*Notes to a video lecture on http://www.unizor.com*

__Impulse__

In a simple case of a single constant force

**acting in the direction of motion on an object of mass**

*F***that moves with constant acceleration**

*m***the Newton's Second Law states that**

*a*

*F = m·a*Let's assume that at moment of time

**the speed of this object was**

*t***and at moment of time**

*v***Δ**

*t+***the speed of this object was**

*t***Δ**

*v+***.**

*v*Since an objects moved with constant acceleration

**, the average increment of speed during this time should be equal to the acceleration multiplied by increment of time**

*a***Δ**

*(v+***[**

*v) − v = a·***Δ**

*(t+***] or**

*t) − t*Δ

**Δ**

*v = a·*

*t*Multiplying both sides of the above equation by mass

**, we obtain**

*m***Δ**

*m·***Δ**

*v = m·a·***or**

*t***Δ**

*m·***Δ**

*v = F·*

*t*Considering

**is constant, we can make the following manipulations with the left side of this equation:**

*m***Δ**

*m·***[**

*v = m·***Δ**

*v(t+***]**

*t)−v(t)***Δ**

*=*

= m·v(t+= m·v(t+

**Δ**

*t)−m·v(t) =*

*(m·v)*Therefore,

Δ

**Δ**

*(m·v) = F·*

*t*The expression on the left side represents an increment of the

*momentum of motion*. The expression on the right side is called

**impulse**of the force

**acting on an object during interval of time Δ**

*F***.**

*t*If the time interval Δ

**is**

*t**infinitesimal*, this can be applied to variable force

**and written in terms of differentials**

*F(t)**d*(

**)**

*m·v*

*= F(t)**·d*

**t**Integrating this equality by time

**, we obtain on the left side the total momentum change during time from moment of time**

*t**to moment*

**t=t**_{1}*:*

**t=t**_{2}

*P = ∫*_{t1}^{t2}*d*(

**)**

*m·v***(**

*=*

= m·v= m·v

**)**

*t*_{2}**(**

*− m·v***)**

*t*_{1}The total impulse exhorted by a variable force

**during this time from**

*F(t)**to*

**t**_{1}*is*

**t**_{2}

*J = ∫*_{t1}^{t2}F(t)*d*

**t**Hence, our qualitative observation, that the force acting on an object

changes its velocity, can be quantitatively characterized as

**impulse exhorted by a force during certain time period equals to change of momentum during this period**.

Obviously, for a simple case of constant force

**this is equivalent to**

*F***(**

*m·v***)**

*t*_{2}**(**

*− m·v***)**

*t*_{1}**(**

*= F·***)**

*t*_{2}−t_{1}which fully corresponds to the Newton's Second Law for constant force

*and, therefore, constant acceleration*

**F***since*

**a**[

**(**

*v***)**

*t*_{2}**(**

*−v***)]**

*t*_{1}**) =**

*/(t*_{2}−t_{1}

*a = F/m*Generally speaking,

*force*is a vector. So is

*velocity*and, therefore,

*momentum*of motion. All the definitions discussed above, considering forces and motions in our three-dimensional space, are for vectors.

Hence, for constant force

*in three-dimensional space the definition of*

**F***impulse*of this force acting during time

*is:*

**t***.*

**J = F·t**For variable force

*acting on object from moment of time*

**F(t)***to moment of time*

**t=t**_{1}*we define the*

**t=t**_{2}*impulse*in the integral form

*J = ∫*_{t1}^{t2}F(t)*d*

**t**where integral of vector is a vector of integrals of its X-, Y- and Z-components.

Using the approach based on the concept of

*impulse*of the force and

*momentum*of motion, we can simplify solutions to some problems.

Consider a case when an object of mass

**, staying at rest in some inertial reference frame, is pushed forward with constant force**

*m***during time**

*F*_{1}**, then with force**

*t*_{1}**in the same direction during time**

*F*_{2}**.**

*t*_{2}What would be its final speed

*?*

**v**_{fin}The impulse given by the first force is

**. It caused an increase in speed from**

*F*_{1}·t_{1}**to**

*0***.**

*v*_{1}Therefore,

**(**

*F*_{1}·t_{1}= m·**)**

*v*_{1}−0Then the impulse given by the second force is

**. It caused an increase in speed from**

*F*_{2}·t_{2}**to**

*v*_{1}*(the final speed).*

**v**_{fin}Therefore,

**(**

*F*_{2}·t_{2}= m·*)*

**v**_{fin}**−v**_{1}Adding them together, we obtain

**(**

*F*

= m·v_{1}·t_{1}+ F_{2}·t_{2}== m·v

_{1}+ m·*)*

**v**_{fin}**−v**_{1}**(**

*=*

= m·= m·

*)*

**v**_{fin}−**0**

*= m·v*

_{fin}which gives the final speed

*v**[*

_{fin}**=***]*

**F**_{1}·t_{1}+ F_{2}·t_{2}

**/m**But the easiest way to solve this is to use combined impulse given to an

object as the cause of increased speed from zero to its final value

*by adding*

**v**_{fin}*impulses*

*F*_{1}·t_{1}+ F_{2}·t_{2}*momentum*

*m·v**with the same result for final speed.*

_{fin}**SUMMARY**

Any action of force on an object during certain time interval exhorts an

Each consecutively or simultaneously applied impulse contributes to this

change of momentum, so the final momentum of an object is the combined

effect of all impulses acting on it in an integrated fashion.

Any action of force on an object during certain time interval exhorts an

*impulse*that causes to change the object's*momentum*.Each consecutively or simultaneously applied impulse contributes to this

change of momentum, so the final momentum of an object is the combined

effect of all impulses acting on it in an integrated fashion

*Example*

Liquid fuel is pumped into combustion chambers of airplane engines at the rate

*(kg/sec).*

**μ**Burned gases are exhausted with speed

*. An airplane is in uniform motion along a straight line with constant velocity.*

**v**_{out}What is the force of air resistance acting against its motion?

Assume that the mass of an airplane is significantly greater than the

mass of exhausted gases, so we can ignore the loss of mass during

engine's work.

*Solution*

Airplane is in uniform motion, which allows us to use a reference frame

associated with it as the inertial frame, where the Newton's Laws are

held and all calculations can be done.

Since an airplane's motion is uniform, the sum of the vector of air

resistance force, directed against its movement, and the vector of the

reaction force from the gases, exhausted by its engines and directed

towards its movement, should balance each other and be equal in

magnitude since their directions are opposite.

During time interval Δ

*the engines exhaust burned gases of mass Δ*

**t***=*

**m***·Δ*

**μ***, accelerating them from speed zero to*

**t***and correspondingly increasing their momentum.*

**v**_{out}Therefore, from the equality between the increment of

*momentum of motion*of burned gases and the

*impulse of force*applied to them by the engines, we have the following equality:

Δ

*·Δ*

**m·v**_{out}**= F·**

**t**Using the expression for Δ

*above, we derive from this*

**m***Δ*

**μ·***·Δ*

**t·v**_{out}**= F·**

**t**And the value for the force produced by an engine applied to burned gases pushing them backward

**F = μ·v**_{out}According to Newton's Third Law, this is the same force, with which burned gases push an airplane forward.

Since the airplane movement is uniform, the air resistance is also equal in magnitude to this value.

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