Head-on Collision of Equal Masses

Work in progress by mtikhonovsky3

Main Idea

A collision is a brief interaction between large forces. A head-on collision can be between two carts rolling or sliding on a track with low fricton or billiard balls, hockey pucks, or vehicles hitting each other head-on.

In terms of the two carts of equal masses example, the two carts are the system. The Momentum Principle tells us that after the collision the total final x momentum p_1xf + p_2xf must equal the initial total x momentum p_1xi. Before the collision, nonzero energy terms included the kinetic energy of cart 1, K_1i, and the internal energies of both carts. After the collision there is internal energy of both carts and kinetic energy of both carts, K_1f + K_2f.

A Mathematical Model

Head-on Collisions of Equal Masses can be based off the Fundamental Principle of Momentum:

[math]\displaystyle{ {\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net}{Δt} }[/math]

where p is the momentum of the system, F is the net force from the surroundings, Δt is the change in time for the process. Energy Principle:

[math]\displaystyle{ {E}={Q}+{W} }[/math] where E is the total energy, Q is the heat given off, and W is the work done.

If we input the various types of energy in for the total energy such as kinetic, potential, and internal we get

Based off of the Momentum Principle and the Energy Principle, we will explore Head-on Collisions of Equal Masses in two different scenarios: elastic and maximally inelastic.

1. Elastic Head-on Collisions of Equal Masses

Momentum Principle

[math]\displaystyle{ {p_{xf}}={p_{xi}}+{F_{net,x}}{Δt} }[/math]

[math]\displaystyle{ {p_{1xf}}+{p_{2xf}}={p_{1xi}}+{0} }[/math]

Energy Principle

[math]\displaystyle{ {E_f}={E_i}+{W}+{Q} }[/math]

[math]\displaystyle{ ({K_{1f}}+{E_{int1f}})+({K_{1f}}+{E_{int2f}})=({K_{1i}}+{E_{int1i}})+({K_{2i}}+{E_{int2i}})+{0}+{0} }[/math]

[math]\displaystyle{ {K_{1f}}+{K_{2f}}={K_{1i}}+{0} }[/math]

Since K=p²/2m, we can combine the momentum and energy equations:

[math]\displaystyle{ {\frac{p^{2}_{1xf}}{2m}}+{\frac{p^{2}_{2xf}}{2m}}= {\frac{(p_{1xf}+{p}_{2xf})^2}{2m}} }[/math]

[math]\displaystyle{ {p^2_{1xf}}+{p^2_{2xf}}={p^2_{1xf}}+{2p_{1xf}p_{2xf}}+{p^2_{2xf}} }[/math]

[math]\displaystyle{ {2p_{1xf}p_{2xf}}={0} }[/math]

There are two possible solutions: p_1xf =0 or p_2xf =0. If p_1xf =0, then object 1 came to a full stop. Based on the momentum equation p_1xf + p_2xf = p_2xf = p_1xi, then object 2 now has the same momentum that object 1 used to have. There is a complete transfer of momentum from object 1 to object 2, and so, there is also a complete transfer of kinetic energy from object 1 to object 2.

A Computational Model

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

Examples

Be sure to show all steps in your solution and include diagrams whenever possible

Connectedness

1. How is this topic connected to something that you are interested in?
2. How is it connected to your major?
3. Is there an interesting industrial application?

History

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

See also

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

Further reading

Books, Articles or other print media on this topic

External links

Internet resources on this topic

References

This section contains the the references you used while writing this page