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AP Physics Chapter 9
Center of Mass and
Linear Momentum
1
AP Physics
 Lecture:

Chapter 9
Q&A
2
Standards:
(AP Physics C: Mechanics)
Newtonian Mechanics
D. Systems of particles, linear momentum
(12%)
1. Center of mass
2. Impulse and momentum
3. Conservation of linear momentum, collisions
3
System
• A system is a collection of objects.
• A system includes the objects, but not
the space in between them.
• In the system


All objects that belong to the system
Not necessary the space enclosed by the system
4
Center of Mass
Center of mass (of a body or a system of
bodies): the point that moves as though all of
the mass were concentrated there and all
external forces were applied there.
Position of Center of Mass:
1D: xcm
ˆ
ˆ
ˆ
r

x
,
y
,
z

x
i

y
j

z
k


3D: cm
cm
cm
cm
cm
cm
cm
5
Position of Center of Mass: xcm
(2 objects, 1D)
xcm
m1 x1  m2 x2

m1  m2
xcm: position of center of mass of two objects
m1: mass of Object 1
m2: mass of Object 2
x1: position of (center of mass of) Object 1
x2: position of (center of mass of) Object 2
•
x1
m1
m2
•
x2
6
Position of Center of Mass: rcm
(n objects, 3D)
1

M
n
m x

i 1


1
1 n
ycm 
mi yi   rcm 

M i 1
M


1 n
zcm 
mi zi 


M i 1
xcm
i i
M  m1  m2 
n
m r
i 1
i i
 mn  mi  Total Mass
7
Position of Center of Mass:
(Continuous Mass Distribution, 3D)

1
xcm 
xdm 

M


1
ycm 
ydm 

M

1

zcm 
zdm 

M

uniform
density
1

 xcm  V  xdV

1

 ycm   ydV
V

1

 zcm  V  zdV
V: volume of object
8
Example: Pg238-102
The distance between the centers of the carbon (C) and oxygen (O)
atoms in a carbon monoxide (CO) gas molecule is 1.131  10-10 m.
Locate the center of mass of a CO molecule relative to the carbon atom.
(Find the masses of C and O in Appendix D.)
C
O
x
0
Let C = 1, O = 2, then m1 = 12.01115u, m2 = 15.9994u.
Also, let x = 0 at C, then x1 = 0, and x2 = 1.131  10-10m.
xcm = ?
xcm 
m1 x1  m2 x2

m1  m2
15.9994u  1.131  1010 m 
12.01115u  15.9994u
 6.46  1011 m
The center of mass is 6.46  10-11 m from the carbon atom, in
between the two atoms.

Center of mass is closer to the object with larger mass.
9
Example: Pg229-3
In Fig. 9-37, three uniform thin rods, each of length L = 22 cm, form
an inverted U. The vertical rods each have a mass of 14 g; the
horizontal rod has a mass of 42 g.
a) What are the x coordinate and
b) the y coordinate of the system’s center of mass?
Set up coordinates, then
m1 = 14g at (0, 11cm)
m2 = 14g at (22cm, 11cm)
m3 = 42g at (11cm, 22cm)
m3
y

m1
•
cm
m
2
x
m1 x1  m2 x2  m3 x3 14 g   0   14 g   22cm   42  g 11cm 

 11cm
m1  m2  m3
14 g  14 g  42 g
m y  m2 y2  m3 y3 14 g  11cm   14 g  11cm   42 g   22cm 

 1 1
 17.6cm
m1  m2  m3
14 g  14 g  42 g
xcm 
ycm
 rcm  11cm,17.6cm 
Symmetric?
10
Newton’s Second Law for a System
of Particles
Fext  Macm
• Fext: external force, force from object not included in
the system
• M: total mass of system
• acm: acceleration of center of mass
 Fext , x  Macm , x

3D    Fext , y  Macm , y

  Fext , z  Macm , z
11
Special Cases and Application
What if the total external force is zero?
acm = 0  vcm = constant
If vcm, i = 0  vcm remains 0  CM does not move.
12
Velocity of Center of Mass, vcm
xcm
m1 x1  m2 x2

m1  m2
 vcm
m1v1  m2 v2

m1  m2
13
Example: Pg238-104
An old Chrysler with mass 2400 kg is moving
along a straight stretch of road at 80 km/h. It
is followed by a Ford with mass 1600 kg
moving at 60 km/h. How fast is the center of
mass of the two cars moving?
Let m1 = 2400kg, v1 = 80km/h, m2 = 1600kg, v2 = 60 km/h
vcm = ?
vcm 
 km 
 km 
2400kg  80
  1600kg  60

m1v1  m2 v2
h
h



  72 km

m1  m2
2400kg  1600kg
h
14
Practice: Pg240-138
Two particles P and Q are initially at rest 1.0 m apart. P has a mass of
0.10 kg and Q a mass of 0.30 kg. P and Q attract each other with a
constant force of 1.0  10-2 N. No external forces act on the system.
a) Describe the motion of the center of mass.
b) At what distance from P’s original position do the particles collide?
Let x1 = 0, then x2 = 1m. Also, m1 = 0.10 kg,
m2 = 0.30kg.
a) Fext = 0  CM stays at rest.
b) What are we actually looking for?
What happens to the center of mass?
xcm  ?
m1 x1  m2 x2 0.10kg  0   0.30kg 1m 
xcm 

 0.75m
m1  m2
0.10kg  0.30kg
15
Newton’s Second Law
d  mv 
dP
dv
dP
 F 

 F  ma  m 
dt
dt
dt
dt
Define: (Linear) Momentum:
P  mv
P is a vector.

Unit:
Same direction as v.
 P   mv  kg 
m
s
16
Momentum of System of Particles
P  Mvcm  m1v1  m2v2  ...
F
ext
 Macm
dP

dt
17
Practice: Pg230-18
A 0.70 kg ball is moving horizontally with a speed
of 5.0 m/s when it strikes a vertical wall. The
ball rebounds with a speed of 2.0 m/s. What is
the magnitude of the change in linear
momentum of the mass?
Let toward the wall be the positive direction, then
m = 0.70kg, vi = 5.0m/s, vf = -2.0 m/s
|P| = ?
P  Pf  Pi  mv f  mvi  m  v f  vi 
m
m
m

 0.70kg  2.0  5.0   4.9kg 
s
s
s

m
P  4.9kg 
s
18
Closed, isolated System

Closed System: n = constant



n = constant
No particles leave or enter the system.
Isolated System:



Fext = 0
The sum of the external forces acting on the system
of particles is zero.
There could be external forces acting on the system,
but the net external force is zero.
19
Law of Conservation of Linear
Momentum
In closed and isolated system:


P = constant
Pi = Pf
P: total momentum of system
P1i + P2i = P1f + P2f
If a component of the net external force on a closed system is
zero along an axis, then the component of the linear
momentum of the system along that axis cannot change.
20
Example: Pg232-35
A 91 kg man lying on a surface of negligible friction shoves
a 68 g stone away from himself, giving it a speed of 4.0
m/s. What speed does the man acquire as a result?
m1  91kg , m2  68 g  0.068kg , v1i  v2i  0, v2 f
m
 4.0
s
v1 f  ?
Pi  Pf
 P1i  P2i  P1 f  P2 f  0  m1v1 f  m2 v2 f
 v1 f  
m2 v2 f
m1
 
0.068kg  4.0
91kg
m
s   2.9  103 m
s
Notice the negative sign. When we define v2 = 4.0 m/s, we
already define the direction of v2f to be the positive direction.
21
Practice: Pg232-37
A space vehicle is traveling at 4300 km/h relative to the Earth when the
exhausted rocket motor is disengaged and sent backward with a speed
of 82 km/h relative to the command module. The mass of the motor is
four times the mass of the module. What is the speed of the command
module relative to earth after the separation?
Motor = 1, Module = 2, v1i =v2i= 4300 km/h, m1 = 4m2,
v1f – v2f = -82 km/h  v1f = v2f – 82 km/h = v2f - vR,
v2f = ?
Pi  Pf
  m1  m2  vi  m1v1 f  m2v2 f
  m1  m2  vi  m1  v2 f  vR   m2 v2 f   m1  m2  v2 f  m1vR
  m1  m2  v2 f   m1  m2  vi  m1vR
 v2 f
5m2 vi  4m2 vR
m1  m2  vi  m1vR
4m2  m2  vi  4m2 vR





5m2
 m1  m2 
 4m2  m2 
5  4300km / h   4 82km / h 
5v  4vR
 i

 4.4  103 km / h
5
5
22
Practice
A railroad flatcar of mass M can roll without
friction along a straight horizontal track.
Initially, a man of mass m is standing on the
car, which is moving to the right with speed
v0. What is the change in velocity of the car
if the man runs to the left so that his speed
relative to the car is vrel?
Man’s motion
Flatcar’s
motion
23
Solution
m1 = M, m2 = m, v1i = v2i = vo, v2f = v1f - vrel,
v1=v1f - v1i = ?
Pi  Pf   m1  m2  vi  m1v1 f  m2v2 f
  m1  m2  vo  m1v1 f  m2  v1 f  vrel    m1  m2  v1 f  m2vrel
v1 f
m1  m2  vo  m2 vrel


 m1  m2 

 M  m  vo   m  vrel
 M  m

mvrel 
  v0 
v1  v1 f  v1i  vo 
 M  m 

 vo 
mvrel
 M  m
mvrel
 M  m
24
Practice: Pg236-85
The last stage of a rocket is traveling at a speed of
7600 m/s. This last stage is made up of two parts
that are clamped together, namely, a rocket case
with a mass of 290.0 kg and a payload capsule with
a mass of 150.0 kg. When the clamp is released, a
compressed spring causes the two parts to separate
with a relative speed of 910.0 m/s.
a) What are the speeds of the two parts after they
have separated? Assume that all velocities are along
the same direction.
b) Find the total kinetic energy of the two parts before
and after they separate; account for any difference.
25
Solution: Pg236-85
case 1, capsule 2, m1 = 290.0kg, m2 = 150.0kg,
v1i=v2i=vi = 7600 m/s, v2f – v1f = vR = 910.0 m/s
a )v1 f  ?, v2 f  ?
Pi  Pf 
 m1  m2  vi  m1v1 f
v1 f
 m2v2 f  m1v1 f  m2  v1 f  vR  
 m1  m2  v1 f
 m2vR
m1  m2  vi  m2 vR


 m1  m2 
m
m


290.0
kg

150.0
kg
7600

150.0
kg
910.0



 

s
s



  7290 m

s
 290.0kg  150.0kg 
v2 f
m
m
m
 v1 f  vR  7290  910.0  8200
s
s
s
26
Solution: Pg236-85 (2)
b) Ki = ?, Kf = ?
2
1
m
1

2
K i   m1  m2  vi   290.0kg  150.0kg   7600   1.271 1010 J
2
s
2

1
1
K f  K1 f  K 2 f  m1v1 f 2  m2 v2 f 2
2
2
2
2
1
m
1
m




  290.0kg   7290   150.0kg   8200   1.275  1010 J
2
s 2
s


K f  Ki
The increase in the KE comes from the spring, when the spring
does positive work to separate the two parts.
27
System with Varying Mass: Rocket
 Rocket
accelerates forward by burning
and ejecting fuel backward.
 Mass of rocket keeps changing.
 Newton’s Second Law and Conservation
of Momentum still apply, though more
complicated.
28
First Rocket Equation
Rvrel  Ma
R = |dM/dt|, rate at which rocket losing mass (fuel)
vrel: exhaust speed, backward speed of fuel relative to the
rocket
M: mass of rocket at that moment
a: acceleration of rocket at that moment
29
Second Rocket Equation
Mi
v f  vi  vrel ln
Mf
 vf: final speed of rocket after some fuel




consumption
vi: initial speed of rocket
vrel: exhaust speed of fuel
Mi: initial mass of rocket
Mf: final mass of rocket after ejecting some fuel
30
Example: Pg238-112
A rocket is moving away from the solar system at a speed of 6.0  103 m/s.
It fires its engine, which ejects exhaust with a velocity of 3.0  103 m/s
relative to the rocket. The mass of the rocket at this time is 4.0  104 kg,
and its acceleration is 2.0 m/s2.
a) What is the thrust of the engine?
b) At what rate, in kilograms per second, was exhaust ejected during the
firing?
vi = 6.0  103 m/s, vrel = 3.0  103 m/s, M = 4.0  104 kg,
a = 2.0 m/s2.
a) F  ?
m

4
ma

4.0

10
kg
2.0

8.0

10
N

  s 2 
F
4
b) R  ?
Rvrel  Ma
m

4.0  104 kg  2.0 2 
Ma
kg
s 

R

 27
m
vrel
s
3.0  103 2
s
31
Practice: P238-114
During a lunar mission, it is necessary to increase the speed of a
spacecraft by 2.2 m/s when it is moving at 400 m/s. The exhaust
speed of rocket engine is 1000 m/s. What fraction of the initial mass
of the spacecraft must be burned and ejected for the increase?
vi = 400 m/s,  v = 2.2 m/s,vrel = 1000 m/s
|M/Mi| = ?
Mi
v
Mi
ln


v  v f  vi  vrel ln
Mf
vrel
Mf
v
2.2 m / s
Mi
vrel

 e  e1000 m / s  1.0022  M i  1.0022M f
Mf
M f  1.0022M f 0.0022M f
M f  Mi
M


 0.0022

Mi
1.0022M f
1.0022M f
Mi

M
 0.0022
Mi
32
Impulse
Newton’s Second Law
v mv   mv  P



F  ma  m
t
t
t
t
 F t  P
Area under curve of F-t
Define Impulse:
J  F t  Ft 
J  P  Pf  Pi
 F  dt
t
Impulse-Momentum Theorem
33
Example:
A cue stick strikes a stationary pool ball, exerting an
average force of 50 N over a time of 10 ms. If the ball
has mass 0.20 kg, what speed does it have after impact?
F  50 N , t  10  103 s, m  0.20kg, vi  0
vf  ?
J  P  Pf  Pi
 Ft  mv f  mvi  mv f
 vf 
F t

m
50 N  10  103 s 
0.20kg
 2.5
m
s
34
Practice:
A ball of mass m and speed v strikes a wall perpendicularly and
rebounds in the opposite direction with the same speed.
a) If the time of collision is t, what is the average force exerted by
the ball on the wall?
b) Evaluate this average force numerically for a rubber ball with mass
140 g moving at 7.8 m/s; the duration of the collision is 3.8 ms.
m, vi  v, v f  v
a ) t , F  ?
Ft  Pf  Pi  mv f  mvi  m  v   mv  2mv
2mv
2mv

F
t
t
b)m  0.140kg , v  7.8m / s, t  3.8  103 s
F ?
m

2  0.140kg   7.8 
2mv
s

F

 570 N
3
t
3.8  10 s
35
Practice: Pg231-26
A 1.2 kg ball drops vertically onto a floor, hitting with a speed of 25
m/s. It rebounds with an initial speed of 10 m/s.
a) What impulse acts on the ball during the contact?
b) If the ball is in contact with the floor for 0.020 s, what is the
average force exerted on the floor?
Upward  , m  1.2kg , vi  25
a) J  ?
m
m
, v f  10
s
s
J  P  Pf  Pi  mv f  mvi  m  v f  vi 
 m 
m 
m
 1.2kg 10   25    42kg 
s 
s
 s 
b)t  0.020s, F  ?
m
42kg 
J
s  2100 N
J  Ft  F 

t
0.020s
36
Series of Collisions
m
F 
v
t
F:
Impulse-Momentum Theorem
Average force on body being hit
m
: Rate at which mass collides with the body
t
v :
Change in velocity of hitting bodies
37
Example: Pg237-99
A pellet gun fires ten 2.0 g pellets per second with a
speed of 500 m/s. The pellets are stopped by a rigid
wall.
a) What is the momentum of each pellet?
b) What is the kinetic energy of each pellet?
c) What is the average force exerted by the stream of
pellets on the wall?
d) If each pellet is in contact with the wall for 0.6 ms,
what is the average force exerted on the wall by each
pellet during contact? Why is this average force so
different from the average force calculated in (c)?
38
Solution: Pg237-99
n  10, m  2.0  103 kg, t  1s, vi  500m / s, v f  0
m
m

a ) P  mv  2.0  103 kg  500   1.0kg 
s
s

2
1 2 1
m


b) K  mv   2.0  103 kg   500   250 J
2
2
s

m

10  2.0  103 kg   500 
nmvi
s
m
nm


 10 N
c) F  
v  
v f  vi  

t
1s
t
t
d )t  0.6  103 s, F  ?
J  P  Ft  Pf  Pi  mv f  mvi  mvi
m

2.0  103 kg  500 
mvi
s
3

3

1.7

10
N
F



1.7

10
N
3
t
0.6  10 s
Most of the time no bullet is in contact with the wall, average is much smaller.
39
Practice:
A movie set machine gun fires 50 g bullets at a speed of
1000 m/s. An actor, holding the machine gun in his hands,
can exert an average force of 180 N against the gun.
Determine the maximum number of bullets he can fire per
minute while still holding the gun steady?
m  0.050kg , vi  0, v f  1000
m
, F  180 N , t  60s
s
n?
nm
nm
m

v

v


vf
F 
v 

f
i
t
t
t
F t

n 
mv f
180 N  60s 
 216
m

0.050kg 1000 
s

40
Collisions

Total momentum is always conserved.


Pi = Pf
Total kinetic energy:
 Conserved  Elastic Collision: Ki = Kf
 Not conserved  Inelastic Collision

Extreme case: Stick and move together  Completely
Inelastic Collision (No other collision loses more portion of
the kinetic energy than this.)
41
Elastic Collisions in 1D
Conservation of P:
Conservation of K:

m1v1i  m2 v2i  m1v1 f  m2 v2 f
1
1
1
1
m1v1i 2  m2 v2i 2  m1v1 f 2  m2 v2 f 2
2
2
2
2
m1  m2
2m2

v

v

 1 f m  m 1i m  m v2i

1
2
1
2
if v2i = 0

v  2m1 v  m2  m1 v
2f
1i
2i

m

m
m

m

1
2
1
2
m1  m2

v1 f  m  m v1i

1
2

v  2m1 v
 2 f m1  m2 1i
42
Special Cases (of v2i = 0)
v1 f  0

v2 f  v1i

Equal masses: m1 = m2

v1 f  v1i
Massive target: m2 >> m1  
v2 f  0

v1 f  v1i

v2 f  2v1i

Light target: m2 << m1
Motion of CM: vcm 
m1v1i  m2 v2i
m1  m2
m1

v1i
m1  m2
m1  m2

v1 f  m  m v1i

1
2

v  2m1 v
 2 f m1  m2 1i
P
 vcm 
m
43
Example:
A hovering fly is approached by an enraged elephant charging at 2.1
m/s. Assuming that the collision is elastic, at what speed does the fly
rebound? Note that the projectile (the elephant) is much more
massive than the stationary target (the fly).
m
Elephant  1, Fly  2, v1i  2.1 , m1  m2
s
v2 f  ?
v2 f
m
m
 2v1i  2  2.1  4.2
s
s
Notice that the elephant is chosen as mass 1 because of the
formula we have.
44
Practice: Pg234-55
A cart with mass 343 g moving on a frictionless
linear air track at an initial speed of 1.2 m/s
strikes a second cart of unknown mass at
rest. The collision between the carts is
elastic. After the collision, the first cart
continues in its original direction at 0.66 m/s.
a) What is the mass of the second cart?
b) What is its speed after impact?
c) What is the speed of the two-cart center of
mass?
45
Solution: Pg234-55
m1  0.340kg , v1i  1.2
m
m
, v2i  0, v1 f  0.66
s
s
a)m2  ?
m1  m2
v1 f 
v1i  m1v1 f  m2 v1 f  m1v1i  m2 v1i
m1  m2
 m2 v1 f  m2 v1i  m1v1i  m1v1 f  m2  v1i  v1 f   m1  v1i  v1 f 
m2 
b)v2 f
m1  v1i  v1 f
v1i  v1 f

m
 m
0.340kg 1.2  0.66 
s
s


 0.099kg  99 g
m
m
1.2  0.66
s
s
2m1
2  0.340kg
m
 m
v1i 

1.2   1.9
m1  m2
0.340kg  0.099kg 
s
s
 m
0.340kg 1.2 
m1v1i
m
s


 0.93
c)vcm 
m1  m2 0.340kg  0.099kg
s
46
Practice: Pg234-59
A body of mass 2.0 kg makes an elastic collision
with another body at rest and continues to
move in the original direction but with onefourth of its original speed.
a) What is the mass of the struck body?
b) What is the speed of the two-body center of
mass if the initial speed of the 2.0 kg body
was 4.0 m/s?
47
Solution: Pg234-59
m1  2.0kg, v1i  v, v1 f  v / 4, v2i  0
a)m2  ?
v1 f m1  m2 v / 4 1
m1  m2

v1 f 
v1i 


v
4
m1  m2
v1i m1  m2
 4  m1  m2   m1  m2  4m1  4m2  m1  m2
3
3
 4m1  m1  m2  4m2  3m1  5m2  m2  m1   2.0kg   1.2kg
5
5
m
b)v1i  4.0 , vcm  ?
s
2.0kg
m
m
m1

4.0

2.5
vcm 
v1i 


2.0kg  1.2kg 
s
s
m1  m2
48
Practice
A platform scale is calibrated to indicated the
mass in kilograms of an object placed on it.
Particles fall from a height of 3.5 m and
collide with the platform of the scale. The
collisions are elastic; the particles rebound
upward with the same speed they had before
hitting the pan. If each particle has a mass
of 110 g and collisions occur at the rate of 42
s-1, what is the average scale reading?
49
Solution:
up  , yground  0, y  3.5m, v f  vi , v2 f  0, m  0.110kg , n
t
 42s 1 ,
F ?
1
m
m

mvi 2  vi   2 gy   2  9.8 2   3.5m   8.28
2
s 
s

m
v f  8.28
s
y=0
m 
m
m
v  v f  vi  8.28   8.28   16.56
s 
s
s
m
nm
n
1
m
F 
v  
v   mv  42  0.110kg  16.56   76.5 N
t
t
t
s
s

mgy 
But that is force on scale. By Newton’s Third Law, the force of scale on
particle is 76.5N, and this is a normal force. Remember that scale reading
(apparent weight) is the normal force.
mread 
N 76.5 N

 7.8kg
m
g 9.8
s2
50
Inelastic Collision

P is conserved.

K is not conserved (lost). Where does the
energy go?
–
–
–
Sound
Heat
Deformation, change in shape
51
Completely Inelastic Collision in 1D
If m1, initially moving with v1i, collides with m2,
initially moving at v2i, and the two stick and
move together with a combined velocity of Vf,
then
Pi  Pf  P1i  P2i  P1 f  P2 f
m1v1i  m2 v2i
 m1v1i  m2v2i   m1  m2 V f  V f 
m1  m2
Reverse CIC: Decay and explosion
Similar analysis
52
Example:
A 6.0 kg box sled is coasting across frictionless
ice at a speed of 9.0 m/s when a 12 kg
package is dropped into it from above. What is
the new speed of the sled?
Motion on horizontal direction only.
m1  6.0kg , v1i  9.0
m
, m2  12kg , v2i  0
s
vf  ?
m

6.0kg  9.0 
m
s
m1v1i


3.0
vf 

6.0kg  12kg
s
m1  m2
53
Practice: Pg240-136
A 3000 kg weight falls vertically through 6.0 m and then collides with
a 500 kg pile, driving it 3.0 cm into bedrock. Assuming that the
weight-pile collision is completely inelastic, find the magnitude of the
average force on the pile by the bedrock during the 3.0 cm descent.
down  , m1  3000kg , y1  6.0m, m2  500kg , v2i  0, d  0.030m,
1
F ?
m
m
1

2
E1  E2  mgy1  mv2  v2  2 gy1  2  9.8 2   6.0m   10.8
s 
s
2

m

3000kg 10.8 
m1v1i
m
s

V3 

 9.26
m1  m2 3000kg  500kg
s
W  K  K4  K3   K3   m1  m2  g d  F d  
m1  m2  V32

F

2d
1
 m1  m2 V32
2
y=0
2
3
4
 m1  m2  g
2
m

3000
kg

500
kg
9.26



m
s



  3000kg  500kg   9.8 2   5.0  106 N
2  0.030m 
s 

54
Practice: Pg234-54
In Fig. 9-62, block 2 (mass 1.0 kg) is at rest on a
frictionless surface and touching the end of an
unstretched spring of spring constant 200 N/m.
The other end of the spring is fixed to a wall.
Block 1 (mass 2.0 kg), traveling at speed v1 =
4.0 m/s, collides with block 2, and the two
blocks stick together. When the blocks
momentarily stop, by what distance is the spring
compressed?
1
v1
2
55
Solution: Pg234-54
m1  2.0kg , m2  1.0kg , k  200 N / m, v1i  4.0
xmax  ?
CIC:
m
, v2i  0
s
m

2.0kg  4.0 
m1v1
m
s

vf 

 2.67
m1  m2 1.0kg  2.0kg
s
Conservation of E during compression of spring:
Ei  E f
1
1
Mv f 2  kxmax 2
2
2
xmax
M

vf 
k
1.0kg  2.0kg   2.67 m  
200 N / m



s
0.33m
56
Decay

Reverse process of completely inelastic collision.
–
Same analysis using conservation of momentum.

Momentum is conserved.

Kinetic energy is not conserved, but total massenergy is conserved. Gain in KE is given by loss
in mass.
–
–
–
Energy released: Q = ΔK = -Δm c2
atomic mass unit: u = 1.66  10-27 kg
1 eV = 1.60  10-19 J
57
Example: Pg237-75E
A particle called - (sigma minus) is initially at rest and
decays spontaneously into two other particles
according to
-  - + n.
The masses are
m = 2340.5 me, m = 273.2me, mn = 1838.65me,
Where me (9.1110-31 kg or 0.511 MeV/c2) is the electron
mass.
a) How much energy is transferred to kinetic energy in
this process?
b) How do the linear momenta of the decay products (and n) compare?
c) Which product gets the larger share of the kinetic
58
energy?
Solution: Pg237-75E
m  2340.5me , m  273.2me , mn  1838.65me
a )K  ?
2
2
2
K  mc   m  m  mn  c   2340.5me  273.2me  1838.65me  c
 228.65me c 2  228.65  0.511MeV / c 2  c 2  117MeV
b) The two particles should have momentum equal in magnitude
and opposite in direction.
c)
mn 183865me
P2
K
1 2  mv 


 6.730
K  mv 


2m
m
273.2me
2
Kn
2m
2
Smaller particle has larger kinetic energy.
59
Collision in 2D
Glancing collision (as compared to head-on
collision.)
Scattering (Not really hitting each other)
60
Collision in 2D (only for v2i = 0)

Conservation of linear momentum
Px: m1v1i  m1v1 f cos 1  m2v2 f cos 2

Py:  0  m1v1 f sin 1  m2 v2 f sin  2

If kinetic energy is also conserved.
1
1
1
2
2
2
m1v1i  m1v1 f  m2v2 f
2
2
2
v2f
y
m2

If m1 = m2:
1  2  90o
2
v1i
m1
m2
x
1
m1
v1f
61
Example:
Two vehicles A and B are traveling west and south, respectively, toward the
same intersection, where they collide and lock together. Before the collision,
A (total weight 2700 lb) is moving with a speed of 40 MPH and B (total
weight 3600 lb) has a speed of 60 MPH. Find the magnitude and direction
of the velocity of the (interlocked) vehicles immediately after the collision.
East  x, North  y, mA  2700lb, mB  3600lb, v Ai  40mphiˆ, vBi  60mphˆj
vf  ?
Pix  Pfx  mAvAix   mA  mB  v f x v f x
2700lb  40mph 
mAvAix


 17.1mph
mA  mB
2700lb  3600lb
Piy  Pfy  mB vBiy   mA  mB  v f y v f y 
vf 
v fx  v fy 
  tan
2
1
vf y
vf x
2
 17.1mph 
2
3600lb  60mph 

 34.3mph
mA  mB
2700lb  3600lb
mB vBiy
  34.3mph   38mph
34.3mph
 tan
 64o ,South of West
17.1mph
1
y
2
B
A

vf
x
62
External Force and Internal Energy
Change
Eint   Fext d cm cos
 Eint: change in internal energy of system that
causes the external force
 Fext: external force that is caused by system
 dcm: displacement of center of mass of system
 : angle between Fext and dcm
In this case, instead of thinking Fextdcmcos  as the work done by the
external force, it is better to understand it as the work done by the system.
If this work is positive, the Eint is negative, and the system loses internal
energy.
63
then kinetic energy of CM is changed:
Kcm  Fext d cm cos 
If there is also potential energy of any object
within the system involved, then the
mechanical energy of the system (which is
expanded to include the earth and/or
spring) will be changed:
E  Kcm  Ucm  Fext d cm cos
64
Collision
 An
isolated event in which two or more
bodies exert relatively strong forces on
each other for a relatively short time.

If not isolated, not applicable, except when


Fext << Fint
Fext = 0 at the direction of consideration
65