8.6 Collisions of Point Masses in Two Dimensions

Collision Analysis in Two Dimensions

When objects collide at angles rather than head-on along a single line, you need to work with momentum as a vector quantity in two dimensions. This is where your vector decomposition skills really pay off: you'll apply conservation of momentum separately to the x and y directions, turning one complicated 2D problem into two manageable 1D problems.

Point masses are an idealization where you treat objects as having mass but no physical size. This simplifies collision analysis because you don't need to worry about rotation or where exactly on the object the contact happens. During the collision itself, the interaction forces are so large and brief that external forces like gravity and friction are negligible by comparison.

Conservation of momentum in each direction

The core principle: total momentum before the collision equals total momentum after, and this holds independently for each axis.

x-direction:

$m_1v_{1x} + m_2v_{2x} = m_1v'_{1x} + m_2v'_{2x}$

y-direction:

$m_1v_{1y} + m_2v_{2y} = m_1v'_{1y} + m_2v'_{2y}$

• $v_{1x}, v_{2x}, v_{1y}, v_{2y}$ are the initial velocity components (before collision)
• $v'_{1x}, v'_{2x}, v'_{1y}, v'_{2y}$ are the final velocity components (after collision)

A common setup: one object moves along the x-axis and strikes a stationary object. In that case, $v_{2x} = v_{2y} = 0$ and $v_{1y} = 0$, which simplifies the equations considerably.

Elastic Collisions and Scattering Angles

Elastic collisions of equal masses

In an elastic collision, both momentum and kinetic energy are conserved. For two objects of equal mass ($m_1 = m_2 = m$), the mass cancels from every term, which makes the algebra cleaner.

Conservation of momentum becomes:

$\vec{v}_1 + \vec{v}_2 = \vec{v}'_1 + \vec{v}'_2$

Conservation of kinetic energy becomes:

$v_1^2 + v_2^2 = v'^2_1 + v'^2_2$

A key result for equal-mass elastic collisions: if one object is initially stationary, the two objects always scatter at 90° relative to each other after the collision. Think of billiard balls or air hockey pucks. This 90° rule is a useful check on your answers.

Velocities and angles in 2D collisions

Scattering angles $\theta_1$ and $\theta_2$ describe how each object's trajectory deflects relative to the original direction of motion. You connect these angles to velocity components using trigonometry:

• Object 1: $v'_{1x} = v'_1 \cos{\theta_1}$, $v'_{1y} = v'_1 \sin{\theta_1}$
• Object 2: $v'_{2x} = v'_2 \cos{\theta_2}$, $v'_{2y} = v'_2 \sin{\theta_2}$

To solve a 2D elastic collision problem, follow these steps:

1. Choose a coordinate system. Typically, align the x-axis with the initial velocity of one object.
2. Break all initial velocities into x and y components.
3. Write the conservation of momentum equation for x and for y.
4. Write the conservation of kinetic energy equation.
5. Substitute the trigonometric expressions for final velocity components.
6. Solve the system of equations for the unknowns (final speeds and angles).

You get three equations (momentum-x, momentum-y, kinetic energy) and four unknowns ($v'_1, v'_2, \theta_1, \theta_2$). That means you typically need one quantity given (like a scattering angle) to solve the rest.

Types of collisions by geometry:

• Head-on collisions produce scattering angles of 0° (forward) or 180° (backward), reducing to the 1D case.
• Glancing collisions produce scattering angles between 0° and 180°, where both objects deflect off the original line of motion.

Additional Collision Concepts

Impulse and relative motion

Impulse is the change in momentum of an object during a collision: $\vec{J} = \Delta \vec{p} = \vec{F}_{avg} \Delta t$. In 2D, impulse is also a vector, so it has x and y components just like momentum.

The relative velocity between two objects before and after the collision helps characterize what type of collision occurred. For a perfectly elastic collision, the relative speed of separation equals the relative speed of approach. For an inelastic collision, the objects separate more slowly (or not at all, in the perfectly inelastic case).