Impulse (physics): Difference between revisions

Impulse
Impulse
Common symbols	J, Imp
SI unit	newton-second (N⋅s) (kg⋅m/s in SI base units)
Other units	pound⋅s
Conserved?	yes
Dimension

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In classical mechanics, impulse (symbolized by $J$ or Imp) is the change in momentum of an object. If the initial momentum of an object is p₁, and a subsequent momentum is p₂, the object has received an impulse $J$ :

{\vec {J}}={\vec {p_{2}}}-{\vec {p_{1}}}

Momentum is a vector quantity, so impulse is also a vector quantity.

Newton’s second law of motion states that the rate of change of momentum of an object is equal to the resultant force $F$ acting on the object:

{\vec {F}}={\frac {{\vec {p_{2}}}-{\vec {p_{1}}}}{\Delta t}}

so the impulse $J$ delivered by a steady force $F$ acting for time Δt is:

{\vec {J}}={\vec {F}}\Delta t

The impulse delivered by a varying force is the integral of the force $F$ with respect to time: ${\vec {J}}=\int {\vec {F}}\,\mathrm {d} t$

The SI unit of impulse is the newton second (N⋅s), and the dimensionally equivalent unit of momentum is the kilogram metre per second (kg⋅m/s). The corresponding English engineering unit is the pound-second (lbf⋅s), and in the British Gravitational System, the unit is the slug-foot per second (slug⋅ft/s).

Mathematical derivation in the case of an object of constant mass

Impulse $J$ produced from time $t 1$ to $t 2$ is defined to be^[1] $\mathbf {J} =\int _{t_{1}}^{t_{2}}\mathbf {F} \,\mathrm {d} t,$ where $F$ is the resultant force applied from $t 1$ to $t 2$ .

From Newton's second law, force is related to momentum $p$ by $\mathbf {F} ={\frac {\mathrm {d} \mathbf {p} }{\mathrm {d} t}}.$

Therefore, ${\begin{aligned}\mathbf {J} &=\int _{t_{1}}^{t_{2}}{\frac {\mathrm {d} \mathbf {p} }{\mathrm {d} t}}\,\mathrm {d} t\\&=\int _{\mathbf {p} _{1}}^{\mathbf {p} _{2}}\mathrm {d} \mathbf {p} \\&=\mathbf {p} _{2}-\mathbf {p} _{1}=\Delta \mathbf {p} ,\end{aligned}}$ where $Δ p$ is the change in linear momentum from time $t 1$ to $t 2$ . This is often called the impulse-momentum theorem^[2] (analogous to the work-energy theorem).

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: $\mathbf {J} =\int _{t_{1}}^{t_{2}}\mathbf {F} \,\mathrm {d} t=\Delta \mathbf {p} =m\mathbf {v_{2}} -m\mathbf {v_{1}} ,$

where

$F$ is the resultant force applied,
$t 1$ and $t 2$ are times when the impulse begins and ends, respectively,
$m$ is the mass of the object,
$v 2$ is the final velocity of the object at the end of the time interval, and
$v 1$ is the initial velocity of the object when the time interval begins.

Impulse has the same units and dimensions (MLT⁻¹) as momentum. In the International System of Units, these are kg⋅m/s = N⋅s. In English engineering units, they are slug⋅ft/s = lbf⋅s.

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). Additionally, in rocketry, the term "total impulse" is commonly used and is considered synonymous with the term "impulse".

Variable mass

The application of Newton's second law for variable mass allows impulse and momentum to be used as analysis tools for jet- or rocket-propelled vehicles. In the case of rockets, the impulse imparted can be normalized by unit of propellant expended, to create a performance parameter, specific impulse. This fact can be used to derive the Tsiolkovsky rocket equation, which relates the vehicle's propulsive change in velocity to the engine's specific impulse (or nozzle exhaust velocity) and the vehicle's propellant-mass ratio.

Notes

^ Hibbeler, Russell C. (2010). Engineering Mechanics (12th ed.). Pearson Prentice Hall. p. 222. ISBN 978-0-13-607791-6.
^ See, for example, section 9.2, page 257, of Serway (2004).

Bibliography

Serway, Raymond A.; Jewett, John W. (2004). Physics for Scientists and Engineers (6th ed.). Brooks/Cole. ISBN 0-534-40842-7.
Tipler, Paul (2004). Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics (5th ed.). W. H. Freeman. ISBN 0-7167-0809-4.

External links

Dynamics

[1] Hibbeler, Russell C. (2010). Engineering Mechanics (12th ed.). Pearson Prentice Hall. p. 222. ISBN 978-0-13-607791-6.

[2] See, for example, section 9.2, page 257, of Serway (2004).

[1]

[2]

@@ Line 13: / Line 13: @@
 {{Classical mechanics|cTopic=Fundamental concepts}}
+In [[classical mechanics]], '''impulse''' (symbolized by {{mvar|J}} or '''Imp''') is the change in [[momentum]] of an object. If the initial momentum of an object is p<sub>1</sub>, and a subsequent momentum is p<sub>2</sub>, the object has received an impulse {{mvar|J}}:
-In [[classical mechanics]], '''impulse''' (symbolized by {{mvar|J}} or '''Imp''') is the [[integral]] of a [[force]], {{mvar|F}}, over the [[time]] interval, {{mvar|t}}, for which it acts. Since force is a [[Vector (physics)|vector]] quantity, impulse is also a vector quantity. Impulse applied to an object produces an equivalent vector [[Mathematics#Calculus and analysis|change]] in its linear [[momentum]], also in the resultant direction. The [[International System of Units|SI]] unit of impulse is the [[newton second]] (N⋅s), and the [[dimensional analysis|dimensionally equivalent]] unit of momentum is the kilogram meter per second (kg⋅m/s). The corresponding [[English engineering unit]] is the [[pound (force)|pound]]-second (lbf⋅s), and in the [[British Gravitational System]], the unit is the [[Slug (unit)|slug]]-foot per second (slug⋅ft/s).
+:<math>\vec{J}=\vec{p_2} - \vec{p_1}</math>
+Momentum is a [[Vector (physics)|vector]] quantity, so impulse is also a vector quantity.
-A non-zero [[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 linear momentum than the same force applied briefly: the change in momentum is equal to the product of the average force and duration. Conversely, a small force applied for a long time produces the same change in momentum—the same impulse—as a larger force applied briefly.
+[[Newton’s second law of motion]] states that the rate of change of momentum of an object is equal to the resultant force {{mvar|F}} acting on the object:
-<math display=block>J = F_{\text{average}} (t_2 - t_1).</math>
+:<math>\vec{F}=\frac{\vec{p_2} - \vec{p_1}}{\Delta t}</math>
-The impulse is the [[integral]] of the resultant force ({{mvar|F}}) with respect to time:
+so the impulse {{mvar|J}} delivered by a steady force {{mvar|F}} acting for time Δt is:
-<math display=block>J = \int F \,\mathrm{d}t.</math>
+:<math>\vec{J}=\vec{F} \Delta t </math>
+The impulse delivered by a varying force is the [[integral]] of the force {{mvar|F}} with respect to time:
+<math display=block>\vec{J} = \int \vec{F} \,\mathrm{d}t</math>
+The [[International System of Units|SI]] unit of impulse is the [[newton second]] (N⋅s), and the [[dimensional analysis|dimensionally equivalent]] unit of momentum is the kilogram metre per second (kg⋅m/s). The corresponding [[English engineering unit]] is the [[pound (force)|pound]]-second (lbf⋅s), and in the [[British Gravitational System]], the unit is the [[Slug (unit)|slug]]-foot per second (slug⋅ft/s).
 ==Mathematical derivation in the case of an object of constant mass==