Derivation: Head-on Elastic Collision and its 4 Special Cases

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In the following we want to consider two bodies (e.g. two spherical masses) which elastically collide with each other. The first body has the mass $m_1$ and moves with the velocity $v_1$ to the right (in the positive $x$ direction). The second body has the mass $m_2$ and moves with the velocity $v_2$ (in the opposite direction, towards the first body).

Our goal is to find a formula for the velocity $v'_1$ of the first body and the velocity $v'_2$ of the second body after the collision.

The first body has the momentum $p_1 ~=~ m_1 \, v_1$.
The second body has the momentum $p_2 ~=~ m_2 \, v_2$.

According to the momentum conservation, the sum of the two momenta before the collision must be equal to the sum of the momenta after the collision:

Conservation of momentum for the collision of two bodies

$$ \begin{align} m_1 \, \class{red}{v_1} ~+~ m_2 \, \class{blue}{v_2} ~=~ m_1 \, \class{red}{v'_1} ~+~ m_2 \, \class{blue}{v'_2} \end{align} $$

Rearrange the formula for the conservation of momentum into a more suitable form. Put the terms with equal masses on one side of the equation:

$$ \begin{align} m_1 \, v_1 ~-~ m_1 \, v'_1 ~=~ m_2 \, v'_2 ~-~ m_2 \, v_2 \end{align} $$

Factor out the masses on both sides:

$$ \begin{align} m_1 \, ( v_1 ~-~ \, v'_1) ~=~ m_2 \, (v'_2 ~-~ v_2) \end{align} $$

Next, we bring the law of conservation of energy into play. We may assume conservation of energy of the kinetic energy here, since we have assumed that the collision is elastic. This means: The total energy (here exclusively in the form of kinetic energy of the collision partners) remains the same after the collision.

The two bodies should have only kinetic energy before and after the collision. According to the law of conservation of energy, this means that the sum of the kinetic energies of both bodies before the collision must be equal to the sum of the kinetic energies after the collision:

Law of conservation of energy of two bodies with kinetic energy

$$ \begin{align} \frac{1}{2}\,m_1\,{v_1}^2 ~+~ \frac{1}{2}\,m_2\,{v_2}^2 ~=~ \frac{1}{2}\,m_1\,{v'_1}^{2} ~+~ \frac{1}{2}\,m_2\,{v'_2}^{2} \end{align} $$

Cancel the factor $\frac{1}{2}$ and bring the same masses to one side as in the conservation of momentum law 3:

$$ \begin{align} m_1 \, \left( {v_1}^2 - {v'_1}^2 \right) ~=~ m_2 \, \left( {v'_2}^2 - {v_2}^2 \right) \end{align} $$

Now use the binomial formula $(a^2 - b^2) = (a-b)(a+b)$. In our case, $ a = v'_1$ and $b = v'_2$:

$$ \begin{align} m_1 \, (v_1 - v'_1) \, (v_1 + v'_1) ~=~ m_2\, (v'_2 - v_2) \, (v'_2 + v_2) \end{align} $$

According to the rearranged law of conservation of momentum 3, the terms $m_1 \, (v_1 - v'_1)$ and $m_2 \, (v'_2 - v_2)$ are equal. These terms occur in the law of conservation of energy 6 and can be cancelled. What remains is:

$$ \begin{align} v_1 + v'_1 ~=~ v'_2 + v_2 \end{align} $$

Usually, we only know the situation before the collision. We know the masses of the two collision partners and the velocities with which we let them collide. Next, let's figure out an equation for the velocity $ v'_1 $ of the first body after the collision, which depends just on the masses and the initial velocities. Then, analogously, we find an equation for the velocity $ v'_2 $ of the other body after the collision.

Velocity of the first body after the collision

To derive the velocity $ v'_1 $ of the first body after the collision, rearrange the equation 7 for $v'_2$:

$$ \begin{align} v'_2 ~=~ v_1 ~+~ v'_1 ~-~ v_2 \end{align} $$

Now you can substitute Eq. 8 into the conservation of momentum 1 and thus eliminate $v'_2$:

$$ \begin{align} m_1 \, v_1 ~+~ m_2 \, v_2 ~=~ m_1 \, v'_1 ~+~ m_2 \, (v_1 ~+~ v'_1 ~-~ v_2) \end{align} $$

Thus we have eliminated the velocity $v'_2$. Now we just have to rearrange Eq. 9 for $v'_1$. Multiply out the parenthesis on the right side:

$$ \begin{align} m_1 \, v_1 ~+~ m_2 \, v_2 ~=~ m_1 \, v'_1 ~+~ m_2 \, v_1 ~+~ m_2 \, v'_1 ~-~ m_2 \, v_2 \end{align} $$

Put all the summands containing $v'_1$ on the left side and all the other summands on the right side of the equation:

$$ \begin{align} m_1 \, v'_1 ~+~ m_2 \, v'_1 ~=~ m_1 \, v_1 ~-~ m_2 \, v_1 ~+~ m_2 \, v_2 ~+~ m_2 \, v_2 \end{align} $$

Factor out the velocity $ v'_1$ on the left side. Also factor out $v_1$ on the right side. On the right side, we can also simplify this: $m_2 \, v_2 + m_2 \, v_2 = 2 m_2 \, v_2$. After these steps you get the following equation:

$$ \begin{align} v'_1 \, (m_1 ~+~ m_2) ~=~ v_1 \, (m_1 ~-~ m_2) ~+~ 2m_2 \, v_2 \end{align} $$

Divide both sides by $(m_1 ~+~ m_2)$ and you get a formula for the unknown velocity $v'_1$ of the first body after the head-on elastic collision with another body:

Formula: Velocity of the first body after the elastic head-on collision

$$ \begin{align} v'_1 ~=~ v_1 \, \left(\frac{m_1 ~-~ m_2}{m_1 ~+~ m_2}\right) ~+~ v_2 \, \left(\frac{2m_2}{m_1 ~+~ m_2}\right) \end{align} $$

Velocity of the second body after the collision

To derive the velocity $v'_2$ of the second body after the head-on collision depending only on the initial conditions, we proceed analogously as with $v'_1$. To do this, rearrange Eq. 7 for $v'_1$:

$$ \begin{align} v'_1 ~=~ v'_2 ~+~ v_2 ~-~ v_1 \end{align} $$

Substitute Eq. 14 into conservation of momentum 1 to eliminate $ v'_1$:

$$ \begin{align} m_1 \, v_1 ~+~ m_2 \, v_2 ~=~ m_1 \, (v'_2 ~+~ v_2 ~-~ v_1) ~+~ m_2 \, v'_2 \end{align} $$

Multiply out the parenthesis:

$$ \begin{align} m_1 \, v_1 ~+~ m_2 \, v_2 ~=~ m_1 \, v'_2 ~+~ m_1 \, v_2 ~-~ m_1 \, v_1 ~+~ m_2 \, v'_2 \end{align} $$

Put all the summands containing $v'_2$ on the left-hand side and all the other summands on the right-hand side:

$$ \begin{align} m_1 \, v'_2 ~+~ m_2 \, v'_2 ~=~ m_1 \, v_1 ~+~ m_1 \, v_1 ~+~ m_2 \, v_2 ~-~ m_1 \, v_2 \end{align} $$

Factor out $ v'_2$ on the left side. Also factor out $v_2 $ on the right side. On the right side we can simplify a bit: $m_1 \, v_1 + m_1 \, v_1 = 2 m_1 \, v_1$. Then you get the following equation:

$$ \begin{align} v'_2 \, (m_1 ~+~ m_2) ~=~ 2m_1 \, v_1 ~+~ v_2 \, (m_2 ~-~ m_1) \end{align} $$

Divide both sides by $(m_1 ~+~ m_2)$. Then you get the formula you are looking for:

Formula: Velocity of the second body after the elastic head-on collision

$$ \begin{align} v'_2 ~=~ v_1 \, \left( \frac{2m_1}{ m_1 + m_2 } \right) ~+~ v_2 \, \left( \frac{m_2 - m_1}{ m_1 + m_2 } \right) \end{align} $$

In these two chapters we have derived two formulas 13 and 19, which allow us to calculate the velocity of the first and the second body after an elastic head-on collision. Next, let us consider some important special cases of these formulas.

Special case #1: Both collision partners have the same mass

If the two colliding bodies have equal masses: $ m_1 = m_2$ $ m_1 = m_2$, then the velocity formulas 13 and 19 simplify. Let us denote the mass of the body as $m$ and insert it into Eqs. 13 and 19. Then we get:

Velocity of the first body after the collision of two equal masses

$$ \begin{align} v'_1 ~=~ v_2 \end{align} $$

Velocity of the second body after the collision of two equal masses

$$ \begin{align} v'_2 ~=~ v_1 \end{align} $$

The two results state that after the collision the two equal masses exchange their velocities.

Hover the image!

Two equal masses collide elastically and exchange their velocities.

Special case #2: One of the bodies is at rest before the collision

In this special case, we assume that the first body is at rest. That is, it has no velocity before the collision: $v_1 = 0$. The second body now elastically collides with the resting first body. So put $v_1 = 0$ in 13 and in 19 and you get simplified collision formulas:

Velocity of the 1st body after the collision (before that it was at rest)

$$ \begin{align} v'_1 ~=~ v_2 \, \left(\frac{2m_2}{m_1 ~+~ m_2}\right) \end{align} $$

Velocity of the 2nd body after the collision with a stationary body

$$ \begin{align} v'_2 ~=~ v_2 \, \left(\frac{m_2 ~-~ m_1}{m_1 ~+~ m_2}\right) \end{align} $$

Special case #3: Collision of a heavy body with a light body at rest

For this special case, we assume that the first body is at rest: $v_1 = 0$. If we inser this velocity into Eqs. 13 and 19, then we get the result we already got out in Eqs. 22 and 23.

The second simplification we want to make here is that the second body is much heavier than the resting first body: $ m_2 \gg m_1 $. The mass $m_2$ is very large compared to the mass $m_1$. So we can neglect the first mass, so set it equal to zero: $ m_1 \approx 0 $. Insert the first mass into equations 22 and 23 and you get:

Velocity of the previously stationary body after collision with a heavy body

$$ \begin{align} v'_1 ~\approx~ 2 v_2 \end{align} $$

Velocity of the heavy body after the collision with a stationary body

$$ \begin{align} v'_2 ~\approx~ v_2 \end{align} $$

From the equation 24 you can read that the light first body has twice the velocity of the second body after the collision.
From the equations 25 you can read that the heavy second body maintains its velocity after the collision.

Hover the image!

A heavy ball collides with a light ball at rest.

Special case #4: Collision of a light body with a heavy body at rest

In this simplification, we again assume that the first body is at rest: $v_1 = 0$. Substituted into equations 13 and 19 we get the equations 22 and 23. Nothing new so far.

What is new is the assumption that the first body should be much heavier than the second body: $ m_1 \gg m_2 $. The mass $m_1$ is so large compared to the mass $m_2$ that we can set the second mass to zero: $ m_2 \approx 0$. So insert the second mass into equations 22 and 23 and you get:

Velocity of the heavy body at rest after the collision

$$ \begin{align} v'_1 ~\approx~ 0 \end{align} $$

Velocity of the light body after the collision with a heavy body at rest

$$ \begin{align} v'_2 ~=~ -v_2 \end{align} $$

Equation 26 states that the heavy body at rest continues to be at rest after the collision.
The equation 27 states that the light second body keeps its velocity after the collision, but reverses its direction (because of the minus sign). The second body is reflected at the heavy resting mass.

Hover the image!

A light ball bounces off the heavy ball at rest.

Now you should know how the law of conservation of momentum and energy is used to find the velocities of two particles after the collision.