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 could be between two carts rolling or sliding on a track with low friction, 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 momentum p_1xf + p_2xf must equal the initial total 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 an increase in the 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.

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: 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. If p_2xf =0, then p_1xf + p_2xf = p_1xf = p_1xi, 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 Inelastic 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]

The final speed of the stuck-together carts is half the initial speed:

[math]\displaystyle{ {v_{f}}={\frac{1}{2}}{v_{i}} }[/math]

Final translational kinetic energy:

[math]\displaystyle{ ({K_{1f}}+{K_{2f}})={2}({\frac{1}{2}}{m}{v^2_{f}}) }[/math]

[math]\displaystyle{ ({K_{1f}}+{K_{2f}})={2}({\frac{1}{2}}{m}({\frac{1}{2}}{v_{i}})^2)={\frac{1}{4}}{m}{v^2_{i}} }[/math]

[math]\displaystyle{ ({K_{1f}}+{K_{2f}})={\frac{K_{1i}}{2}} }[/math]

Energy Principle:

[math]\displaystyle{ {K_{1f}}+{K_{2f}}+{E_{int,f}}={K_{1i}}+{E_{int,i}} }[/math]

[math]\displaystyle{ {E_{int,f}}-{E_{int,i}}={K_{1i}}-({K_{1f}}+{K_{2f}}) }[/math]

[math]\displaystyle{ {ΔE_{int}}={K_{1i}}-{\frac{K_{1i}}{2}} }[/math]

[math]\displaystyle{ {ΔE_{int}}={\frac{K_{1i}}{2}} }[/math]

The Momentum Principle is still valid even though the collision is inelastic, and fundamental principles apply in all situations. The final kinetic energy of the system is only half of the original kinetic energy, which mean that the other half of the original kinetic energy has been dissipated into increased internal energy of the two carts.

From both examples, we know:

1. If the collision is elastic, object 1 stops and object 2 moves with the speed object 1 used to have.

2. If the collision is maximally inelastic, the carts stick together and move with half the original speed. Half of the original kinetic energy is dissipated into increased internal energy.

A Computational Model

https://trinket.io/embed/glowscript/37540ee8e0

Examples

Toyota is testing the crash-safety of their new car. They crash two cars in a head on collision, one moving 30 miles per hour and one moving 45 miles per hour. The cars are they same model, so they both weigh 3,500 pounds. Assuming the cars stick together, what is their speed immediately after the collision?

This is a maximally inelastic collision since the cars stick together. Therefore, we can use conservation of momentum to calculate the final speed.

First, it is useful to convert all given measurements into the metric system. This isn't a totally necessary step, but I find it makes the problem easier and saves time in the end. It is necessary to convert the given weight measurement to a mass measurement, however.

[math]\displaystyle{ {30 mph} * (1609.34 meters / 1 mile) * (1 hour / 3600 seconds) = 13.41 m/s }[/math]

[math]\displaystyle{ {45 mph} * (1609.34 meters / 1 mile) * (1 hour / 3600 seconds) = 20.12 m/s }[/math]

[math]\displaystyle{ {3500 pounds} * (4.448 newtons / 1 pound) = 15568.78 N }[/math]

[math]\displaystyle{ {15568.78 newtons} * (1 / 9.81 m/s) = 1,587 kg }[/math]

[math]\displaystyle{ {m}{v_{1i}}+{m}{v_{2i}}={m}{v_{f}} }[/math]

[math]\displaystyle{ {1587}{13.41}+{1587}{20.12}={3174}{v_{f}} }[/math]

[math]\displaystyle{ {21281.67}+{31930.44}={3174}{v_{f}} }[/math]

[math]\displaystyle{ {53212.11}={3174}{v_{f}} }[/math]

[math]\displaystyle{ {v_{f}}={16.77m/s} }[/math]

Simple

Question

A 8 kg mass traveling at speed 18 m/s strikes a stationary 8 kg mass head-on, and the two masses stick together. What are the final speeds?

Answer

[math]\displaystyle{ {m}{v_{1i}}+{m}{v_{2i}}={m}{v_{f}}+{m}{v_{f}} }[/math]

[math]\displaystyle{ {m}{v_{1i}}+{m}({0})={2m}{v_{f}} }[/math]

[math]\displaystyle{ {8 kg}({18m/s})={2}({8kg}){v_{f}} }[/math]

[math]\displaystyle{ {144kg}{m/s}={16 kg}{v_{f}} }[/math]

[math]\displaystyle{ {v_{f}}={9m/s} }[/math]

Middling

Question

Answer

Difficult

Question

Answer

Connectedness

A real world application of head-on collisions of equal mass are billiard balls (playing pool). All the balls have equal mass, and one ball is shot towards another, which is stationary. When the two collide, the ball that is hit will move at a 90 angle from the direction the striking ball comes from. Keeping this mind and with a little practice one's level of playing billiards should increase.

Head-on Collision of Equal Masses

Contents

Main Idea

A Mathematical Model

A Computational Model

Examples

Simple

Middling

Difficult

Connectedness

See also

Further reading

External links

References

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