Notes to a video lecture on http://www.unizor.com
Impulse
In a simple case of a single constant force F acting in the direction of motion on an object of mass m that moves with constant acceleration a the Newton's Second Law states that
F = m·a
Let's assume that at moment of time t the speed of this object was v and at moment of time t+Δt the speed of this object was v+Δv.
Since an objects moved with constant acceleration a, the average increment of speed during this time should be equal to the acceleration multiplied by increment of time
(v+Δv) − v = a·[(t+Δt) − t] or
Δv = a·Δt
Multiplying both sides of the above equation by mass m, we obtain
m·Δv = m·a·Δt or
m·Δv = F·Δt
Considering m is constant, we can make the following manipulations with the left side of this equation:
m·Δv = m·[v(t+Δt)−v(t)] =
= m·v(t+Δt)−m·v(t) = Δ(m·v)
Therefore,
Δ(m·v) = F·Δt
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 F acting on an object during interval of time Δt.
If the time interval Δt is infinitesimal, this can be applied to variable force F(t) and written in terms of differentials
d(m·v) = F(t)·dt
Integrating this equality by time t, we obtain on the left side the total momentum change during time from moment of time t=t1 to moment t=t2:
P = ∫t1t2d(m·v) =
= m·v(t2) − m·v(t1)
The total impulse exhorted by a variable force F(t) during this time from t1 to t2 is
J = ∫t1t2F(t)dt
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 F this is equivalent to
m·v(t2) − m·v(t1) = F·(t2−t1)
which fully corresponds to the Newton's Second Law for constant force F and, therefore, constant acceleration a since
[v(t2)−v(t1)]/(t2−t1) = 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 F in three-dimensional space the definition of impulse of this force acting during time t is:
For variable force F(t) acting on object from moment of time t=t1 to moment of time t=t2 we define the impulse in the integral form
J = ∫t1t2F(t)dt
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 m, staying at rest in some inertial reference frame, is pushed forward with constant force F1 during time t1, then with force F2 in the same direction during time t2.
What would be its final speed vfin?
The impulse given by the first force is F1·t1. It caused an increase in speed from 0 to v1.
Therefore,
F1·t1 = m·(v1−0)
Then the impulse given by the second force is F2·t2. It caused an increase in speed from v1 to vfin (the final speed).
Therefore,
F2·t2 = m·(vfin−v1)
Adding them together, we obtain
F1·t1 + F2·t2 =
= m·v1 + m·(vfin−v1) =
= m·(vfin−0) = m·vfin
which gives the final speed
vfin = [F1·t1 + F2·t2] /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 vfin by adding impulses
SUMMARY
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
Burned gases are exhausted with speed vout. An airplane is in uniform motion along a straight line with constant velocity.
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 Δt the engines exhaust burned gases of mass Δm=μ·Δt, accelerating them from speed zero to vout and correspondingly increasing their momentum.
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:
Δm·vout = F··Δt
Using the expression for Δm above, we derive from this
μ·Δt·vout = F··Δt
And the value for the force produced by an engine applied to burned gases pushing them backward
F = μ·vout
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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