# PHYS 1441 � Section 004 Lecture #14

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```							                  PHYS 1441 – Section 004
Lecture #14
Monday, Mar. 22, 2004
Dr. Jaehoon Yu
•        Linear Momentum
•        Linear Momentum Conservation
•        Impulse
•        Collisions: Elastic and Inelastic collisions
•        Center of Mass

Monday, Mar. 22, 2004          PHYS 1441-004, Spring 2004      1
Dr. Jaehoon Yu
Announcements
• Quiz results
– Average: 52/ 90 58/100
– Top score: 80/90  89/100
– How did we do compared to the other quizzes?
• 1st quiz: 38.5/100
• 2nd quiz: 41.8/100
– Marked improvement
• Second term exam next Monday, Mar. 29
–   In the class, 1:00 – 2:30pm in SH101
–   Sections 5.6 – 8.2
–   Mixture of multiple choices and numeric problems
–   Will give you exercise test problems Wednesday
Monday, Mar. 22, 2004              PHYS 1441-004, Spring 2004   2
Dr. Jaehoon Yu
Linear Momentum
The principle of energy conservation can be used to solve problems
that are harder to solve just using Newton’s laws. It is used to
describe motion of an object or a system of objects.
A new concept of linear momentum can also be used to solve physical
problems, especially the problems involving collisions of objects.
Linear momentum of an object whose mass is m
and is moving at a velocity of v is defined as                            p  mv
What can you tell from this         1.     Momentum is a vector quantity.
definition about momentum?          2.     The heavier the object the higher the momentum
3.     The higher the velocity the higher the momentum
4.     Its unit is kg.m/s
What else can use see from the            The change of momentum in a given time interval

      
r r
definition? Do you see force?           ur   r    r
 p mv  mv 0   m v  v0
r
v   r     r
u
                    m    ma   F
t      t         t        t
Monday, Mar. 22, 2004           PHYS 1441-004, Spring 2004                           3
Dr. Jaehoon Yu
Linear Momentum and Forces
r
u
ur  p           What can we learn from this Force-momentum
 F  t           relationship?

•       The rate of the change of particle’s momentum is the same as
the net force exerted on it.
•       When net force is 0, the particle’s linear momentum is
constant as a function of time.
•       If a particle is isolated, the particle experiences no net force,
therefore its momentum does not change and is conserved.
Something else we can do                  The relationship can be used to study
with this relationship. What              the case where the mass changes as a
do you think it is?                       function of time.

Can you think of a          Motion of a meteorite           Motion of a rocket
few cases like this?
Monday, Mar. 22, 2004            PHYS 1441-004, Spring 2004                         4
Dr. Jaehoon Yu
Linear Momentum Conservation

p1i  p 2i  m1 v1  m2 v 2

'           '
p1 f  p 2 f  m v  m2 v
1 1           2

Monday, Mar. 22, 2004   PHYS 1441-004, Spring 2004                        5
Dr. Jaehoon Yu
Conservation of Linear Momentum in a Two
Particle System
Consider a system with two particles that does not have any external
forces exerting on it. What is the impact of Newton’s 3rd Law?
If particle#1 exerts force on particle #2, there must be another force that
the particle #2 exerts on #1 as the reaction force. Both the forces are
internal forces and the net force in the SYSTEM is still 0.
Now how would the momenta             Let say that the particle #1 has momentum
of these particles look like?         p1 and #2 has p2 at some point of time.
r
u                       ur
Using momentum-         ur    p                ur    p
force relationship      F 21  1       and      F 12  2
t                       t
ur     r
u

 p 2  p1   u  u

And since net force                                               r r
of this system is 0      F      F 12  F 21   
t

t    t
p 2 p1       0

Therefore p 2  p1  const The total linear momentum of the system is conserved!!!
Monday, Mar. 22, 2004          PHYS 1441-004, Spring 2004                          6
Dr. Jaehoon Yu
More on Conservation of Linear Momentum in
a Two Particle System
From the previous slide we’ve learned that the total
momentum of the system is conserved if no external                           p p      2     p1  const
forces are exerted on the system.

What does this mean?                     As in the case of energy conservation, this means
that the total vector sum of all momenta in the
system is the same before and after any interaction
u
r      u
r     r
u       r
u
Mathematically this statement can be written as                              p 2i  p1i  p 2 f  p1 f

P
system
xi       P
system
xf      P
system
yi      P
system
yf       P
system
zi      P
system
zf

This can be generalized into                           Whenever two or more particles in an
conservation of linear momentum in                     isolated system interact, the total
many particle systems.                                 momentum of the system remains constant.
Monday, Mar. 22, 2004                      PHYS 1441-004, Spring 2004                                          7
Dr. Jaehoon Yu
Example for Linear Momentum Conservation
Estimate an astronaut’s resulting velocity after he throws his book to a
direction in the space to move to a direction.
vA                           vB   From momentum conservation, we can write
p i  0  p f  mA v A  mB v B
Assuming the astronaut’s mass if 70kg, and the book’s
mass is 1kg and using linear momentum conservation
mB v B     1
vA                vB
mA        70

Now if the book gained a velocity
of 20 m/s in +x-direction, the                   vA  
1
70
 
20i   0.3 i m / s 
Astronaut’s velocity is
Monday, Mar. 22, 2004              PHYS 1441-004, Spring 2004                8
Dr. Jaehoon Yu
Impulse and Linear Momentum
ur
Net force causes change of momentum                               u p
r      r
u u  r
Newton’s second law                                                F     p  Ft
t
By summing the above                         r r r
u u u                   ur            r u r      ur
equation in a time interval ti to           p  p f  pi          F t          I  Ft   p
tf, one can obtain impulse I.

So what do you                  Impulse of the force F acting on a particle over the time
think an impulse is?            interval t=tf-ti is equal to the change of the momentum of
the particle caused by that force. Impulse is the degree of
which an external force changes momentum.
The above statement is called the impulse-momentum theorem and is equivalent to Newton’s second law.
What are the          Defining a time-averaged force     Impulse can be rewritten    If force is constant
dimension and
unit of Impulse?          ur 1    ur
What is the
direction of an
F    F i t
t i                             I  Ft                  I  F t
impulse vector?
It is generally assumed that the impulse force acts on a
Monday, Mar. 22, 2004                 PHYS 1441-004, Spring 2004                                 9
short time but much greater than any other forces present.
Dr. Jaehoon Yu
Example 7-5
(a) Calculate the impulse experienced when a 70 kg person lands on firm ground
after jumping from a height of 3.0 m. Then estimate the average force exerted on
the person’s feet by the ground, if the landing is (b) stiff-legged and (c) with bent
legs. In the former case, assume the body moves 1.0cm during the impact, and in
the second case, when the legs are bent, about 50 cm.

We don’t know the force. How do we do this?
Obtain velocity of the person before striking the ground.
mv  mg  y  yi   mgyi
1 2
KE  PE
2
Solving the above for velocity v, we obtain
v  2 gyi  2  9.8  3  7.7 m / s
Then as the person strikes the ground, the
momentum becomes 0 quickly giving the impulse

I  F t  p  p f  pi  0  mv 
 70kg  7.7m / s  540 N  s
Monday, Mar. 22, 2004             PHYS 1441-004, Spring 2004                             10
Dr. Jaehoon Yu
Example 7-5 cont’d
In coming to rest, the body decelerates from 7.7m/s to 0m/s in a distance d=1.0cm=0.01m.
0  vi    7.7
The average speed during this period is    v                 3.8m / s
2       2
d      0.01m
The time period the collision lasts is  t                  2.6 103 s
v     3.8m / s
Since the magnitude of impulse is        I  F t  540N  s
The average force on the feet during        I        540
F                   2.1105 N
this landing is                            t 2.6 103
How large is this average force? Weight  70kg  9.8m / s  6.9 10 N
2          2

F  2.1105 N  304  6.9 102 N  304 Weight
If landed in stiff legged, the feet must sustain 300 times the body weight. The person will
likely break his leg.

Monday, Mar. 22, 2004             PHYS 1441-004, Spring 2004                        11
Dr. Jaehoon Yu
Example 7-5 cont’d
What if the knees are bent in coming to rest? The body decelerates from 7.7m/s to 0m/s in a
distance d=50cm=0.5m.
0  vi      7.7
The average speed during this period is still the same v                    3.8m / s
2       2
The time period the collision lasts         d      0.5m
t                 1.3 101 s
changes to                                  v    3.8m / s
Since the magnitude of impulse is      I  F t  540N  s
The average force on the feet during      I       540
F                  4.1103 N
this landing is                          t 1.3 101
How large is this average force? Weight  70kg  9.8m / s  6.9 10 N
2          2

F  4.1103 N  5.9  6.9 102 N  5.9 Weight
It’s only 6 times the weight that the feet have to sustain! So by bending the knee
you increase the time of collision, reducing the average force exerted on the knee,
and will avoid injury!
Monday, Mar. 22, 2004               PHYS 1441-004, Spring 2004                      12
Dr. Jaehoon Yu
Example for Impulse
In a crash test, an automobile of mass 1500kg collides with a wall. The initial and
final velocities of the automobile are vi=-15.0i m/s and vf=2.60i m/s. If the collision
lasts for 0.150 seconds, what would be the impulse caused by the collision and the
average force exerted on the automobile?

Let’s assume that the force involved in the collision is a lot larger than any other
forces in the system during the collision. From the problem, the initial and final
momentum of the automobile before and after the collision is
p i  mvi  1500 15.0i  22500i kg  m / s

p f  mv f  1500 2.60i  3900i kg  m / s
Therefore the impulse on the       I   p  p  p i  3900  22500 i kg  m / s
f

automobile due to the collision is    26400i kg  m / s  2.64 104 i kg  m / s
p   2.64  10 4
The average force exerted on the           F  t    0.150
automobile during the collision is
 1.76 105 i kg  m / s 2  1.76 105 i N
Monday, Mar. 22, 2004             PHYS 1441-004, Spring 2004                             13
Dr. Jaehoon Yu
Collisions
Generalized collisions must cover not only the physical contact but also the collisions
without physical contact such as that of electromagnetic ones in a microscopic scale.
The collisions of these ions never involves a
Consider a case of a collision
physical contact because the electromagnetic
between a proton on a helium ion.
repulsive force between these two become great
as they get closer causing a collision.
F                 F12
Assuming no external forces, the force                      r r
u u
t   exerted on particle 1 by particle 2, F21,                  p1  F 21t
changes the momentum of particle 1 by
F21                                                                          r r
u u
Likewise for particle 2 by particle 1
 p2  F12t
r
u                   r
u
Using Newton’s      3rd     law we obtain         p 2  F 12 t           F 21t   p1

So the momentum change of the system in the                         p   p1   p 2  0
collision is 0 and the momentum is conserved
p system    p1  p 2     constant
Monday, Mar. 22, 2004                   PHYS 1441-004, Spring 2004                               14
Dr. Jaehoon Yu
Example for Collisions
A car of mass 1800kg stopped at a traffic light is rear-ended by a 900kg car, and the
two become entangled. If the lighter car was moving at 20.0m/s before the collision
what is the velocity of the entangled cars after the collision?
Before collision              The momenta before and after the collision are
m2                    p i  m1 v1i  m2 v2i  0  m2 v 2i
20.0m/s
m1
p f  m1 v1 f  m2 v2 f  m1  m2 v f
After collision
Since momentum of the system must be conserved
m2         pi  p f                      m1  m2 v f    m2 v 2i
m1
vf                                            m2 v 2i   900  20 .0i
vf                                 6.67 i m / s
m1  m2  900  1800
What can we learn from these equations      The cars are moving in the same direction as the lighter
on the direction and magnitude of the       car’s original direction to conserve momentum.
velocity before and after the collision?    The magnitude is inversely proportional to its own mass.
Monday, Mar. 22, 2004               PHYS 1441-004, Spring 2004                               15
Dr. Jaehoon Yu
Elastic and Inelastic Collisions
Momentum is conserved in any collisions as long as external forces negligible.
Collisions are classified as elastic or inelastic by the conservation of kinetic
energy before and after the collisions.

Elastic              A collision in which the total kinetic energy and momentum
Collision            are the same before and after the collision.
Inelastic            A collision in which the total kinetic energy is not the same
Collision            before and after the collision, but momentum is.
Two types of inelastic collisions:Perfectly inelastic and inelastic
Perfectly Inelastic: Two objects stick together after the collision
moving at a certain velocity together.
Inelastic: Colliding objects do not stick together after the collision but
some kinetic energy is lost.
Note: Momentum is constant in all collisions but kinetic energy is only in elastic collisions.
Monday, Mar. 22, 2004                PHYS 1441-004, Spring 2004                         16
Dr. Jaehoon Yu
Elastic and Perfectly Inelastic Collisions
In perfectly Inelastic collisions, the objects stick
m1 v1i  m2 v2i  (m1  m2 )v f
together after the collision, moving together.
Momentum is conserved in this collision, so the                               m1 v1i  m2 v 2i
vf 
final velocity of the stuck system is                                          (m1  m2 )

How about elastic collisions?                       m1 v1i  m2 v2i  m1 v1 f  m2 v2 f
1             1             1             1
In elastic collisions, both the           m1v12i  m2v2i  m1v12f  m2v2 f
2                            2

2             2             2             2
momentum and the kinetic energy
are conserved. Therefore, the          m1 v12i  v12f   m2 v2i  v2 f 
2      2

final speeds in an elastic collision   m1 v1i  v1 f v1i  v1 f   m2 v2i  v2 f v2i  v2 f 
can be obtained in terms of initial From momentum
m1 v1i  v1 f   m2 v2i  v2 f 
speeds as                            conservation above
 m  m2        2m2                       2m1         m  m2 
v1 f   1
m m           m  m v2i
v1i                     v2 f  
m m v1i   1
 m  m v2i

 1    2        1    2                    1  2        1    2 

Monday, Mar. 22, 2004                PHYS 1441-004, Spring 2004                                17
Dr. Jaehoon Yu
Two dimensional Collisions
In two dimension, one can use components of momentum to apply
momentum conservation to solve physical problems.
m1 v 1i  m2 v 2 i  m1 v1 f  m2 v 2 f
v1i
m1                              x-comp.      m1v1ix  m2 v2ix  m1v1 fx  m2 v2 fx
m2
y-comp.      m1v1iy  m2 v2iy  m1v1 fy  m2 v2 fy

q            Consider a system of two particle collisions and scatters in
two dimension as shown in the picture. (This is the case at
f                  fixed target accelerator experiments.) The momentum
conservation tells us:
m1 v 1i  m2 v 2i  m1 v1i
m1v1ix  m1v1 fx  m2v2 fx  m1v1 f cosq  m2 v2 f cos f
m1v1iy  0  m1v1 fy  m2v2 fy  m1v1 f sin q  m2 v2 f sin f
What do you think
And for the elastic conservation,    1         1        1
m1v12i  m1v12f  m2v2 f
2
we can learn from
the kinetic energy is conserved:     2         2        2                 these relationships?
Monday, Mar. 22, 2004           PHYS 1441-004, Spring 2004                                   18
Dr. Jaehoon Yu
Example of Two Dimensional Collisions
Proton #1 with a speed 3.50x105 m/s collides elastically with proton #2 initially at
rest. After the collision, proton #1 moves at an angle of 37o to the horizontal axis and
proton #2 deflects at an angle f to the same axis. Find the final speeds of the two
protons and the scattering angle of proton #2, f.
v1i                             Since both the particles are protons m1=m2=mp.
m1
m2               Using momentum conservation, one obtains
x-comp. m p v1i  m p v1 f cosq  m p v2 f cos f
q                   y-comp.     m p v1 f sin q  m p v2 f sin f  0
f                       Canceling mp and put in all known quantities, one obtains
v1 f cos37  v2 f cosf  3.50 105 (1)

From kinetic energy                       v1 f sin 37  v2 f sin f (2)
conservation:                                                  v1 f  2.80 105 m / s
Solving Eqs. 1-3                                        Do this at
3.50 10 
5 2
 v v
2
1f
2
2f   (3) equations, one gets        v2 f  2.1110 m / s
5
home
f  53 .0 
Monday, Mar. 22, 2004                   PHYS 1441-004, Spring 2004                            19
Dr. Jaehoon Yu
Center of Mass
We’ve been solving physical problems treating objects as sizeless
points with masses, but in realistic situation objects have shapes
with masses distributed throughout the body.
Center of mass of a system is the average position of the system’s mass and
represents the motion of the system as if all the mass is on the point.

What does above               The total external force exerted on the system of
total mass M causes the center of mass to move at
statement tell you
an acceleration given by a   F / M as if all
concerning forces being
the mass of the system is concentrated on the
exerted on the system?        center of mass.
Consider a massless rod with two balls attached at either end.
m1                  m2
x1                  x2    The position of the center of mass of this system is
the mass averaged position of the system
xCM
m x  m2 x2        CM is closer to the
xCM  1 1
m1  m2         heavier object
Monday, Mar. 22, 2004            PHYS 1441-004, Spring 2004                         20
Dr. Jaehoon Yu

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