Short Version : 9. Systems of Particles

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 .

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

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

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

Example 9.4. 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 ).

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

9.2. Momentum Total momentum: M constant 

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

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.

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 :

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:

9.3. Kinetic Energy of a System

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 )

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

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

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

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)

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 = 4 ( 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.

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

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

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

Example 9.11. 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: 

Center of Mass Frame