A perfectly elastic collision is a collision in which both total momentum and total kinetic energy are conserved.
In one dimension, all motion occurs along one straight line, so signs show direction.
Sign convention
Choose one direction as positive. In this calculator, positive velocity means motion to the right, and negative velocity
means motion to the left.
Momentum conservation
Total momentum before equals total momentum after.
\[
\begin{aligned}
m_1u_1+m_2u_2 &= m_1v_1+m_2v_2
\end{aligned}
\]
Here \(u_1\) and \(u_2\) are the initial velocities, while \(v_1\) and \(v_2\) are the final velocities.
Kinetic energy conservation
In a perfectly elastic collision, kinetic energy is also conserved:
Total kinetic energy before equals total kinetic energy after.
\[
\begin{aligned}
\frac12m_1u_1^2+\frac12m_2u_2^2
&=
\frac12m_1v_1^2+\frac12m_2v_2^2
\end{aligned}
\]
This is what separates a perfectly elastic collision from an inelastic collision. In an inelastic collision, momentum is
conserved, but kinetic energy is not.
Final velocity formulas
Combining momentum conservation and kinetic energy conservation gives the standard one-dimensional elastic-collision formulas:
Final velocity of object 1.
\[
\begin{aligned}
v_1
&=
\frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2}
\end{aligned}
\]
Final velocity of object 2.
\[
\begin{aligned}
v_2
&=
\frac{2m_1u_1+(m_2-m_1)u_2}{m_1+m_2}
\end{aligned}
\]
Relative velocity reversal
A useful property of a perfectly elastic collision in 1D is that the relative velocity reverses. The speed of approach
before collision equals the speed of separation after collision.
Relative velocity relation.
\[
\begin{aligned}
u_1-u_2 &= v_2-v_1
\end{aligned}
\]
This is the same as saying the coefficient of restitution is \(e=1\).
Special cases
Some common elastic-collision cases are easy to recognize:
-
Equal masses, target initially at rest: the first object stops and the second object takes its velocity.
-
Light object hitting a heavy target: the light object rebounds strongly, while the heavy target moves slowly.
-
Very heavy wall: the small object reverses velocity approximately as \(v_1\approx -u_1\).
Worked example
Suppose \(m_1=2\ \mathrm{kg}\), \(u_1=10\ \mathrm{m\,s^{-1}}\), \(m_2=4\ \mathrm{kg}\), and
\(u_2=0\ \mathrm{m\,s^{-1}}\).
Final velocity of object 1.
\[
\begin{aligned}
v_1
&=
\frac{(2-4)(10)+2(4)(0)}{2+4}\\
&=
\frac{-20}{6}\\
&=
-3.33\ \mathrm{m\,s^{-1}}
\end{aligned}
\]
Final velocity of object 2.
\[
\begin{aligned}
v_2
&=
\frac{2(2)(10)+(4-2)(0)}{2+4}\\
&=
\frac{40}{6}\\
&=
6.67\ \mathrm{m\,s^{-1}}
\end{aligned}
\]
The negative sign for \(v_1\) means object 1 rebounds to the left. Object 2 moves to the right.
Check momentum
\[
\begin{aligned}
p_i &= (2)(10)+(4)(0)=20\ \mathrm{kg\,m\,s^{-1}}\\
p_f &= (2)(-3.33)+(4)(6.67)\approx 20\ \mathrm{kg\,m\,s^{-1}}
\end{aligned}
\]
Check kinetic energy
\[
\begin{aligned}
K_i &= \frac12(2)(10)^2+\frac12(4)(0)^2=100\ \mathrm{J}\\
K_f &= \frac12(2)(-3.33)^2+\frac12(4)(6.67)^2\approx100\ \mathrm{J}
\end{aligned}
\]
Formula summary
| Concept |
Formula |
Meaning |
| Momentum conservation |
\(m_1u_1+m_2u_2=m_1v_1+m_2v_2\) |
Total momentum before equals total momentum after |
| Kinetic energy conservation |
\(\tfrac12m_1u_1^2+\tfrac12m_2u_2^2=\tfrac12m_1v_1^2+\tfrac12m_2v_2^2\) |
Total kinetic energy before equals total kinetic energy after |
| Final velocity 1 |
\(v_1=\frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2}\) |
Velocity of object 1 after collision |
| Final velocity 2 |
\(v_2=\frac{2m_1u_1+(m_2-m_1)u_2}{m_1+m_2}\) |
Velocity of object 2 after collision |
| Relative velocity reversal |
\(u_1-u_2=v_2-v_1\) |
Speed of approach equals speed of separation |
| Center-of-mass velocity |
\(v_{cm}=\frac{m_1u_1+m_2u_2}{m_1+m_2}\) |
Constant velocity of the center of mass |
Perfectly elastic collisions conserve both momentum and kinetic energy. This makes them ideal models for hard-sphere
collisions, gas molecules, and many introductory mechanics problems.
