Interaction and Normals Intro to Programming in 3D Applications Lecture 20

Collision handling 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 detection response

Collision Detection One part of the physics generally necessary in today’s game environments One part of the physics generally necessary in today’s game environments Basics 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 Separate from Collision Response

Ray Basics Segment-Plane Intersection Segment-Plane Intersection – Intersect movement segment with plane of polygon – Segment is defined by start and end points Intersection Point Intersection Point – Find the exact 3D location of the intersection Point in Polygon Check Point in Polygon Check – Easy for simple polygons Calculate dot product of edge normals to vector Calculate dot product of edge normals to vector Fast, Minimum Storage Ray-Triangle Intersection (best) Fast, Minimum Storage Ray-Triangle Intersection (best) – – More difficult for concave polygons Sum angles between vectors to vertices Sum angles between vectors to vertices Divide polygon into quadrants, sum edge crossings Divide polygon into quadrants, sum edge crossings Scanline +/- Scanline +/- If the point is in the polygon the movement vector went through the surface and we have a collision If the point is in the polygon the movement vector went through the surface and we have a collision

Collision detection: point-plane N

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 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 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. 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 We often deal with various different ‘primitives’ that we describe our geometry with. Objects are constructed from these primitives Examples 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 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 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 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 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 Colliding against moving objects requires some additional modifications that we will also look at

Segment vs. Triangle Does segment ab intersect triangle v 0 v 1 v 2 ? Does segment ab intersect triangle v 0 v 1 v 2 ?

Segment vs. Triangle First, compute signed distances of a and b to plane First, compute signed distances of a and b to plane Reject if both are above or both are below triangle Reject if both are above or both are below triangle Otherwise, find intersection point x Otherwise, find intersection point x

Segment vs. Triangle Is point x inside the triangle? Is point x inside the triangle? (x-v 0 )·((v 2 -v 0 )×n) > 0 (x-v 0 )·((v 2 -v 0 )×n) > 0 Test all 3 edges Test all 3 edges x v0v0 v1v1 v2v2 v 2 -v 0 (v 2 -v 0 )×n x-v 0

Faster Way Reduce to 2D: remove smallest dimension Reduce to 2D: remove smallest dimension Compute barycentric coordinates Compute barycentric coordinates x' =x-v 0 e 1 =v 1 -v 0 e 2 =v 2 -v 0 α =(x'×e 2) /(e 1 ×e 2 ) β =(x'×e 1) /(e 1 ×e 2 ) Reject if α 1 Reject if α 1 x v0v0 v1v1 v2v2 α β

Segment vs. Mesh To test a line segment against a mesh of triangles, simply test the segment against each triangle 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. 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 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 M 0 is the object’s matrix at time t 0 M 0 is the object’s matrix at time t 0 M 1 is the matrix at time t 1 M 1 is the matrix at time t 1 Compute delta matrix: Compute delta matrix: M 1 =M 0 ·M Δ M Δ = M 0 -1 ·M 1 Transform a by M Δ Transform a by M Δ a'=a·M Δ a'=a·M Δ Test segment a'b against object with matrix M 1 Test segment a'b against object with matrix M 1

Triangle vs. Triangle Given two triangles: T 1 (u 0 u 1 u 2 ) and T 2 (v 0 v 1 v 2 ) Given two triangles: T 1 (u 0 u 1 u 2 ) and T 2 (v 0 v 1 v 2 ) u0u0 u2u2 u1u1 v0v0 v1v1 v2v2 T1T1 T2T2

Triangle vs. Triangle Step 1: Compute plane equations n 2 =(v 1 -v 0 )×(v 2 -v 0 ) d 2 =-n 2 ·v 0 v0v0 v1v1 v2v2 v 1 -v 0 v 2 -v 0 n

Triangle vs. Triangle Step 2: Compute signed distances of T 1 vertices to Step 2: Compute signed distances of T 1 vertices to plane of T 2 : d i =n 2 ·u i +d 2 (i=0,1,2) Reject if all d i 0 Reject if all d i 0 Repeat for vertices of T 2 against plane of T 1 Repeat for vertices of T 2 against plane of T 1 d0d0 u0u0

Triangle vs. Triangle Step 3: Find intersection points Step 3: Find intersection points Step 4: Determine if segment pq is inside triangle or intersects triangle edge Step 4: Determine if segment pq is inside triangle or intersects triangle edge p q

Mesh vs. Mesh Geometry: points, edges, faces Geometry: points, edges, faces Collisions: p2p, p2e, p2f, e2e, e2f, f2f Collisions: p2p, p2e, p2f, e2e, e2f, f2f Relevant ones: p2f, e2e (point to face & edge to edge) Relevant ones: p2f, e2e (point to face & edge to edge) Multiple simultaneous collisions Multiple simultaneous collisions

Sphere-Sphere Intersection Two objects are said to have collided if their bounding spheres intersect. 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. 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. If the distance is greater than the sum of the two sphere radii, they don’t intersect. Otherwise they intersect. r1r1 r2r2 d = r 1 + r 2 d > r 1 + r 2

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). 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) 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 The sphere and the plane intersect if |(n.C) – k| < r

Dot Product Let U and V be vectors such that U = (U x, U y, U z ), and V = (V x, V y, V z ) Let U and V be vectors such that U = (U x, U y, U z ), and V = (V x, V y, V z ) Then, the dot product U.V = U x V x + U y V y + U z V z Then, the dot product U.V = U x V x + U y V y + U z V z U.V is also equal to |U||V| cos q where q is the angle between U and V. U.V is also equal to |U||V| cos q where q is the angle between U and V. U V q

Collision of Fast-Moving Objects We need a different method to detect collision of fast- moving, and often small, 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. 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. 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: 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 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 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 Suppose the line and sphere intersect at point X, then | P + tD – C | 2 – r 2 = 0 – Let M = P – C. Then, | tD + M | 2 – r 2 = 0 – Expanding, t 2 + 2D.Mt + | M | 2 – r 2 = 0 See Note 1 on next page See Note 1 on next page – Solving for t, t = -D.M +/- sqrt((D.M) 2 – ( | M | 2 – r 2 )) See Note 2 on next page See Note 2 on next page – The discriminant d is (D.M) 2 – ( | M | 2 – r 2 ) If d > 0, the line and sphere intersect at two points. 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 intersect at one point. If d < 0, the line and sphere don’t intersect. 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 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 |a+b| 2 = |a| 2 + |b| 2 + 2a.b a b a+b axax byby bxbx ayay Proof: |a+b| 2 = (a x +b x ) 2 + (a y +b y ) 2 = a x 2 + 2a x b x + b x 2 + a y 2 + 2a y b y + b y 2 But, |a| 2 = a x 2 + a y 2 and |b| 2 = b x 2 + b y 2 Therefore, |a+b| 2 = |a| 2 + |b| 2 + 2(a x b x + a y b y ) |a+b| 2 = |a| 2 + |b| 2 + 2a.b Note 1 Note 2 Let A, B and C be coefficients of the quadratic equation: Ax 2 + Bx + C = 0 Then, x = -B +/- sqrt(B 2 -4AC) 2A

Line-Triangle Intersection Once again, let a point on a line be X(t) = P + tD Once again, let a point on a line be X(t) = P + tD Let a triangle be defined by its three corner points P 0, P 1 and P 2. Let a triangle be defined by its three corner points P 0, P 1 and P 2. Strategy: 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 e 0 be P 1 – P 0. Let edge e 1 be P 2 – P 1. Then n = e 0 x e 1 In other words, the normal of the plane is the cross product of two edges. Step 2: Find k k = P 0.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 P0P0 P1P1 P2P2 R e0e0 e1e1 e2e2 Therefore, Point R is inside the triangle if: (e 0 x (R – P 0 )). n > 0 and (e 1 x (R – P 1 )). n > 0 and (e 2 x (R – P 2 )). n > 0 Remember that the cross product of consecutive vectors going counter-clockwise will always be of the same sign. R is inside the triangle if it is always to the left side of each edge.

Speeding up Collision Detection Spatial subdivision method Spatial subdivision method Divide the space into different regions. Divide the space into different regions. At each step, determine which region each object is in. At each step, determine which region each object is in. Only test objects in the same region for collision. Only test objects in the same region for collision.

Collision detection: polyhedra 2. test for vertex of one object inside of other object 1. test bounding volumes for overlap 3. test for edge of one object intersecting face of other object Order tests according to computational complexity and power of detection

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?

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 2. test for vertex of one object inside of other object 1. test bounding volumes for overlap 3. test for edge of one object intersecting face of other object

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. 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." 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." 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: p x = mv x p y = mv y p z = mv z 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 velocity. Name an object with large momentum but small mass Name an object with large momentum but small mass

Recall that 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 ConservationofMomentum In that case, p i = p f.

Conservation of Momentum In the absence of external forces, the total momentum of a given system remains constant. 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: m 1 = 90 kg m 2 =80 kg v 1 = 6 m/s v 2 = -7 m/s m1v1 + m 2 v 2 = (m 1 + m 2 ) (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 m 1 and m 2. Consider a system consisting of just the two masses m 1 and m 2. Mass m 1 exerts a force F 21 on mass m 2. Mass m 1 exerts a force F 21 on mass m 2. Mass m 2 exerts a force F 12 on mass m 1. Mass m 2 exerts a force F 12 on mass m 1. Force on m 1 = rate of change of momentum of m 1 Force on m 1 = rate of change of momentum of m 1 – F 12 =Dp 1 / Dt Force on m 2 = rate of change of momentum of m 2 Force on m 2 = rate of change of momentum of m 2 – F 21 =Dp 2 / Dt Dp 1 / Dt + Dp 2 / Dt = F 12 + F 21 = 0 (Newton’s Third Law). Dp 1 / Dt + Dp 2 / Dt = F 12 + F 21 = 0 (Newton’s Third Law). D(p 1 +p 2 )/ Dt = 0 D(p 1 +p 2 )/ Dt = 0 Rate of change of total momentum is zero. Rate of change of total momentum is zero. Total Momentum does not change if net external force 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 system 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. 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. 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 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 Impulse is a vector quantity The magnitude of the impulse is equal to the area under the force-time curve The magnitude of the impulse is equal to the area under the force-time curve Dimensions of impulse are M L / T 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 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. 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 The impulse can also be found by using the time averaged force I = Dt I = Dt This would give the same impulse as the time-varying force does 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 v n ’ would be the recoil velocity. Recoil is the term that describes the backward movement of an object that has propelled another object forward. In the nuclear decay example, the v n ’ would be the recoil velocity.

Conservation of Momentum, Archer Example The archer is standing on a frictionless surface (ice) The archer is standing on a frictionless surface (ice) Approaches: 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) 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 x- direction There are no external forces in the x-direction, so it is isolated in terms of momentum in the x- direction Total momentum before releasing the arrow is 0 Total momentum before releasing the arrow is 0 The total momentum after releasing the arrow is p 1f + p 2f = 0 The total momentum after releasing the arrow is p 1f + p 2f = 0

Archer Example, final The archer will move in the opposite direction of the arrow after the release 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 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 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 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 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 m 2 =F(1  2) Force on m 1 =F(2  1)

Collisions In general, a “collision” is an interaction in which two objects strike one another two objects strike one another the net external impulse is zero or negligibly small (momentum is conserved) 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: 1 2 m 1 m 2 v1v1 v2v2 V 1 V

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 elastic collisions: K is conserved inelastic collisions: K is not conserved inelastic collisions: K is not conserved perfectly inelastic: objects stick together after colliding perfectly inelastic: objects stick together after colliding

Collisions – Example 1 Collisions may be the result of direct contact Collisions may be the result of direct contact The impulsive forces may vary in time in complicated ways The impulsive forces may vary in time in complicated ways – This force is internal to the system Momentum is conserved Momentum is conserved

Perfectly Inelastic Collisions After a perfectly inelastic collision the two objects stick together and move with the same final velocity: p i = p f m 1 v 1,i + m 2 v 2,i = (m 1 + m 2 )v f This gives the maximum possible loss of kinetic energy. In non-relativistic collisions, the total mass is conserved

From the conservation of momentum: p i = p f m 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,f v 2,f v 2,i v 1,f v 1,i

Perfectly Inelastic Collisions Since the objects stick together, they share the same velocity after the collision Since the objects stick together, they share the same velocity after the collision m 1 v 1i + m 2 v 2i = m 1 v 1i + m 2 v 2i = (m 1 + m 2 ) v f (m 1 + m 2 ) v f

Elastic Collisions Kinetic energy is conserved and momentum is conserved: p i = p f K i = K f

Elastic Collisions Both momentum and kinetic energy are conserved Both momentum and kinetic energy are conserved

Elastic Collisions in 1- dimension Kinetic energy is conserved in addition to momentum: p i = p f K i = K 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 rear- ends 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. 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. 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. 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. 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 s, what is the average force exerted on the ball by the wall?

Two-Dimensional Collisions The momentum is conserved in all directions The momentum is conserved in all directions Use subscripts for 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 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 v 1i and particle 2 is at rest Particle 1 is moving at velocity v 1i and particle 2 is at rest In the x-direction, the initial momentum is m 1 v 1i In the x-direction, the initial momentum is m 1 v 1i In the y-direction, the initial momentum is 0 In the y-direction, the initial momentum is 0

Two-Dimensional Collision, example cont After the collision, the momentum in the x- direction is m 1 v 1f cos q + m 2 v 2f cos f After the collision, the momentum in the x- direction is m 1 v 1f cos q + m 2 v 2f cos f After the collision, the momentum in the y- direction is m 1 v 1f sin q + m 2 v 2f sin f After the collision, the momentum in the y- direction is m 1 v 1f sin q + m 2 v 2f sin f

Two-Dimensional Collision Example Before the collision, the car has the total momentum in the x- direction and the van has the total momentum in the y- direction Before the collision, the car has the total momentum in the x- direction and the van has the total momentum in the y- direction After the collision, both have x- and y- components After the collision, both have x- and y- components

Elastic Collisions Involving an Angle Momentum is conserved in both the x- direction and in the y-direction. Momentum is conserved in both the x- direction and in the y-direction. Before: Before:

Elastic Collisions Involving an Angle After: After:

Elastic Collisions Involving an Angle Directions for the velocities before and after the collision must include the positive or negative sign. Directions for the velocities before and after the collision must include the positive or negative sign. The direction of the x-components for v 1 and v 2 do not change and therefore remain positive. The direction of the x-components for v 1 and v 2 do not change and therefore remain positive. The directions of the y-components for v 1 and v 2 do change and therefore one velocity is positive and the other velocity is negative. The directions of the y-components for v 1 and v 2 do change and therefore one velocity is positive and the other velocity is negative.

Elastic Collisions Involving an Angle p x before = p x after p x before = p x after p y before = p y after p y before = p y after Velocity after collision: Velocity after collision:

Elastic Collisions Perfectly elastic collisions do not have to be head-on. Perfectly elastic collisions do not have to be head-on. Particles can divide or break apart. 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) Example: nuclear decay (nucleus of an element emits an alpha particle and becomes a different element with less mass)

Elastic Collisions m n = mass of nucleus m n = mass of nucleus m p = mass of alpha particle m p = mass of alpha particle v n = velocity of nucleus before event v n = velocity of nucleus before event v n ’ = velocity of nucleus after event v n ’ = velocity of nucleus after event v p = velocity of particle after event v p = 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. Head-on collisions occur when all of the motion, before and after the collision, is along one straight line. Glancing collisions involve an angle. Glancing collisions involve an angle. A vector diagram can be used to represent the momentum for a glancing collision. A vector diagram can be used to represent the momentum for a glancing collision.

Vector Diagrams Use the three vectors and construct a triangle. Use the three vectors and construct a triangle.

Vector Diagrams Use the appropriate expression to determine the unknown variable. Use the appropriate expression to determine the unknown variable.

Vector Diagrams Total vector momentum is conserved. You could break each momentum vector into an x and y component. Total vector momentum is conserved. You could break each momentum vector into an x and y component. p x before = p x after p y before = p y after You would use the x and y components to determine the resultant momentum for the object in question You would use the x and y components to determine the resultant momentum for the object in question Resultant momentum = Resultant momentum =

Vector Diagrams Right triangle trigonometry can be used to solve this type of problem: Right triangle trigonometry can be used to solve this type of problem:

Vector Diagrams Pythagorean theorem: Pythagorean theorem: 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 v T if the angle is 25°. 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 v T if the angle is 25°.

Vector Diagrams To determine the angle at which the cars go off together after the impact: To determine the angle at which the cars go off together after the impact:

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. 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. m a = m b m a = m b v a = 0 m/s v a = 0 m/s

Special Condition Use the three vectors to construct a triangle. Use the three vectors to construct a triangle.

Special Condition Use the appropriate expression to determine the unknown variable. Use the appropriate expression to determine the unknown variable.

Elastic Collision Example Example: mass 1 and mass 2 collide and bounce off of each other Example: mass 1 and mass 2 collide and bounce off of each other Momentum equation: Momentum equation: Kinetic energy equation: Kinetic energy equation: v 1 and v 2 = velocities before collision v 1 and v 2 = velocities after collision Velocities are + or – to indicate directions. Velocities are + or – to indicate directions.

Elastic Collision Example Working with kinetic energy: Working with kinetic energy: 0.5 cancels out. 0.5 cancels out.

Elastic Collision Example The velocity terms are perfect squares and can be factored: The velocity terms are perfect squares and can be factored: a 2 -b 2 = (a – b)·(a + b) a 2 -b 2 = (a – b)·(a + b) We will use this equation later. We will use this equation later.

Elastic Collision Example Momentum equation: Momentum equation:

Elastic Collision Example Both the kinetic energy and momentum equations have been solved for the ratio of m 1 /m 2. Both the kinetic energy and momentum equations have been solved for the ratio of m 1 /m 2. Set m 1 /m 2 for kinetic energy equal to m 1 /m 2 for momentum: Set m 1 /m 2 for kinetic energy equal to m 1 /m 2 for momentum:

Elastic Collision Example Get all the v 1 terms together and all the v 2 terms together: Get all the v 1 terms together and all the v 2 terms together: Cancel the like terms: Cancel the like terms:

Elastic Collision Example Rearrange to get the initial and final velocities back together on the same side of the equation: Rearrange to get the initial and final velocities back together on the same side of the equation: This equation can be solved for one of the two unknowns, then substituted back into the conservation of momentum equation. This equation can be solved for one of the two unknowns, then substituted back into the conservation of momentum equation.

Collision response: kinematic N k – damping factor =1 indicates no energy loss No forces involved! Negate component of velocity in direction of normal

Collision response – penalty method

Collision reaction Coefficient of restitution N k – coefficient of restitution But now want to add angular velocity contribution to separation velocity

Impulse response 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 forceSeparation velocity

Update linear and angular velocities as a result of impulse force

Velocities of points of contact