# Impulse (physics)

 Common symbol(s): J, Imp SI unit: N · s = kg · m/s

In classical mechanics, the impulse of force (noted as J or Imp[1]) is defined as the product of the average force multiplied by the time it is exerted.[2] Impulse is a vector quantity since force is a vector quantity. The SI unit of impulse is the newton second (N·s) or, in base units, the kilogram meter per second (kg·m/s). Impulse is also defined as the change in the linear momentum of a body.

A resultant force causes acceleration and a change in the velocity of the body for as long as it acts. A resultant force applied over a longer time therefore produces a bigger change in momentum than the same force applied briefly: the change in momentum is equal to the product of the average force and time. Conversely, a small force applied for a long time can produce the same change in momentum—the same impulse—as a large force applied briefly.

The quantity of an impulse is average force × time interval, or in shorthand notation:

$J = F_{average} (t_2 - t_1)$

If a force is not of constant magnitude over time, the impulse is the integral of the magnitude of the resultant force (F) with respect to time:

$J = \int F dt$

## Mathematical derivation in the case of an object of constant mass

Impulse J produced from time t1 to t2 is defined to be[3]

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

where F is the resultant force applied from t1 to t2.

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

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

Therefore

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

where Δp is the change in linear momentum from time t1 to t2. This is often called the impulse-momentum theorem.[4]

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

$\mathbf{J} = \int_{t_1}^{t_2} \mathbf{F}\, dt = \Delta\mathbf{p} = m \mathbf{v_2} - m \mathbf{v_1}$

where

F is the resultant force applied,
t1 and t2 are times when the impulse begins and ends, respectively,
m is the mass of the object,
v2 is the final velocity of the object at the end of the time interval, and
v1 is the initial velocity of the object when the time interval begins.

The term "impulse" is also used to refer to a fast-acting force or impact. 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. However, this is a useful model for computing the effects of ideal collisions (such as in game physics engines).

The longer the club remains in contact with the ball the greater the impulse

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

## 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:[5]

$d\mathbf{p} = d(m\mathbf{v})\,+\,(dm)\mathbf{v} = (dt)\mathbf{F}\,+\,(dm)\mathbf{v} = d\mathbf{J}\,+\,(dm)\mathbf{v} = (dm)\mathbf{v}_e\,+\,(dm)\mathbf{v} = (dm)(\mathbf{v}_e + \mathbf{v})$

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[6]
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
dJ is the infinitesimal impulse exerted on the rocket

## Notes

1. ^ Beer, F.P., E.R. Johnston, Jr., D.F. Mazurek, P.J. Cornwell, and E.R. Eisenberg. (2010). Vector Mechanics for Engineers; Statics and Dynamics. 9th ed. Toronto: McGraw-Hill.
2. ^ Impulse of Force, Hyperphysics
3. ^ Hibbeler, Russell C. (2010). Engineering Mechanics (12th ed.). Pearson Prentice Hall. p. 222. ISBN 0-13-607791-9.
4. ^ See, for example, section 9.2, page 257, of Serway (2004).
5. ^ 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.
6. ^ 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

• Serway, Raymond A.; Jewett, John W. (2004). Physics for Scientists and Engineers (6th ed.). Brooks/Cole. ISBN 0-534-40842-7.
• Tippler, Paul (2004). Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics (5th ed.). W. H. Freeman. ISBN 0-7167-0809-4.
1. ^ Beer, F.P., E.R. Johnston, Jr., D.F. Mazurek, P.J. Cornwell, and E.R. Eisenberg. (2010). Vector Mechanics for Engineers; Statics and Dynamics. 9th ed. Toronto: McGraw-Hill.
2. ^ Impulse of Force, Hyperphysics
3. ^ Hibbeler, Russell C. (2010). Engineering Mechanics (12th ed.). Pearson Prentice Hall. p. 222. ISBN 0-13-607791-9.
4. ^ See, for example, section 9.2, page 257, of Serway (2004).
5. ^ 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.
6. ^ 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