model the cars as particles located at the center of mass The collision is perfectly inelastic M is the total mass of the system The cars stick together
5 Center of Mass, Extended Center of Mass, Coordinates Object
The coordinates of the Similar analysis can be center of mass are done for an extended mxii x i object CM M Consider the extended my ii object as a system y i CM M containing a large mzii number of particles i zCM M Since particle separation M is the total mass of the is very small, it can be system considered to have a Use the active figure to observe effect of different constant mass masses and positions distribution
Finding Center of Mass, Center of Mass, position Irregularly Shaped Object
The center of mass in three dimensions can Suspend the object from one point be located by its position vector, rCM For a system of particles, The suspend from 1 another point rr m CMM i i i The intersection of the r is the position of the ith particle, defined by resulting lines is the i ˆˆˆ center of mass rijkiixyz i i For an extended object, 1 rr dm CM M
Center of Gravity Center of Mass, Rod
Each small mass element of an extended Conceptualize Find the center of mass object is acted upon by the gravitational force of a rod of mass M and The net effect of all these forces is equivalent length L The location is on the x- to the effect of a single force M g acting axis (or yCM = zCM = 0) through a point called the center of gravity Categorize If g is constant over the mass distribution, the Analysis problem center of gravity coincides with the center of mass Analyze
Use equation for xcm xCM = L / 2
6 Velocity and Momentum of a Acceleration of the Center of System of Particles Mass
The velocity of the center of mass of a system of The acceleration of the center of mass can particles is be found by differentiating the velocity with dr 1 vvCM m CM ii respect to time dt M i The momentum can be expressed as dv 1 aa CM m CM ii MmvvppCMii i tot dt M i ii
The total linear momentum of the system equals the total mass multiplied by the velocity of the center of mass
Newton’s Second Law for a Forces In a System of Particles System of Particles
The acceleration can be related to a force Since the only forces are external, the net external force equals the total mass of the system multiplied MaF CM i by the acceleration of the center of mass: i If we sum over all the internal forces, they Fa M cancel in pairs and the net force on the ext CM system is caused only by the external forces The center of mass of a system of particles of combined mass M moves like an equivalent particle of mass M would move under the influence of the net external force on the system
Impulse and Momentum of a Motion of the Center of Mass, System of Particles Example
The impulse imparted to the system by A projectile is fired into the air and suddenly explodes external forces is With no explosion, the IFdt M d vp projectile would follow the ext CM tot dotted line The total linear momentum of a system of After the explosion, the center of mass of the particles is conserved if no net external force fragments still follows the is acting on the system dotted line, the same parabolic path the projectile MvpCM tot constant when F ext 0 would have followed with no explosion Use the active figure to observe a variety of explosions
7 Deformable System (Spring) Deformable Systems Example
To analyze the motion of a deformable Conceptualize system, use Conservation of Energy and the See figure Impulse-Momentum Theorem Push on left block, it moves to right, spring ETKU0 compresses system At any given time, the blocks are generally Ip Fdt m v tot ext moving with different velocities If the force is constant, the integral can be easily The blocks oscillate back evaluated and forth with respect to the center of mass
Spring Example, cont Spring Example, final
Categorize Analyze, cont. Non isolated system Find energies Work is being done on it by the applied force Finalize It is a deformable system Answers do not depend on spring length, spring The applied force is constant, so the acceleration of the center of mass is constant constant, or time interval Model as a particle under constant acceleration Analyze Apply impulse-momentum
Solve for vcm
Rocket Propulsion Rocket Propulsion, 2
The operation of a rocket depends upon the The initial mass of the law of conservation of linear momentum as rocket plus all its fuel is M + m at time t and applied to a system of particles, where the i speed v system is the rocket plus its ejected fuel The initial momentum of the system is pvi = (M + m)
8 Rocket Propulsion, 3 Rocket Propulsion, 4
At some time t + t, the Because the gases are given some momentum rocket’s mass has been when they are ejected out of the engine, the rocket reduced to M and an receives a compensating momentum in the opposite amount of fuel, m has direction been ejected Therefore, the rocket is accelerated as a result of the “push” from the exhaust gases The rocket’s speed has In free space, the center of mass of the system increased by v (rocket plus expelled gases) moves uniformly, independent of the propulsion process
Rocket Propulsion, 5 Thrust
The basic equation for rocket propulsion is The thrust on the rocket is the force exerted on it by the ejected exhaust gases Mi vvvfieln Mf dv dM thrust M ve The increase in rocket speed is proportional to the dt dt
speed of the escape gases (ve) The thrust increases as the exhaust speed So, the exhaust speed should be very high increases The increase in rocket speed is also proportional to the natural log of the ratio Mi/Mf The thrust increases as the rate of change of mass So, the ratio should be as high as possible, meaning the mass of increases the rocket should be as small as possible and it should carry as The rate of change of the mass is called the burn rate much fuel as possible
9