9.4 Two-Dimensional Collisions

1. 9.4 Two-Dimensional Collisions

2. 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

3. 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

4. Two-Dimensional Collision, 3 • For a collision of two particles in two dimensions implies that the momentum in each direction xand yis 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

5. Two-Dimensional Collision, 4 • Particle 1 is moving at velocity v1iand particle 2 is at rest • In the x-direction, the initial momentum is m1v1i • In the y-direction, the initial momentum is 0

6. Two-Dimensional Collision, final • After the glancing collision, the conservation of momentum in the x-direction is m1v1i≡m1v1fcosq+m2v2f cosf (9.24) • After the collision, the conservation of momentum in the y-direction is 0 ≡ m1v1fsinq+m2v2f sinf(9.25)

7. Active Figure 9.13

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

9. 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)

10. Example 9.8 Collision at an Intersection, final • Since (mc + mv ) = 400kg.  3.75x104 kg·m/s = 4000vf 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

11. 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= Σmiis the total mass of the system • The coordinates of the center of mass are (9.28) (9.29)

12. 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

13. Active Figure 9.16

14. 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)

15. 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

16. Active Figure 9.17

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

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

19. 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

20. 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

21. 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

22. 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:

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

24. Material for the Final • Examples to Read!!! • Example 9.12(page 269) • Example 9.13(page 272) • Example 9.16(page 276) • Homework to be solved in Class!!! • Problems: 43