In two-dimensional collisions, momentum is conserved as a vector. This means the x-component and y-component of total
momentum must each be conserved, provided the net external impulse on the system is negligible during the collision.
Vector momentum conservation
Main vector equation.
\[
\begin{aligned}
m_1\vec v_{1i}+m_2\vec v_{2i}
&=
m_1\vec v_{1f}+m_2\vec v_{2f}
\end{aligned}
\]
This one vector equation is equivalent to two scalar equations:
x- and y-component equations.
\[
\begin{aligned}
m_1v_{1ix}+m_2v_{2ix}
&=
m_1v_{1fx}+m_2v_{2fx} \\
m_1v_{1iy}+m_2v_{2iy}
&=
m_1v_{1fy}+m_2v_{2fy}
\end{aligned}
\]
Speed-angle form
A velocity entered as a speed and angle can be converted into components using:
Component conversion.
\[
\begin{aligned}
v_x &= v\cos\theta \\
v_y &= v\sin\theta
\end{aligned}
\]
Angles are measured counter-clockwise from the positive x-axis. A negative angle means clockwise from the positive x-axis.
Perfectly inelastic 2D collision
If the objects stick together, they share a common final velocity vector:
Common final vector.
\[
\begin{aligned}
\vec v_f
&=
\frac{m_1\vec v_{1i}+m_2\vec v_{2i}}{m_1+m_2}
\end{aligned}
\]
Momentum is conserved, but kinetic energy usually decreases.
Momentum-only vector solve
If one final velocity vector is known, the other final velocity vector is found directly from vector momentum conservation.
Solving for object 2.
\[
\begin{aligned}
\vec v_{2f}
&=
\frac{m_1\vec v_{1i}+m_2\vec v_{2i}-m_1\vec v_{1f}}{m_2}
\end{aligned}
\]
Solving for object 1.
\[
\begin{aligned}
\vec v_{1f}
&=
\frac{m_1\vec v_{1i}+m_2\vec v_{2i}-m_2\vec v_{2f}}{m_1}
\end{aligned}
\]
Known final directions
If the final directions are known but the final speeds are unknown, the two component equations can be solved as a
two-by-two linear system.
Final direction equations.
\[
\begin{aligned}
m_1s_1\cos\theta_{1f}+m_2s_2\cos\theta_{2f} &= p_{ix} \\
m_1s_1\sin\theta_{1f}+m_2s_2\sin\theta_{2f} &= p_{iy}
\end{aligned}
\]
Here \(s_1=|\vec v_{1f}|\) and \(s_2=|\vec v_{2f}|\). The directions must not be parallel; otherwise the system
does not have a unique solution.
Elastic 2D angle solve
In an elastic collision, kinetic energy is also conserved:
Kinetic energy conservation.
\[
\begin{aligned}
\frac12m_1|\vec v_{1i}|^2+\frac12m_2|\vec v_{2i}|^2
&=
\frac12m_1|\vec v_{1f}|^2+\frac12m_2|\vec v_{2f}|^2
\end{aligned}
\]
If the final direction of object 1 is known, write
\(\vec v_{1f}=s_1(\cos\theta_{1f},\sin\theta_{1f})\), use momentum to express
\(\vec v_{2f}\), then use kinetic energy conservation to solve for \(s_1\).
Central impact with restitution
For a frictionless central impact, the collision impulse acts along the impact normal direction \(\hat n\). The tangential
components are unchanged, while the normal components follow a one-dimensional restitution law.
Impulse along the normal.
\[
\begin{aligned}
J
&=
\frac{(1+e)\left[(\vec v_{1i}-\vec v_{2i})\cdot\hat n\right]}
{\frac{1}{m_1}+\frac{1}{m_2}}
\end{aligned}
\]
Final vectors after the normal impulse.
\[
\begin{aligned}
\vec v_{1f} &= \vec v_{1i}-\frac{J}{m_1}\hat n \\
\vec v_{2f} &= \vec v_{2i}+\frac{J}{m_2}\hat n
\end{aligned}
\]
When \(e=1\), the normal part of the impact is elastic. When \(e=0\), the normal relative speed after collision is zero.
Worked example
Suppose \(m_1=0.5\ \mathrm{kg}\), \(|\vec v_{1i}|=8\ \mathrm{m\,s^{-1}}\) at \(0^\circ\),
and \(m_2=0.3\ \mathrm{kg}\) is initially at rest. If object 1 leaves at \(30^\circ\) and the collision is elastic,
then the calculator uses both momentum and kinetic energy to solve the final speeds.
Initial momentum.
\[
\begin{aligned}
\vec p_i
&=
m_1\vec v_{1i}+m_2\vec v_{2i} \\
&=
(0.5)(8,0)+(0.3)(0,0) \\
&=
(4,0)\ \mathrm{kg\,m\,s^{-1}}
\end{aligned}
\]
With \(\theta_{1f}=30^\circ\), momentum gives \(\vec v_{2f}\) in terms of \(|\vec v_{1f}|\), and kinetic energy
closes the system.
Formula summary
| Case |
Formula |
Meaning |
| Vector momentum |
\(m_1\vec v_{1i}+m_2\vec v_{2i}=m_1\vec v_{1f}+m_2\vec v_{2f}\) |
Total momentum vector is conserved |
| x-component |
\(m_1v_{1ix}+m_2v_{2ix}=m_1v_{1fx}+m_2v_{2fx}\) |
Horizontal momentum conservation |
| y-component |
\(m_1v_{1iy}+m_2v_{2iy}=m_1v_{1fy}+m_2v_{2fy}\) |
Vertical momentum conservation |
| Stick together |
\(\vec v_f=(m_1\vec v_{1i}+m_2\vec v_{2i})/(m_1+m_2)\) |
Perfectly inelastic final velocity |
| Energy check |
\(\Delta K=K_f-K_i\) |
Shows energy loss, conservation, or added energy |
Momentum conservation in 2D is vector conservation: both the x-component and the y-component must balance.
