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 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

5. 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

6. 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)

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
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

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 = 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

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 = Σmi is 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 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)

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 m m)/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)
Homework to be solved in Class!!!
Problems: 43