Head-on Collision of Unequal Masses

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Main Idea

The two main types of collisions are Elastic Collisions and Inelastic Collisions, but these are very broad as there are many much more specific types of collisions under these umbrella terms. One of the specific types of collisions is head-on collisions of unequal masses. This is exactly what it sounds like - two objects of different masses collide with each other head-on, and this causes some changes in kinetic energy and speed. This can be thought of as two different types of cars colliding with each other, but to make the visualization a bit easier, think about a ping pong ball colliding with a bowling ball.

Elastic head-on collision between a car and truck[1]

The Equations Behind It

The most common type of head-on collision of unequal masses studied is an elastic collision, and if this is the case kinetic energy is conserved. What this means is that the total final kinetic energy of the system is equal to the total initial kinetic energy of the system. In equations, it looks like this: [math]\displaystyle{ {1 \over 2}m_1v_1i^2 + {1 \over 2}m_2v_2i^2 = {1 \over 2}m_1v_1f^2+ {1 \over 2}m_2v_2f^2 }[/math].

If the equation is inelastic, the idea of conservation of momentum can be used because momentum is always conserved in collisions. The equation for the conservation of momentum is: [math]\displaystyle{ m_1v_1i^2 + m_2v_2i^2 = m_1v_1f^2 + m_2v_2f^2 }[/math]

[2]

Let's talk about a specific example to truly understand what happens in a head-on collision of unequal masses. As referenced to before, the ping pong and bowling ball example clearly expresses what happens. Assuming the bowling ball is at rest when the ping pong ball hits it, the final momentum of the bowling ball is twice the initial momentum of the ping pong ball. This can be seen by using the conservation of momentum principle: [math]\displaystyle{ p_1f + p_2f = p_1i + p_2i }[/math]


Knowing the speed of the ping pong ball remains about the same, the new equation is [math]\displaystyle{ -p_1i + p_2f = p_1i }[/math] which gives [math]\displaystyle{ p_2f = 2p_1i }[/math] showing that the final momentum of the bowling ball is twice the initial momentum of the ping pong ball.

A Computational Model

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References

http://www.physicsclassroom.com/mmedia/momentum/cthoe.cfm