24/08/2013
The second law states that the net force on an object is equal to the rate of change (that is, the derivative) of its linear momentum p in an inertial reference frame:
\mathbf{F} = \frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t} = \frac{\mathrm{d}(m\mathbf v)}{\mathrm{d}t}.
The second law can also be stated in terms of an object's acceleration. Since the law is valid only for constant-mass systems,[16][17][18] the mass can be taken outside the differentiation operator by the constant factor rule in differentiation. Thus,
\mathbf{F} = m\,\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = m\mathbf{a},
where F is the net force applied, m is the mass of the body, and a is the body's acceleration. Thus, the net force applied to a body produces a proportional acceleration. In other words, if a body is accelerating, then there is a force on it.
Consistent with the first law, the time derivative of the momentum is non-zero when the momentum changes direction, even if there is no change in its magnitude; such is the case with uniform circular motion. The relationship also implies the conservation of momentum: when the net force on the body is zero, the momentum of the body is constant. Any net force is equal to the rate of change of the momentum.
Any mass that is gained or lost by the system will cause a change in momentum that is not the result of an external force. A different equation is necessary for variable-mass systems (see below).
Newton's second law requires modification if the effects of special relativity are to be taken into account, because at high speeds the approximation that momentum is the product of rest mass and velocity is not accurate.
Impulse
An impulse J occurs when a force F acts over an interval of time Δt, and it is given by[19][20]
\mathbf{J} = \int_{\Delta t} \mathbf F \,\mathrm{d}t .
Since force is the time derivative of momentum, it follows that
\mathbf{J} = \Delta\mathbf{p} = m\Delta\mathbf{v}.
This relation between impulse and momentum is closer to Newton's wording of the second law.[21]
Impulse is a concept frequently used in the analysis of collisions and impacts.[22]
Variable-mass systems
Main article: Variable-mass system
Variable-mass systems, like a rocket burning fuel and ejecting spent gases, are not closed and cannot be directly treated by making mass a function of time in the second law;[17] that is, the following formula is wrong:[18]
\mathbf{F}_\mathrm{net} = \frac{\mathrm{d}}{\mathrm{d}t}\big[m(t)\mathbf{v}(t)\big] = m(t) \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} + \mathbf{v}(t) \frac{\mathrm{d}m}{\mathrm{d}t}. \qquad \mathrm{(wrong)}
The falsehood of this formula can be seen by noting that it does not respect Galilean invariance: a variable-mass object with F = 0 in one frame will be seen to have F ≠ 0 in another frame.[16]
The correct equation of motion for a body whose mass m varies with time by either ejecting or accreting mass is obtained by applying the second law to the entire, constant-mass system consisting of the body and its ejected/accreted mass; the result is[16]
\mathbf F + \mathbf{u} \frac{\mathrm{d} m}{\mathrm{d}t} = m {\mathrm{d} \mathbf v \over \mathrm{d}t}
where u is the relative velocity of the escaping or incoming mass as seen by the body. From this equation one can derive the Tsiolkovsky rocket equation.
Under some conventions, the quantity udm/dt on the left-hand side, known as the thrust, is defined as a force (the force exerted on the body by the changing mass, such as rocket exhaust) and is included in the quantity F. Then, by substituting the definition of acceleration, the equation becomes F = ma.
