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Two blocks of masses and velocities $(m_1, \textbf{v}_{1,\:i})$ and $(m_2, \textbf{0})$ composed of identical material move towards each other on a frictionless plane in vacuum.

A turqoise block with a velocity approaches a purple block on a grey plane

They collide. It is always observed that the quantity of motion respective to direction is conserved, such that $$ m_1 \cdot \textbf{v}_{1,\:i} = m_1 \cdot \textbf{v}_{1,\:f} + m_2 \cdot \textbf{v}_{2,\:f}.$$ Put a better way, the amount of mass going in any direction before and after the collision stays the same. We call the quantity momentum. This is quite palatable.

Experimentally, it is also observed that as more and more "elastic" materials, like steel, are used for the composition of the two blocks, a certain quantity approaches conservation, such that $$m_1 \cdot ||\textbf{v}_{1,\:i}||^2 = m_1 \cdot ||\textbf{v}_{1,\:f}||^2 + m_2 \cdot ||\textbf{v}_{2,\:f}||^2.$$ Firstly, I notice that this quantity is scalar, rather than vector, as the quantity of motion is. The only practical difference between is that we square the velocity. And yet, experimentally, the data tend towards this conservation equation. We call the quantity kinetic energy.

Usually, explanations point to black-box theorems and formulas that aren't helpful in terms of intuition, use circular reasoning, or create elaborate setups that, although helpful in mathematically understanding the conservation, are also not intuitive to me. So, intuitively, why is the quantity of $\frac{1}{2}m \cdot ||\textbf{v}||^2$ conserved in elastic collisions?

Makemake
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2 Answers2

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So, intuitively, why is kinetic energy conserved in elastic collisions?

Because purely elastic collisions only involve conversions between mechanical kinetic and elastic potential energy without the loss of energy in the form of heat, light, sound, etc.. An example is a collision between ideal springs.

Hope this helps.

Bob D
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The definition of elastic collision is one where kinetic energy is conserved, but that is not the intuitive answer you are asking for.

The blocks approach one another until they become very close, $M_1$ at $x_1$, $M_2$ at $x_1 + \Delta$, and exert force on one another for a short period of time. Then, when $M_1$ is at $x_2$, and $M_2$ is at $x_2 + \Delta$, the force stops. The force is caused by them being less than $\Delta$ apart.

Notice that each block has traveled the same exact distance during the collision period, and that at each instant, $\vec{F}_{1,2} =-\vec{F}_{2,1}$. If the force in question was constant $F$ during this collision time period, then the energy leaving $M_1$ would be force times distance $F (x2 - x1)$, and the energy gained by $M_2$ also would be $F (x2 - x1)$, resulting in conserved energy. The integral of $\int_0^{v_x}{F_x dx}$ works out to $mv_x^2/2$ .

The force during the collision is unlikely to be constant, but if we modeled the force as a spring at distances less than $\Delta$, we again find that the integral of the energy lost by $M_1$ equals the energy gained by $M_2$. We see the spring absorb energy as the blocks compress it, and see the spring transfer energy out as the blocks separate. A 'perfect' spring would release all its energy to its compressors.

Actually, both blocks act as deforming springs, with different spring constants.

David Graham
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