9.4 Two-Dimensional Collisions

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, 2 Qualitative Analysis Physical Principles: The same as in One-Dimension 1. Conservation of VECTOR momentum: P1x + P2x = P1x + P2x & P1y + P2y= P1y + P2y 2. Conservation of Kinetic Energy ½m1v12 + ½m2v22 = ½m1v’12 + ½m2v’22

Two-Dimensional Collision, 3 For a collision of two particles in two dimensions implies that the momentum in each direction x and y is conserved The game of billiards is an example for such two dimensional collisions The equations for conservation of momentum are: m1v1ix + m2v2ix ≡ m1v1fx+ m2v2fx m1v1iy + m2v2iy ≡ m1v1fy+ m2v2fy Subscripts represent: (1,2) Objects (i,f) Initial and final values (x,y) Component direction

Two-Dimensional Collision, 4 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, final After the glancing collision, the conservation of momentum in the x-direction is m1v1i ≡ m1v1f cosq + m2v2f cosf (9.24) After the collision, the conservation of momentum in the y-direction is 0 ≡ m1v1f sinq + m2v2f sinf (9.25)

Active Figure 9.13

Example 9.8 Collision at an Intersection (Example 9.10 Text Book) Mass of the car mc = 1500kg Mass of the van mv = 2500kg Find vf if this is a perfectly inelastic collision (they stick together). Before collision The car’s momentum is: Σpxi = mcvc  Σpxi = (1500)(25) = 3.75x104 kg·m/s The van’s momentum is: Σpyi = mvvv  Σpyi = (2500)(20) = 5.00x104 kg·m/s

Example 9.8 Collision at an Intersection, 2 After collision, both have the same x- and y-components: Σpxf = (mc + mv )vf cos Σpyf = (mc + mv )vf sin Because the total momentum is both directions is conserved: Σpxf = Σpxi  3.75x104 kg·m/s = (mc + mv )vf cos (1) Σpyf = Σpyi  5.00x104 kg·m/s = (mc + mv )vf sin (2)

Example 9.8 Collision at an Intersection, final Since (mc + mv ) = 400kg.  3.75x104 kg·m/s = 4000 vf cos (1) 5.00x104 kg·m/s = 4000vf sin (2) Dividing Eqn (2) by (1) 5.00/3.75 =1.33 = tan   = 53.1° Substituting  in Eqn (2) or (1)  5.00x104 kg·m/s = 4000vf sin53.1°  vf = 5.00x104/(4000sin53.1° )  vf = 15.6m/s

9.5 The Center of Mass There is a special point in a system or object, called the center of mass (CM), that moves as if all of the mass of the system is concentrated at that point The system will move as if an external force were applied to a single particle of mass M located at the CM M = Σmi is the total mass of the system The coordinates of the center of mass are (9.28) (9.29)

Center of Mass, position The center of mass can be located by its position vector, rCM (9.30) ri is the position of the i th particle, defined by

Active Figure 9.16

Center of Mass, Example Both masses are on the x-axis The center of mass (CM) is on the x-axis One dimension, x-axis xCM = (m1x1 + m2x2)/M M = m1+m2 xCM ≡ (m1x1 + m2x2)/(m1+m2)

Center of Mass, final The center of mass is closer to the particle with the larger mass xCM ≡ (m1x1 + m2x2)/(m1+m2) If: x1 = 0, x2 = d & m2 = 2m1 xCM ≡ (0 + 2m1d)/(m1+2m1)  xCM ≡ 2m1d/3m1  xCM = 2d/3

Active Figure 9.17

Center of Mass, Extended Object Up to now, we’ve been mainly concerned with the motion of single (point) particles. To treat extended bodies, we’ve approximated the body as a point particle & treated it as if it had all of its mass at a point! How is this possible? Real, extended bodies have complex motion, including: translation, rotation, & vibration! Think of the extended object as a system containing a large number of particles

Center of Mass, Extended Object, Coordinates The particle separation is very small, so the mass can be considered a continuous mass distribution: The coordinates of theCM of the object are: (9.31) (9.32)

Center of Mass, Extended Object, Position The position of CM can also be found by: (9.33) The CM of any symmetrical object lies on an axis of symmetry and on any plane of symmetry An extended object can be considered a distribution of small mass elements, Dm The CM is located at position rCM

Example 9.9 Three Guys on a Raft A group of extended bodies, each with a known CM and equivalent mass m. Find the CM of the group. xCM = (Σmixi)/Σmi xCM = (mx1 + mx2+ mx3)/(m+m+m)  xCM = m(x1 + x2+ x3)/3m = (x1 + x2+ x3)/3  xCM = (1.00m + 5.00m + 6.00m)/3 = 4.00m

Example 9.10 Center of Mass of a Rod (Example 9.14 Text Book) Find the CM position of a rod of mass M and length L The location is on the x-axis (yCM = zCM = 0) (A). Assuming the road has a uniform mass per unit length λ = M/L (Linear mass density) From Eqn 9.31

Example 9.10 Center of Mass of a Rod, 2 But λ = M/L  (B). Assuming now that the linear mass density of the road is no uniform: λ = x The CM will be:

Example 9.10 Center of Mass of a Rod, final But mass of the rod and  are related by:  The CM will be:

Material for the Final Examples to Read!!! Example 9.12 (page 269) Homework to be solved in Class!!! Problems: 43