Thursday, May 17, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Momentum of Motion

Notes to a video lecture on

Momentum of Motion

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 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

Δv = m·a·Δt or

Δv = F·Δt or

v/Δt) = F

If the time interval Δt is infinitesimal, this can be written in terms of derivative of speed v by time t as

(dv/dt) = F

or, since mass m is constant,

d(m·v)/dt = F

So, the force F equals to a derivative by time of the value of a product of mass by speed m·v, called momentum of motion.

Since we have reduced the time interval Δt to
infinitesimal value, it's no longer important that the force and
acceleration are constant. In the next moment of time it can be
different, but at any moment of time the dependency between the force
and a derivative of the momentum above will take place.

Let's consider a more complicated case of motion in the three-dimensional space.

The Newton's Second Law states that the vector of force F(t), as a function of time t, equals to a product of constant mass m and the vector of acceleration a(t), which, in turn, also depends on time t:

F(t) = m·a(t)

An immediate consequence from this, considering the vector of acceleration is a derivative of vector of velocity v(t), is the following chain of equalities:

F(t) = m·a(t) = m·dv(t)/dt =


from which we conclude that the force F(t), acting on an object, is a derivative by time of the characteristic of motion called momentum, that is equal to v(t):

F(t) = [v(t)]I

The same equality can be expressed in term of infinitesimal increments:

F(t)·dt = d(m·v(t))

The left side of this equation is an increment in impulse of the force F(t) during the time from t to t+dt, while the right side represents an increment of the momentum of object of mass m moving with velocity v(t) during the same time interval.

Let's return to a simple case of a constant force acting in the direction of movement with an equation that connects the force with a derivative of momentum:

d(m·v)/dt = F

The first consequence from this equation is that in the absence of force (F=0)
the derivative of momentum is zero and, therefore, momentum is
constant. With constant mass it is equivalent to the Principle of
Inertia because it implies the constant velocity.

But, expressed as a constant momentum, it covers a more general case of complex systems of objects with no external forces.

Assume that a two-stage rocket of mass m flies in open space far from any gravity fields along a straight line with speed v
with no engine working. According to the Principle of Inertia (and the
Newton's First Law), it will move like this indefinitely.

Then at some moment the first stage of mass m1 is separated from the second stage of mass m2=m−m1. This separation is achieved by pushing the first stage back along the same straight line with a force F during time t.

The resulting speed of the first stage is

v1 = v − (F/m1)·t

The resulting momentum of the first stage is

m1·v1 = m1·v − F·t

Meanwhile, according to the Newton's Third Law, the same by magnitude
force should push the second stage forward during the same time. Its new
speed is

v2 = v + (F/m2)·t

The momentum of the second stage is

m2·v2 = m2·v + F·t

Let's add the momentums of two stages after the separation:

m1·v1 + m2·v2 =

= m1·v − F·t + m2·v + F·t =

= (m1 + m2)·v = m·v

As we see, even in case of a complex system of objects with no external forces, the momentum of an entire system is constant.

Our example with a rocket can be generalized into a system of any number
of objects and not only straight line motion. We will address this
issue further in the course as the Law of Conservation of Momentum.

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