KAHIBARO
Table of Contents
Extending Collisions into a Plane
In one dimensional collisions, all motion happens along a single line. In two dimensional collisions, objects can move in different directions after impact, so both the size and the direction of momentum matter. This makes two dimensional collisions a natural application of vectors.
The key idea remains the same, momentum is conserved for an isolated system. The difference is that momentum must now be conserved separately in each coordinate direction, usually the $x$ and $y$ directions.
If two objects collide in a plane, the total momentum before the collision equals the total momentum after the collision:
$$
\vec{p}_{1i} + \vec{p}_{2i} = \vec{p}_{1f} + \vec{p}_{2f}
$$
Because momentum is a vector, this single vector equation becomes two scalar equations:
$$
p_{x,\text{before}} = p_{x,\text{after}}
$$
$$
p_{y,\text{before}} = p_{y,\text{after}}
$$
In a two dimensional collision, conservation of momentum must be applied in each perpendicular direction separately.
Choosing Axes
The first step in solving a two dimensional collision is to choose a coordinate system. A common choice is to place the initial motion of one object along the $x$ axis. This often makes the initial $y$ momentum zero, which simplifies the equations.
Suppose object 1 moves initially with speed $v_{1i}$ along the $x$ axis, and object 2 is initially at rest. After the collision, object 1 moves at angle $\theta_1$ and speed $v_{1f}$, while object 2 moves at angle $\theta_2$ and speed $v_{2f}$.
Then the momentum equations are
$$
m_1 v_{1i} = m_1 v_{1f}\cos\theta_1 + m_2 v_{2f}\cos\theta_2
$$
and
$$
0 = m_1 v_{1f}\sin\theta_1 + m_2 v_{2f}\sin\theta_2
$$
The sign of each term depends on the direction chosen for positive axes. If one object moves below the $x$ axis, its $y$ component is negative.
Breaking Momentum into Components
In two dimensions, each momentum vector is written in terms of components. If an object of mass $m$ moves with speed $v$ at angle $\theta$ from the positive $x$ axis, then
$$
p_x = mv\cos\theta
$$
$$
p_y = mv\sin\theta
$$
This means the vector momentum can be written as
$$
\vec{p} = (mv\cos\theta)\hat{i} + (mv\sin\theta)\hat{j}
$$
Using components avoids confusion and turns the vector conservation law into algebra.
Elastic and Inelastic Cases in Two Dimensions
The momentum equations alone are often not enough to determine all unknown speeds and angles. In many collision problems, extra information is needed.
For an elastic collision, kinetic energy is also conserved:
$$
\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2
=
\frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2
$$
For an inelastic collision, kinetic energy is not generally conserved, even though momentum still is.
If the collision is perfectly inelastic, the objects stick together and move as one combined mass after the collision. In two dimensions, the final velocity of the combined object is found from
$$
(m_1 + m_2)\vec{v}_f = m_1\vec{v}_{1i} + m_2\vec{v}_{2i}
$$
So,
$$
\vec{v}_f = \frac{m_1\vec{v}_{1i} + m_2\vec{v}_{2i}}{m_1 + m_2}
$$
Momentum is always conserved in an isolated collision, but kinetic energy is conserved only in elastic collisions.
A Common Special Situation
A very common example is an object striking another object initially at rest, after which both move off at angles. This often appears in billiard ball type problems.
If object 2 is initially at rest, the initial momentum is entirely due to object 1. After collision, the final momentum vectors must add to give the original momentum vector. Geometrically, the two final momentum vectors combine head to tail to reproduce the initial one.
This picture shows that the vector sum of the final momenta equals the initial momentum.
Solving Strategy
A two dimensional collision problem is usually solved by translating the physical situation into component equations. The steps are systematic.
First, identify the masses, initial velocities, and known directions. Next, choose axes, often aligned with an initial velocity. Then write conservation of momentum in the $x$ and $y$ directions. If the collision is elastic, also write the kinetic energy equation. Finally, solve the resulting system for the unknown speeds or angles.
The following table summarizes this structure.
| Collision type | Always use | Sometimes also use |
|---|---|---|
| Any isolated 2D collision | $p_x$ conservation, $p_y$ conservation | Given angle or speed information |
| Elastic 2D collision | $p_x$ conservation, $p_y$ conservation | Kinetic energy conservation |
| Perfectly inelastic 2D collision | Vector momentum conservation | Common final velocity relation |
Example Structure
Suppose a mass $m_1$ moving with velocity $\vec{v}_{1i}$ hits a stationary mass $m_2$. After collision, the two masses move away with unknown speeds and known angles $\theta_1$ and $\theta_2$.
Then the unknowns are usually $v_{1f}$ and $v_{2f}$. The two momentum component equations are enough if both final angles are known:
$$
m_1 v_{1i} = m_1 v_{1f}\cos\theta_1 + m_2 v_{2f}\cos\theta_2
$$
$$
0 = m_1 v_{1f}\sin\theta_1 - m_2 v_{2f}\sin\theta_2
$$
Here the minus sign appears because the second object is taken to move below the $x$ axis.
If the collision is elastic and one of the angles is unknown, the kinetic energy equation can provide the extra relation.
Equal Mass Elastic Collisions
A famous result occurs for elastic collisions between equal masses when one mass is initially at rest. In that case, the two final velocities are perpendicular to each other.
This result comes from combining momentum conservation with kinetic energy conservation. It is a special result, not a general rule for all two dimensional collisions.
For an elastic collision of equal masses, with one object initially at rest, the two final velocity directions are at right angles.
Direction of the Final Motion in a Perfectly Inelastic Collision
If two objects stick together after colliding, the final direction is determined by the total initial momentum vector. If the initial momentum components are
$$
p_x = m_1 v_{1x} + m_2 v_{2x}
$$
$$
p_y = m_1 v_{1y} + m_2 v_{2y}
$$
then the direction angle $\phi$ of the combined motion satisfies
$$
\tan\phi = \frac{p_y}{p_x}
$$
and the magnitude of the final velocity is
$$
v_f = \frac{\sqrt{p_x^2 + p_y^2}}{m_1 + m_2}
$$
This is especially useful in crash type problems where two vehicles lock together and slide off in a new direction.
Common Difficulties
Students often make mistakes by treating momentum as if it were just a number. In two dimensional collisions, momentum has direction, so writing only one conservation equation is not enough.
Another common mistake is to use angles incorrectly. Always check whether the angle is measured from the $x$ axis, from the $y$ axis, or below the axis. The trigonometric components depend on this choice.
It is also important not to assume that kinetic energy is conserved unless the problem states or implies an elastic collision.
Do not add speeds directly when directions differ. Add momentum vectors by components.
Physical Interpretation
Two dimensional collisions show clearly why momentum is a vector quantity. After impact, the motion can spread into different directions, but the total momentum vector remains unchanged. The system redistributes momentum among the objects while still preserving the total in both horizontal and vertical directions.
This is why collision analysis in a plane is such an important step beyond one dimensional motion. It connects conservation laws with vector methods and gives a more realistic description of impacts in the real world.