Impulse (physics)

Classical mechanics
\mathbf{F} = m \mathbf{a}
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In classical mechanics, an impulse (abbreviated I or J) is defined as the integral of a force with respect to time. When a force is applied to a rigid body it changes the momentum of that body. A small force applied for a long time can produce the same momentum change as a large force applied briefly, because it is the product of the force and the time for which it is applied that is important. The impulse is equal to the change of momentum.

Contents

Mathematical derivation in the case of an object of constant mass

Impulse I produced from time t1 to t2 is defined to be[1]

\mathbf{I} = \int_{t_1}^{t_2} \mathbf{F}\, dt

where F is the force applied from t_1 to t_2.

From Newton's second law, force is related to momentum p by

\mathbf{F} = \frac{d\mathbf{p}}{dt}.

Therefore

\begin{align} \mathbf{I} &= \int_{t_1}^{t_2} \frac{d\mathbf{p}}{dt}\, dt \\ &= \int_{p_1}^{p_2} d\mathbf{p} \\ &= \Delta \mathbf{p}, \end{align}

where Δp is the change in momentum from time t1 to t2. This is often called the impulse-momentum theorem.[2]

As a result, an impulse may also be regarded as the change in momentum of an object to which a force is applied. The impulse may be expressed in a simpler form when both the force and the mass are constant:

\mathbf{I} = \mathbf{F}\,\Delta t = m \,\Delta \mathbf{v} = \Delta\mathbf{p}

It is often the case that not just one but both of these two quantities vary.

In the technical sense, impulse is a physical quantity, not an event or force. The term "impulse" is also used to refer to a fast-acting force. This type of impulse is often idealized so that the change in momentum produced by the force happens with no change in time. This sort of change is a step change, and is not physically possible. This is a useful model for computing the effects of ideal collisions (such as in game physics engines).

Impulse has the same units (in the International System of Units, kg·m/s = N·s) and dimensions (M L T−1) as momentum.

Impulse can be calculated using the equation

\mathbf{F}\,\Delta t = \Delta p = m\mathbf{v}_1 - m\mathbf{v}_0 \,

where

F is the constant total net force applied,
t is the time interval over which the force is applied,
m is the constant mass of the object,
v1 is the final velocity of the object at the end of the time interval, and
v0 is the initial velocity of the object when the time interval begins.

Variable mass

When a system expels mass in one direction, the force the expelled mass applies to the system is called thrust; the force the system applies to the mass being expelled is of equal magnitude but opposite direction.

Consider for example a rocket. The momentum of the rocket (including the remaining fuel) changes due to two effects: one is the applied thrust, the other one is the reduction of mass:[3]

dp = d(mv) = mdv + (dm)v = (dt)F + (dm)v = dI + (dm)v = (dm)ve + (dm)v = (dm)(ve + v)

where

p is the momentum of the rocket including the remaining fuel
dp is the infinitesimal change of the momentum of the rocket including the remaining fuel; it is the negative of the momentum of the mass being expelled, just after expulsion (the total momentum does not change)
m is the mass of the rocket including the remaining fuel (it decreases when mass is expelled)
dm is the infinitesimal change of the mass of the rocket including the remaining fuel, so the negative of the mass being expelled[4]
v is the velocity of the rocket
ve is the velocity of the just expelled mass relative to the rocket (effective exhaust velocity), hence:
ve + v is the velocity of the just expelled mass
F is the thrust
dI is the infinitesimal impulse exerted on the rocket

See also

Notes

  1. ^ Hibbeler, Russell C. (2010), EngineEring Mechanics, 12th edition, Pearson Prentice Hall, p. 222, ISBN 0-13-607791-9 
  2. ^ See, for example, section 9.2, page 257, of Serway (2004).
  3. ^ Scalar multiplication is written here in the regular notation, with the scalar first. Alternatively the vector is sometimes written first (like above), thus avoiding the need of parentheses in some cases.
  4. ^ With this definition dm is non-positive. Alternatively, in this context it can be convenient to define dm as a non-negative number: as the infinitesimal mass being expelled, so the negative of the infinitesimal change of the mass of the rocket including the remaining fuel

Bibliography

External links