Game Development Reference
In-Depth Information
x , in Equations (6.19) and (6.20), indicate the post-collision velocities.
The velocity normal to the line of action, v n , is not the post-collision value because that velocity
component doesn't change due to the collision.
You might be asking yourself why you need to bother with rotating the coordinate system
and then later rotating it back again. Why not just use conservation of momentum in the x-
and y-directions and be done with it? The problem with that approach is there would be more
unknowns than there would be equations. You need to use Equation (6.13) to solve for the post-
collision velocities, and Equation (6.13) only applies along the line of action of the collision.
The accent marks, as in v
Two-Dimensional Collision Example
Let's go through a sample problem that will demonstrate the process of analyzing a two-
dimensional, frictionless collision. In Figure 6-8, two spheres collide. Sphere 1 has a mass of
10 kg and is traveling horizontally with a velocity v 1 x = 8 m/s . Sphere 2 is stationary and has a
mass of 5 kg . The line of action for the collision is at an angle of 30 degrees with respect to the
x-axis. The coefficient of restitution between the two spheres is 0.9. What will be the post-collision
velocities of the two spheres?
m 2 = 5
kg
m 1 = 10
kg
v 1x = 8
m/s
y
θ = 30
x
Figure 6-8. A sample two-dimensional collision
This problem, like all linear collision problems, can be broken down into four steps:
1.
Determine the line-of-action vector for the collision.
2.
Use a rotation matrix to determine the velocity components along the line of action and
normal to it.
3.
Compute the post-collision velocities from Equation (6.14).
4.
Rotate the post-collision velocities back to the original Cartesian coordinate system.
The line-of-action vector has already been provided by the problem statement, so we can
move on to step 2. The velocity components parallel and normal to the line of action can be
computed from Equations (6.16) and (6.18). Sphere 2 is initially not moving, so its velocity
components will be zero.
 
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