Head-on Collision of Equal Masses: Difference between revisions

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If we input the various types of energy in for the total energy such as kinetic, potential, and internal we get
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.
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 (objects become stuck together).


'''1. Elastic Head-on Collisions of Equal Masses'''
'''1. Elastic Head-on Collisions of Equal Masses'''
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If ''p<sub>1xf</sub> =0'', then object 1 came to a full stop. Based on the momentum equation ''p<sub>1xf</sub> + p<sub>2xf</sub> = p<sub>2xf</sub> = p<sub>1xi</sub>'', 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.
If ''p<sub>1xf</sub> =0'', then object 1 came to a full stop. Based on the momentum equation ''p<sub>1xf</sub> + p<sub>2xf</sub> = p<sub>2xf</sub> = p<sub>1xi</sub>'', 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.
If ''p<sub>2xf</sub> =0'', then ''p<sub>1xf</sub> + p<sub>2xf</sub> = p<sub>1xf</sub> = p<sub>1xi</sub>'', and so object 1 keeps going, missing object 2. This won't happen if the carts are on the same track. It is not possible for both final momenta to be zero, since the total final momentum of the system must equal the nonzero total initial momentum of the system.
If ''p<sub>2xf</sub> =0'', then ''p<sub>1xf</sub> + p<sub>2xf</sub> = p<sub>1xf</sub> = p<sub>1xi</sub>'', and so object 1 keeps going, missing object 2. This won't happen if the carts are on the same track. It is not possible for both final momenta to be zero, since the total final momentum of the system must equal the nonzero total initial momentum of the system.
'''2. Maximally Inelast Collision of Equal Masses'''
Momentum Principle
::<math>{p_{1xf}}+{p_{2xf}}={p_{1xi}}</math>
::<math>{2p_{1xf}}={p_{1xi}}</math>
::<math>{p_{1xf}}={\frac{1}{2}}{p_{1xi}}</math>
Energy Principle
:: <math>{E_f}={E_i}+{W}+{Q} </math>
:: <math>({K_{1f}}+{E_{int1f}})+({K_{1f}}+{E_{int2f}})=({K_{1i}}+{E_{int1i}})+({K_{2i}}+{E_{int2i}})+{0}+{0} </math>
:: <math>{K_{1f}}+{K_{2f}}={K_{1i}}+{0} </math>
Since ''K=p<sup>2</sup>/2m'', we can combine the momentum and energy equations:
::<math>{\frac{p^{2}_{1xf}}{2m}}+{\frac{p^{2}_{2xf}}{2m}}= {\frac{(p_{1xf}+{p}_{2xf})^2}{2m}}</math>
:: <math>{p^2_{1xf}}+{p^2_{2xf}}={p^2_{1xf}}+{2p_{1xf}p_{2xf}}+{p^2_{2xf}} </math>
:: <math>{2p_{1xf}p_{2xf}}={0} </math>


===A Computational Model===
===A Computational Model===

Revision as of 15:23, 3 December 2015

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 p1xf + p2xf must equal the initial total x momentum p1xi. Before the collision, nonzero energy terms included the kinetic energy of cart 1, K1i, and the internal energies of both carts. After the collision there is internal energy of both carts and kinetic energy of both carts, K1f + K2f.


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 (objects become stuck together).

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: p1xf =0 or p2xf =0. If p1xf =0, then object 1 came to a full stop. Based on the momentum equation p1xf + p2xf = p2xf = p1xi, 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. If p2xf =0, then p1xf + p2xf = p1xf = p1xi, and so object 1 keeps going, missing object 2. This won't happen if the carts are on the same track. It is not possible for both final momenta to be zero, since the total final momentum of the system must equal the nonzero total initial momentum of the system.

2. Maximally Inelast Collision of Equal Masses

Momentum Principle

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

A Computational Model

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