10.6 Collisions of Extended Bodies in Two Dimensions

Collisions of Extended Bodies in Two Dimensions

When extended bodies collide, you can't just track where they go; you also have to account for how they spin. Unlike point-particle collisions, these interactions involve both linear and angular momentum changing simultaneously.

This matters because most real objects (bats, cars, spinning discs) have size and shape. Understanding how impulse location affects both translation and rotation is central to analyzing these collisions.

Collisions of Extended Bodies in Two Dimensions

Collisions of extended bodies

Two conservation laws govern these collisions, and you need both.

Conservation of linear momentum still holds in two dimensions. The total linear momentum of the system before the collision equals the total after:

$m_1 \vec{v}_1 + m_2 \vec{v}_2 = m_1 \vec{v'}_1 + m_2 \vec{v'}_2$

• $m_1, m_2$ are the masses of the two objects
• $\vec{v}_1, \vec{v}_2$ are their velocities before the collision
• $\vec{v'}_1, \vec{v'}_2$ are their velocities after

Since this is two dimensions, remember that this equation applies independently to both the $x$ and $y$ components.

Conservation of angular momentum is what makes extended-body collisions different from point-particle ones. If no external torques act on the system, total angular momentum is conserved:

$I_1 \omega_1 + I_2 \omega_2 = I_1 \omega'_1 + I_2 \omega'_2$

• $I_1, I_2$ are the moments of inertia of each object (about their respective rotation axes)
• $\omega_1, \omega_2$ and $\omega'_1, \omega'_2$ are the angular velocities before and after

Connecting impulse to both types of momentum: The impulse $\vec{J}$ during the collision equals the change in linear momentum of an object:

$\vec{J} = \Delta \vec{p} = m(\vec{v'} - \vec{v})$

That same impulse also changes the angular momentum, but the effect depends on where the force is applied. The change in angular momentum is:

$\Delta \vec{L} = \vec{r} \times \vec{J}$

Here $\vec{r}$ is the position vector from the rotation axis to the point where the impulse acts. This cross product means that an impulse applied far from the axis produces a larger change in angular momentum than the same impulse applied close to it. The magnitude works out to $|\Delta L| = rJ\sin\theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{J}$.

Percussion point applications

The center of percussion (or percussion point) is the specific point on an extended body where you can apply an impulse without producing a reaction force at the pivot or axis of rotation. Hit an object at this point, and the pivot feels no jolt.

It sits at a distance $R$ from the pivot, given by:

$R = \frac{I}{mh}$

• $I$ is the moment of inertia about the pivot axis
• $m$ is the object's mass
• $h$ is the distance from the pivot to the center of mass

Why does this matter in practice?

• Baseball bats: The "sweet spot" is near the center of percussion. Hitting the ball there minimizes the stinging vibration you feel in your hands (the pivot point) and transfers energy to the ball most efficiently.
• Tennis rackets: Striking near the center of percussion reduces the twisting torque on your wrist, giving you a cleaner, more controlled shot.
• Golf clubs: The percussion point on the clubface helps produce a straight shot by minimizing unwanted rotation of the clubhead at impact.

In each case, the physics is the same: hitting at the percussion point means the impulse produces rotation and translation that are perfectly balanced, so the pivot point stays still.

Energy changes in rotating collisions

For extended bodies, total kinetic energy has two parts. You need to track both.

Rotational kinetic energy:

$K_r = \frac{1}{2} I \omega^2$

Linear (translational) kinetic energy:

$K_l = \frac{1}{2} m v^2$

The total change in kinetic energy during a collision is the sum of changes in both:

$\Delta K = \Delta K_r + \Delta K_l = \frac{1}{2} I (\omega'^2 - \omega^2) + \frac{1}{2} m (v'^2 - v^2)$

This total determines the collision type:

• Elastic collisions: $\Delta K = 0$. Total kinetic energy (rotational + translational) is conserved. No energy is lost to deformation or heat.
• Inelastic collisions: $\Delta K < 0$. Some kinetic energy is converted to heat, sound, or permanent deformation. A car crash is a classic example.

The coefficient of restitution $e$ quantifies how elastic a collision is. It ranges from 0 (perfectly inelastic, objects stick together) to 1 (perfectly elastic). For extended bodies, you need to be careful: even if translational kinetic energy decreases, rotational kinetic energy might increase, so always check the total.

Additional considerations in extended body collisions

A few factors can complicate the analysis beyond the core conservation laws:

• Center of mass motion: The center of mass of the entire system moves at a constant velocity throughout the collision (assuming no external forces). Analyzing the collision in the center-of-mass frame can simplify the math considerably.
• Friction at the contact point: If the colliding surfaces aren't frictionless, tangential impulses arise during the collision. These can spin up or slow down rotation and redirect linear motion, adding another unknown to your equations.
• Coupling of rotation and translation: For extended bodies, angular momentum conservation and linear momentum conservation aren't independent problems. The point of contact determines how the impulse splits its effect between changing $\vec{v}$ and changing $\omega$, which is exactly why the percussion point concept is so useful.