Newtonian

Momentum is a vector quantity: it has both magnitude and direction. Since momentum has a direction, it can be used to predict the resulting direction and speed of motion of objects after they collide. Below, the basic properties of momentum are described in one dimension. The vector equations are almost identical to the scalar equations (see multiple dimensions).

Single particle

The momentum of a particle is conventionally represented by the letter p. It is the product of two quantities, the particle's mass (represented by the letter m) and its velocity (v):

${\displaystyle p=mv.}$

The unit of momentum is the product of the units of mass and velocity. In SI units, if the mass is in kilograms and the velocity is in meters per second then the momentum is in kilogram meters per second (kg⋅m/s). In cgs units, if the mass is in grams and the velocity in centimeters per second, then the momentum is in gram centimeters per second (g⋅cm/s).

Being a vector, momentum has magnitude and direction. For example, a 1 kg model airplane, traveling due north at 1 m/s in straight and level flight, has a momentum of 1 kg⋅m/s due north measured with reference to the ground.

Many particles

The momentum of a system of particles is the vector sum of their momenta. If two particles have respective masses m₁ and m₂, and velocities v₁ and v₂, the total momentum is

The momenta of more than two particles can be added more generally with the following:

${\displaystyle p=\underset{i}{\sum <!--\; \sum \; -->}{m}_i{v}_i.}$

A system of particles has a center of mass, a point determined by the weighted sum of their positions:

${\displaystyle r_{\text{cm}}=\frac{m_1{r}_1+m_2{r}_2+\cdots <!--\; \cdots \; -->}{m_1+m_2+\cdots <!--\; \cdots \; -->}=\frac{\underset{i}{\sum <!--\; \sum \; -->}{m}_i{r}_i}{\underset{i}{\sum <!--\; \sum \; -->}{m}_i}.}$

If one or more of the particles is moving, the center of mass of the system will generally be moving as well (unless the system is in pure rotation around it). If the total mass of the particles is ${\displaystyle m}$, and the center of mass is moving at velocity v_cm, the momentum of the system is:

${\displaystyle p=m{v}_{\text{cm}}.}$

This is known as Euler's first law.

Relation to force

If the net force F applied to a particle is constant, and is applied for a time interval Δt, the momentum of the particle changes by an amount

${\displaystyle \mathrm{\Delta <!--\; \Delta \; -->}p=F\mathrm{\Delta <!--\; \Delta \; -->}t.}$

In differential form, this is Newton's second law; the rate of change of the momentum of a particle is equal to the instantaneous force F acting on it,

${\displaystyle F=\frac{dp}{dt}.}$

If the net force experienced by a particle changes as a function of time, F(t), the change in momentum (or impulse J ) between times t₁ and t₂ is

${\displaystyle \mathrm{\Delta <!--\; \Delta \; -->}p=J={\int <!--\; \int \; -->}_{t_1}^{t_2}F(t)dt.}$

Impulse is measured in the derived units of the newton second (1 N⋅s = 1 kg⋅m/s) or dyne second (1 dyne⋅s = 1 g⋅cm/s)

Under the assumption of constant mass m, it is equivalent to write

${\displaystyle F=\frac{d(mv)}{dt}=m\frac{dv}{dt}=ma,}$

hence the net force is equal to the mass of the particle times its acceleration.

Example: A model airplane of mass 1 kg accelerates from rest to a velocity of 6 m/s due north in 2 s. The net force required to produce this acceleration is 3 newtons due north. The change in momentum is 6 kg⋅m/s due north. The rate of change of momentum is 3 (kg⋅m/s)/s due north which is numerically equivalent to 3 newtons.

Conservation

In a closed system (one that does not exchange any matter with its surroundings and is not acted on by external forces) the total momentum is constant. This fact, known as the law of conservation of momentum, is implied by Newton's laws of motion. Suppose, for example, that two particles interact. Because of the third law, the forces between them are equal and opposite. If the particles are numbered 1 and 2, the second law states that F₁ = dp₁/dt and F₂ = dp₂/dt. Therefore,

${\displaystyle \frac{d{p}_1}{dt}=-<!--\; -\; -->\frac{d{p}_2}{dt},}$

with the negative sign indicating that the forces oppose. Equivalently,

${\displaystyle \frac{d}{dt}\left(p_1+p_2\right)=0.}$

If the velocities of the particles are u₁ and u₂ before the interaction, and afterwards they are v₁ and v₂, then

${\displaystyle m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2.}$

This law holds no matter how complicated the force is between particles. Similarly, if there are several particles, the momentum exchanged between each pair of particles adds up to zero, so the total change in momentum is zero. This conservation law applies to all interactions, including collisions and separations caused by explosive forces. It can also be generalized to situations where Newton's laws do not hold, for example in the theory of relativity and in electrodynamics.

Dependence on reference frame

Momentum is a measurable quantity, and the measurement depends on the motion of the observer. For example: if an apple is sitting in a glass elevator that is descending, an outside observer, looking into the elevator, sees the apple moving, so, to that observer, the apple has a non-zero momentum. To someone inside the elevator, the apple does not move, so, it has zero momentum. The two observers each have a frame of reference, in which, they observe motions, and, if the elevator is descending steadily, they will see behavior that is consistent with those same physical laws.

Suppose a particle has position x in a stationary frame of reference. From the point of view of another frame of reference, moving at a uniform speed u, the position (represented by a primed coordinate) changes with time as

${\displaystyle x^{\prime }=x-<!--\; -\; -->ut.}$

This is called a Galilean transformation. If the particle is moving at speed dx/dt = v in the first frame of reference, in the second, it is moving at speed

${\displaystyle v^{\prime }=\frac{d{x}^{\prime }}{dt}=v-<!--\; -\; -->u.}$

Since u does not change, the accelerations are the same:

${\displaystyle a^{\prime }=\frac{d{v}^{\prime }}{dt}=a.}$

Thus, momentum is conserved in both reference frames. Moreover, as long as the force has the same form, in both frames, Newton's second law is unchanged. Forces such as Newtonian gravity, which depend only on the scalar distance between objects, satisfy this criterion. This independence of reference frame is called Newtonian relativity or Galilean invariance.

A change of reference frame, can, often, simplify calculations of motion. For example, in a collision of two particles, a reference frame can be chosen, where, one particle begins at rest. Another, commonly used reference frame, is the center of mass frame – one that is moving with the center of mass. In this frame, the total momentum is zero.

Application to collisions

By itself, the law of conservation of momentum is not enough to determine the motion of particles after a collision. Another property of the motion, kinetic energy, must be known. This is not necessarily conserved. If it is conserved, the collision is called an elastic collision; if not, it is an inelastic collision.

Elastic collisions

An elastic collision is one in which no kinetic energy is absorbed in the collision. Perfectly elastic "collisions" can occur when the objects do not touch each other, as for example in atomic or nuclear scattering where electric repulsion keeps them apart. A slingshot maneuver of a satellite around a planet can also be viewed as a perfectly elastic collision. A collision between two pool balls is a good example of an almost totally elastic collision, due to their high rigidity, but when bodies come in contact there is always some dissipation.

A head-on elastic collision between two bodies can be represented by velocities in one dimension, along a line passing through the bodies. If the velocities are u₁ and u₂ before the collision and v₁ and v₂ after, the equations expressing conservation of momentum and kinetic energy are:

A change of reference frame can simplify analysis of a collision. For example, suppose there are two bodies of equal mass m, one stationary and one approaching the other at a speed v (as in the figure). The center of mass is moving at speed v/2 and both bodies are moving towards it at speed v/2. Because of the symmetry, after the collision both must be moving away from the center of mass at the same speed. Adding the speed of the center of mass to both, we find that the body that was moving is now stopped and the other is moving away at speed v. The bodies have exchanged their velocities. Regardless of the velocities of the bodies, a switch to the center of mass frame leads us to the same conclusion. Therefore, the final velocities are given by

In general, when the initial velocities are known, the final velocities are given by

${\displaystyle v_1=\left(\frac{m_1-<!--\; -\; -->m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2}$

${\displaystyle v_2=\left(\frac{m_2-<!--\; -\; -->m_1}{m_1+m_2}\right)u_2+\left(\frac{2m_1}{m_1+m_2}\right)u_1.}$

If one body has much greater mass than the other, its velocity will be little affected by a collision while the other body will experience a large change.

Inelastic collisions

In an inelastic collision, some of the kinetic energy of the colliding bodies is converted into other forms of energy (such as heat or sound). Examples include traffic collisions, in which the effect of loss of kinetic energy can be seen in the damage to the vehicles; electrons losing some of their energy to atoms (as in the Franck–Hertz experiment); and particle accelerators in which the kinetic energy is converted into mass in the form of new particles.

In a perfectly inelastic collision (such as a bug hitting a windshield), both bodies have the same motion afterwards. A head-on inelastic collision between two bodies can be represented by velocities in one dimension, along a line passing through the bodies. If the velocities are u₁ and u₂ before the collision then in a perfectly inelastic collision both bodies will be travelling with velocity v after the collision. The equation expressing conservation of momentum is:

If one body is motionless to begin with (e.g. ${\displaystyle u_2=0}$), the equation for conservation of momentum is

${\displaystyle m_1{u}_1=\left(m_1+m_2\right)v,}$

so

${\displaystyle v=\frac{m_1}{m_1+m_2}{u}_1.}$

In a different situation, if the frame of reference is moving at the final velocity such that ${\displaystyle v=0}$, the objects would be brought to rest by a perfectly inelastic collision and 100% of the kinetic energy is converted to other forms of energy. In this instance the initial velocities of the bodies would be non-zero, or the bodies would have to be massless.

One measure of the inelasticity of the collision is the coefficient of restitution C_R, defined as the ratio of relative velocity of separation to relative velocity of approach. In applying this measure to a ball bouncing from a solid surface, this can be easily measured using the following formula:

${\displaystyle C_{\text{R}}=\sqrt{\frac{\text{bounce height}}{\text{drop height}}}.}$

The momentum and energy equations also apply to the motions of objects that begin together and then move apart. For example, an explosion is the result of a chain reaction that transforms potential energy stored in chemical, mechanical, or nuclear form into kinetic energy, acoustic energy, and electromagnetic radiation. Rockets also make use of conservation of momentum: propellant is thrust outward, gaining momentum, and an equal and opposite momentum is imparted to the rocket.

Multiple dimensions

Real motion has both direction and velocity and must be represented by a vector. In a coordinate system with x, y, z axes, velocity has components v_x in the x-direction, v_y in the y-direction, v_z in the z-direction. The vector is represented by a boldface symbol:

${\displaystyle \mathbf{v}=\left(v_x,v_y,v_z\right).}$

Similarly, the momentum is a vector quantity and is represented by a boldface symbol:

${\displaystyle \mathbf{p}=\left(p_x,p_y,p_z\right).}$

The equations in the previous sections, work in vector form if the scalars p and v are replaced by vectors p and v. Each vector equation represents three scalar equations. For example,

${\displaystyle \mathbf{p}=m\mathbf{v}}$

represents three equations:

The kinetic energy equations are exceptions to the above replacement rule. The equations are still one-dimensional, but each scalar represents the magnitude of the vector, for example,

${\displaystyle v^2=v_x^2+v_y^2+v_z^2.}$

Each vector equation represents three scalar equations. Often coordinates can be chosen so that only two components are needed, as in the figure. Each component can be obtained separately and the results combined to produce a vector result.

A simple construction involving the center of mass frame can be used to show that if a stationary elastic sphere is struck by a moving sphere, the two will head off at right angles after the collision (as in the figure).

Objects of variable mass

The concept of momentum plays a fundamental role in explaining the behavior of variable-mass objects such as a rocket ejecting fuel or a star accreting gas. In analyzing such an object, one treats the object's mass as a function that varies with time: m(t). The momentum of the object at time t is therefore p(t) = m(t)v(t). One might then try to invoke Newton's second law of motion by saying that the external force F on the object is related to its momentum p(t) by F = dp/dt, but this is incorrect, as is the related expression found by applying the product rule to d(mv)/dt:

${\displaystyle F=m(t)\frac{dv}{dt}+v(t)\frac{dm}{dt}.}$ (incorrect)why?

This equation does not correctly describe the motion of variable-mass objects. The correct equation is

${\displaystyle F=m(t)\frac{dv}{dt}-<!--\; -\; -->u\frac{dm}{dt},}$

where u is the velocity of the ejected/accreted mass as seen in the object's rest frame. This is distinct from v, which is the velocity of the object itself as seen in an inertial frame.

This equation is derived by keeping track of both the momentum of the object as well as the momentum of the ejected/accreted mass (dm). When considered together, the object and the mass (dm) constitute a closed system in which total momentum is conserved.

${\displaystyle P(t+dt)=(m-<!--\; -\; -->dm)(v+dv)+dm(v-<!--\; -\; -->u)=mv+mdv-<!--\; -\; -->udm=P(t)+mdv-<!--\; -\; -->udm}$