1. Short Version : 9. Systems of Particles

2. 9.1. Center of Mass N particles: 
= total mass
= Center of mass
= mass-weighted average position
with
3rd law 
Cartesian coordinates:
Extension: “particle” i may stand for an extended object with cm at ri .

3. Example 9.2. Space Station 2: 2m x H L 1: m 3:m y
A space station consists of 3 modules arranged in an equilateral triangle, connected by struts of length L & negligible mass.
2 modules have mass m, the other 2m.
Find the CM.
Coord origin at m2 = 2m & y points downward.
2: 2m x
30 H L CM
obtainable by symmetry
1: m
3:m y

4. Continuous Distributions of Matter
Discrete collection:
Continuous distribution:
Let  be the density of the matter.

5. Example 9.3. Aircraft Wing y w x L
A supersonic aircraft wing is an isosceles triangle of length L, width w, and negligible thickness.
It has mass M, distributed uniformly.
Where’s its CM?
Density of wing = .
Coord origin at leftmost tip of wing.
By symmetry, y dm h w x L

6. Example Circus Train
Jumbo, a 4.8-t elephant, is standing near one end of a 15-t railcar,
which is at rest on a frictionless horizontal track.
Jumbo walks 19 m toward the other end of the car.
How far does the car move?
1 t = 1 tonne
= 1000 kg
Final distance of Jumbo from xc : 
Jumbo walks, but the center of mass doesn’t move (Fext = 0 ).

7. Alternative Solution  relative to ground relative to car 19m + xc

8. 9.2. Momentum
Total momentum:
M constant 

9. Conservation of Momentum
Conservation of Momentum:
Total momentum of a system is a constant if there is no net external force.

10. Conceptual Example 9.1. Kayaking
Jess (mass 53 kg) & Nick (mass 72 kg) sit in a 26-kg kayak at rest on frictionless water.
Jess toss a 17-kg pack, giving it a horizontal speed of 3.1 m/s relative to the water.
What’s the kayak’s speed after Nick catches it?
Why can you answer without doing any calculations ?
Initially, total p = 0.
frictionless water  p conserved
After Nick catches it , total p = 0.
Kayak speed = 0
Simple application of the conservation law.

11. Making the Connection
Jess (mass 53 kg) & Nick (mass 72 kg) sit in a 26-kg kayak at rest on frictionless water.
Jess toss a 17-kg pack, giving it a horizontal speed of 3.1 m/s relative to the water.
What’s the kayak’s speed while the pack is in the air ?
Initially
While pack is in air:
Note: Emech not conserved
After Nick catches it :

12. Example 9.5. Radioactive Decay
A lithium-5 ( 5Li ) nucleus is moving at 1.6 Mm/s when it decays into
a proton ( 1H, or p ) & an alpha particle ( 4He, or  ). [ Superscripts denote mass in AMU ]
 is detected moving at 1.4 Mm/s at 33 to the original velocity of 5Li.
What are the magnitude & direction of p’s velocity?
Before decay:
After decay:

13. 9.3. Kinetic Energy of a System

14. 9.4. Collisions Examples of collision: Balls on pool table.
tennis rackets against balls.
bat against baseball.
asteroid against planet.
particles in accelerators.
galaxies
spacecraft against planet ( gravity slingshot )
Characteristics of collision:
Duration: brief.
Effect: intense
(all other external forces negligible )

15. Momentum in Collisions
External forces negligible  Total momentum conserved
For an individual particle
t = collision time
impulse
More accurately,
Same size
Average
Crash test

16. Energy in Collisions Elastic collision: K conserved.
Inelastic collision: K not conserved.
Bouncing ball: inelastic collision between ball & ground.

17. 9.5. Totally Inelastic Collisions
Totally inelastic collision: colliding objects stick together
 maximum energy loss consistent with momentum conservation.

18. Example 9.9. Ballistic Pendulum
The ballistic pendulum measures the speeds of fast-moving objects.
A bullet of mass m strikes a block of mass M and embeds itself in the latter.
The block swings upward to a vertical distance of h.
Find the bullet’s speed. 
Caution:
(heat is generated when bullet strikes block)

19. 9.6. Elastic Collisions
Momentum conservation:
Energy conservation:
Implicit assumption: particles have no interaction when they are in the initial or final states. ( Ei = Ki )
2-D case:
number of unknowns = 2  2 = ( final state: v1fx , v1fy , v2fx , v2fy )
number of equations = 2 +1 = 3
 1 more conditions needed.
3-D case:
number of unknowns = 3  2 = 6 ( final state: v1fx , v1fy , v1fz , v2fx , v2fy , v2fz )
number of equations = 3 +1 = 4
 2 more conditions needed.

20. Elastic Collisions in 1-D
1-D collision
1-D case:
number of unknowns = 1  2 = ( v1f , v2f )
number of equations = 1 +1 = 2
 unique solution.
This is a 2-D collision 

21.      mom. cons. rel. v reversed (a) m1 << m2 :
(b) m1 = m2 : 
(c) m1 >> m2 : 
Mathematica

22. Elastic Collision in 2-D
Impact parameter b :
additional info necessary to fix the collision outcome.
Mathematica

23. Example Croquet
A croquet ball strikes a stationary one of equal mass.
The collision is elastic & the incident ball goes off 30 to its original direction.
In what direction does the other ball move?
p cons:
E cons: 

24. Center of Mass Frame