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In an elastic collision, the total momentum AND the total kinetic energy $$E_k$$ are conserved.

Kinetic energy ($$E_k$$) is energy stored as movement (that could, for example, be converted to electrical energy in a generator): $$$E_k=\frac{1}{2}mv^2$$$

In an inelastic collision only the total momentum is conserved. Some energy is lost (e.g. as heat) during the collision.

Most real situations are inelastic, but collisions of hard stones or pool balls come close to being elastic. Elastic collisions occur between very rigid materials.

For an elastic collision between two particles $$1$$ and $$2$$, we can write: $$$ \begin{align*} m_1\vecphy{u}_1+m_2\vecphy{u}_2&=m_1\vecphy{v}_1+m_2\vecphy{v}_2\\ E_{k(1i)}+ E_{k(2i)}&=E_{k(1f)}+ E_{k(2f)}\\ \frac{1}{2}m_1\vecphy{u}_1^2+\frac{1}{2}m_2\vecphy{u}_2^2&=\frac{1}{2}m_1\vecphy{v}_1^2+\frac{1}{2}m_2\vecphy{v}_2^2\\ \end{align*}$$$

$$i$$ and $$f$$ indicate initial and final values. $$m=$$ mass; $$\vecphy{u}=$$ velocity before the collision$$=$$speed of approach; $$v=$$velocity after the collision$$=$$speed of separation.

To express the final velocities in terms of the initial velocities in an elastic collision, it is necessary to use the equations for the conservation of kinetic energy and conservation of momentum:

$$$(1) \quad m_1\vecphy{u}_1+m_2\vecphy{u}_2=m_1\vecphy{v}_1+m_2\vecphy{v}_2$$$ $$$(2)\quad\frac{1}{2}m_1\vecphy{u}_1^2+\frac{1}{2}m_2\vecphy{u}_2^2=\frac{1}{2}m_1\vecphy{v}_1^2+\frac{1}{2}m_2\vecphy{v}_2^2$$$

The momentum/kinetic energy lost by body 1 is equal to the momentum/kinetic energy gained by body 2: $$$ \begin{align*} (1) & \iff m_1(\vecphy{u}_1- \vecphy{v}_1)=m_2(\vecphy{v}_2-\vecphy{u}_2)\\ (2) &\ \iff \frac{1}{2}m_1(\vecphy{u}_1^2- \vecphy{v}_1^2)= \frac{1}{2}m_2(\vecphy{v}_2^2- \vecphy{u}_2^2)\\ \end{align*} $$$

Now we derive the intermediary result: $$$ (2) \iff \frac{1}{2}m_1(\vecphy{u}_1+ \vecphy{v}_1) (\vecphy{u}_1- \vecphy{v}_1)= \frac{1}{2}m_2(\vecphy{v}_2+ \vecphy{u}_2) (\vecphy{v}_2- \vecphy{u}_2) $$$ And using $$(1)$$: $$$ \begin{align*} (2) & \iff \vecphy{u}_1+ \vecphy{v}_1=\vecphy{v}_2+ \vecphy{u}_2\\ & \iff \vecphy{u}_1- \vecphy{u}_2=\vecphy{v}_2 - \vecphy{v}_1\\ & \Rightarrow \text{ relative speed of approach=relative speed of separation} \end{align*} $$$

$$\vecphy{v}_1$$ and $$\vecphy{v}_2$$ can be obtained using: $$$ \begin{align*} & m_1(\vecphy{u}_1- \vecphy{v}_1)= m_2(\vecphy{v}_2- \vecphy{u}_2)\\ \iff & \frac{m_2}{m_1}\vecphy{v}_2+ \vecphy{v}_1= \frac{m_2}{m_1}\vecphy{u}_2+ \vecphy{u}_1 \end{align*} $$$ Using $$\vecphy{u}_1- \vecphy{u}_2=\vecphy{v}_2- \vecphy{v}$$: $$$ \begin{align*} & (\frac{m_2}{m_1}+1) \vecphy{v}_2=2\vecphy{u}_1+(\frac{m_2}{m_1}-1) \vecphy{u}_2\\ \iff & \vecphy{v}_2=(\frac{2m_1}{m_1+m_2})\vecphy{u}_1+(\frac{m_2-m_1}{m_1+m_2})\vecphy{u}_2 \end{align*} $$$ A similar result holds for $$\vecphy{v}_1$$: $$$\vecphy{v}_1=(\frac{m_1-m_2}{m_1+m_2})\vecphy{u}_1+(\frac{2m_2}{m_1+m_2})\vecphy{u}_2$$$

An elastic collision between two objects with equal mass moving in opposite directions (a frontal collision) has some special properties. The two objects simply exchange velocity: $$$ \begin{align*} & \vecphy{v}_1=\vecphy{u}_2\\ & \vecphy{v}_2=\vecphy{u}_1 \end{align*} $$$

When the mass of the first body is much bigger than the mass of the second, the bigger mass does not change its velocity noticeably. Only the smaller mass noticeably changes its direction and speed. The following is an approximation: $$$\vecphy{v}_1=\vecphy{u}_1 \quad \quad \vecphy{v}_2=2\vecphy{u}_1+\vecphy{u}_2$$$

Two pool balls have the same mass. When they collide head-on, they simply exchange their velocities.
Two pool balls have the same mass. When they collide head-on, they simply exchange their velocities.

In an inelastic collision, some of the kinetic energy is converted to other forms of energy. This energy can take the form of heat (thermal energy), sound or deformation of the structure of one of the objects.

Most real collisions are inelastic. The degree to which energy is converted does, however, vary widely.

While some kinetic energy is converted to other forms of energy, total energy is conserved.

The principle of conservation of momentum is maintained: $$$m_1\vecphy{u}_1+m_2\vecphy{u}_2= m_1\vecphy{v}_1+m_2\vecphy{v}_2$$$

In a perfectly inelastic collision, the two objects stick together after the collision.

Here, the principle of conservation of momentum takes the form: $$$m_1\vecphy{u}_1+m_2\vecphy{u}_2=(m_1+m_2) \vecphy{v}$$$ $$\vecphy{v}$$ is the final velocity of the new body.

A car crash is an example of an inelastic collision, as a large amount of kinetic energy is converted into the deformation of material.

Balls for squash (a racquet sport) are not very elastic. They convert a lot of kinetic energy into heat following collision with the wall and get quite hot during intense play.