Moméntum

Dinafisika,moméntum(dina hijianSIkg·m/s, atawa, nu satara,N·s) mangrupa hasil pakalianmassajeunglajuhiji obyék (p=mv{\displaystyle p=mv}). Inyana kalan-kalan disebut salakumoméntum liniérpikeun ngabédakeun jeung jejer nu pakait ngeunaanmoméntum sudutatawamoméntum puter. Moméntum liniér mangrupa hiji kuantitasvéktorlantaran inyana boga arah tujuan sarta gedé. Moméntum sudut mangrupa hiji kuantitasvéktor palsulantaran inyana
Momentum is aconservedquantity, méaning that the total momentum of anyclosed system(one not affected by external forces) cannot change. This concept is extended in thesecond law of thermodynamicsandspecial relativity.
The word for the general concept ofmōmentumwas used in theRoman Republicprimarily to méan "a movement, motion (as an indwelling force ...)." A fish was able to change velocity (velocitas) through themōmentumof its tail.The word is formed by an accretion ofsufficeson the stem ofLatinmovēre, "to move." Amovi-men-is the result of themovērejust asfrag-men-is the result offrangere, "to break." Extension by -to- obtainsmōvimentumandfragmentum, the former contracting tomōmentum.
Themōmentumwas not méré ly the motion, which wasmōtus, but was the power residing in a moving object, captured by today's mathematical definitions. Amōtus, "movement", was a stage in any sort of change,whilevelocitas, "swiftness", captured onlyspeed. The Romans, due to limitations inherent in theRoman numeralsystem,Citakan:Clarifymewere unable to go further with the perception.
The concept of momentum in classical mechanics was originated by a number of gréat thinkers and experimentalists. The first of these wasIbn Sina(Avicenna)circa1000, who referred toimpetusas proportional toweighttimesvelocity.René Descarteslater referred tomass times velocityas thefundamental force of motion.Galileoin hisTwo New Sciencesused theItalianword "impeto."
The question has been much debated as to what SirIsaac Newton's contribution to the concept was. Apparently nothing, except to state more fully and with better mathematics what was alréady known. The first and second ofNewton's Laws of Motionhad alréady been stated byJohn Wallisin his 1670 work,Mechanica slive De Motu, Tractatus Geometricus: "the initial state of the body, either of rest or of motion, will persist" and "If the force is greater than the resistance, motion will result...."Wallis usesmomentumandvisfor force.
Newton's "Mathematical Principles of Natural Philosophy" when it first came out in 1686 showed a similar casting around for words to use for the mathematical momentum. His Definition IIdefinesquantitas motus, "quantity of motion," as "arising from the velocity and quantity of matter conjointly", which identifies it as momentum.Thus when in Law II he refers tomutatio motus, "change of motion," being proportional to the force impressed, he is generally taken to méan momentum and not motion.
It remained only to assign a standard term to the quantity of motion. The first use of "momentum" in its proper mathematical sense is not cléar but by the time of Jenning'sMiscellaneain 1721, four yéars before the final edition of Newton'sPrincipia Mathematica, momentum M or "quantity of motion" was being defined for students as "a rectangle", the product of Q and V where Q is "quantity of material" and V is "velocity", s/t.
If an object is moving in anyreference frame, then it has momentum in that frame. It is important to note that momentum isframe dependent. That is, the same object may have a certain momentum in one frame of reference, but a different amount in another frame. For example, a moving object has momentum in a reference frame fixed to a spot on the ground, while at the same time having 0 momentum in a reference frame attached to the object'scenter of mass.
The amount of momentum that an object has depends on two physical quantities: themassand thevelocityof the moving object in theframe of reference. In physics, the usual symbol for momentum is a small boldp(bold because it is avector); so this can be written:
where:
Example: a modél airplane of 1 kg travelling due north at 1 m/s in straight and level flight has a momentum of 1 kg m/s due north méasured from the ground. To the dummy pilot in the cockpit it has a velocity and momentum of zero.
According toNewton's second lawthe rate of change of the momentum of a particle is proportional to the resultant force acting on the particle and is in the direction of that force. In the case of constant mass, and velocities much less than the speed of light, this definition results in the equation
or just simply
whereFis understood to be theresultant.
Example: a modél airplane of 1 kg accelerates from rest to a velocity of 1 m/s due north in 1 sec. The thrust required to produce this acceleration is 1newton. The change in momentum is 1 kg m/s. To the dummy pilot in the cockpit there is no change of momentum. Its pressing backward in the séat is a réaction to the unbalanced thrust, shortly to be balanced by the drag.
The linéar momentum of a system of particles is thevector sumof the momenta of all the individual objects in the system.
where
It can be shown that, in thecenter of mass framethe momentum of a system is zero. Additionally, the momentum in a frame of reference that is moving at a velocityvcmwith respect to that frame is simply:
where:
The linéar momentum of a system of particles can also be defined as the product of the total massM{\displaystyle \ M}of the system times the velocity of the center of massvcm{\displaystyle \mathbf {v} _{cm}}
This is commonly known asNewton's second law.
For a more general derivation using tensors, we consider a moving body (see Figure), assumed as acontinuum, occupying a volumeV{\displaystyle \ V}at a timet{\displaystyle \ t}, having a surface aréaS{\displaystyle \ S}, with defined traction or surface forcesTi(n){\displaystyle \ T_{i}^{(n)}}acting on every point of the body surface, body forcesFi{\displaystyle \ F_{i}}per unit of volume on every point within the volumeV{\displaystyle \ V}, and a velocity fieldvi{\displaystyle \ v_{i}}prescribed throughout the body. Following the previous equation, The linéar momentum of the system is:
By definition thestressvector isTi(n)=σijnj{\displaystyle \ T_{i}^{(n)}=\sigma _{ij}n_{j}}, then
Using theGauss's divergency theoremto convert a surface integral to a volume integral gives
For an arbitrary volume the integrand itself must be zero, and we have theCauchy's equation of motion
If a system is in equilibrium, the change in momentum with respect to time is equal to 0, as there is no acceleration.
or using tensors,
These are the equilibrium equations which are used insolid mechanicsfor solving problems oflinear elasticity.
In engineering notation, the equilibrium equations are expressed inCartesian coordinatesas
The law ofconservation of linéar momentumis a fundamental law of nature, and it states that the total momentum of a closed system of objects (which has no interactions with external agents) is constant. One of the consequences of this is that thecenter of massof anysystemofobjectswill always continue with the same velocity unless acted on by a force from outside the system.
Conservation of momentum is a mathematical consequence of thehomogeneity(shiftsymmetry) of space (position in space is thecanonical conjugatequantity to momentum). So, momentum conservation can be philosophically stated as "nothing depends on location per se".
In an isolated system (one where external forces are absent) the total momentum will be constant: this is implied by Newton'sfirst law of motion. Newton's third law of motion, thelaw of reciprocal actions, which dictates that the forces acting between systems are equal in magnitude, but opposite in sign, is due to the conservation of momentum.
Since position in space is a vector quantity, momentum (being thecanonical conjugateof position) is a vector quantity as well - it has direction. Thus, when a gun is fired, the final total momentum of the system (the gun and the bullet) is the vector sum of the momenta of these two objects. Assuming that the gun and bullet were at rest prior to firing (méaning the initial momentum of the system was zero), the final total momentum must also equal 0.
In an isolated system with only two objects, the change in momentum of one object must be equal and opposite to the change in momentum of the other object. Mathematically,
Δp1=−Δp2{\displaystyle \Delta \mathbf {p} _{1}=-\Delta \mathbf {p} _{2}}
Momentum has the special property that, in aclosed system, it is always conserved, even incollisionsand separations caused by explosive forces.Kinetic energy, on the other hand, is not conserved in collisions if they are inelastic. Since momentum is conserved it can be used to calculate an unknown velocity following a collision or a separation if all the other masses and velocities are known.
A common problem in physics that requires the use of this fact is the collision of two particles. Since momentum is always conserved, the sum of the momenta before the collision must equal the sum of the momenta after the collision:
where:
Usually, we either only know the velocities before or after a collision and would like to also find out the opposite. Correctly solving this problem méans you have to know what kind of collision took place. There are two basic kinds of collisions, both of which conserve momentum:
A collision between twoPoolballs is a good example of analmosttotally elastic collision; a totally elastic collision exists only in théory. In addition to momentum being conserved when the two balls collide, the sum of kinetic energy before a collision must equal the sum of kinetic energy after:
Since the 1/2 factor is common to all the terms, it can be taken out right away.
In the case of two objects colliding héad on we find that the final velocity
which can then éasily be réarranged to
Special Case: m1>>m2
Now consider the case when the mass of one body, say m1, is far gréater than that of the other, m2(m1>>m2).
In that case m1+m2is approximately equal to m1and m1-m2is approximately equal to m1.
Using these approximations, the above formula forv2,f{\displaystyle v_{2,\mathrm {f} }}reduces tov2,f=2v1,i−v2,i{\displaystyle v_{2,\mathrm {f} }=2v_{1,\mathrm {i} }-v_{2,\mathrm {i} }}.
Its physical interpretation is that in the case of a collision between two bodies, one of which is much more massive than the other, the lighter body ends up moving in the opposite direction with twice the original speed of the more massive body.
Special Case: m1=m2
Another special case is when the collision is between two bodies of equal mass.
Say body m1 moving at velocity v1strikes body m2that is at rest (v2).
Putting this case in the equation derived above we will see that after the collision, the body that was moving (m1) will start moving with velocity v2and the mass m2will start moving with velocity v1. So there will be an exchange of velocities.
Now suppose one of the masses, say m2, was at rest. In that case after the collision the moving body, m1, will come to rest and the body that was at rest, m2, will start moving with the velocity that m1had before the collision.
Note that all of these observations are for anelastic collision.
This phenomenon is demonstrated byNewton's cradle, one of the best known examples of conservation of momentum, a réal life example of this special case.
In the case of objects colliding in more than one dimension, as in oblique collisions, the velocity is resolved into orthogonal components with one component perpendicular to the plane of collision and the other component or components in the plane of collision. The velocity components in the plane of collision remain unchanged, while the velocity perpendicular to the plane of collision is calculated in the same way as the one-dimensional case.
For example, in a two-dimensional collision, the momenta can be resolved intoxandycomponents. We can then calculate éach component separately, and combine them to produce a vector result. The magnitude of this vector is the final momentum of the isolated system.
See theelastic collisionpage for more details.x=2a{\displaystyle x=2a}
A common example of a perfectly inelastic collision is when two snowballs collide and thensticktogether afterwards. This equation describes the conservation of momentum:
It can be shown that a perfectly inelastic collision is one in which the maximum amount ofkinetic energyis converted into other forms. For instance, if both objects stick together after the collision and move with a final common velocity, one can always find a reference frame in which the objects are brought to rest by the collision and 100% of the kinetic energy is converted. This is true even in the relativistic case and utilized inparticle acceleratorsto efficiently convert kinetic energy into new forms ofmass-energy(i.e. to créate massive particles).
In case of Inelastic collision, there is a paraméter attached called coefficient of restitution (denoted by small 'e' or 'c' in many text books). It is defined as the ratio of relative velocity of separation to relative velocity of approach. It is a ratio hence it is a dimensionless quantity.
When we have an elastic collision the value of e (= coefficient of restitution) is 1, i.e. the relative velocity of approach is same as the relative velocity of separation of the colliding bodies. In an elastic collision the Kinetic energy of the system is conserved.
When a collision is not elastic (e<1) it is an inelastic collision. In case of a perfectly inelastic collision the relative velocity of separation of the centre of masses of the colliding bodies is 0. Hence after collision the bodies stick together after collision. In case of an inelastic collision the loss of Kinetic energy is maximum as stated above.
In all types of collision if no external force is acting on the system of colliding bodies, the momentum will always be preserved.
An explosion occurs when an object is divided into two or more fragments due to a reléase of energy. 
Note that kinetic energy in a system of explosion is not conserved because it involves energy transformation. (i.e. kinetic energy changes into héat and sound energy)
http://www.glenbrook.k12.il.us/gbssci/phys/Class/momentum/u4l2e.htmlArchived2008-09-15 diWayback Machine
In the exploding cannon demonstration, 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 momenta of the two objects is 0. Total system momentum is conserved.
See theinelastic collisionpage for more details.
In relativistic mechanics, in order to be conserved, momentum must be defined as:
where
Relativistic momentum can also be written as invariant mass times the object'sproper velocity, defined as the rate of change of object position in the observer frame with respect to time elapsed on object clocks (i.e. objectproper time).  Relativistic momentum becomes Newtonian momentum:mv{\displaystyle m\mathbf {v} }at low speed(v/c→0){\displaystyle {\big (}\mathbf {v} /c\rightarrow 0{\big )}}.
Relativisticfour-momentumas proposed byAlbert Einsteinarises from the invariance offour-vectorsunder Lorentzian translation.  The four-momentum is defined as:
where
The "length" of the vector is the mass times the speed of light, which is invariant across all reference frames:
Momentum of massless objects
Objects without a rest mass, such asphotons, also carry momentum. The formula is:
where
Generalization of momentum
Momentum is theNoether chargeof translational invariance. As such, even fields as well as other things can have momentum, not just particles. However, incurved space-timewhich is not asymptoticallyMinkowski, momentum isn't defined at all.
Inquantum mechanics, momentum is defined as anoperatoron thewave function. TheHeisenberguncertainty principledefines limits on how accurately the momentum and position of a single observable system can be known at once. In quantum mechanics, position and momentum areconjugate variables.
For a single particle with noelectric chargeand nospin, the momentum operator can be written in the position basis as
where:
This is a commonly encountered form of the momentum operator, though not the most general one.
Electric and magnetic fields possess momentum regardless of whether they are static or
they change in time. It is a gréat surprise for freshmen who are introduced
to the well known fact of the pressureP{\displaystyle {P}}of an electrostatic (magnetostatic) field upon a metal sphere,
cylindrical capacity or ferromagnetic bar:
whereW{\displaystyle {W}},E{\displaystyle {\mathbf {E} }},B{\displaystyle {\mathbf {B} }},
are electromagnetic energy density , electric and magnetic fields respectively.
The electromagnetic pressureP=W{\displaystyle {P}={W}}may be sufficiently high to explode capacity.
Thus electric and magnetic fields do carry momentum.
Light (visible, UV, radio) is an electromagnetic wave and also has momentum.  Even thoughphotons(the particle aspect of light) have no mass, they still carry momentum.  This léads to applications such as thesolar sail.
Momentum is conserved in an electrodynamic system (it may change from momentum in the fields to mechanical momentum of moving parts).  The tréatment of the momentum of a field is usually accomplished by considering the so-calledenergy-momentum tensorand the change in time of thePoynting vectorintegrated over some volume. This is a tensor field which has components related to the energy density and the momentum density.
The definition canonical momentum corresponding to the momentum operator of quantum mechanics when it interacts with the electromagnetic field is, using theprinciple of least coupling:
instéad of the customary
where: