Short Version : 9. Systems of Particles

9.1. Center of Mass

i i imF a1

N

total ii

F F2

21

Ni

ii

dm

dt

r 2

21

N

i ii

dm

dt

r

2

2cmd

Mdt

r

1

1 N

cm i ii

mM

r r

2

2cm

cm

d

dt

ra

1

1 N

cm i ii

x m xM

1

1 N

cm i ii

y m yM

1

1 N

cm i ii

z m zM

Cartesian coordinates:

1

N

ii

M m

= total mass

= Center of mass = mass-weighted average position

total cmMF a with

Extension: “particle” i may stand for an extended object with cm at ri .

net cmMF a

1

Next int

total i ii

F F F1

Nexti

i

F1

0N

inti

i

F3rd law netF

N particles:

Example 9.2. Space Station

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.

1 1 10

4 2 2cmx L

2 4M m m m m

1 1

1, , cos30

2x y L L

1 3,

2 2L

2 2, 0 , 0x y

1 3 30

4 2 2cmy L

3

4L 0.43L

0

1

1 N

cm i ii

mM

r r

obtainable by symmetry

2: 2m

1: m 3:m

L

x

y

CM

30

3 3

1, , cos30

2x y L L

1 3,

2 2L

1

2H

H

Continuous Distributions of Matter

1

1 N

cm i ii

mM

r r

01

1lim

i

N

cm i im

i

mM

r r

1

N

ii

M m

Continuous distribution:

Discrete collection:

01

limi

N

im

i

M m

dm

1dm

M r

Let be the density of the matter.

M dV r 1cm dV

M r r r dm dV r

Example 9.3. Aircraft WingA 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 = .

dm dx dy

1

2w L

By symmetry, 0cmy

2

3L

L

wx

y

dm

Coord origin at leftmost tip of wing.

h

2h w

x L

0

L h x

h xM dx dy

02

Ldx h x

0

L wdx x

L

0

L h x

cm h xy dx dy y

M

2 2

00

2

Ldx h x h x

M

0

L h x

cm h xx dx dy x

M

2

0

L wdx x

M 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

Jumbo walks, but the center of mass doesn’t move (Fext = 0 ).

J J i c cicm i

m x m xx

M

J cM m m

J J f c c fcm f

m x m xx

M

19f Jic f iJ cmx xx x

cm i cm fx x

4.6 m 19J

c f ciJ c

m mx x

m m

Final distance of Jumbo from xc : 19 ci Jim x x

Alternative Solution

J J c ccm

m x m xx

M

J cM m m

J J c ccm

m x m xx

M

0

Jc J

c

mx x

m 19J

cc

mm x

m 19J

c J

mm

m m

4.6 m

relative to ground relative to car

xc

19m

19m + xc

9.2. Momentum

Total momentum:i

i

P p ii

i

dm

dt r

i ii

dm

dt

r cm

dM

dt r

cm

dM

d

t

rPM constant cmM v

cmd

dt

dM

dt

vPcmM a net extF

Conservation of Momentum

net ext

d

dt

PF

0net ext F constP

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 0 0( ) 0J N k pp m m m m v

While pack is in air:

1 1 0( ) 0J N k p pp m m m v m v p

1p

pJ N k

mv v

m m m

173.1 /

55 72 26

kgm s

kg kg kg

0.35 /m s

Note: Emech not conserved

2 2( ) 0J N k pp m m m m v 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: 0 Li LimP v , 0Li Lim v

After decay:1 p pm m P v v

1 cos , sinp p x p p ym v m v m v m v P

cosLi Li p p xm v m v m v 0 sinp p ym v m v

1cosp x Li Li

p

v m v m vm 1

5 1.6 / 4 1.4 / cos331.0

u Mm s u Mm su

3.3 /Mm s

sinp yp

mv v

m

4 1.4 / cos33

1.0

u Mm s

u

3.05 /Mm s

2 2p p x p yv v v 4.5 /Mm s

1tan p yp

p x

v

v 43

p LiK K K

p LiU U U

9.3. Kinetic Energy of a System

ii

K K 21

2 i ii

m v 1

2 i cm i rel cm i reli

m v v v v

2 21 1

2 2i cm i cm i rel i i reli i i

m m m v v v v

2 21

2

1

2 cm i i reli

mM v vi

i

M m

i cm i rel cm i i reli i

m m v v v v

intcmKK K

21

2 mcm cK M v

2int

1

2 i i reli

K m v

cm i i cmi

dm

dt v r r

cm i i cmi

dm

dt v r r

0cm cm cm

dM M

dt v r r

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 p F t = collision time

J impulse

More accurately, t dt J p F

Average

Same size

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.

1 1 2 2initial m m P v v 1 2final fm m P v

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.

m M

m

v V

21

2embE m M V finalE m M g h

init mP v emb m M P V

2 2V g h 2m M

v g hm

Caution: 21

2init finalE m E v (heat is generated when bullet strikes block)

9.6. Elastic Collisions

Momentum conservation: 1 1 2 2init i im m P v v 1 1 2 2final f fm m P v v

Energy conservation: 2 21 1 2 2

1 1

2 2init i iE m m v v 2 21 1 2 2

1 1

2 2final f fE m m v v

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.

Implicit assumption: particles have no interaction when they are in the initial or final states. ( Ei = Ki )

Elastic Collisions in 1-D

1 1 2 2init i ip m v m v 1 1 2 2final f fp m v m v

2 21 1 2 2

1 1

2 2init i iE m v m v 2 21 1 2 2

1 1

2 2final f fE m v m v

1-D case:

number of unknowns = 1 2 = 2 ( v1f , v2f )

number of equations = 1 +1 = 2

unique solution.

1 1 1 2 2 2f i f im v v m v v

2 2 2 21 1 1 2 2 2f i f im v v m v v

1 1 2 2f i f iv v v v

1 2 1 2i i f fv v v v i fv v

This is a 2-D collision

1-D collision

1 2p p

1 2E E

1 1 1 2 2 2f i f im v v m v v

1 1 2 2f i f iv v v v

1 2 1 2 21

1 2

2i if

m m v m vv

m m

1 1 2 2 1 1 2 2f f i im v m v m v m v

1 2 1 2f f i iv v v v

1 1 2 1 2

21 2

2 i if

m v m m vv

m m

(a) m1 << m2 : 1 1 22f i iv v v 2 2f iv v

2 0iv 1 1f iv v 2 0fv

(b) m1 = m2 : 1 2f iv v 2 1f iv v

2 0iv 1 0fv 2 1f iv v

(c) m1 >> m2 : 1 1f iv v 2 1 22f i iv v v

2 0iv 1 1f iv v 2 12f iv v

mom. cons.

rel. v reversed

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?

1 1 2i f f v v vp cons:

2 2 21 1 2i f fv v v E cons:

2 2 21 1 1 2 22i f f f f v v v v v

2 2 21 1 1 2 22 cos 30i f f f fv v v v v

1 22 cos 30 0f fv v

9030 60

Center of Mass Frame

0i f P P