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To say the momentum of a BODY is constant is to say "at the initial instant of observation" the BODY had a certain velocity which it continued to have thereafter. Since the mass of a BODY is constant, its momentum, mV, is constant provided its velocity is constant. For constant, non-zero velocity, the event of the BODY is said to be "uniform motion." The special (sometimes called "trivial") case has velocity equal to zero and is called "rest."
The trivial case of constant momentum is zero initial momentum meaning zero initial velocity. The non-trivial case corresponds to the body having a non-zero initial velocity. Below the momentum equation with conditions and a schematic with full notation is presented. Then the momentum equation is solved step by step.
Newton stated this case in the words:
A BODY in a state of motion continues in a straight line at
constant speed unless acted upon by a non-zero sum of forces."
Straight-line constant speed motion is "Constant Momentum Motion." The differential equation that describes such events of a BODY is written below center. The consequence attendant to the motion, zero sum of forces, is expressed below right.
A minor complication is that zero velocity is a case of constant momentum of a body. Inspect the equations then read below.
Initial Conditions (pairs) |
Constant Momentum (differential equation) |
Constant Momentum (sum of forces) |
i) P(t = 0+) = P_{o} with either ii) V(t = 0+) = 0 or iii) V(t = 0+) = V_{o} |
||
The time domain extends from the initial time of observation (t = 0+) until any time thereafter, t |
Solutions:
Regarding Forces (table right-column): By the table entry the set of forces acting on the BODY sums to zero. The expression above right is solved . (A small note here: this equation is written with a "0" left of equality. This is not an "algebraic" zero. It is a physical zero; it means there is a system and its momentum is constant therefore the derivative if its momentum is zero. Also, "0" has units - what are they?)
Regarding Constant Momentum (table above-center):
CASE II: The solution of this mathematical model is important because very many physical events can be expressed mathematically by a single first order equation (and other physical events are expressed as a sets of first-order equations). Solution steps are classical. But first confirm the ideas, space, time frame initial condition (sufficiency ) and all (above) are correct. These ideas are needed to start:
Step 1: SEPARATE VARIABLES: As written, the left side of the differential equation has two variables. The differential of momentum, d[mV(t)] has the dependent variable, V(t). The independent variable is time, "t." When the equation is multiplied by "dt" the variables are separated.
(1) 1 |
Step 2: INTEGRATE the differential equation.
i) First apply integration symbols to both sides of the separated equation:
(2) 2 |
ii) Next it must be decided what the limits of integration should be. When dealing with equations, a good approach is to search among all terms to find the easiest. Deal with that term first. Here, the "right-side" integral is easier. The limits are time (upper limit) and the starting time (lower limit) is the definite time, zero-positive. The upper limit is generally either an ending time or as in this case just "t" meaning all times after start. Specify time limits (right side integral) first then set the limits of the (left-side) differential to correspond time-wise. The right side integration is indefinite (has a variable for one limit - upper), hence so also must be the left side integral. At time 0+, the momentum is mV_{o}. Later, at indefinite time, the momentum is indefinite: mV(t). Write the limits with their integrals:
(3) 3 |
iii) Perform the integrations (often symbolic) and do a little algebra:
(4) 4 |
Step 3: INTERPRET RESULTS: The above result states that the velocity was V_{o} at t = 0+ and it remained constant at V_{o} for all times thereafter.
One Step Further: This result can be taken one step further. The steps are the same as above. We just write the resulting equations:
(5) 5 |
The last statement above says the position of the body is a distance from where it was originally, the distance being the initial speed times the time and the direction (from the initial position) being in the direction of the initial velocity.
Regarding Constant Momentum (table above-left):
CASE I: The initial conditions of this case state that velocity is zero. Consequently the solution of Case I equals the solution of Case II with velocity set to zero. Specifically, set V equal to zero in the last equation of the above solution. In words, the solution says: "The Body was at position P_{o} initially and it stayed there for all times thereafter.