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 . It is the product of two quantities, the particle's
mass (represented by the letter ) and its
velocity (): p = m v. 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 and , and velocities and , the total momentum is \begin{align} p &= p_1 + p_2 \\ &= m_1 v_1 + m_2 v_2\,. \end{align} The momenta of more than two particles can be added more generally with the following: p = \sum_{i} m_i v_i . A system of particles has a
center of mass, a point determined by the weighted sum of their positions: r_\text{cm} = \frac{m_1 r_1 + m_2 r_2 + \cdots}{m_1 + m_2 + \cdots} = \frac{\sum_{i}m_ir_i}{\sum_{i}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 m, and the center of mass is moving at velocity , the momentum of the system is: p= mv_\text{cm}. This is known as
Euler's first law. Suppose, for example, that two particles interact. As explained by the third law, the forces between them are equal in magnitude but opposite in direction. If the particles are numbered 1 and 2, the second law states that and . Therefore, \frac{\text{d}p_1}{\text{d}t} = - \frac{\text{d}p_2}{\text{d}t}, with the negative sign indicating that the forces oppose. Equivalently, \frac{\text{d}}{\text{d} t} \left(p_1 + p_2\right)= 0. If the velocities of the particles are and before the interaction, and afterwards they are and , then m_A v_{A1} + m_B v_{B1} = m_{A} v_{A2} + m_B v_{B2}. 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 to zero, so the total change in momentum is zero. The conservation of the total momentum of a number of interacting particles can be expressed as 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 If two particles, each of known momentum, collide and coalesce, the law of conservation of momentum can be used to determine the momentum of the coalesced body. If the outcome of the collision is that the two particles separate, the law is not sufficient to determine the momentum of each particle. If the momentum of one particle after the collision is known, the law can be used to determine the momentum of the other particle. Alternatively if the combined
kinetic energy after the collision is known, the law can be used to determine the momentum of each particle after the collision. Kinetic energy is usually not 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 transformed into heat or some other form of energy. 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 the objects 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 and before the collision and and after, the equations expressing conservation of momentum and kinetic energy are: \begin{align} m_{A} v_{A1} + m_{B} v_{B1} &= m_{A} v_{A2} + m_{B} v_{B2}\\ \tfrac{1}{2} m_{A} v_{A1}^2 + \tfrac{1}{2} m_{B} v_{B1}^2 &= \tfrac{1}{2} m_{A} v_{A2}^2 + \tfrac{1}{2} m_{B} v_{B2}^2\,.\end{align} A change of reference frame can simplify analysis of a collision. For example, suppose there are two bodies of equal mass , one stationary and one approaching the other at a speed (as in the figure). The center of mass is moving at speed and both bodies are moving towards it at speed . 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 . 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 \begin{align} v_{A2} &= \left( \frac{m_{A} - m_{B}}{m_{A} + m_{B}} \right) v_{A1} + \left( \frac{2 m_{B}}{m_{A} + m_{B}} \right) v_{B1} \\ v_{B2} &= \left( \frac{m_{B} - m_{A}}{m_{A} + m_{B}} \right) v_{B1} + \left( \frac{2 m_{A}}{m_{A} + m_{B}} \right) v_{A1}\,. \end{align} 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 and before the collision then in a perfectly inelastic collision both bodies will be travelling with velocity after the collision. The equation expressing conservation of momentum is: \begin{align} m_A v_{A1} + m_B v_{B1} &= \left( m_A + m_B \right) v_2\,.\end{align} If one body is motionless to begin with (e.g. v_{B1} = 0 ), the equation for conservation of momentum is m_A v_{A1} = \left( m_A + m_B \right) v_2\,, so v_2 = \frac{m_{A}}{m_{A}+m_{B}} v_{A1}\,. In a different situation, if the frame of reference is moving at the final velocity such that v_2 = 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 , 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: 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 axes, velocity has components in the -direction, in the -direction, in the -direction. The vector is represented by a boldface symbol: \mathbf{v} = \left(v_x,v_y,v_z \right). Similarly, the momentum is a vector quantity and is represented by a boldface symbol: \mathbf{p} = \left(p_x,p_y,p_z \right). The equations in the previous sections, work in vector form if the scalars and are replaced by vectors and . Each vector equation represents three scalar equations. For example, \mathbf{p}= m \mathbf{v} represents three equations:
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: . The momentum of the object at time is therefore . One might then try to invoke Newton's second law of motion by saying that the external force on the object is related to its momentum by , but this is incorrect, as is the related expression found by applying the product rule to : F = m(t) \frac{\text{d}v}{\text{d}t} + v(t) \frac{\text{d}m}{\text{d}t}. \text{(incorrect)} This equation does not correctly describe the motion of variable-mass objects. The correct equation is F = m(t) \frac{\text{d}v}{\text{d}t} - u \frac{\text{d}m}{\text{d}t}, where is the velocity of the ejected/accreted mass ''as seen in the object's rest frame''. This is distinct from , 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 (). When considered together, the object and the mass () constitute a closed system in which total momentum is conserved. P(t+\text{d}t) = ( m - \text{d}m ) ( v + \text{d}v ) + \text{d}m ( v - u ) = mv+m \text{d}v - u \text{d}m = P(t) +m \text{d}v - u \text{d}m ==Generalized==