Impulse (physics)

In classical mechanics, an impulse 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.

Mathematical derivation

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

t 2

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

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

Therefore

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

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:

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

where

F is the constant total net force applied,

Δt is the time interval over which the force is applied,

m is the constant mass of the object,

Δv is the change in velocity produced by the force in the considered time interval, and

m Δv = Δ(mv) is the change in linear momentum.

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

$\mathbf{F}\,\Delta t = \Delta p = mv_1 - mv_0 \,$

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.

Specific impulse in seconds

General definition

For all vehicles specific impulse (impulse per unit weight-on-Earth of propellant) in seconds can be defined by the following equation^[7]:

$\mathrm{F_{\rm thrust}}=I_{\rm sp} \cdot \frac{\Delta m} {\Delta t} \cdot g_{\rm 0} \,$

where:

F thrust

is the thrust obtained from the engine, in newtons (or poundals).

I sp

is the specific impulse measured in seconds.

$\frac {\Delta m} {\Delta t}$ is the mass flow rate in kg/s (lb/s), which is negative the time-rate of change of the vehicle's mass since propellant is being expelled.

g 0

is the acceleration at the Earth's surface, in m/s² (or ft/s²).

(When working with English units, it is conventional to divide both sides of the equation by g₀ so that the left hand side of the equation has units of lbs rather than expressing it in poundals.)
This I_sp in seconds value is somewhat physically meaningful—if an engine's thrust could be adjusted to equal the initial weight of its propellant (measured at one standard gravity), then I_sp is the duration the propellant would last.
The advantage that this formulation has is that it may be used for rockets, where all the reaction mass is carried onboard, as well as aeroplanes, where most of the reaction mass is taken from the atmosphere. In addition, it gives a result that is independent of units used (provided the unit of time used is the second).

The specific impulse of various hydrocarbon fuelled jet engines

[edit] Rocketry

In rocketry, where the only reaction mass is the propellant, an equivalent way of calculating the specific impulse in seconds is also frequently used. In this sense, specific impulse is defined as the change in momentum per unit weight-on-Earth of the propellant:

$I_{\rm sp}=\frac{v_{\rm e}}{g_{\rm 0}}$

where
I_sp is the specific impulse measured in seconds

v e

is the average exhaust speed along the axis of the engine in (ft/s or m/s)
g₀ is the acceleration at the Earth's surface (in ft/s² or m/s²)
In rockets, due to atmospheric effects, the specific impulse varies with altitude, reaching a maximum in a vacuum. It is therefore most common to see the specific impulse quoted for the vehicle in a vacuum; the lower sea level values are usually indicated in some way (e.g. 'sl').

[edit] Specific impulse as a speed (effective exhaust velocity)

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Because of the geocentric factor of g₀ in the equation for specific impulse, many prefer to define the specific impulse of a rocket (in particular) in terms of thrust per unit mass flow of propellant (instead of per unit weight flow). This is an equally valid (and in some ways somewhat simpler) way of defining the effectiveness of a rocket propellant. For a rocket, the specific impulse defined in this way is simply the effective exhaust velocity relative to the rocket, v_e. The two definitions of specific impulse are proportional to one another, and related to each other by:

$v_{\rm e} = g_0 I_{\rm sp} \,$

where

$I_{\rm sp} \,$ - is the specific impulse in seconds

$v_{\rm e} \,$ - is the specific impulse measured in m/s, which is the same as the effective exhaust velocity measured in m/s (or ft/s if g is in ft/s²)

$g_0 \,$ - is the acceleration due to gravity at the Earth's surface, 9.81 m/s² (in English units 32.2 ft/s²).

This equation is also valid for airbreathing jet engines, but is rarely used in practice.
(Note that different symbols are sometimes used; for example, c is also sometimes seen for exhaust velocity. While the symbol

I sp

might logically be used for specific impulse in units of N•s/kg, to avoid confusion it is desirable to reserve this for specific impulse measured in seconds.)
It is related to the thrust, or forward force on the rocket by the equation:

$\mathrm{F_{\rm thrust}}=v_{\rm e} \cdot \frac {\Delta m} {\Delta t} \,$

where
$\frac {\Delta m} {\Delta t}$ is the propellant mass flow rate, which is the rate of decrease of the vehicle's mass
A rocket must carry all its fuel with it, so the mass of the unburned fuel must be accelerated along with the rocket itself. Minimizing the mass of fuel required to achieve a given push is crucial to building effective rockets. Using Newton's laws of motion it is not difficult to verify that for a fixed mass of fuel, the total change in velocity (in fact, momentum) it can accomplish can only be increased by increasing the effective exhaust velocity.
A spacecraft without propulsion follows an orbit determined by the gravitational field. Deviations from the corresponding velocity pattern (these are called Δv) are achieved by sending exhaust mass in the direction opposite to that of the desired velocity change.

[edit] Actual exhaust speed versus effective exhaust speed

Note that effective exhaust velocity and actual exhaust velocity can be significantly different, for example when a rocket is run within the atmosphere, atmospheric pressure on the outside of the engine causes a retarding force that reduces the specific impulse and the effective exhaust velocity goes down, whereas the actual exhaust velocity is largely unaffected. Also, sometimes rocket engines have a separate nozzle for the turbopump turbine gas, and then calculating the effective exhaust velocity requires averaging the two mass flows as well as accounting for any atmospheric pressure.
For airbreathing jet engines, particularly, turbofans, the actual exhaust velocity and the effective exhaust velocity are different by orders of magnitude. This is because a good deal of additional momentum is obtained by using air as reaction mass. This allows for a better match between the airspeed and the exhaust speed which saves energy/propellant and enormously increases the effective exhaust velocity while reducing the actual exhaust velocity.

Momentum

Mōmentum was not merely the motion, which was mōtus, but was the power residing in a moving object, captured by today's mathematical definitions. A mōtus, "movement", was a stage in any sort of change,^[1] while velocitas, "swiftness", captured only speed. The concept of momentum in classical mechanics was originated by a number of great thinkers and experimentalists. The first of these was Byzantine philosopher John Philoponus, in his commentary to Aristotle´s Physics. As regards the natural motion of bodies falling through a medium, Aristotle's verdict that the speed is proportional to the weight of the moving bodies and indirectly proportional to the density of the medium is disproved by Philoponus through appeal to the same kind of experiment that Galileo was to carry out centuries later.^[2] This idea was refined by the European philosophers Peter Olivi and Jean Buridan. Buridan referred to impetus being proportional to the weight times the speed.^[3][4] Moreover, Buridan´s theory was different to his predecessor´s in that he did not consider impetus to be self dissipating, asserting that a body would be arrested by the forces of air resistance and gravity which might be opposing its impetus.^[5]
René Descartes believed that the total "quantity of motion" in the universe is conserved, where the quantity of motion is understood as the product of size and speed. This should not be read as a statement of the modern law of momentum, since he had no concept of mass as distinct from weight and size, and more importantly he believed that it is speed rather than velocity that is conserved. So for Descartes if a moving object were to bounce off a surface, changing its direction but not its speed, there would be no change in its quantity of motion.^[6] Galileo, later, in his Two New Sciences, used the Italian word "impeto."
The question has been much debated as to what Isaac Newton contributed to the concept. The answer is apparently nothing, except to state more fully and with better mathematics what was already known. Yet for scientists, this was the death knell for Aristotelian physics and supported other progressive scientific theories (i.e., Kepler's laws of planetary motion). Conceptually, the first and second of Newton's Laws of Motion had already been stated by John Wallis in his 1670 work, Mechanica sive De Motu, Tractatus Geometricus: "the initial state of the body, either of rest or of motion, will persist" and "If the force is greater than the resistance, motion will result".^[7] Wallis uses momentum and vis for force. Newton's Philosophiæ Naturalis Principia Mathematica, when it was first published in 1687, showed a similar casting around for words to use for the mathematical momentum. His Definition II^[8] defines quantitas motus, "quantity of motion", as "arising from the velocity and quantity of matter conjointly", which identifies it as momentum.^[9] Thus when in Law II he refers to mutatio motus, "change of motion", being proportional to the force impressed, he is generally taken to mean momentum and not motion.^[10] It remained only to assign a standard term to the quantity of motion. The first use of "momentum" in its proper mathematical sense is not clear but by the time of Jenning's Miscellanea in 1721, four years before the final edition of Newton's Principia Mathematica, momentum M or "quantity of motion" was being defined for students as "a rectangle", the product of Q and V, where Q is "quantity of material" and V is "velocity", s/t.^[11]
Some languages, such as French still lack a single term for momentum, and use a phrase such as the literal translation of "quantity of motion".

Linear momentum of a particle

If an object is moving in any reference frame, then it has momentum in that frame. It is important to note that momentum is frame dependent. That is, the same object may have a certain momentum in one frame of reference, but a different amount in another frame. For example, a moving object has momentum in a reference frame fixed to a spot on the ground, while at the same time having 0 momentum in a reference frame attached to the object's center of mass.

The amount of momentum that an object has depends on two physical quantities: the mass and the velocity of the moving object in the frame of reference. In physics, the usual symbol for momentum is a bold p (bold because it is a vector); so this can be written

$\mathbf{p}= m \mathbf{v}\,,$

where p is the momentum, m is the mass and v is the velocity.

Example: a model airplane of 1 kg traveling due north at 1 m/s in straight and level flight has a momentum of 1 kg m/s due north measured from the ground. To the dummy pilot in the cockpit it has a velocity and momentum of zero.

According to Newton's second law, the rate of change of the momentum of a particle is proportional to the resultant force acting on the particle and is in the direction of that force. The derivation of force from momentum is given below, however because mass is constant the second term of the derivative is 0 so it is ignored.

$\sum{\mathbf{F}} = {\mathrm{d}\mathbf{p} \over \mathrm{d}t} = m{\mathrm{d}\mathbf{v} \over \mathrm{d}t} + v{\mathrm{d}\mathbf{m} \over \mathrm{d}t} = m\mathbf{a}\,$

(if mass is constant)

or just simply

$\mathbf{F}= m \mathbf{a}\,,$

where F is understood to be the resultant.

Example: a model airplane of 1 kg accelerates from rest to a velocity of 1 m/s due north in 1 s. The thrust required to produce this acceleration is 1 newton. The change in momentum is 1 kg m/s. To the dummy pilot in the cockpit there is no change of momentum. Its pressing backward in the seat is a reaction to the unbalanced thrust, shortly to be balanced by the drag.

Linear momentum of a system of particles

Relating to mass and velocity

The linear momentum of a system of particles is the vector sum of the momenta of all the individual objects in the system:

$\sum{\mathbf{F}} = {\mathrm{d}\mathbf{P} \over \mathrm{d}t}= M \frac{\mathrm{d}\mathbf{v}_{cm}}{\mathrm{d}t}=M\mathbf{a}_{cm}\,.$

This is a special case of Newton's second law. (If mass is constant)

For a more general derivation using tensors, we consider a moving body (see Figure), assumed as a continuum, occupying a volume $V\,$ at a time $t\,$ , having a surface area $S\,$ , with defined traction or surface forces per unit area represented by the stress vector $T_i^{(n)}\,$ acting on every point of every body surface (external and internal), body forces $F_i\,$ per unit of volume on every point within the volume $V\,$ , and a velocity field $v_i\,$ prescribed throughout the body. Following the previous equation, the linear momentum of the system is:

$\int_S T_i^{(n)}dS + \int_V F_i dV = \frac{d}{dt}\int_V \rho\, v_i\, dV\,.$

By definition the stress vector is defined as $T_i^{(n)} \equiv \sigma_{ij}n_j\,$ , then

$\int_S \sigma_{ij}n_j\, dS + \int_V F_i\, dV = \frac{d}{dt}\int_V \rho\, v_i\, dV\,.$

Using the Gauss's divergence theorem to convert a surface integral to a volume integral gives (we denote $\partial_j \equiv \frac{\partial}{\partial x_j}\,$ as the differential operator):

$\int_V \partial_j\sigma_{ij}\, dV + \int_V F_i\, dV = \frac{d}{dt}\int_V \rho\,v_i\, dV\,.$

Now we only need to take care of the right side of the equation. We have to be careful, since we cannot just take the differential operator under the integral. This is because while the motion of the continuum body is taking place (the body is not necessarily solid), the volume we are integrating on can change with time too. So the above integral will be:

$\frac{d}{dt}\int \rho\,v_i\, dV=\int \frac{\partial (\rho v_i)}{\partial t}\, dV +\oint \rho v_i v_k n_k dA\,.$

Performing the differentiation in the first part, and applying the divergence theorem on the second part we obtain:

$\frac{d}{dt}\int \rho\,v_i\, dV =\int \left[ \left(\rho\frac{\partial v_i}{\partial t}+v_i\frac{\partial \rho}{\partial t}\right)+\partial_k (\rho v_i v_k)\right]\, dV\,.$

Now the second term inside the integral is: $\partial_k (\rho v_i v_k)=\rho v_k \cdot \partial_k v_i +v_i\partial_k(\rho v_k)\,.$ Plugging this into the previous equation, and rearranging the terms, we get:

$\frac{d}{dt}\int \rho\,v_i\, dV=\int\rho\left[\frac{\partial}{\partial t}+v_k\partial_k\right]v_i\,dV +\int\left[\frac{\partial\rho}{\partial t}+\partial_k(\rho v_k)\right]v_i\,dV\,.$

We can easily recognize the two integral terms in the above equation. The first integral contains the Convective derivative of the velocity vector, and the second integral contains the change and flow of mass in time. Now lets assume that there are no sinks and sources in the system, that is mass is conserved, so this term is zero. Hence we obtain:

$\frac{d}{dt}\int \rho\,v_i\, dV=\int \rho\,\frac{Dv_i}{Dt}\, dV\,$

putting this back into the original equation:

$\int_V \left[ \partial_j\sigma_{ij} + F_i - \rho \frac{D v_i}{Dt}\right]\, dV = 0\,.$

For an arbitrary volume the integrand itself must be zero, and we have the Cauchy's equation of motion

$\partial_j\sigma_{ij} + F_i = \rho \frac{D v_i}{Dt}\,.$

As we see the only extra assumption we made is that the system doesn't contain any mass sources or sinks, which means that mass is conserved. So this equation is valid for the motion of any continuum, even for that of fluids. If we are examining elastic continuums only then the second term of the convective derivative operator can be neglected, and we are left with the usual time derivative, of the velocity field.

If a system is in equilibrium, the change in momentum with respect to time is equal to 0, as there is no acceleration

$\sum{\mathbf{F}} = {\mathrm{d}\mathbf{P} \over \mathrm{d}t}=\ M\mathbf{a}_{cm}= 0\,$

or using tensors,

$\partial_j\sigma_{ij} + F_i = 0\,.$

These are the equilibrium equations which are used in solid mechanics for solving problems of linear elasticity. In engineering notation, the equilibrium equations are expressed in Cartesian coordinates as

$\frac{\partial \sigma_x}{\partial x} + \frac{\partial \tau_{yx}}{\partial y} + \frac{\partial \tau_{zx}}{\partial z} + F_x = 0\,$

$\frac{\partial \tau_{xy}}{\partial x} + \frac{\partial \sigma_y}{\partial y} + \frac{\partial \tau_{zy}}{\partial z} + F_y = 0\,$

$\frac{\partial \tau_{xz}}{\partial x} + \frac{\partial \tau_{yz}}{\partial y} + \frac{\partial \sigma_z}{\partial z} + F_z = 0\,.$

SOAL FISIKA IMPLUS

1. Hitunglah besar momentum serangga yang massanya 22 gram yang tengah terbang

dengan laju 80 m/s.

2. Sebuah bola tenis bermassa 60 gram dipukul hingga mencapai kecepatan 144 km/jam,

Hitunglah impuls bola tenis tersebut.

3. Bola kasti bermassa 145 gram dilempar dengan kecepatan 39 m/s ternyata dapat dipukul

balik hingga mencapai kecepatan 52 m/s. Hitunglah impuls yang terjadi pada bola kasti.

4. Gerbong bermassa 6500 kg bergerak dengan kecepatan 5 m/s menumbuk gerbong yang
massanya sama yang sedang diam, setelah tubukan kedua-duanya bergerak bersama-sama.
Hitunglah kecepatan kedua gerbong itu setelah tumbukan.

5. Peluru bermassa 20 gram ditembakkan dengan kecepatan 230 m/s mengenai balok yang
diam di atas lantai. Massa balok 2 kg. Ternyata peluru melewati bagian dalam balokdan
setelah keluar dari balok kecepatan peluru berkurang menjadi 160 m/s. Hitunglah
kecepatan balok setelah ditembus peluru demikian.

6. Sebuah pasak panjangnya 40 cm menancap bagian depannya di atas tanah ditumbuk
dengan martil yang dilemparkan jatuh bebas dari ketinggian 5 meter. Massa martil 10 kg.
Jika gaya tahan tanah terhadap martil 100 KN(1KN=1000 N). Perhitungkan bagaimana
pasak bisa masuk ke dalam tanah.

1. Diketahui:

massa(m) = 22 gram = 0,022 kg
kecepatan(v) = 80 m/s
momentum(p) = ...
p= m.v

= 0,022.80

= 1,76 kg m/s

2. Diketahui:

massa = 60 gram = 0,060 kg

kecepatan(v) = 144 km/jam = 40 m/s

http://htmlimg4.scribdassets.com/9ubuq2g9x955hq8/images/2-c9f1cb5be8/000.jpg

http://html.scribd.com/9ubuq2g9x955hq8/images/2-c9f1cb5be8/000.jpg

impuls(I) = ...
I=perubahan momentum(p)
p= m.v

= 0,060.40

= 2,4 kg m/s

I= 2,4 Ns

3. Diketahui:

massa = 145 gram = 0,145 kg

kecepatan(v) = 39 m/s

v’ = 52 m/s

I = ...
I= perubahan momentum(p)
p1 = 0, 145.39

= 5,655 m/s

p2 = 0,145.52

= 7,54 m/s

I = p2-p1
= 7,54-5,655
= 1,855 Ns

4. Diketahui:

m1 = 6500 kg
v1 = 5 m/s
m2 = 6500 kg
v1 = 0 m/s(diam)
kecepatan setelah tumbukan(v’) = ...

m1v1+ m2v2

v’=

m1+m2

6500.5 + 6500.0

6500 + 6500

=32500

13000

v’= 2,5 m/s

5. Diketahui:

m1 = 20 gram = 0,02 kg

v1 = 230 m/s v1’ = 160 m/s m2 = 2 kg

v2 = 0 m/s(diam)

v2’ = ...

m1.v1 + m2.v2 = m1.v1’ + m2.v2’
v2 = 0, maka m1.v1 = m1.v1’ + m2.v2’
0,02 . 230 = 0,02 . 160 + 2.v2’
4,6 = 3,2 + 2.v2’
4,6 – 3,2 = 2.v2’
1,4 = 2.v2’

1,4=0,7

v2’ = 0,7 m/s

6. Diketahui:

panjang(l) = 40 cm = 0,4 m

m1 = 10 kg

h =5 m

Vm(kecepatan martil)=...

Ftanah = 100 KN = 100.000 N

Vm= √2.g.h
= √2.10.5
= √100

Vm= 10 m/s

P = m.v

= 10.10
P = 100 Ns
P(momentum) = I(impuls)
100 = F.t
100 = 10000.t

100

t = 10000 = 0,01 s

irwan maolana

Selasa, 04 Januari 2011

tugas fisika

Mathematical derivation

Specific impulse in seconds

General definition

[edit] Rocketry

[edit] Specific impulse as a speed (effective exhaust velocity)

[edit] Actual exhaust speed versus effective exhaust speed

Momentum

Linear momentum of a particle

Linear momentum of a system of particles

Relating to mass and velocity