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Explain Conservation Of Momentum With Examples

Referencing Styles : APA | Pages : 1

The law of conservation of momentum is explained in the following example. When it comes to any collision occurring among two objects – object 1 and object 2 in a completely isolated environment, the net momentum of these two objects prior to their collision will are to equal the overall momentum of these two objects after their collision. Conservation of momentum, thus becomes an important chapter in physics in line with law of conservation of energy as also the law of conservation of mass. Momentum gets defined as the product of an object’s mass with the associated velocity of that same object. What the law of conservation for momentum states is that among some problem domain, the magnitude of momentum stays constant that is to say momentum can neither be created nor be destroyed and can only be changed by means of forces acting on the body according to the laws of motion given by Isaac Newton. To Deal with momentum becomes much more complex than getting to deal with mass or energy since momentum happens to be a vector quantity containing both magnitude as well as direction. Momentum can be conserved across all the three physical directions simultaneously. It gets even harder to deal with gases as forces from one direction can cause the momentum be generated in another direction due to collisions of all the molecules involved. Here a very, very simplified flow problem will be presented where properties tend to change in a single direction. This problem can be more simplified if a steady flow is considered that do not change time to time as also by placing a limit for forces to act only on those associated with pressure. It should be noted that actual flow problems can be more complex such a simple example.

Let the flow of the gas be considered to be through a domain where properties of the can change in single direction only, which can be called ‘x’. This gas first enters the domain at station 1 with velocity u having pressure p and exits through station 2 having differing values of pressure and velocity. Simply put, it is to be assumed that density r stays constant in this domain as also the area A along which this gas is flowing. Locations of stations 1 and 2 remain separated along a distance ‘d’ of x or dx, where d or delta represents a varying length given by the Greek alphabet ‘d’. Change in distance is called gradient for avoiding possible confusions with changes in time called rate. Gradient of velocity is represented as du/dx, that is changes in velocity with every change of distance. Hence, in station 2, velocity can be written as velocity at 1 + the gradient * the distance.

Or u2 = u1 + (du / dx) * dx

Similarly, the following expression provides the pressure at exit:

p2 = p1 + (dp / dx) * dx

The second law of motion by Newton states that a force ‘F’ will always equal the change in momentum per every intervals of time. For objects having constant mass ‘m’ this gets reduced to the mass ‘m’ multiplied by acceleration ‘a’. Acceleration refers to the changes in velocity with changes of intervals of time that is du / dt. This can be given by the following equation:  

F = (m) * (a) = (m * (du / dt))

The force in the above problem gets derived from pressure gradient. As pressure is the force per a unit of area, the total force on the fluid domain becomes product of pressure with area towards the exit subtracted by the product of pressure and the area towards the entrance. The resultant equation is:

F = - [(((p) * (A))2) – (((p) * (A))1)] = (m) * [((u2) – (u1)) / dt]

This minus sign at the start of this equation has been used to denote gases move from regions of higher pressure to regions of lower pressure, if pressure level increases by x, the associated velocity will definitely decrease. Substitution of these equations for and pressure and velocity are the following:

- [{((p) + (dp / dx) * dx} * (A)) - ((p) * (A))] = (m) * [((u) + (du / dx) * dx – (u)) / dt]

Simplified:

- (dp / dx) * dx * A = m * (du / dx) * dx / dt

To note, that (dx / dt) is velocity and the mass is density r multiplied by the volume (that is product of area with dx):

- (dp / dx) * dx * A = r * dx * A * (du / dx) * u

Simplify:

- (dp / dx) = r * u * (du / dx)

The dp / dx and du / dx represents the gradients for pressure and velocity respectively. If it is shrunk to this domain down to differentials of sizes, such gradients can become the differentials:

- (dp / dx) = r * u * (du / dx)

This is an example of one-dimensional steady form of Euler's Equation. This is interesting for noting the corresponding pressure drops of the fluid where the term to the left is proportional to both value of velocity as well as gradient of velocity. Thus, Conservation of momentum becomes a fundamental law in physics that ends up stating that resultant momentum of the system is always constant as long as no external forces are acting on the objects present in the system. This got mentioned in first law of Newton which talks about law of inertia.

Supposing there exists two different interacting particles p1 and p2 with different masses. All forces among them are equal in magnitude and opposite in direction. As per second law of Newton, the force can be given by product of rate of changes in momentum, therefore it can be concluded that rate of changes in momentum p1 is equivalent to minus of rate of change in momentum for p2.

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