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Chapter 6: Many-body Interactions
Chapter 6 Goals:
To appreciate the ubiquity of action-reaction force pairs
• To understand and utilize Newton’s Third (and final) law
• To define a system of massive particles
• To identify the critical importance of the center-of-mass
point in a system
• To derive general principles for conservation of energy,
momentum, and angular momentum, for a system
• To define and utilize the notion of the impulse of an
interaction: the change in momentum
• To apply these principles to two-body collisions
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Newton’s Third law
• ‘to every action, there is
an equal and opposite [sic]
reaction’
• by ‘action’ we mean a
force (and therefore a
momentum change)
• the actions occur on/to
DIFFERENT bodies: the
bodies are said to interact
• How? strings…
contact… springs…
The 2 forces on the monitor, and their
COLLISIONS
Newton’s-third-law counterparts
• FBA = ─ FAB
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Example
The pair of blocks
accelerates at 5 m/s2.
Find all of the forces
on both blocks.
F
5 kg
N1
F
15 kg
N2
R21 R
12
W1
W2
Note that R12 = ─ R21
FBD 1  N1 = m1g =50 N
 F ─ R21 = m1a = 25 N
FBD 2  N2 = m2g = 150 N
 R12 = m2a = 75 N
But R12 = R21 = 75N and so
from above, F = 100N
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5 kg
Example
15 kg
N1
FK
T21
W1
T12
The pair of blocks
accelerate. The coefficient of
kinetic friction is 0.40. Find a and
T. String and pulley are ideal.
Note that T12 = ─ T21
FBD 1  N1 = m1g = 50 N
 ─ FK + T21 = m1a
and note FK = mKN1 = 20 N
so T21 = 20 N + 5 kg a
FBD 2  ─ T12 + m2g = m2a
So T12 = 150 N ─ 15 kg a
But T12 = T21 so we get
20 N + 5 kg a = 150 N ─ 15 kg a
W2  a = 130 N/ 20 kg = 6.5 m/s2
T = 20 N + (5 kg)( 6.5 m/s2)
= 52.5 N
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A system of masses interacting with one
another
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A system of masses and their felt forces
• a system is a collection of N masses {mi} i = 1, 2, 3,…, N
• each one feels both internal and external forces
N
dpi
 the net force on mass i is Fi  (Fi ) ext 
f j i 
dt
N
N
j i

 define
system mass M :
 mi and system momentum P :  pi
i 1
• Take the time derivative of P, using sum rule
dP

:
dt
N

dpi

dt
i 1
N
N
N
 Fi ext   f ji
i 1
i 1
Messy double sum...
i 1 j i
• first term: total external force on system
N
 Fi ext : FEXT
i 1
• second term: messy double sum that contains all pairs of
internal forces added up , like f12 + f21 +…
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Newton’s action-reaction pairs for two bodies
• the forces may be central
(like gravity or electric)
• but they don’t have to be
(like magnetism)
• they may be gigantic and
short-lived (like collisions)
• for this situation we will
introduce the useful notion of
the impulse of a force
• there are exceptions in
advanced electricity due to the
finite speed of light
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A system of masses and the forces on it
• by N3, f12 = ─ f21 : the entire double sum adds to zero!!
• in English: the internal forces in a system add to zero and
thereforce cannot affect the momentum of the system!!

dP
we are left with N2 for a system :
 FEXT
dt
• important special case: if FEXT = 0 (either because there
are no external forces, of because they happen to add to 0)
dP
we get that
 0 and so system momentum is conserved!!
dt
• an extremely powerful conservation law that relies
on the truth of N3 and permits one to not worry too
much about the internal doings of a system or
extended body
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The center of mass of a system M = ∑mi
• Let the positions of the bodies be given by {ri}

dri d  N
P
pi  mi v i  mi

mi ri  (constant masses)

dt dt  i 1
i 1
i 1
i 1

1 N
 define the center ofmass position R C :
mi ri
M i 1
N

N

N



N
 or, equivalent ly,
 miri : MR C
i 1
dR C
dR C
d N

P 
 MVC where VC :
  mi ri   M
dt  i 1
dt
dt

• very cool!! The system momentum is the simply the
system mass times the velocity of center of mass!!
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Further implications, and where is it?

dVC
dVC
dP N
  Fi ext  FEXT  M
 MA C where A C :
dt i 1
dt
dt
• very cool!! The rate of change of system momentum
is the simply the system mass times the acceleration of
the center of mass!!
• And if FEXT = 0, it’s N1 time!! No matter what the
system is doing (spinning, writhing, exploding… the
center of mass just cruises along without accelerating
• but.. where is the center of mass??
• example: two bodies. Where is c.o.m.?
• answer: along the line joining the
1 N
mi ri
bodies, between them, and closer to the R C :

M i 1
more massive body
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Where is it for the two-body problem?
1
R C :
M
N
m r
i 1
i i
define r '1 and r '2 via ri  r 'i  R C
• the primed positions have the c.o.m. as their origin
r '1  r1  R C so Mr '1  Mr1  MR C  m1  m2 r1  m1r1  m2r2
 Mr '1  m2r1  m2r2  m2 r1  r2 ; easy to get Mr '2  m1 r2  r1 
• both primed positions are vectors parallel to the vector
that starts on one body and ends on the other
• they are both shorter than that vector and proportional
in length to the other mass/M
• they are in opposite directions
• Thus, the c.o.m. lies where we said it did
• For N > 2, the c.o.m. is a ‘weighted’ position average
and visualizable at least: see figure next slide
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m2
r1  r2 
r '1 
M
m1
r2  r1 
r '2 
M
r’1
Three bodies much
more complicated
r’2
r1
C
r1–r2
r1
C
r2
r3
r2
O
O
{show Active Figure 09_14}
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Not in book: the idea of the impulse
• The impulse J delivered to
any body in an interaction is
the change in momentum of
that body Dp
• it is the net force Fnet(t) that
one graphs
• interaction duration Dt
• useful for very rapid highforce situations
Fnet, x
dpx

 Dp x : J x 
dt
t
f
 Fnet, x (t ) dt
t
i Fnet(t) graph
• The impulse is the area under
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More about the impulse
• if the graph is a simple
shape (triangle, rectangle)
one gets the area easily
• if a triangle, J = bh/2 =
(Dt)(Fnet,max)/2 so we see Fnet,max
that Fnet,max = 2J/ Dt
• for any shape graph,
the average net force is
given by < Fnet>  J/ Dt
from the rectangle’s area
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Example
Assume the collision lasts for .25 s.
The car’s mass is 800 kg and Fnet(t)
has a triangular shape. The net
force is essentially the collision
force since it is dominant.
Find the impulse delivered to the
car, and <F> and Fmax .
J  Dp  D(mv )  mDv
 800 (2.6  (15))  14,100 kg  m/s
 F >  J  14,080
Dt
.25
 56,400 N (wow!!)
Fmax  2  F >  113,000 N (WOW!!!)
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The reduced mass of a two-body system
• a way to replace the motions of the pair of bodies,
as seen from a fixed origin, with the motion of one of
the bodies (here, #1), as seen relative to a moving
origin located at the other body (here, #2)
• of course, the moving origin is probably
accelerating so it is not an inertial frame
• the kinematics (motion) is straightforward
• the dynamics (forces) may require fictitious forces
to explain the kinematics
• we start with the origin at the c.o.m. and we define
r to be the vector from #2 to #1: r := r1 ─ r2
• thus, r is the position of r1 relative to r2
• recall that the primed positions are parallel to r
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Some remarks about the notation
Relative to the fixed origin O
• positions relative to O are ri
• velocities relative to O are vi := dri/dt
• accelerations relative to O are ai := dvi/dt
Relative to the c.o.m. origin
• positions relative to c.o.m. are r’i := ri – RC
• velocities relative to c.o.m. are ui := vi – VC
• accelerations relative to c.o.m. are a’i := ai – AC
For the two-body problem
• position of m1 relative to m2 is r = r1 – r2 [= r’1 – r’2 ]
• velocity of m1 relative to m2 is u = u1 – u2 [= v1 – v2]
• acceleration of m1 relative to m2 is a = a1 – a2 [=a’1 – a’2]
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The reduced mass II
• let the internal net forces be F21 = ─ F12 := F
• a1 = F/m1 ; a2 = ─ F/m2  a = a1 ─ a2 = F/m1 + F/m2
• common denominator: a = F(m1+m2)/m1m2
2
m1m2
m1m2 d r
 so F 
a
M
M dt 2
m1m2
 and we get F  m a where m 
M
• m is called reduced mass; m < m1 and m < m2
• the motion of m1 as seen from an origin on m2 is
controlled by F, but it is as if there is only one body
moving, and that body has the reduced mass!!
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The reduced mass III
• rephrase K1: when 2 bodies orbit, they both move on
(different) ellipses, with c.om. at common focus
• here, the two stars have comparable masses so the c.om.
is near the center, and the reduced mass is about half of
either mass taken alone
• in general, m is less than and closer to the smaller mass
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The reduced mass IV
• obviously, VC = dRC/dt ; again vi = dri /dt
• then u = dr/dt = v1 ─ v2 since r = r1 ─ r2
• now consider VC ─ vi , which is the velocity of c.o.m.
as seen from body i (we’ll do #1 and #2)
m1v1  m2 v 2
m1v1  m2 v 2  m1  m2 v1
 VC  v1 
 v1 
M
M
m2 v 2  m2 v1
m2u
m2


 v1  VC 
u
M
M
M
m1
 similarly, by processing VC  v 2  v 2  VC 
u
M
• you can get the same stuff just by taking d/dt of our
previous r1 and r2 equations much more easily
• note that we have re-expressed velocities in terms of
c.o.m velocity VC and relative velocity u
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What happens to system kinetic energy?
• definition is straightforward; the interpretation less so
N
N
1
1
 K :
Ki 
mi vi2 
mi VC  u i
2
2
i 1
i 1
i 1

N


 
 
N
mi 2

VC  2VC  u i  ui2 
2
i 1
N
MVC2

 VC  mi u i 
2
i 1

N
2
miVC2

2
i 1
[used u i : v i  VC ]
N

N

1
mi VC  u i 
mi ui2
2
i 1
i 1
N

1
mi ui2
2
i 1
• term 2 has velocity of c.o.m. relative to c.o.m.  = 0
two terms for K: kinetic energy of the
MVC2 N 1
2
K 
  mi ui center of mass (as if system were a
2
2
i 1
point mass M moving at c.o.m.
velocity VC) plus kinetic energy
relative to the c.o.m.
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What happens to two-body kinetic energy?
• look at the second term and re-express in terms of u
m
2
M
 recall r '1 
N
1
2

m
u
2 i i
i 1

r and r '2 

m
1

2
m

2 1 M
2
m
 M1 r
 u 2 

m
 u1  M2 u and u 2
2
m
1
  1  u 2
m
2 2 M 

m
 M2
 m1m22  m2m12  2  m1m2  2 1 2
u  
 
u  2 μu
2
2
M


 2M

K 
1
2
MVC2
1
2
 μu 2
• two terms for K: kinetic energy a body of mass M
moving at VC, plus kinetic energy of a body of mass m
moving at u: the two-body problem is effectively now a
one-body problem, if you observe from the c.o.m.
• recall
system momentum: much simpler P = MVC
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u
What happens to the system potential energy?
What about the mechanical energy conservation?
• If definable, U is a function of position only
•U = U(r1 , r2) could include internal and external forces
• If the forces are only internal and central and vary only
with separation distance (common!) we say more:
• U(r1 , r2) = U(|r1 ─ r2|) = U(|r|) = U(r) so we have
MVC2 m u 2
 E : K  U 

 U (r )
2
2
• but there is no external force: the first term can’t change
MVC2
 EC :
 a constant (N1situati on for M )
2
E

m u2
2
 U (r )  a constant (N2 situation for m )
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What happens to system angular momentum?
• proceed as we have been doing, writing positions ri as
c.o.m. position plus positions referred to c.o.m [and for vi]
L :

N
N
N
i 1
N
i 1
i 1
 Li   miri  vi   mi R C  r'i  VC  ui 
N
N
N
i 1
i 1
i 1
 mi R C  VC    mi R C  ui   mir'i VC   mir'i ui
i 1
 N

 M R C  VC   R C  mi u i   mi r 'i   VC 


i 1
 i 1

N


N
 mir'i ui
i 1
• term 2 contains velocity of c.o.m. as seen from c.o.m = 0
• term 3contains position of c.o.m. as seen from c.o.m  = 0
L : M R C  VC  
N
 mir'i ui
= ang mom of c.o.m + ang
i 1
mom relative to c.o.m
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How does system angular momentum change?
• take the time derivative of the system ang mom
dL

:
dt
N

dL i

dt
i 1
N

d
ri  pi  
dt
i 1
N
dr
dv 

i
mi  dt  v i  ri  dti 


i 1

N



 mi v i  v i  ri  a i    ri   Fi ext  f j i  



i 1
i 1 
j i


N 
N
N N

  M i ext  ri  f j i  : M EXT 
ri  f j i


i 1 
j i
i 1 j i

N
N






• term 2: assume N3 so there are terms like (ri ─ ri) x fji
= vector from j to i crossed into fji
• further assume central forces, so those two vectors are
parallel: entire double sum is zero!!
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How does system angular momentum change?
dL

: M EXT N2 for angular momentum for system
dt
• if there are no external torques (or if they add to zero) then
angular momentum of the system is conserved
Process L to express in terms of reduced mass
• look at the second term and re-express in terms of u
N

m  m
m   m 



2
2
mi r 'i u i  m1  r    u   m2   1 r     1 u 
M   M 
M   M 


i 1

  m2  2

m 2 
m m2  m m2 

  m1    m2   1  r  u    1 2 2 2 1 r  u 
M
 M 
M


  
mm 

  1 2 r  u   μr  u 
 M 
 L  M R C  VC   μr  u   R C  P  μr  u 
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Summary of the system quantities
• system momentum is P = MVC
• it can only change if (F)EXT ≠ 0
N
2
• system kinetic energy K  MVC   1 mi ui2
2
2
i 1
N
• system angular momentum L : R  P  m r' u
C
K
i i
i
i 1
•it can only change if (M)EXT ≠ 0
• two-body kinetic energy

1
2
MVC2

1
2
μu 2
• two-body angular momentum L  R C  P  μr  u 
• if c.o.m. obeys N1 there is an inertial frame, called the
center-of-mass frame, from which the physics is
particularly simple (since the c.o.m. is stationary!!)
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Important scenario: the two-body collision in 1d
• m1 and m2 collide and all motion is in 1d
• velocities before collision are u1 and u2 (known)
• velocities after collision are v1 and v2 (not known)
• momenta before collision are p1 and p2
• momenta after collision are q1 and q2
• during collision, all important forces are internal
• therefore system momentum is conserved!
u1 ; p1
u2 ; p2
v1 ; q1
v2 ; q2
P  Q  p1  p2  q1  q2  m1 (v1  u1 )  m2 (u2  v2 )
 u 2  v2 
m
1
m
2
(v1  u1 ) [statement about velo city changes]
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The elastic collision in 1d
• if DE = 0  DK = 0 too (DU = 0 too because there is
no height change)  elastic collision
• conventional approach is to derive expressions for the
final velocities v1 and v2 in terms of the masses m1 and
m2 and the initial velocities u1 and u2
• there are two conditions to impose: system
momentum conservation, and kinetic energy
conservation  two equations in two unknowns
• trouble is, one of them is ‘quadratic’ statement so the
arithmetic is very tricky
• Watch carefully…
 we will
make use of u2  v2 
m
1
m
2
(v1  u1 )
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The final velocities for the elastic 1d collision
1
2
m
u
2 1 1
1
1
1



 2 m2u 22  2 m1v12  2 m2 v22  m2 u 22  v22  m1 v12  u12
m2 u 2  v2 u 2  v2   m1 v1  u1 v1  u1  but u 2  v2 
m
1
m
2

(v1  u1 )
 m1 
 m2  v1  u1 u 2  v2   m1 v1  u1 v1  u1 
 m2 
 u 2  v2  v1  u1 add this to the above
 m1 

 2u 2  u1  1  m   v1  1 
2


 1
u1 m2  m1   v1 m2  m1 

 m2
m1  m2
2m2
 2m2u 2  u1 m2  m1   v1M  v1 
u1 
u2
M
M
2m
m  m1
and we also get by solving for v2  v2  1 u1  2
u2
M
M
m
1
m
2
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The totally inelastic collision in 1d
• if maximum energy is lost  totally inelastic collision
• the bodies stick together, so velocities v1 = v2 := v
• there is only one condition to impose: system
momentum conservation conservation  one equation
 since u 2
v 
 m1  m2 
v m m  
 1 2 
m
1
m
2
m 
m

1
(v  u1 )  v1  m   u 2  m1 u1
2
2

m u m u
2 2 11
m
 1m2
v
m u m u
2 2 11
M
• proving that maximum energy is lost if the bodies
stick together is not so easy unless one ‘goes to’ the
center of mass frame of reference
• this is alluded to in the book… we will illustrate!
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The totally inelastic collision in 1d as expressed
relative to the c.o.m
1
m1u1  m2u2  and now define primed velocitie s :
 VC :
m1  m2
 u '1 : u1
 in
 VC
u '2 : u 2  VC
' relative to c.o.m.' velocitie s
this ' frame of reference'P'  Q'  m1u '1  m2u '2  Mv'
 m1
u1  VC   m2 u2  VC   Mv'




m
m1u1  m2u2  m2u2  m2 M1 m1u1  m2u2  Mv'
 m1u1 
 Mm1u1  m12u1  m1m2u2  Mm2u2  m1m2u1  m22u 2  M 2 v'
m
1
M
 everythingon left cancels so v'  0
• result is obvious: c.o.m. is motionless in c.o.m frame
• there is NO kinetic energy left as seen in this frame
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Important case: initially stationary m2 (u2 = 0)
• elastic collision is worked out in example 6.3 (p 135)
2m1
m1  m2
u1
u1 v2 
 v1 
m1  m2
m1  m2
2r
r 1
u1
u1 v2 
 v1 
m2
r 1
r 1
 some cases for r are interestin g (book has entire table!!0
m
 now book defines r  1
r  1 (puck collides with bus) v1  u1 v2  0
r  1 (puck collides with puck) v1  0 v2  u1
r >> 1 (bus collides with puck) v1  u1 v2  2u1
• totally inelastic collision is simplicity itself
m1
1
v 
u1 
u1 {show active figures 09_05-07}
m1  m2
r 1
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example
Problem 6.3 Two crates (masses 300 kg and 100 kg), joined by a
light rope, are being pulled along a factory floor by a constant
force exerted on the heavier crate at an angle of 25° to the
horizontal. The coefficient of kinetic friction between the
heavier crate and the floor is 0.11 and that between the lighter
crate and the floor is 0.18. What should the magnitude of the
applied force be to move the crates at constant speed? What
applied force would be required to give the system an
acceleration of 0.10 m/s2? What would the tension in the rope be
in both cases?
100 kg
300 kg
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example 6.6
A glider of mass 180 g travelling on a horizontal airtrack at a
speed of 3.0 m/s catches up with another glider of mass 360 g
moving in the same direction at 0.9 m/s. If 10% of the energy is
lost in the collision what are the speeds of the gliders after the
collision?
P  m1u1  m2u2  (.18)(3.0)  (.36)(.90)  .540  .324  .864
 Q  .864  m1v1  m2v2  .180 v1  .360 v2 one equation
 v2  2.400  .500 v1
solved for v2
K init  12 m1u12  12 m2u22  12 (.18)(3.0) 2  12 (.36)(.90) 2  .810  .146  .956
 K fin  (.90)(.956)  .860  12 m1v12  12 m2v22 second equation
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example cont’d
this becomes, when squaring the above expression for v2

.860  .090 v12  (.180 ) 5.76  2.40v1  .250 v12

combining terms  .135v12  .432 v1  .177  0
 v1 
.432 
.432 2  4.135 .177  .432  .187  .096

2.135 
2.135 
.432  .302 2.72 m/s


not sure which to choose...
.270
.481 m/s
Plug both of these into the v2 equation:
 2.40  .50(2.72)  1.04 m/s makes sense
v2  
2.40  .50(.481)  2.16 m/s makes sense
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example cont’d
P  .864  Q  (.180 ) (2.72)  (.360)(2.16)  1.27
 Q  (.180 ) (.481)  (.360)(2.16)  .864 yes!!
The other answer would have come from the 360 g mass moving
in the opposite direction initially. This would have slowed
down the 180 g mass a lot more, but would have corresponded to
the same kinetic energy intially. So the quadratic nature of the
kinetic energy introduces subtleties!
Example 5.17
A 200 g block is constrained to move on a smooth horizontal
table by a string that passes downward through a hole in the
centre of the table. Its speed is 0.5 m/s and radius is 0.4 m. The
string is pulled to reduce the radius to 0.3 m.
What is new speed? How much work is done by the puller?
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example 5.17 cont’d
• find the net torque on the body as seen
TOP VIEW
from the hole as origin
• MT = r x T  MT = rT sin 180° = 0
MW = r x W  MW = rmg sin 90° = rmg
r
• direction is anti-parallel to p
T
• clearly, N = – W so MN = rmg sin 90° too
O
• but direction is parallel to p
• net torque Mnet = 0  L is conserved
• L = r x p  L = mvr sin 90° = mvr; direction is ‘down’
L init  L fin  Linit  L fin  mvinit rinit  mv fin r fin
 v fin
 rinit 
 vinit  r   .54 3   0.67 m/s
fin 

DK  tension force's work  0  0  K fin  K init 


.20
2 .50 2  .019 J
.
67
2 Education, Inc., publishing as Pearson Addison-Wesley.
Copyright © 2008 Pearson
 DK 
m
2
v
2
fin
2
 vinit

p
Example 5.17 cont’d
• find the net torque on the body as seen
from the hole as origin
• MT = r x T  MT = rT sin 180° = 0
• direction is immaterial!
• MW = r x W  MW = rmg sin 90° = rmg
• direction is anti-parallel to p
• clearly, N = – W so MN = rmg sin 90° too
• but direction is parallel to p
• net torque Mnet = 0  L is conserved
• L = r x p  L = mvr sin 90° = mvr
• direction is into table: ‘down’
L init  L fin  Linit  L fin  mvinit rinit  mv fin r fin
 v fin
 rinit 
 vinit  r   .54 3   6.7 m/s
fin 

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example of a two-body system I
A mass m1 = 500 g moves at 4.0 m/s to the right, and collides
elastically with a mass m2 = 1500 g moving to the left at 2.0 m/s.
a) Find system P and K in this frame
b) Find the c.o.m velocity and the reduced mass
c) Find the quantities in (a) using M and m
d) Find the relative-to-c.o.m velocities and the same 4
quantities relative to (“in”) the c.o.m. frame
e) confirm that P and K are conserved in the original frame
P  m1u1  m2u 2  (.50)(4.0)  (1.50)(2.0)  1.0 kg - m/s
1
1
2
K  2 m1u1  2 m2u 22  4.0  3.0  7.0 J
1
P
1.0 kg-m/s
VC  M
pi  M  2.0 kg  0.50 m/s

m
mm
1 2
m m
1 2

.75 kg2
2.0 kg
 .375 kg (less than either mass)
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example of a two-body system II
c) Find the quantities in (a) using M and m
d) Find the relative-to-c.o.m velocities and the same
4 quantities relative to (“in”) the c.o.m. frame
u '1  u1  VC  4.0  (0.50)  4.5 m/s
u '2  u2  VC  2.0  (0.50)  1.5 m/s
P'  m1u '1  m2u '2  (.50)(4.5)  (1.50)(1.5)  0.0 kg - m/s
1
1
K '  2 m1u '12  2 m2u '22  5.06  1.69  6.75 J
V 'C 
1
M

p'i 
P'
M
 0.0 m/s
P  MVC  (2.0)(0.50)  1.0 kg - m/s
u  u1  u2  4.0  (2.0)  6.0 m/s
K
1
2
MVC
2

1
2
mu
2
 0.25  6.75  7.0 J
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example of a two-body system III
e) confirm that P and K are conserved in the original frame
v1 
m m
1 2u
1
M

2m
2
M
u2 
.51.5
2(1.5)
(4.0)  2.0 (2.0)
2.0
 2.0  3.0  5.0 m/s
v2 
2m
1u
M 1

m m
2 1u
2
M

2(.5)
1.5.5
(
4
.
0
)

(

2
.
0
)
2.0
2.0
 2.0  1.0  1.0 m/s
Q  m1v1  m2v2  (.50)(5.0)  (1.50)(1.0)  1.0 kg - m/s
1
1
K  2 m1v12  2 m2v22  6.25  0.75  7.0 J
one can confirm that P and K are conserved in the c.o.m. frame…..
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.