<p>When two objects interact in a collision, momentum is conserved where momentum (p) is defined as the product of mass (m) and velocity (v). However, if the kinetic energy also depends on mass and velocity (1/2 mv²) then shouldn’t it also be conserved? Nope. Let’s work through the whole thing.</p>
<h1>Simple Collision Between Two Point Particles</h1>
<p>Let’s start with the simplest example possible (maybe). Imagine that there is a particle with mass m moving in the x-direction with a velocity v (called particle A). It collides with another particle with mass m, but that’s stationary (particle B). Oh, it’s a head on collision so that every stays in 1 dimension (the x-direction).</p>
<p><img alt="" src="https://miro.medium.com/v2/resize:fit:700/1*UlBT1EIi690vvy3tcG7B1w.png" style="height:352px; width:700px" /></p>
<p>The key thing idea is that during the collision object A pushes on B and B pushes back on A. But these are the SAME interaction such that the magnitude of A on B is equal to B on A. Now let’s consider the momentum principle. It says the following:</p>
<p><img alt="" src="https://miro.medium.com/v2/resize:fit:638/1*
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<p>We can do the same thing for particle B, but remember the two forces are equal and opposite. This means that the change in momentum for A must be the opposite of mass B (assuming the time intervals are the same — but why wouldn’t they be).</p>
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