Interaction and Normals
Intro to Programming in 3D
Applications
Lecture 20
Collision handling
detection & response
detection
response
Particle-plane collision detection
Polyhedron-polyhedron collision detection
overlap of Bounding volumes
Vertex inside polyhedron test
Concave case
Convex case
Edge-face intersection test
kinematic response
Penalty method
Impulse force of collision
Collision Detection
• One part of the physics generally necessary in today’s
•
game environments
Basics
–
–
–
–
–
–
Ray-Polygon Intersection
Object motion vector is the ray
Wall or other object is the polygon(s)
Simple to implement
Polygon-Polygon Intersection
Can be expensive to calculate
• Separate from Collision Response
Ray Basics
• Segment-Plane Intersection
– Intersect movement segment with plane of polygon
– Segment is defined by start and end points
• Intersection Point
– Find the exact 3D location of the intersection
• Point in Polygon Check
– Easy for simple polygons
• Calculate dot product of edge normals to vector
• Fast, Minimum Storage Ray-Triangle Intersection (best)
– http://www.acm.org/jgt/papers/MollerTrumbore97/
– More difficult for concave polygons
• Sum angles between vectors to vertices
• Divide polygon into quadrants, sum edge crossings
• Scanline +/-
• If the point is in the polygon the movement vector went
through the surface and we have a collision
Collision detection: point-plane
E( p)  ax  by  cz  d  N  p  d
E ( p)  0
N
E ( p)  0
E ( p)  0
Collision detection: time of impact
2 options
Consider collision at next time step
Compute fractional time at which collision actually occurred
Tradeoff: accuracy v. complexity
Intersection Testing
• General goals: given two objects with current
•
•
and previous orientations specified, determine if,
where, and when the two objects intersect
Alternative: given two objects with only current
orientations, determine if they intersect
Sometimes, we need to find all intersections.
Other times, we just want the first one.
Sometimes, we just need to know if the two
objects intersect and don’t need the actual
intersection data.
Primitives
• We often deal with various different ‘primitives’
•
that we describe our geometry with. Objects are
constructed from these primitives
Examples
–
–
–
–
–
Triangles
Spheres
Cylinders
AABB = axis aligned bounding box
OBB = oriented bounding box
• At the heart of the intersection testing are
various primitive-primitive tests
Particle Collisions
• mainly be concerned with the problem of testing
•
•
•
if particles collide with solid objects
A particle can be treated as a line segment from
it’s previous position to it’s current position
If we are colliding against static objects, then we
just need to test if the line segment intersects
the object
Colliding against moving objects requires some
additional modifications that we will also look at
Segment vs. Triangle
• Does segment ab intersect triangle v0v1v2 ?
a
v2
v0
•x
v1
b
Segment vs. Triangle
• First, compute signed distances of a and b to plane
d a  a  v 0   n
n
a
d b  b  v 0   n
da
x
•
v0
• Reject if both are above or both are below triangle
• Otherwise, find intersection point x
d ab  d ba
x
d a  db
db
b
Segment vs. Triangle
• Is point x inside the triangle?
(x-v0)·((v2-v0)×n) > 0
• Test all 3 edges
v2
v2-v0
x-v0
v0
•x
v1
(v2-v0)×n
Faster Way
• Reduce to 2D: remove smallest dimension
v2
• Compute barycentric coordinates
•
x' =x-v0
e1=v1-v0
β
e2=v2-v0
v0
α=(x'×e2)/(e1×e2)
α
β=(x'×e1)/(e1×e2)
Reject if α<0, β<0 or α+β >1
x
v1
Segment vs. Mesh
• To test a line segment against a mesh of
•
•
triangles, simply test the segment against each
triangle
Sometimes, we are interested in only the ‘first’
hit along the segment from a to b. Other times,
we want all intersections. Still other times, we
just need any intersection.
Testing against lots of triangles in a large mesh
can be time consuming. We will look at ways to
optimize this later
Segment vs. Moving Mesh
• M0 is the object’s matrix at time t0
• M1 is the matrix at time t1
• Compute delta matrix:
•
•
M1=M0·MΔ
MΔ= M0-1·M1
Transform a by MΔ
a'=a·MΔ
Test segment a'b against object with matrix M1
Triangle vs. Triangle
•
Given two triangles: T1 (u0u1u2) and T2 (v0v1v2)
v2
u2
v0
T2
u0
T1
u1
v1
Triangle vs. Triangle
Step 1: Compute plane equations
n2=(v1-v0)×(v2-v0)
v2
d2=-n2·v0
v2-v0
n
v0
v1-v0
v1
Triangle vs. Triangle
• Step 2: Compute signed distances of T1 vertices to
•
•
plane of T2:
di=n2·ui+d2
(i=0,1,2)
Reject if all di<0 or all di>0
Repeat for vertices of T2 against plane of T1
u0
d0
Triangle vs. Triangle
• Step 3: Find intersection points
• Step 4: Determine if segment pq is inside
triangle or intersects triangle edge
p
q
Mesh vs. Mesh
• Geometry: points, edges, faces
• Collisions: p2p, p2e, p2f, e2e, e2f, f2f
• Relevant ones: p2f, e2e (point to face &
edge to edge)
• Multiple simultaneous collisions
Sphere-Sphere Intersection
• Two objects are said to have collided if their bounding
•
•
spheres intersect.
To determine if two spheres intersect, simply calculate
the distance between the centers of the two spheres.
If the distance is greater than the sum of the two sphere
radii, they don’t intersect. Otherwise they intersect.
r1
r2
d = r1 + r2
d > r1 + r2
Sphere-Plane Intersection
• Sometimes, it’s necessary to find the
•
intersection between a sphere and a plane, for
example, the bounding sphere of an object with
a wall (or slope).
Given that the equation of the plane is n.p = k
(where n is the unit normal of the plane, p is any
point on the plane, and k is a number)
– Then, given the center coordinates C of the sphere,
and the radius r of the sphere,
• The sphere and the plane intersect if |(n.C) – k| < r
Dot Product
• Let U and V be vectors such that U = (Ux, Uy,
•
•
Uz), and V = (Vx, Vy, Vz)
Then, the dot product U.V = UxVx + UyVy + UzVz
U.V is also equal to |U||V| cos q where q is the
angle between U and V.
U
q
V
Collision of Fast-Moving Objects
• We need a different method to detect collision of fast•
•
•
moving, and often small, objects.
Example, a bullet is fired, and we want to see if it
intersects a wall. However, if we examine every time
frame, because the bullet moves very fast, even though
at some point in time it intersects the wall, we may only
sample it in front of the wall and behind it, but on at the
point of intersection.
Therefore, we need to consider the path of the bullet,
and determine if that path intersects the wall.
We use a line to represent the path of the bullet. We
then test for line-object intersection. We consider:
– Line-Sphere intersection, and
– Line-Triangle intersection
Line-Sphere Intersection
• Let a point on a line be X(t) = P + tD
– Here X(t) is a function of t, and gives the point
on the line, P is the starting point of the line,
and D is a unit vector in the direction of the
line.
• Let a point on a sphere satisfy | X – C | = r
– Here, X is a point on the sphere, C is the
center of the sphere, and r is the radius of the
sphere.
Line-Sphere Intersection
• Suppose the line and sphere intersect at point X, then
| P + tD – C |2 – r2 = 0
– Let M = P – C. Then, | tD + M |2 – r2 = 0
– Expanding, t2 + 2D.Mt + | M |2 – r2 = 0
• See Note 1 on next page
– Solving for t, t = -D.M +/- sqrt((D.M)2 – ( | M |2 – r2 ))
• See Note 2 on next page
– The discriminant d is (D.M)2 – ( | M |2 – r2 )
• If d > 0, the line and sphere intersect at two points.
• If d = 0, the line and sphere intersect at one point.
• If d < 0, the line and sphere don’t intersect.
– If (d>0) or (d=0), we can solve for t. Assuming that P is the
position of the fast-moving object at the beginning of the game
loop, and D is the vector that it will travel during a game loop,
then the objects intersect during this game loop if 0<t<1.
Notes on Line-Sphere Intersection
Note 1
|a+b|2 = |a|2 + |b|2 + 2a.b
a+b
b
by
Note 2
Let A, B and C be
coefficients of the
quadratic equation:
Ax2 + Bx + C = 0
ay
a
ax
bx
Proof: |a+b|2 = (ax+bx)2 + (ay+by)2
= ax2 + 2axbx + bx2 + ay2+ 2ayby + by2
But, |a|2 = ax2 + ay2 and
|b|2 = bx2 + by2
Therefore, |a+b|2 = |a|2 + |b|2 + 2(axbx + ayby)
|a+b|2 = |a|2 + |b|2 + 2a.b
Then,
x = -B +/- sqrt(B2-4AC)
2A
Line-Triangle Intersection
• Once again, let a point on a line be X(t) =
P + tD
• Let a triangle be defined by its three
corner points P0, P1 and P2.
• Strategy:
– First, find the intersection between the line
and the plane containing the triangle.
– Then, find out if this point is within the triangle
Line-Triangle Intersection
The Plane Containing the Triangle
Equation of a plane:
A point p on the plane will satisfy the equation p.n = k
where n is the normal of the plane.
Step 1: Find the normal n of the plane
Let edge e0 be P1 – P0.
Let edge e1 be P2 – P1.
Then n = e0 x e1
In other words, the normal of the plane is the cross product of two edges.
Step 2: Find k
k = P0.n
Line-Triangle Intersection
Line-Plane Intersection
1. Substitute equation of the line into equation of the plane.
(P + tD) . n = k
2. Find t.
Re-arranging, t = (k – P.n)/(D.n)
3. Substitute t back to get intersection point.
Intersection point R = P + tD.
Line-Triangle Intersection
Check if point is within triangle
Remember that the cross product of
consecutive vectors going counter-clockwise
will always be of the same sign.
P2
e2
P0
R is inside the triangle if it is always to the
left side of each edge.
e1
R
e0
P1
Therefore, Point R is inside the triangle if:
(e0 x (R – P0)) . n > 0
and
(e1 x (R – P1)) . n > 0
and
(e2 x (R – P2)) . n > 0
Speeding up Collision Detection
• Spatial subdivision method
• Divide the space into different regions.
• At each step, determine which region each
object is in.
• Only test objects in the same region for
collision.
Collision detection: polyhedra
Order tests according to
computational complexity and power
of detection
1. test bounding volumes for overlap
2. test for vertex of one object inside of other object
3. test for edge of one object intersecting face
of other object
Collision detection: bounding
volumes
Don’t do vertex/edge intersection testing if
there’s no chance of an intersection between
the polyhedra
Want a simple test to remove easy cases
Tradeoff complexity of test with power
to reject non-intersecting polyhedra
(goodness of fit of bounding volume)
Bounding Spheres
Compute bounding sphere of vertices
Compute in object space and transform with
object
1.Find min/max pair of points in each
dimension
2. use maximally separated pair – use to create
initial bounding sphere (midpoint is center)
3. for each vertex adjust sphere to include
point
Bounding Boxes
Axis-aligned (AABB): use min/max in each
dimension
Oriented (OBB): e.g., use AABB in object space
and transform with object. Vertex is inside of
OBB if on inside of 6 planar equations
Bounding Slabs
For better fit bounding polyhedron: use
arbitrary (user-specified) collection of
bounding plane-pairs
Is a vertex between each pair?
d 2  N  P  d1
Convex Hull
Best fit convex polyhedron to concave
polyhedron but takes some (one-time)
computation
1. Find highest vertex, V1
2. Find remaining vertex that minimizes
angle with horizontal plane through point.
Call edge L
3. Form plane with this edge and horizontal
line perpendicular to L at V1
4. Find remaining vertex that for triangle that
minimizes angle with this plane. Add this
triangle to convex hull, mark edges as
unmatched
5. For each unmatched edge, find remaining
vertex that minimizes angle with the plane
of the edge’s triangle
Collision detection: polyhedra
1. test bounding volumes for overlap
2. test for vertex of one object inside of other object
3. test for edge of one object intersecting face of other obje
Collision detection: polyhedra
Intersection = NO
For each vertex, V, of object A
if (V is inside of B) intersection = YES
For each vertex, V, of object B
if (V is inside of A) intersection = YES
A vertex is inside a convex
polyhedron if it’s on the ‘inside’
side of all faces
A vertex is inside a cancave
polyhedron if a semi-infinite ray
from the vertex intersects an odd
number of faces
Collision detection: polyhedra
Edge intersection face test
Finds ALL polyhedral intersections
But is most expensive test
If vertices of edges are on opposite side of plane
of face
Calculate intersection of edge with
plane
Test vertex for inside face (2D test in plane of
face)
Collision detection: swept volume
Time & relative direction of travel sweeps out a volume
Only tractable in simple cases (e.g. linear translation)
If part of an object is in the volume,
it was intersected by object
Laws of Motion
• First law simplified into the sentence "A body
•
•
continues to maintain its state of rest or of
uniform motion unless acted upon by an
external unbalanced force." This law is known as
the law of inertia.
Second law is often stated as "F = ma: the net
force on an object is equal to the mass of the
object multiplied by its acceleration."
Third law Whenever a particle A exerts a force on
another particle B, B simultaneously exerts a force on
A with the same magnitude in the opposite direction.
This law is often simplified as "To every action
there is an equal and opposite reaction."
Linear Momentum and Collisions
Linear momentum is defined as:
p = mv
Momentum is given by mass times velocity.
Momentum is a vector.
The units of momentum are (no special unit):
[p] = kg·m/s
Since p is a vector, we can also consider
the components of momentum:
px = mvx
py = mvy
pz = mvz
Note: momentum is “large” if m and/or v is
large. (define large, meaning hard for
you to stop).
• Name an object with large momentum but
•
small velocity.
Name an object with large momentum but
small mass
Recall that
F  ma and
v
a
t
mv p
F 

t
t
Another way of writing Newton’s Second Law is
F = Dp/Dt= rate of change of momentum
This form is valid even if the mass is changing.
This form is valid even in Relativity and Quantum
Mechanics.
Impulse
We can rewrite F = Dp/Dt as:
FDt = Dp
I = FDt is known as the impulse.
The impulse of the force acting on an object equals
the change in the momentum of that object.
If there are no external forces on a
system, then the total momentum of that
system is constant. This is known as:
The Principle of
Conservation
of
Momentum
In that case, pi = pf.
Conservation of Momentum
• In the absence of external forces, the total
momentum of a given system remains
constant.
A 90 kg hockey player traveling with a velocity of 6 m/s collides head-on with an 80
kg player traveling a 7 m/s. If the two players entangle and continue traveling
together as a unit following the collision, what is their combined velocity?
Known: m1= 90 kg m2=80 kg v1= 6 m/s v2= -7 m/s
m1v1 + m2v2 = (m1 + m2) (v)
(90 kg) (6 m/s) + (80 kg) (-7 m/s) = (90 kg + 80 kg) (v)
540 kg m/s – 560 kg m/s = (170 kg) (v)
- 20 kg m/s = (170 kg) (v)
v = 0.12 m/s in the direction of the 80 kg player’s original direction of travel
Conservation of Momentum and
Newton’s Third Law
• Consider a system consisting of just the two masses m1
•
•
•
•
•
•
•
•
and m2.
Mass m1 exerts a force F21 on mass m2.
Mass m2 exerts a force F12 on mass m1.
Force on m1 = rate of change of momentum of m1
– F12 =Dp1 / Dt
Force on m2 = rate of change of momentum of m2
– F21 =Dp2 / Dt
Dp1 / Dt + Dp2 / Dt = F12 + F21 = 0 (Newton’s Third Law).
D(p1+p2 )/ Dt = 0
Rate of change of total momentum is zero.
Total Momentum does not change if net external force
is zero
– Composite objects can be treated like point particles
Internal vs. External Forces
Here the system is just the box and table.
Any forces between those two objects are
internal. Example: The normal forces
between the table and the box are internal
forces. Internal forces on the system sum
to zero.
system
External forces do not necessarily sum to zero.
Something outside the circle is pushing or pulling
something inside the circle.
Example: gravity is an external force.
Impulse and Bouncing
• Impulses are greater
•
when bouncing takes
place.
The impulse required
to bring an object to a
stop and then throw it
back again is greater
than the impulse
required to bring an
object to a stop.
The Forces
Equal and Opposite
More About Impulse: F-t The
Graph
• Impulse is a vector
•
•
•
quantity
The magnitude of the
impulse is equal to the
area under the force-time
curve
Dimensions of impulse
are M L / T
Impulse is not a property
of the particle, but a
measure of the change in
momentum of the particle
Impulse
• If your car runs into a
brick wall and you come
to rest along with the
car, there is a significant
change in momentum.
If you are wearing a seat
belt or if the car has an
air bag, your change in
momentum occurs over
a relatively long time
interval. If you stop
because you hit the
dashboard, your change
in momentum occurs
over a very short time
interval.
Impulse
• The impulse can also
•
•
be found by using the
time averaged force
I = F Dt
This would give the
same impulse as the
time-varying force
does
Recoil
• Recoil is the term that describes the backward
movement of an object that has propelled
another object forward. In the nuclear decay
example, the vn’ would be the recoil velocity.
Conservation of Momentum,
Archer Example
• The archer is standing
•
on a frictionless surface
(ice)
Approaches:
– Newton’s Second Law –
no information about F or
a
– Energy approach – no
information about work or
energy
– Momentum – yes
Archer Example, 2
• Let the system be the archer with bow (particle
•
•
•
1) and the arrow (particle 2)
There are no external forces in the x-direction,
so it is isolated in terms of momentum in the xdirection
Total momentum before releasing the arrow is 0
The total momentum after releasing the arrow is
p1f + p2f = 0
Archer Example, final
• The archer will move in the opposite
direction of the arrow after the release
– Agrees with Newton’s Third Law
• Because the archer is much more massive
than the arrow, his acceleration and
velocity will be much smaller than those of
the arrow
Overview: Collisions –
Characteristics
• We use the term collision to represent an event
•
•
during which two particles come close to each
other and interact by means of forces
The time interval during which the velocity
changes from its initial to final values is assumed
to be short
The interaction force is assumed to be much
greater than any external forces present
– This means the impulse approximation can be used
What goes on during the collision?
Force on m2
=F(12)
Force on m1
=F(21)
Collisions
In general, a “collision” is an interaction in
which
• two objects strike one another
• the net external impulse is zero or
negligibly small (momentum is conserved)
Examples: car crash; billiard balls
Collisions can involve more than 2 objects
Consider two particles:
v2
v1
m1
m2
1
1
V1
2
2
V2
What about conservation of energy?
The total energy of an isolated system is
conserved, but the total kinetic energy
may change.
• elastic collisions: K is conserved
• inelastic collisions: K is not conserved
• perfectly inelastic: objects stick together
after colliding
Collisions – Example 1
• Collisions may be the
•
result of direct contact
The impulsive forces
may vary in time in
complicated ways
– This force is internal to
the system
• Momentum is
conserved
Perfectly Inelastic Collisions
After a perfectly inelastic collision the two
objects stick together and move with the
same final velocity:
pi = pf
m1 v1,i + m2 v2,i = (m1+ m2)vf
This gives the maximum possible loss of kinetic energy.
In non-relativistic collisions, the total mass is conserved
From the conservation of momentum:
v1,i
v1,f
v2,i
v2,f
pi = pf
m1v1,i + m2v2,i = m1v1,f + m2v2,f
Perfectly Inelastic Collisions
• Since the objects
•
stick together, they
share the same
velocity after the
collision
m1v1i + m2v2i =
(m1 + m2) vf
Elastic Collisions
Kinetic energy is conserved and
momentum is conserved:
pi = pf
m1 v1,i  m2 v 2,i  m1 v1, f  m2 v 2, f
Ki = Kf
1
1
1
1
2
2
2
m1v1,i  m2v2,i  m1v1, f  m2 v22, f
2
2
2
2
Elastic Collisions
• Both momentum and
kinetic energy are
conserved
m1v1i  m2 v 2i 
m1v1 f  m2 v 2 f
1
1
2
m1v1i  m2 v 22i 
2
2
1
1
2
m1v1 f  m2 v 22 f
2
2
Elastic Collisions in 1dimension
Kinetic energy is conserved in addition to
momentum:
m1v1,i  m2 v2,i  m1v1, f  m2 v2, f
pi = pf
m1 v1,i  v1, f   m2 v2, f  v2,i 
1
1
1
1
m1v12,i  m2 v22,i  m1v12, f  m2 v22, f
2
2
2
2
Ki = Kf


 
m1 v12,i  v12, f  m2 v22, f  v22,i

 


m1 v1,i  v1, f v1,i  v1, f  m2 v2, f  v2,i v2, f  v2,i
Divide :
v

1, i  v1, f  v2, f  v2,i


v1,i  v2,i  v2, f  v1, f
Relative velocity of approach before collision = relative velocity of separation after collision
***Inelastic head-on collision between a
car and a truck…
A 3000-kg truck moving with a velocity of 20 m/s rearends a 1000-kg parked car. The impact causes the
1000-kg car to be set in motion at 15 m/s. Are the two
vehicles stuck together after the collision?
***The animation below portrays the elastic collision
between a 1000-kg car and a 3000-kg truck.
Two Dimensional Collisions
Do you know how to play
pool?
Rebound
• When objects/bodies separate (move apart) after a
collision or impact occurs.
• Angle of incidence and angle of reflection/rebound
measured with respect to the vertical.
• Coefficient of elasticity/restitution refers to the degree
(amount) of recoil/bounce that objects have. The greater
the bounce the greater the coefficient (value between 0
and 1) with 0 signifying a completely inelastic object and
1 signifying a completely elastic object.
• Affected by temperature and rebounding surface. Heat
causes balls to bounce more while artificial turf also will
cause a greater bounce.
Angle of Reflection/Rebound
Incidence
Rebound
A 3.00-kg steel ball strikes a wall with a speed of 10.0 m/s at
an angle of 60.0° with the surface. It bounces off with the
same speed and angle (Fig. P9.9). If the ball is in contact
with the wall for 0.200 s, what is the average force exerted
on the ball by the wall?
Two-Dimensional Collisions
• The momentum is conserved in all directions
• Use subscripts for
– identifying the object
– indicating initial or final values
– the velocity components
• If the collision is elastic, use conservation of
kinetic energy as a second equation
– Remember, the simpler equation can only be used for
one-dimensional situations
Two-Dimensional Collision,
example
• Particle 1 is moving at
•
•
velocity v1i and
particle 2 is at rest
In the x-direction, the
initial momentum is
m1v1i
In the y-direction, the
initial momentum is 0
Two-Dimensional Collision,
example cont
• After the collision, the
•
momentum in the xdirection is m1v1f cos
q + m2v2f cos f
After the collision, the
momentum in the ydirection is m1v1f sin q
+ m2v2f sin f
Two-Dimensional Collision
Example
• Before the collision,
•
the car has the total
momentum in the xdirection and the van
has the total
momentum in the ydirection
After the collision,
both have x- and ycomponents
Elastic Collisions Involving an
Angle
• Momentum is conserved in both the xdirection and in the y-direction.
• Before:
v 1 x  v 1  cos 1
positive
v 1y  v 1  sin 1
negative
v 2 x  v 2  cos 2
positive
v 2 y  v 2  sin 2
positive
Elastic Collisions Involving an
Angle
• After:
v1x '  v1 ' cos θ 3
positive
v1y '  v1 ' sin θ 3
positive
v 2 x '  v 2 ' cos θ 4
positive
v 2 y '  v 2 ' sin θ 4
negative
Elastic Collisions Involving an Angle
• Directions for the velocities before and
after the collision must include the
positive or negative sign.
• The direction of the x-components for
v1 and v2 do not change and therefore
remain positive.
• The directions of the y-components for
v1 and v2 do change and therefore one
velocity is positive and the other
velocity is negative.
Elastic Collisions Involving an Angle
• px before = px after
m1  v1x  m2  v 2 x  m1  v1x 'm2  v 2 x '
• py before = py after
m1  v1y  m2  v 2 y  m1  v1y 'm2  v 2 y '
• Velocity after collision:
2
v1 '  v1x '  v1 y '
2
v 2 '  v 2 x '2  v 2 y ' 2
Elastic Collisions
• Perfectly elastic collisions do not
have to be head-on.
• Particles can divide or break apart.
• Example: nuclear decay (nucleus
of an element emits an alpha
particle and becomes a different
element with less mass)
Elastic Collisions


mn  v n  mn  mp  v n 'mp  v p
•
•
•
•
•
mn = mass of nucleus
mp = mass of alpha particle
vn = velocity of nucleus before event
vn’ = velocity of nucleus after event
vp = velocity of particle after event
Head-on and Glancing Collisions
• Head-on collisions occur when all of the
motion, before and after the collision, is
along one straight line.
• Glancing collisions involve an angle.
• A vector diagram can be used to represent
the momentum for a glancing collision.
Vector Diagrams
• Use the three vectors and construct a
triangle.
Vector Diagrams
mB  v B m A  v A

•
sin 115 sin30
mB  v B

sin 115
Use the
appropriate
mB  v B ' expression to
determine the
sin 35 unknown
variable.
mR  v R mB  v B '

sin 30 sin 35
Vector Diagrams
• Total vector momentum is conserved. You
•
•
could break each momentum vector into an x
and y component.
px before = px after
py before = py after
You would use the x and y components to
determine the resultant momentum for the object
in question
Resultant momentum =
2
px  py
2
Vector Diagrams
• Right triangle trigonometry can be used
to solve this type of problem:
Vector Diagrams
• Pythagorean theorem:
ma  va 
2
 mb  vb   ma  mb   v T 
2
2
• If the angle  for the direction in which the
cars go in after the collision is known, you can
use sin, cos, or tan to determine the unknown
quantity. Example: determine final velocity vT
if the angle is 25°.
ma  v a
sin 25 
ma  m b   v T
mb  v b
cos 25 
ma  m b   v T
Vector Diagrams
• To determine the angle at which the cars go off
together after the impact:
1  ma
 va
θ  tan 
 mb  v b




Special Condition
• When a moving ball strikes a stationary
ball of equal mass in a glancing collision,
the two balls move away from each other
at right angles.
• ma = mb
• va = 0 m/s
Special Condition
• Use the three vectors to construct a
triangle.
Special Condition
mB  v B m A  v A

• Use the
sin 90
sin 50 appropriate
mB  v B mB  v B ' expression to

determine the
sin 90
sin 40
m A  v A ' mB  v B '

sin 50
sin 40
unknown
variable.
Elastic Collision Example
• Example: mass 1 and mass 2 collide
•
and bounce off of each other
Momentum equation:
m1  v1  m2  v 2  m1  v1'm2  v 2 '
• Kinetic energy equation:
0.5  m1  v12  0.5  m2  v22  0.5  m1  v1' 2 0.5  m2  v2 ' 2
v1 and v2 = velocities before collision
v1 and v2 = velocities after collision
• Velocities are + or – to indicate
directions.
Elastic Collision Example
• Working with kinetic energy:
0.5  m1  v12  0.5  m2  v22  0.5  m1  v1'2 0.5  m2  v2 '2
• 0.5 cancels out.
m1  v12  m2  v 22  m1  v1'2 m2  v 2 '2
m1  v12  m1  v1'2  m2  v 2 '2 m2  v 22
m1 

2
v1
2
 v1'
2
  m  v ' v 
2
2
v 2 ' v 2
m1
 2
m2 v1  v1'2
2
2
2
2
Elastic Collision Example
• The velocity terms are perfect squares and can
be factored:
a2-b2 = (a – b)·(a + b)

v 2 ' v 2   v 2 ' v 2 
m1

m 2 v1  v1 '  v1  v1 '
• We will use this equation later.
Elastic Collision Example
• Momentum equation:
m1  v1  m2  v 2  m1  v1'm2  v 2 '
m1  v1  m1  v1'  m2  v 2 'm2  v 2
m1  v1  v1'  m2  v 2 'v 2 
v 2 'v 2
m1

m2 v1  v1'
Elastic Collision Example
• Both the kinetic energy and momentum
•
equations have been solved for the ratio of
m1/m2.
Set m1/m2 for kinetic energy equal to m1/m2 for
momentum:
v 2 'v 2   v 2 ' v 2   v 2 'v 2
v1  v1'  v1  v1' v1  v1'
Elastic Collision Example
• Get all the v1 terms together and all the v2 terms
together:
v 2 'v 2   v 2 ' v 2   v1  v1'  v1  v1'
v 2 ' v 2
v1  v1 '
• Cancel the like terms:
v 2 ' v 2  v1  v1'
Elastic Collision Example
• Rearrange to get the initial and final
velocities back together on the same
side of the equation:
v 2  v1  v1'v 2 '
• This equation can be solved for one of
the two unknowns, then substituted
back into the conservation of
momentum equation.
Collision response: kinematic
v(ti 1 )  v(ti )  N  v(ti )  k ( N  v(ti ))
v(ti )  N  v(ti )
v(ti )
N
N  v(ti )
 v(ti )  (1  k ) N  v(ti )
k – damping factor
=1 indicates no energy loss
Negate component of velocity in
direction of normal
No forces involved!
Collision response – penalty method
Collision reaction
Coefficient of restitution
v(ti 1 )  v(ti )  N  v(ti )  k ( N  v(ti ))
v(ti )  N  v(ti )
N
v(ti )
N  v(ti )
 v(ti )  (1  k ) N  v(ti )
k – coefficient of restitution
But now want to add angular velocity
contribution to separation velocity
Impulse response
A (t )
x A (t )
vB (t )
B (t )
xB (t )
v A (t )
How to compute the collision
response of two rotating rigid
objects?
Impulse response
Given
Separation velocity is to be negative of colliding velocity
Compute
Impulse force that produces sum of linear and angular
velocities that produce desired separation velocity
Rigid body simulation
Impulse force
Separation velocity

vrel  vrel
j  ft
A (t )
x A (t )
vB (t )
pA
pB
xB (t )
v A (t )
B (t )

Update linear and angular velocities
as a result of impulse force
A (t )

vA  vA

vB  vB



jn

MA
jn

MB

1
 A   A  I A (t )(rA  jn)
 B    B   I B 1 (t )(rB  jn)
x A (t )
vB (t )
pA
pB
xB (t )
v A (t )
B (t )
Velocities of points of contact
A (t )
x A (t )
rA
pA
pB
rA  p A  x A (t )
rB  p B  x B (t )
 A (t )  p
 B (t ))  n
vrel  ( p
 A (t )  v A (t )   A (t )  rA
p
 B (t )  v B (t )   B (t )  rB
p
vB (t )
rB
xB (t )
v A (t )
B (t )