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Collision Response - PowerPoint PPT Presentation

Collision Response. CS 134 Soon Tee Teoh. Collision Response. Sometimes, it is necessary to calculate what happens after two objects collide. Where do the objects move? Example 1: What is the velocity of a billiard ball after it collides against the edge?

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Collision Response

CS 134

Soon Tee Teoh

• Sometimes, it is necessary to calculate what happens after two objects collide.

• Where do the objects move?

• Example 1: What is the velocity of a billiard ball after it collides against the edge?

• Example 2: What are the velocities of two billiard balls after they collide?

• We will consider these two cases, which serve as generalizations of collisions.

• Example: Billiard ball against edge of table

• What happens is like simple reflection.

• Let the unit vector in the direction that the object hits the rigid surface be V.

• Let the unit normal of the surface be N.

• Then, the vector after collision R = V + 2N((-V).N)

N

q

q

V

R

surface

• Collision involves two steps.

• 1. Period of compression: Two objects collide and deform

• 2. Period of restitution: After compression, they bounce off each other.

• If during restitution, entire deformation recedes and none of the mechanical energy is lost (changed to heat for example), this is called elastic collision.

• If there is no restitution (that is, the two objects remain stuck to each other after collision), this is called inelastic collision.

• Real-life situations are in between.

• Suppose that two spheres collide.

• As we model objects as spheres, a sphere is just a generalization of an object.

• We assume elastic collision. Elastic collision means that no energy is lost during the collision. Therefore, both momentum and kinetic energy are preserved.

• The velocity of the spheres after collision depends on their initial velocities, angle of impact, and mass.

• In elastic collision, these two objects will bounce off each other and continue moving.

• Let v0 and v1 be the velocity vectors of the two spheres S0 and S1 respectively.

• Let vc be the vector from the center of S0 to the center of S1.

• Let v0c be the projection of v0 onto vc, and let v1c be the projection of v1 onto vc.

• Let vp be the vector perpendicular to vc. Note that vp lies on the plane containing v0 and v1.

• Let v0p be the projection of v0 onto vp, and let v1p be the projection of v1 onto vp.

• Let M0 and M1 be the masses of S0 and S1 respectively.

v1

v1p

vp

v1

v0

vp

v0p

v0

vc

v0c

v1c

• Let the new velocities after collision be r0 and r1 respectively for S0 and S1.

• Then, r0 and r1 are given by the following formulas:

r0c = v0c*(m0-m1)/(m0+m1) + v1c*2m1/(m0+m1)

r0p = v0p

r1c = v0c*2*m0/(m0+m1) + v1c*(m1-m0)/(m0+m1)

r1p = v1p

r0 = r0c + r0p

r1 = r1c + r1p

• But, how to find v0c, v0p, v1c and v1p in the first place?

• The unit vector joining the center of the spheres, vc = (centerS1 – centerS0)/ ( | centerS1 – centerS0 |)

• Then,

multiply

dot product

v0c = vc * (vc.v0)

v0p = v0 – v0c

v1c = -vc * (-vc.v1)

v1p = v1 – v1c

Perfectly (continued)inelastic collision

• Now, suppose we have inelastic collision.

• Happens when two pieces of clay collide and stick together.

• Momentum is conserved, but kinetic energy is not conserved.

• From conservation of momentum,

u = (m1v1 + m2v2)/(m1+m2)

• where u is the velocity of the (combined) object after collision, m1 and v1 are the mass and velocity respectively of object 1 before collision, and m2 and v2 are the mass and velocity of object 2 before collision.