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9. Systems of Particles
1.
2.
3.
4.
5.
6.
Center of Mass
Momentum
Kinetic Energy of a System
Collisions
Totally Inelastic Collisions
Elastic Collisions
As the skier flies through the air,
most parts of his body follow complex trajectories.
But one special point follows a parabola.
What’s that point, and why is it special?
Ans. His center of mass (CM)
Rigid body: Relative particle positions fixed.
9.1. Center of Mass
N particles:

Fi  mi ai

d 2 ri
d 2 rcm
d2  N
  Fi   m i
M
m r
2 
2
2  i i 
dt
dt
dt  i  1
i 1
i 1

N
Ftotal
N
N
M   mi
1
rcm 
M
= total mass
i 1
N
m r
i 1
i i
= Center of mass
= mass-weighted average position
Ftotal  M acm
Ftotal    F  F
N
i 1
ext
i
int
i

with
N
 F
i 1
ext
i
acm
d 2 rcm

dt 2
 Fnet
3rd
law 
N
F
i 1
int
i
0
Fnet  M acm
Cartesian
coordinates:
1
xcm 
M
N
m x
i 1
i
i
1
ycm 
M
N
m y
i 1
i
i
1
zcm 
M
N
m z
i 1
Extension: “particle” i may stand for an extended object with cm at ri .
i i
Example 9.1. Weightlifting
Find the CM of the barbell consisting of 50-kg & 80-kg weights
at opposite ends of a 1.5 m long bar of negligible weight.
xcm 
m1 x1  m2 x2
m1  m2
80 kg 1.5 m 


50 kg  80 kg
 0.92 m
CM is closer to the heavier mass.

m2 x2
m1  m2
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.
1
rcm 
M
2 modules have mass m, the other 2m.
Find the CM.
Coord origin at m2 = 2m & y points downward.
x
m r
i 1
i i
 1
3
1

L , L cos 30   L   ,

2
2
2




 x1 , y1    
 x2 , y2    0
2: 2m
N
, 0
1
  L1 , 3
x
,
y

L
,
L
cos
30



 3 3 

2
2
2




30
L
M  m  2m  m  4m
CM
1: m
3:m
y
1 1
1
xcm     0   L  0
4 2
2
obtainable
by symmetry
3
1 3
3
L  0.43L
ycm  
0
 L 
4
4 2
2 
Continuous Distributions of Matter
Discrete collection:
1
rcm 
M
N
N
m r
i 1
M   mi
i i
i 1
Continuous distribution:
M  lim
m i  0
N
 m
i 1
1
rcm  lim
m i  0 M
N
i
  dm
 mi ri 
i 1
1
r dm
M
Let  be the density of the matter.
d m    r  dV
M     r  dV
rcm 
1
r   r  dV

M
Example 9.3. Aircraft Wing
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.
y
dx
ycm  0
d m   h dx   x
h
W
L
By symmetry,
x
w
M 
L
xcm  

L
0
w
M L
w
dx
L
h w

x L
1
x dx   w L
2

L
0
x 2 dx  
w 1 3
2
L  L
3
M L 3
d m   b dy
b
L w



y

 dy
w/2 2

y
b
ycm
dy
w/2
x
L
w /2  w
2L

M 
2    y  dy
0
w
2

L
w/2
2 L w/2  w


y

y

 dy  0


w
/2
w
2

w /2
2L  0
w

w
 

y

y
dy

y

y



 dy 




w
/2
0
w 
2

2
 
W
w/2
w
 y
2

0
w /2

2 L  w 2 1 3 
w
1

2
3

 y  y  
 y  y 
w  4
3   w/2  4
3 0 
2 L  w3 w3 w3 w3 

 

 

w  16 24 16 24 
 w  2 1  w  2 
L
 2
2       
w
 2  2  2  

0
1
 wL
2
CMfuselage
CMplane
CMwing
A high jumper clears the bar,
but his CM doesn’t.
Got it? 9.1.
A thick wire is bent into a semicircle.
Which of the points is the CM?
Example 9.4. Circus Train
Jumbo, a 4.8-t elephant, is standing near one end of a 15-t railcar,
1 t = 1 tonne
= 1000 kg
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?
xcm i 
mJ xJ i  mc xci
xcm f 
M
M  mJ  mc
mJ xJ f  mc xc f
M
Final distance of Jumbo from xc : 19 m   xci  xJi 
xJ f  xc f  19 m   xci  xJi 
xcm i  xcm f
Jumbo walks, but the center of mass
doesn’t move (Fext = 0 ).

xc f  xci  
mJ 19m 
mJ  mc
 4.6 m
9.2. Momentum
Total momentum:
P   pi   mi
i
M constant 
i
PM
d ri
d
d 


 M rcm 
   mi ri 
dt
dt  i
 dt
d rcm
 M vcm
dt
d v cm
dP
M
dt
dt
 M acm
 Fnet ext
Conservation of Momentum
dP
 Fnet ext
dt
Fnet ext  0

P  const
Conservation of Momentum:
Total momentum of a system is a constant if there is no net external force.
GOT IT! 9.2.
A 500-g fireworks rocket is moving with velocity v = 60 j m/s at the instant it explodes.
If you were to add the momentum vectors of all its fragments just after the explosion,
what would you get?
 0.5 kg   60 ˆj m / s   30 ˆj
kg m / s
K.E. is not conserved.
Emech = K.E. + P.E. grav is not conserved.
Etot = Emech + Uchem is conserved.
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
p0  0
While pack is in air:
p1  (mJ  mN  mk )v1  m p v p  p0  0
v1  

mp
mJ  mN  mk
vp
17 kg
 3.1 m / s   0.35 m / s
55 kg  72 kg  26 kg
Note: Emech not conserved
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?
P0  mLi v Li  mLi  vLi , 0
Before decay:
After decay:

P1  m p v p  m v
P1   m p v p x  m v cos  , m p v p y  m v sin 
mLi vLi  m p v p x  m v cos 
vp x 
1
 mLi vLi  m v cos
mp
vp y  


0  m p v p y  m v sin 
1
 5 u 1.6 Mm / s    4 u 1.4 Mm / s  cos 33  
1.0 u
m
4 u 1.4 Mm / s  cos 33 
 3.05 Mm / s
v sin     
mp
1.0 u
vp 
v v
2
px
2
py
 4.5 Mm / s

 p  tan 1
vp y
vp x
 43
 3.3 Mm / s
Example 9.6. Fighting a Fire
A firefighter directs a stream of water to break the window of a burning building.
The hose delivers water at a rate of 45 kg/s, hitting the window horizontally at 32 m/s.
After hitting the window, the water drops horizontally.
What horizontal force does the water exert on the window?
Momentum transfer to a plane  stream:
dP dm

v   45 kg / s  32 m / s   1400 N
dt
dt
= Rate of momentum transfer to window
= force exerted by water on window
GOT IT? 9.3.
Two skaters toss a basketball back & forth on frictionless ice.
Which of the following does not change:
(a) momentum of individual skater.
(b) momentum of basketball.
(c) momentum of the system consisting of one skater & the basketball.
(d) momentum of the system consisting of both skaters & the basketball.
Application: Rockets
Ptot  Procket  P fuel  const
Thrust:
F  v exhaust
dM
dt
9.3. Kinetic Energy of a System
K   Ki  
i

i

i
1
1
mi vi2   mi  v cm  v i rel    v cm  v i rel 
2
2
i
1
1
2
mi v cm
  mi v cm  vi rel   mi vi2rel
2
2
i
i
1
1
2
M v cm
  mi vi2rel
2
2
i
M   mi
i
m
i
K  Kcm  Kint
K cm
1
 M v c2m
2
Kint  
i
1
mi v i2rel
2
v cm  vi rel  v cm   mi vi rel
i
i
 v cm   mi
i
d
 ri  rcm 
dt
 v cm 
d
 mi  ri  rcm 
dt i
 v cm 
d
 M rcm  M rcm   0
dt
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
p  F t
J
More accurately,
t = collision time
impulse
J  p   F  t  dt
Same size
Average
Crash
test
Energy in Collisions
Elastic collision: K conserved.
Inelastic collision: K not conserved.
Bouncing ball: inelastic collision between ball & ground.
GOT IT? 9.4.
Which of the following qualifies as a collision?
Of the collisions, which are nearly elastic & which inelastic?
elastic (a) a basketball rebounds off the backboard.
elastic (b) two magnets approach, their north poles facing; they repel & reverse
direction without touching.
(c) a basket ball flies through the air on a parabolic trajectory.
inelastic
(d) a truck crushed a parked car & the two slide off together.
inelastic
(e) a snowball splats against a tree, leaving a lump of snow adhering to the bark.
9.5. Totally Inelastic Collisions
Totally inelastic collision: colliding objects stick together
 maximum energy loss consistent with momentum conservation.
Pinitial  m1v1  m2 v 2  Pfinal   m1  m2  v f
Example 9.7. Hockey
A Styrofoam chest at rest on frictionless ice is loaded with sand to give it a mass of 6.4 kg.
A 160-g puck strikes & gets embedded in the chest, which moves off at 1.2 m/s.
What is the puck’s speed?
Pinitial  m p v p
vp 

m p  mc
mp
 Pfinal   m p  mc  v c
vc
0.16 kg  6.4 kg
1.2 m / s 
0.16 kg
 49 m / s
Example 9.8. Fusion
2H
Consider a fusion reaction of 2 deuterium nuclei
+ 2H  4He .
Initially, one of the 2H is moving at 3.5 Mm/s, the other at 1.8 Mm/s at a 64 angle to the 1st.
Find the velocity of the Helium nucleus.
Pinit  mD  v1  v2   P final  mHe v f
vf 

mD
 v1  v 2 
mHe
2
  3.5 , 0   1.8  cos 64 , sin 64   Mm / s
4
  2.14 , 0.809  Mm / s
vf 
 2.14   0.809
2
 2.3 Mm / s
2
Mm / s
  tan 1
0.809
 21
2.14
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.
Pinit  m v  Pemb   m  M  V

Eemb 
v
1
 m  M  V 2  E final   m  M  g h
2
V2  2 g h
Caution:
Einit 
1
m v 2  E final
2
mM
V
m
v
mM
m
2gh
(heat is generated when bullet strikes block)
9.6. Elastic Collisions
Pinit  m1 v1i  m2 v 2i  Pfinal  m1 v1 f  m2 v 2 f
Momentum conservation:
Energy conservation:
Einit 
1
1
1
1
m1 v12i  m2 v 22 i  E final  m1 v12 f  m2 v 22 f
2
2
2
2
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
number of equations = 3 +1 = 4
 2 more conditions needed.
( final state: v1fx , v1fy , v1fz , v2fx , v2fy , v2fz )
Elastic Collisions in 1-D
pinit  m1v1i  m2v2i  p final  m1v1 f  m2v2 f
Einit 
1
1
1
1
m1 v12i  m2 v22 i  E final  m1 v12 f  m2 v22 f
2
2
2
2
1-D collision
1-D case:
number of unknowns = 1  2 = 2
( v1f , v2f )
number of equations = 1 +1 = 2

This is a 2-D collision
unique solution.
m1  v1 f  v1i   m2  v2 f  v2i 
 p1  p2 
m1  v12f  v12i   m2  v22 f  v22i 
 E1  E2 

v1 f  v1i  v2 f  v2i
v1i  v2i    v1 f  v2 f

vi  v f
m1  v1 f  v1i   m2  v2 f  v2i 

v1 f  v1i  v2 f  v2i

v1 f 
 m1  m2  v1i  2m2v2i
m1  m2
(a) m1 << m2 :
m1v1 f  m2 v2 f  m1v1i  m2v2i
v1 f  v2 f  v1i  v2i
v2 f 
2m1v1i   m2  m1  v2 i
m1  m2
v2i  0 
(b) m1 = m2 :
v1 f  v1i
v1 f  v2i
v2i  0 
(c) m1 >> m2 :
v2 f  v2i
v1 f  v1i  2v2i
v1 f  v1i
v2i  0 
v2 f  0
v2 f  v1i
v1 f  0
v2 f  v1i
v2 f  2v1i  v2i
v1 f  v1i
v2 f  2v1i
Example 9.10. Nuclear Engineering
Moderator slows neutrons to induce fission.
A common moderator is heavy water ( D2O ).
Find the fraction of a neutron’s kinetic energy that’s transferred to an initially
stationary D in a head-on elastic collision.
v1 f 
 m1  m2  v1i  2m2v2i
v2i  0
K1i 
K2 f
v2 f 
m1  m2
v1 f 
m1  m2
v1i
m1  m2
1
  v1i
3
v2 f 
2m1
v1i
m1  m2

2m1v1i   m2  m1  v2 i
m1  m2
m1  1 u
m2  2 u
2
v1i
3
1
m1 v12i
2
1
 m2 v22 f
2
K2 f
K1i

m2 v22 f
m1 v12i

4 m1 m2
 m1  m2 
2

4 1u  2u 
1u  2u 
2

8
 89%
9
GOT IT? 9.5.
One ball is at rest on a level floor.
Another ball collides elastically with it & they move off in the same direction separately.
What can you conclude about the masses of the balls?
1st one is lighter.
Elastic Collision in 2-D
Impact parameter b :
additional info necessary to fix the collision outcome.
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:
v1i  v1 f  v 2 f
E cons:
v12i  v12f  v22 f
v12i  v12 f  2v1 f  v2 f  v22 f
v12i  v12f  2v1 f v2 f cos   30  v22 f
2v1 f v2 f cos   30  0

  30  90
  60
Center of Mass Frame
Pi  P f  0