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9.4: Elastic and Inelastic Collisions

Difficulty Level: At Grade Created by: CK-12
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This device is known as Newton’s cradle. As the balls collide with each other, nearly all the momentum and kinetic energy is conserved. If one ball swings down, exactly one ball will swing up; if three balls swing down, exactly three will swing back up. The collisions between the balls are very nearly elastic. 

Elastic and Inelastic Collisions

For all collisions in a closed system, momentum is conserved. In some collisions in a closed system, kinetic energy is conserved. When both momentum and kinetic energy are conserved, the collision is called an elastic collision. Most collisions are inelastic because some amount of kinetic energy is converted to potential energy, usually by raising one of the objects higher (increasing gravitation PE) or by flexing the object. Any denting or other changing of shape by one of the objects will also be accompanied by a loss of kinetic energy. The only commonly seen elastic collisions are those between billiard balls or ball bearings, because these balls do not compress. And, of course, collisions between molecules are elastic if no damage is done to the molecules.

Much more common are inelastic collisions. These collisions occur whenever kinetic energy is not conserved, primarily when an object's height is increased after the collision or when one of the objects is compressed.

Example Problem: A 12.0 kg toy train car moving at 2.40 m/s on a straight, level train track, collides head-on with a second train car whose mass is 36.0 kg and was at rest on the track. If the collision is perfectly elastic and all motion is frictionless, calculate the velocities of the two cars after the collision.

Solution: Since the collision is elastic, both momentum and KE are conserved. We use the conservation of momentum and conservation of KE equations.

Conservation of momentum: \begin{align*}m_1 v_1+m_2v_2=m_1v{_1}^\prime+m_2v{_2}^\prime\end{align*}m1v1+m2v2=m1v1+m2v2

Conservation of KE: \begin{align*}\frac{1}{2} m_1v{_1}^2+\frac{1}{2}m_2v_2^2=\frac{1}{2}m_1{v{_1}^\prime}^2=\frac{1}{2} m_2{v{_2}^\prime}^2\end{align*}12m1v12+12m2v22=12m1v12=12m2v22

Since \begin{align*}m_1, m_2, v_1\end{align*}m1,m2,v1, and \begin{align*}v_2\end{align*}v2 are known, only \begin{align*}v{_1}^\prime\end{align*}v1 and \begin{align*}v{_2}^\prime\end{align*}v2 are unknown. When the known values are plugged into these two equations, we will have two equations with two unknowns. Such systems can be solved with algebra.

\begin{align*}(12.0 \ kg)(2.40 \ m/s) + (36.0 \ kg)(0 \ m/s) = (12.0 \ kg)(v{_1}^\prime) + (36.0 \ kg)(v{_2}^\prime)\end{align*}(12.0 kg)(2.40 m/s)+(36.0 kg)(0 m/s)=(12.0 kg)(v1)+(36.0 kg)(v2)

\begin{align*}28.8 = 12.0 \ v{_1}^\prime + 36.0 \ v{_2}^\prime\end{align*}28.8=12.0 v1+36.0 v2

Solving this equation for \begin{align*}v{_1}^\prime\end{align*}v1 yields \begin{align*}v{_1}^\prime = 2.4 - 3 \ v{_2}^\prime\end{align*}v1=2.43 v2

\begin{align*}\frac{1}{2} (12.0)(2.40)^2 + \frac{1}{2} (36.0)(0)^2 = \frac{1}{2} (12.0)(v{_1}^\prime)^2 + \frac{1}{2} (36.0)(v{_2}^\prime)^2\end{align*}12(12.0)(2.40)2+12(36.0)(0)2=12(12.0)(v1)2+12(36.0)(v2)2

\begin{align*}69.1 = 12.0 \ {v{_1}^\prime}^2 + 36.0 \ {v{_2}^\prime}^2\end{align*}69.1=12.0 v12+36.0 v22

\begin{align*}5.76 = {v{_1}^\prime}^2 + 3 \ {v{_2}^\prime}^2\end{align*}5.76=v12+3 v22

Substituting the equation for \begin{align*}v{_1}^\prime\end{align*}v1 into this equation yields

\begin{align*}5.76 = (2.4 - 3 \ v{_2}^\prime)^2 + 3 \ {v{_2}^\prime}^2\end{align*}5.76=(2.43 v2)2+3 v22

\begin{align*}5.76 = 5.76 - 14.4 \ v{_2}^\prime + 9 \ {v{_2}^\prime}^2 + 3 \ {v{_2}^\prime}^2\end{align*}5.76=5.7614.4 v2+9 v22+3 v22

\begin{align*}12 \ {v{_2}^\prime}^2 -14.4 \ v{_2}^\prime = 0\end{align*}12 v2214.4 v2=0

\begin{align*}12 \ v{_2}^\prime = 14.4\end{align*}12 v2=14.4

\begin{align*}v{_2}^\prime = 1.2 \ m/s\end{align*}v2=1.2 m/s

Substituting this result back into \begin{align*}v{_1}^\prime = 2.4 - 3 \ v{_2}^\prime\end{align*}v1=2.43 v2 , we get \begin{align*}v{_1}^\prime = - 1.2 \ m/s\end{align*}v1=1.2 m/s.

So, the heavier car is moving in the original direction at 1.2 m/s and the lighter car is moving backward at 1.2 m/s.


  • Elastic collisions are those in which both momentum and kinetic energy are conserved.
  • Inelastic collisions are those in which either momentum or kinetic energy is not conserved.


The following video is a demonstration of elastic and inelastic collisions. Use this resource to answer the questions that follow.

  1. Explain what happened in the first demonstration on elastic collisions.
  2. Explain what happened in the second demonstration on inelastic collisions.
  3. Assuming the first carts started at the same speed in both demonstrations, explain using momentum why the inelastic collision ended slower than the elastic collision.

Practice problems for elastic collision:



  1. A 4.00 kg metal cart is sitting at rest on a frictionless ice surface.  Another metal cart whose mass is 1.00 kg is fired at the cart and strikes it in a one-dimensional elastic collision.  If the original velocity of the second cart was 2.00 m/s, what are the velocities of the two carts after the collision?
  2. Identify the following collisions as most likely elastic or most likely inelastic.
    1. A ball of modeling clay dropped on the floor.
    2. A fender-bender automobile collision.
    3. A golf ball landing on the green.
    4. Two billiard balls colliding on a billiard table.
    5. A collision between two ball bearings.

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Difficulty Level:
At Grade
Date Created:
Feb 05, 2013
Last Modified:
Sep 08, 2016

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