# Determining Kinetic Energy Lost in Inelastic Collisions

A **perfectly inelastic collision** is one in which two objects colliding stick together, becoming a single object. For instance, two balls of sticky putty thrown at each other would likely result in perfectly inelastic collision: the two balls stick together and become a single object after the collision.

Unlike elastic collisions, perfectly inelastic collisions don't conserve energy, but they do conserve momentum. While the total energy of a system is always conserved, the **kinetic energy** carried by the moving objects is not always conserved. In an inelastic collision, energy is **lost** to the environment, transferred into other forms such as heat.

$A$ and $B$ with the same mass $m$ are $15$ m away from each other. Now, $A$ and $B$ are thrown horizontally at the same time at the velocities $20$ m/s and $-10$ m/s, respectively, eventually colliding with each other in the air. If the two objects stick together after a perfectly inelastic collision, what is the speed of the mass at the moment of collision (in m/s)?

In the above figure, two objectsAssume that gravitational acceleration is $g= 10$ m/s$^{2}$.

## General Solution for Perfectly Inelastic Collisions

Consider two particles of mass $m_{1}$ and $m_{2}$ moving at velocities $\vec{v}_{1}$ and $\vec{v}_{2}$, respectively. Before they collide, they have a combined energy of $E_{\text{init}} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{1} v_{2}^{2}$ and a combined momentum of $\vec{p}_{\text{init}} = m_{1} \vec{v}_{1} + m_{2} \vec{v}_{2}$.

Since the collision is perfectly inelastic, after the collision there is a single combined object of mass $m_{1} + m_{2}$. Since momentum is conserved, this object has momentum equal to the total intitial momentum $\vec{p} = (m_{1} + m_{2}) \vec{v}_{f}$. The velocity of the combined object $\vec{v}_f$ is then given by

$\begin{aligned} (m_{1} + m_{2}) \vec{v}_{f} &= m_{1} \vec{v}_{1} + m_{2} \vec{v}_{2}\\ \vec{v}_{f} &= \frac{m_1}{m_1 + m_2} \vec{v}_1 + \frac{m_{2}}{m_1 + m_2} \vec{v}_2. \end{aligned}$

The energy depends on the squared magnitude of $\vec{v}_f$, which is the dot product of $\vec{v}_f$ with itself. If the angle between $\vec{v}_1$ and $\vec{v}_2$ is $\theta$, then this equals

$\| \vec{v}_f \|^2 = \frac{m_1^2}{(m_1 + m_2)^2} v_1^2 + \frac{m_2^2}{(m_1 + m_2)^2} v_2^2 + \frac{m_1 m_2}{(m_1 + m_2)^2} v_1 v_2 \text{cos} \theta.$

The final energy $E_f$ is

$\begin{aligned} E_f &= \frac{1}{2} (m_1 + m_2) \| \vec{v}_f \|^2\\ &= \frac{1}{2} \left[\frac{m_1^2}{(m_1 + m_2)} v_1^2 + \frac{m_2^2}{(m_1 + m_2)} v_2^2 + 2 \frac{m_1 m_2}{(m_1 + m_2)} v_1 v_2 \text{cos} \theta\right]. \end{aligned}$

This equation is the general solution for perfectly inelastic collisions. It's somewhat ugly, but exploring how it works in particular simplified cases can help build intuition for what it says.

What is the energy difference $\Delta E = E_ f - E_ i$ if $m_2$ is much much smaller than $m_1$? Physicists express this with symbols as $m_2 \ll m_1$. In this case, $m_1 + m_2 \approx m_1$. This simplifies the equation to

$\begin{aligned} E_f &= \frac{1}{2} \left[\frac{m_1^2}{m_1} v_1^2 + \frac{m_2}{m_1} v_2^{2} + 2 \frac{m_1 m_2}{m_1} v_1 v_2 \text{cos} \theta\right] \\ &= \frac{1}{2} \left(m_1 v_1^2 + 2 m_2 v_1 v_2 \text{cos} \theta + \frac{m_2}{m_1} m_1 v_2^2\right). \end{aligned}$

Since $m_2 \ll m_1$, $\frac{m_2}{m_1} \ll 1$, the last term is small if in addition $v_2$ is smaller than or not much larger than $v_1$. These combined assumptions allow $E_f$ to be further simplified to

$\begin{aligned} E_f &= \frac{1}{2} \left(m_1 v_1^2 + 2 m_2 v_1 v_2 \text{cos} \theta\right)\\\\ \Delta E &= E_f - E_i \\ &= m_2 v_1 v_2 \text{cos} \theta - \frac{1}{2} m_2 v_2^2. \end{aligned}$

This equation gives a nice interpretation for this limiting case. The second term "eliminates" the energy of the original particle, while the first term "creates" a particle of mass $m_2$ with velocity projected in the direction of the more massive $m_1$, because it's stuck to $m_1$. The energy of the mass $m_1$ is left unchanged.

Take special care that this simplification required that the velocity of the smaller particle was not too high. If it were, then the larger particle would have its energy changed as well, and

**Cite as:**Determining Kinetic Energy Lost in Inelastic Collisions.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/determining-kinetic-energy-lost-in-inelastic/