PHY 113 C General Physics I
11 AM – 12:15 PM MWF Olin 101
Plan for Lecture 10
Chapter 9 -- Linear momentum
1. Impulse and momentum
2. Conservation of linear momentum
3. Examples – collision analysis
4. Notion of center of mass
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PHY 113 C Fall 2013 -- Lecture 10
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PHY 113 C Fall 2013 -- Lecture 10
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Summary of physics “laws”
Newton' s second law :
dv
F  ma  m
dt
Work - kinetic energy theorem :
f
i f
total
W
1 2 1 2
  F  dr  mv f  mvi
2
2
i
For conservative forces :
i f
Wconservati
ve  U r f   U ri 
i f
i f
i f
Wtotal
 Wconservati

W
ve
non  conservative
1 2
1 2
i f
mv f  U r f   mvi  U ri   Wnon
 conservative
2
2
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Another way to look at Newton’s second law:
dv d mv 
F  ma  m

dt
dt
Define " linear momentum":
mv  p
Units of linear momentum : kg  m/s  N  s
iclicker question:
Why would you want to define linear momentum?
A. To impress your friends.
B. To exercise your brain.
C. It might be helpful.
D. To distinguish it from angular momentum.
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Relationship between Newton’s second law and
linear momentum:
dv d mv  dp
F  ma  m


dt
dt
dt
Using a little calculus :
(if m is constant)
Fdt  dp
f
f
 Fdt   dp  p
i
f
 pi
i
f
Define " impulse": I   Fdt
i
f
I   Fdt  p f  p i
i
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PHY 113 C Fall 2013 -- Lecture 10
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f
Example : Suppose the impulse I   Fdt
i
shown in this figure has the value of
1000 Ns which is imparted to an object
of mass m  50 kg, initially at rest.
What is its final velocity?
f
I   Fdt  p f  p i
i
vf 
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I 1000

m / s  20m / s
m
50
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Suppose that a tennis ball with
mass m=0.057 kg approaches a
tennis racket at a speed of 45m/s.
What is the impulse the racket must
exert on the ball to return the ball in
the opposite direction at the same
speed. Assume that the motion is
completely horizontal.
I  p f  p i  2mv  2(0.057)(45)  5.13 Ns
What is the average force exerted by the racket
if the interaction lasted for 0.3 s?
F 
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I 5.13 Ns

 17.1N
t
0 .3 s
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Example: A 1500 kg car collides with a wall, with
vi= -15m/s and vf=2.6m/s. What is the impulse
exerted on the car?
I  p f  p i  mv f  vi 
 1500kg 2.6   15m / s
 26400 Ns
If the collision time is 0.15 s, what is the
average force exerted on the car?
I
26400 Ns
F 

 1.76 105 N
t
0.15s
Energy loss due to collision


1 2 1 2 1
ΔE  mv f  mvi  m v 2f  vi2  1.6368 105 J
2
2
2
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Example of graphical representation of F(t)
Estimate of impulse :
0.0015
1
1
I   F (t )dt  Fmax 0.0015s   18000 N 0.0015s   13.5 Ns
2
2
0.001
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Physics of composite systems
Newton' s second law :
dv i
d mi v i  d 

i Fi  i miai  i mi dt i dt  dt  i pi 
Note that if
 F  0,
i
then :
i
d 

  pi   0
dt  i

  p i  (constant)
i
  p i initial 
i
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p
i final
i
PHY 113 C Fall 2013 -- Lecture 10
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Example – completely inelastic collision; balls moving
on a frictionless surface
p
i initial
i

p
i final
i
m1 v1i  m2 v 2i  m1  m2 v f
m1 v1i  m2 v 2i
vf 
m1  m2
For
m1  0.3kg , m2  0.5kg
v  2m / sˆi , v  1m / sˆi
1i
vf
2i

0.32 ˆi  0.5 1ˆi

m/s
0.3  0.5
 0.125 m/s ˆi
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Energy loss in this example:
2
1
1
2
2
1
E  m1  m2  v f   m1 v1i  m2 v 2i 
2
2
2

For m1  0.3kg , m2  0.5kg
v  2m / sˆi , v  1m / sˆi
1i
2i
v f  0.125 m/s ˆi
1
0.3  0.5 0.125 2   1 0.3 2 2  1 0.51 2 
2
2
2

 0.84 J
E 
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Example – completely elastic collision; balls moving on
a frictionless surface
p

p

i
i initial

i final
i
1

1

2
2
m
v

m
v
i  2 i i initial  i  2 i i final 
m1 v1i  m2 v 2i  m1 v1 f  m2 v 2 f
1
1
1
1
2
2
2
m1v1i  m2 v2i  m1v1 f  m2 v22 f
2
2
2
2
For 1 - d motion, after some algebra :
 m1  m2 
 2m2 
v1i  
v2i
v1 f  
 m1  m2 
 m1  m2 
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 2m1 
 m2  m1 
v1i  
v2i
v2 f  
m1  m2 
m1  m2 


PHY 113 C Fall 2013 -- Lecture 10
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Completely elastic collision; numerical example:
 m1  m2 
 2m2 
v1i  
v2i
v1 f  
 m1  m2 
 m1  m2 
 2m1 
 m2  m1 
v1i  
v2i
v2 f  
 m1  m2 
 m1  m2 
For m1  0.3kg , m2  0.5kg
v  2m / sˆi , v  1m / sˆi
1i
2i
 0.3  0.5 
2m / s  
v1 f  
 0.3  0.5 
 20.5 

 1m / s   1.75m / s
 0.3  0.5 
 20.3 
2m / s  
v1 f  
 0.3  0.5 
 0.5  0.3 

 1m / s   1.25m / s
 0.3  0.5 
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PHY 113 C Fall 2013 -- Lecture 10
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iclicker exercise:
We have assumed that there is no net force acting on the
system. What happens if there are interaction forces
between the particles?
A. Analysis still applies
B. Analysis must be modified
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Example from homework:
p
i initial
i

p
i final
i
mC v Ci  mT v Ti  mC v Cf  mT v Tf
v Tf  ?
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v Tf 
mC v Ci  v Cf 
PHY 113 C Fall 2013 -- Lecture 10
mT
 v Ti
16
Example from homework: -- continued
Change in mechanical energy :
1
1
1
1
2
2
2
K f  K i  mC vCf  mT vTf  mC vCi  mT vTi2
2
2
2
2
iclicker question
Do you expect Kf-Ki to be
A. >0
B. <0
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Another example:
before
vf
after
p
i
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i initial

p
i final
i
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Examples of two-dimensional collision; balls moving on a
frictionless surface
p
i initial

p
i
i final
i
m1v1i  m1v1 f cos   m2 v2 f cos 
0  m1v1 f sin   m2 v2 f sin 
Knowns : m1 , m2 , v1i
Unknowns : v1 f , v2 f ,  , 
Need 2 more equations - -
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Examples of two-dimensional collision; balls moving on a
frictionless surface
Suppose : m1  m2  0.06kg , v1i  2m / s,
v2 f  1m / s,
  20o
m1v1i  m1v1 f cos   m2 v2 f cos 
0  m1v1 f sin   m2 v2 f sin 
v1 f sin   v2 f sin 
 1m / s sin 20o  0.342m / s
v1 f cos   v1i  v2 f cos 


 2m / s   1m / s  cos 20o  1.060m / s
0.342
tan  
   17.88o
1.060
0.342m / s 1.060m / s
v1 f 

 1.11m / s
o
o
sin 17.88
cos17.88
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Example: two-dimensional totally inelastic collision
m1 v1i  m2 v 2i  m1  m2 v f
vf 
m1
m2
v1i 
v 2i
m1  m2
m1  m2
1500
2500
25m / sˆi 
20m / sˆj
4000
4000
 9.375m / sˆi  12.5m / sˆj
vf 
m1=1500kg
Loss of kinetic energy in this case :
2
1
m1  m2  v f   1 m1 v1i 2  1 m2 v 2i 2 
2
2
2

1
2
2
 4000  9.375  12.5
2
1
1
2
2
  150025  250020  
2
2

ΔE 

m2=2500kg

ΔE  4.8  105 J
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PHY 113 C Fall 2013 -- Lecture 10
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iclicker exercise:
Can this analysis be used to analyze a real collision?
A. Of course! The laws of physics must be obeyed.
B. Of course NOT! In physics class we only deal
with idealized situations which never happen.
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Another example of 2-dimensional elastic collision:
In this case, the collision is elastic - - v i  v f  v
vf
m
I  2mv sin ˆi
1
E  m v f
2
2
1
2
 m vi  0
2
vi
m
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Energy analysis of a simple nuclear reaction :
226
88
Ra 
222
86
Rn  He
4
2
Q=4.87 MeV
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PHY 113 C Fall 2013 -- Lecture 10
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Energy analysis of a simple reaction :
226
88
4
Ra  222
Rn

86
2 He
Q=4.87 MeV
p Rn  p He  p
2
2
2
pRn
pHe
p
Q


2mRn 2mHe 2mu
1
 1
 

 222 4 
2
E He
pHe
1/ 4


Q  0.98  Q  4.8MeV
2mHe 1 / 222  1 / 4
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Elastic collision in two dimensions
p
i initial

i
K
i
p
i final
i
i initial

K
i final
i
m1v1i  m1v1 f cos   m2 v2 f cos 
0  m1v1 f sin   m2 v2 f sin 
1
1
1
1
2
2
2
m1v1i  m2 v2i  m1v1 f  m2 v22 f
2
2
2
2
Knowns : m1 , m2 , v1i
Unknowns : v1 f , v2 f ,  , 
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Elastic collision in two dimensions -example of elastic proton-proton scattering
p
i initial

i
K
p
i final
i
i initial

K
i
i final
i
m1v1i  m1v1 f cos   m2 v2 f cos 
0  m1v1 f sin   m2 v2 f sin 
1
1
1
1
m1v12i  m2 v22i  m1v12f  m2 v22 f
2
2
2
2
Suppose : v1i  3.5  105 m / s;
m1  m2  m;   37 o
Unknowns : v1 f , v2 f , 
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Elastic collision in two dimensions -example of elastic proton-proton scattering
v1i  v1 f cos   v2 f cos 
0  v1 f sin   v2 f sin 
v12i  v22i  v12f  v22 f
Given : v1i  3.5  105 m / s;   37 o
After some algebra :
v1 f  2.80  105 m / s
v2 f  2.11 105 m / s
  53o
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The notion of the center of mass and the physics of
composite systems
Newton' s second law :
dv i
d mi v i  d 

i Fi  i miai  i mi dt i dt  dt  i pi 
or
d 2ri
d 2 mi ri 
i Fi  i miai  i mi dt 2 i dt 2
rCM 
Define :
F  F
i
i
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total
 m r 
i i
i
M
d 2rCM
M
dt 2
M   mi 
PHY 113 C Fall 2013 -- Lecture 10
i
29
General relation for center of mass :
rCM 
Define :
F  F
i
total
i
 m r 
i i
i
M
M   mi 
i
d 2rCM
M
dt 2
If Ftotal   Fi  0, then :
i
d 

  pi   0
dt  i

  p i  (constant)
i
  p i initial
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i
drCM
  p i final  M
dt
i
PHY 113 C Fall 2013 -- Lecture 10
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Finding the center of mass
rCM 
 m r 
i i
i
M
M   mi 
i
In this example : m1  m2  1kg ; m3  2kg
m1 x1ˆi  m2 x2 ˆi  m3 y3ˆj
rCM 
m1  m2  m3
(1)(1m)ˆi  12m ˆi  2 2m ˆj
rCM 
4
 0.75mˆi  1.00mˆj
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Example from webassign :
y
m1
m2
m3
m4
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PHY 113 C Fall 2013 -- Lecture 10
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32
Finding the center of mass
For a solid object composed of constant density
material, the center of mass is located at the center
of the object.
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