Chapter 9: Conservation of Momentum

Equations Introduced and Used in this Topic (all equations can be written and solved as both scalar and vector and all equations are generally solved as vectors):

In an Isolated System

[latex]\Sigma\overrightarrow{p} = constant[/latex]

or …         [latex]\underrightarrow{p}_{before} =\underrightarrow{p}_{after} (Vector Equation)[/latex]

[latex]Where ...\underrightarrow{p} =m\underrightarrow{v}[/latex]

For two or more bodies:

p_{1i (x, y, z)} + p_{2i (x, y, z)} + … = p_{1f (x, y, z)} + p_{2f (x, y, z)} + …

Image is of the electron-positron (antimatter) particles being created in the lab. (Image source: lolscience.co.uk)

The Law of the Conservation of Momentum ³ is one of the most powerful and useful laws in the study of mechanics. It belongs to a family of four conservation laws: Conservation of Mass Energy, Conservation of Electric Charge, Conservation of Linear Momentum and Conservation of Angular Momentum ⁴.

The law of conservation of momentum is derived mainly from Newton’s Third Law and can be analyzed as in the following description excerpted from the Physics Classroom:

In every interaction, there is a pair of forces acting on the two interacting objects. The size of the force on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs – equal and opposite action-reaction force pairs.

9.1 Total Momentum: the resolution of multiple momentum vectors

Total Momentum of a System: The Resolutions of multiple momentums acting during an event or interaction

Momentums are Vectors. They are added, subtracted and combined according to the direction they take. Quite often, vectors in different directions, such as North and South, East and West, and up and down, are left as the sum of the three different vector directions and not combined any further.

Vectors in the same directions are added and vectors in the opposite direction are subtracted.

For instance:

100 Ns East + 50 Ns East = 150 Ns East

(The vectors are in the same direction so they are added.)

100 Ns North + 50 Ns South = 50 Ns North

(The vectors are in the opposite direction so they are subtracted.)

Vectors in different directions are generally left as the vector sum in each of those directions.

For instance:

100 Ns North + 50 Ns East = 100 Ns North + 50 Ns East

(The vectors are in the different direction so they are left alone.)

Vectors that are angled between two directions are first broken up into the various components for those directions before they are added or subtracted.

To find the Momentum upwards use the Sine function:

Sin 42° = Momentum up

                    150 Ns

To find the Momentum right use the Cosine function:

Cos 42° = Momentum right

                       150 Ns

9.2 Linear Collisions ^{5, 6, 7}

In an Isolated System

[latex]\Sigma\overrightarrow{p}[/latex] = constant

or …

[latex]\underrightarrow{p}_{before} =\underrightarrow{p}_{after}[/latex](Vector Equation)

Where [latex]\underrightarrow{p} = m\underrightarrow{v}[/latex]

For two or more bodies:

[latex]\underrightarrow{p}_{li(x, y, z)} +\underrightarrow{p}_{2i(x, y, z)} + ... =\underrightarrow{p}_{1f(x, y, z)} +\underrightarrow{p}_{2f(x, y,z)} + ...[/latex]

The Law of Momentum Conservation is one of the foundational laws in Physics. The law is stated as follows:

In the absence of external forces that can change the momentum of objects inside a system from the outside, the vector sum of the momentum of all objects moving within the system will equal the vector sum of momentum of these same objects after an interaction has occurred, be it a collision or explosion.

This absence of external forces acting on the object means that the system being looked at is termed an isolated system^8. Isolated systems require that no external forces outside of the forces acting between the objects within a system can be acting. This means that the forces acting on the objects are in fact being balanced between the objects and not by outside means.

[latex]\overrightarrow{F}orce_{object1} = -\overrightarrow{F}orce_{object2}[/latex]

If two objects collide, the forces acting between the two objects are equal and opposite as defined by Newton’s Third Law.

9.4 Two Dimensional Collisions ^{9, 10, 11}

The Law of the Conservation of Momentum can be expanded from linear one-dimensional to the multi-dimensional universe of 2 or 3 dimensions.  This is due to the fact that momentum must be conserved in its original sum as long as no external forces act to change the momentum of the system.

The grid shown to the left is the common (x, y, z) representation of 3 dimensional space. All that is missing from this would be the time that identifies the spot where an object would be at a specific time.

It is also common to see vector momenta sums to be identified according to the directions defined by

    (x, y, z).

As such, you could see the momentum of an object defined by:

[latex]\overrightarrow{p}_{iobject} =\overrightarrow{p}_{ix} +\overrightarrow{p}_{iy} +\overrightarrow{p}_{iz}[/latex]

When solving two or three dimensional problems using the Law of Conservation of Momentum, you solve as you did before with the one dimensional problems, except now you need to make sure that the momentum in each of the three directions is conserved both before and after an event occurs.

This means that:

[latex]\overrightarrow{p}_{before(x, y, z)} =\overrightarrow{p}_{after(x, y, z})[/latex]

As an example of this, if the vector sum of the momentum of two or more interacting objects is:

[latex]\overrightarrow{p}_{before(x, y, z)}[/latex]= 25 Ns _x – 30 Ns _y + 10 Ns _z

then the vector sum of the momentum of two or more interacting object afterwards is:

[latex]\overrightarrow{p}_{after(x, y, z)}[/latex]= 25 Ns _x – 30 Ns _y + 10 Ns _z

In solving these two or three dimensional problems, you will be using trigonometry and the Pythagorean Theorem extensively, as you can see in the following two examples.

At this point, you will use trigonometry and the Pythagorean Theorem to further analyze this collision.

Now:  The total momentum can be found by using the Pythagorean Theorem.

[latex](\overrightarrow{p}_{total})^{2} = (816 Ns)^{2} + (1020 Ns)^{2}[/latex]

[latex](\overrightarrow{p}-{total})^{2} = 665856 N^{2}s^{2} + 1040400 N^{2}s^{2}[/latex]

[latex](\overrightarrow{p}_{total})^{2} = 1706256 N^{2}s^{2}[/latex]

[latex]\overrightarrow{p} total = 1306 Ns[/latex]

The angle at which these players move off is now found by using trigonometry, specifically the Tangent function:

Tan ø = Opposite Side / Adjacent Side

Tan ø = 1020 Ns / 819 Ns

Tan ø = 1.245

ø = Tan^-1 1.245

ø = 51.3°  specifically ≈ 51° North from East

The velocity at which these players move off can be found by using the equation for momentum:

[latex]\overrightarrow{p} = m\overrightarrow{v}[/latex]

[latex]\text{1306 NS = (120 kg + 85 kg)}\overrightarrow{v}_{f(120+85)}[/latex]

[latex]\overrightarrow{v}_{f(120+85)}\text{= 1306 Ns ÷ 205 kg}[/latex]

                                                                    [latex]\overrightarrow{v}_{f(120+85)}\text{= 6.4 m/s 51º North from East}[/latex]

All that is known are the mass of the two colliding vehicles, original velocity of the car and the angle at which the wreck moved off.

Combining this data, we can calculate the momentum of the car, and by using trigonometry we can calculate the momentum of the truck.  The momentum of the truck allows us to estimate the initial velocity at which it was traveling.

First: The momentum of the car.

[latex]\overrightarrow{p}\text{= (1250 kg)(11.1 m/s East)}[/latex]

[latex]\overrightarrow{p}\text{= 13 900 Ns East}[/latex]

Now:

Using the Tangent function

[latex]\text{Tan 23.5º} =\dfrac{13 900 Ns}{\overrightarrow{p}_{truck}}[/latex]

[latex]\overrightarrow{p}_{truck}=\dfrac{13 900 Ns}{Tan 23.5º}[/latex]

[latex]\overrightarrow{p}_{truck}\text{= 32 00 Ns South}[/latex]

Finally, the velocity of the truck is

[latex]\overrightarrow{p}\text{= m v}[/latex]

[latex]\overrightarrow{v}\text{= (32 000 Ns South) ÷ (1600 kg)}[/latex]

[latex]\overrightarrow{v}\text{= 20 m/s South or 72 km/h}[/latex]

The truck was speeding by 12 km/h and could also be charged with speeding.