Goals for Chapter 7

Chapter 7
• To study impulse and momentum.
Impulse and Momentum
• To understand conservation of momentum.
• To study momentum changes during
collisions.
• To understand center of mass and how forces
act on the c.o.m.
• To apply momentum to rocket propulsion.

When the bat strikes the ball, the magnitude of the force exerted DEFINITION OF IMPULSE
on the ball rises to a maximum value and then returns to zero The impulse of a force is the product of the average
force and the time interval during which the force acts:

 
J  F t

The collision time between a bat The time interval during Impulse is a vector quantity and has the
and a ball is very short, often less which the force acts is t, same direction as the average force.
than a millisecond, but the force and the magnitude of the
can be quite large. average force is F. SI unit: newton  seconds (N  s)

Momentum transfer (collision) ‘timescales’ Force and Impulse

• Collisions typically involve F
interactions that happen quickly. same area

• During this brief time, the forces F

involved can be quite large t t
ti
t big, F small
tf
 
vf
J  F t
vi
t
t
ti tf
F Vf
initial final t small, F big
The balls are in contact for a very short time.
DEFINITION OF LINEAR MOMENTUM Linear Momentum
The linear momentum of an object is the product of the
object’s mass times its velocity:   Momentum of a particle is defined as the product of

p  mv its mass and velocity

 
p  mv
SI unit:
• Momentum components
kilogram  meter/second (kg  m/s)
– px = m vx and py = m vy
– Applies to two-dimensional motion as well

Momentum (magnitude) is related to kinetic energy

1 p2
K mv 2 
2 2m

Relation between Impulse and Momentum (Newton 2nd) IMPULSE-MOMENTUM THEOREM

 
 vf  vo When a net force acts on an object, the impulse of this
a force is equal to the change in the momentum of the object
 t
 final momentum
 F  ma
 
initial momentum
  
impulse  F t  mv f  mv o
 mv f  mv o
 F  t   
J   p  F t Impulse is a vector quantity;

 F  t  mv  mv
SI Unit: N s or kg m / s

f o impulse = change in momentum!

Impulse and average force Example: A Well Hit Ball

A baseball (m = 0.14 kg) has an initial velocity of vo = - 38
• We can use the notion of impulse to define “average m/s as it approaches a bat. The bat applies an average force
force”, which is a useful concept. that is much larger than the weight of the ball, and the ball
departs from the bat with a final velocity of vf = 58 m/s .
(a) Determine the impulse applied to the ball by the bat.
(b) Assuming that the time of contact is t = 1.6 × 10-3 s,
Define average force
 find the average force exerted on the ball by the bat.
such that (even if Fav
is not constant),
impulse is given by

    
J  Fav ( t f  t i )  Fav t p
or: Fav 
t
Example 2: A Rainstorm Conceptual Example: Hailstones versus Raindrops
Rain comes down with a velocity of -15 m/s and hits the roof of a Instead of rain, suppose hail is falling. Unlike rain, hail usually
car. The mass of rain per second that strikes the roof of the car is bounces off the roof of the car.
0.060 kg/s. Assuming that rain comes to rest upon striking the If hail fell instead of rain, would the force be smaller than,
car, find the average force exerted by the rain on the roof. equal to, or greater than that calculated previously ?

 F  t  mv  mv f o
   
J  F t  mv f  mv o
Neglecting the weight of the raindrops,
the net force on a raindrop is simply the
 mv f  mv o
F
force on the raindrop due to the roof.

  
  m 
F    vo
t
F t  mv f  mv o  t 

F   0.060 kg s   15 m s   0.90 N (force on the raindrop)
For a raindrop, the change in velocity is from (downward) to zero.
For a hailstone, the change is from (downward) to (upward).
Fon-the-roof = -0.90 F (Newton’s third law) 
Thus hailstones have a larger F t

Impulse applied to auto collisions Conservation of Linear Momentum

• The most important factor is the collision time or WORK-ENERGY THEOREM 

the time it takes the person to come to a rest CONSERVATION OF ENERGY
– This will reduce the chance of dying in a car crash
Apply the impulse-momentum theorem
• Ways to increase the time to the midair collision
– Seat belts
between two objects….. THEOREM ???
IMPULSE-MOMENTUM
– Air bags

   
p m ( v f  v i )
Fav   Internal forces – Forces that objects
t t within the system exert on each other.

External forces – Forces exerted on

e.g. Weight=W
The air bag increases the time of the collision and objects by agents external to the system.
absorbs some of the energy from the body

Conservation of Linear Momentum Conservation of Linear Momentum




 
 F t  mv f  mv o W  F  t  m v
1 12 1 f1

 m1 v o1
Consider system:
both objects involved
+
OBJECT 1
W  F  t  m v
2 21 2 f2

 m2 vo2

W  F  t  m v
1 12 1 f1

 m1 v o1
W  W  F 
    
 F21 t  m1 v f 1  m 2 v f 2   m1 v o1  m 2 v o 2 
1 2 12

OBJECT 2
 
 
   
  F12   F21 Pf Po
W2  F21 t  m 2 v f 2  m 2 v o 2
The internal forces cancel out.
Principle of Conservation of Linear Momentum Definition of Total Momentum for a System of Particles
 
   
W1  W2 t  Pf  Po  For a system of particles the total  N  N

momentum P is the vector sum of P   pi   m iv i
sum of average external forces
 t  
the individual particle momenta:
   
i 1

i 1

P  p A  p B  p C  ...  m A v A  m B v B  m C v C  ...
 Pf  Po
components of total momentum Px  p A , x  p B,x  ......
If the sum of the external forces is zero, then
    Py  p A , y  p B.y  ......
0  Pf  Po Pf  Po     
 P  (p A  p B  p C  ...)  Fnet t
CONSERVATION OF LINEAR MOMENTUM
The total linear momentum of an isolated system is constant • The momentum of each object will change
(conserved). An isolated system is one for which the sum of the • The total momentum of the system remains
average external forces acting on the system is zero. constant if there are no external forces

Example: Assembling a Freight Train Example: Recoil of a rifle

A freight train is being assembled in a switching yard. Car 1 has a mass A marksman holds a 3.00 kg rifle
of m1 = 65 × 103 kg and moves at a velocity of v01 = +0.80 m/s. Car 2, loosely, allowing it to recoil freely
with a mass of m2 = 92 × 103 kg and a velocity of v02 = +1.3 m/s, when fired, and fires a bullet of
overtakes car 1 and couples to it. Neglecting friction, find the common mass 5.00 g horizontally with a
velocity vf of the cars after they become coupled. speed vB = 300 m/s. What is the
recoil speed of the rifle ?

PT  P R  P B
Before PT  0
=
After PR  PB  0
mB v B  m R v R  0
mB 0.005
vR    vB    300  0.50 m / s.
mR 3.00

Concept Test: Exploding Projectile Example: Momentum Conservation

A model rocket travels as a projectile in a parabolic path after its A box with mass m = 6.0 kg slides with speed v = 4.0 m/s across a
first stage burns out. At the top of its trajectory, where its velocity frictionless floor in the positive direction of an x axis. It suddenly
points horizontally to the right, a small explosion separates it into explodes into two pieces. One piece, with mass m1 = 2.0 kg, moves
two sections with equal masses. One section falls straight down, in the positive x-direction with speed v1 = 8.0 m/s.
with no horizontal motion. What is the direction of the other part What is the velocity of the second piece, with mass m2 = 4.0 kg ?
just after the explosion ?
before
A. Up and to the left
B. Straight up
C. Up and to the right

after
Example: Conservation of Linear Momentum - Ice Skaters Concept Test: Conservation of Momentum
Starting from rest, two skaters push
off against each other on ice where A boy stands at one end of a floating raft that is stationary
friction is negligible. One is a 54-kg relative to the shore. He then walks to the opposite end,
woman and one is a 88-kg man. The towards the shore. Does the raft move (assume no friction)?
woman moves away with a speed of
+2.5 m/s. Find the recoil velocity
  1. No, it will not move at all
P P
of the man.
f o 2. Yes, it will move away from the shore 
m1v f 1  m 2 v f 2  0
3. Yes, it will move towards the shore
m1v f 1
vf 2  
m2

54 kg   2.5 m  Note: Since momentum is conserved in the boy-raft system and neither was

vf 2    s
 1.5
m moving at first, the raft must move in the direction opposite to the
boy’s.
88 kg s

In Collissions Total Momentum is Conserved Types of Collisions

• Momentum is conserved in any collision
In collisions, we assume that external forces either
sum to zero, or are small enough to be ignored. • Elastic collisions
Hence, momentum is conserved in all collisions. – both momentum and kinetic energy
– A collision may be the result of physical contact between are conserved
two objects
– “Contact” may also arise from the electrostatic interactions • Inelastic collisions
of the electrons in the surface atoms of the bodies – Kinetic energy is not conserved
• Mathematically (for two objects): • Some of the kinetic energy is converted
into other types of energy such as heat,
m1 v1i  m 2 v 2i  m1 v1f  m 2 v 2 f
sound, work to permanently deform an
object
– Perfectly inelastic collisions occur when
– Momentum is conserved for the system of objects the objects stick together
– The system includes all the objects interacting with each other • Not all of the KE is necessarily lost
– Assumes only internal forces are acting during the collision Most collisions fall between elastic
– Can be generalized to any number of objects and perfectly inelastic collisions

Elastic Collisions Inelastic and Elastic Collisions

A completely inelastic collision An elastic collision
• Elastic means that kinetic energy is conserved as well
as momentum.
• This gives us more constraints
– We can solve more complicated problems!!
– Billiards (2-D collision)
– The colliding objects
have separate motions
after the collision as
well as before.
Initial Final

• First: simpler 1-D problem

Applying the Principle of Conservation of Momentum Problem Solving for One-Dimensional Collisions
• Set up a coordinate axis and define the velocities with respect
1. Decide which objects are included in the system. to this axis
– It is convenient to make your axis coincide with one of the
2. Relative to the system, identify the internal and initial velocities
external forces. • In your sketch, draw all the velocity vectors with labels
including all the given information
3. Verify that the system is isolated.
• Draw “before” and “after” sketches
• Label each object
4. Set the final momentum of the system equal to its
– include the direction of velocity
initial momentum.
– keep track of subscripts

Remember that momentum is a vector.

Problem Solving for One-Dimensional Collisions Perfectly Inelastic Collisions

Sketch for perfectly inelastic colllision • Suppose, for example, v2i=0.
• The objects stick together Conservation of momentum becomes
• Include all the velocity directions
• The “after” collision combines the masses m1v1i  m 2 v 2i  (m1  m 2 ) v f
• Write the expressions for the momentum
of each object before and after the collision m1v1i  0  (m1  m 2 ) v f
– Remember to include the appropriate signs
E.g., if m1  1000 kg, m 2  1500 kg :
• Write an expression for the total momentum before and after
the collision --- momentum of the system is conserved (1000kg )(50 m s)  0  (2500kg) v f ,
• If the collision is inelastic, solve the momentum equation for 5 10 4 kg  m s
the unknown --- Remember, KE is not conserved vf   20 m s.
2.5 103 kg
• If the collision is elastic, you can use the KE equation to solve
for two unknowns

Perfectly Inelastic Collisions Example: A Ballistic Pendulum

The mass of the block of wood is 2.50-kg
• What amount of KE lost and the mass of the bullet is 0.0100-kg.
during collision? The block swings to a maximum height of
1 1 0.650 m above the initial position.
KE before  m1v12i  m 2 v 22i Find the initial speed of the bullet.
2 2
1 Apply conservation of momentum to the collision:
 (1000 kg )(50 m s) 2  1.25 10 6 J
2
1
KE after  (m1  m 2 ) v f2
m1v f 1  m 2 v f 2  m1v o1  m 2 v o 2
2
1
 ( 2500 kg)(20 m s) 2  0.50 106 J
m1  m 2  v f  m1v o1
2

KElost  0.75  10 6 J v o1 
m1  m 2  v f
m1
lost in heat and sound …
Applying conservation of energy to the swinging motion: Elastic Collisions
• Both momentum and kinetic energy are conserved
mgh  mv 1
2
2

• Typically have two unknowns

m1  m 2  gh f  1
m1  m 2  vf2 • Solve the equations
2
m1v1i  m 2 v 2i  m1v1f  m 2 v 2 f simultaneously
g h f  12 v f2
1 1 1 1
m1v12i  m 2 v 22i  m1v12f  m 2 v 22 f

v f  2gh f  2 9.80 m s 2
0.650 m 2 2 2 2

v o1 
m1  m 2  v f
m1

 0.0100 kg  2.50 kg 
vo1    
 2 9.80 m s 2 0.650 m   896 m s • Incoming and outgoing velocities are very mass dependant
 0.0100 kg 

mass m2 initially at rest Two-dimensional Collisions

• For a general collision
of two objects in three-
dimensional space, the
conservation of
momentum principle

… implies that the total momentum of the system in each

direction is conserved
Elastic Collision
m1 v1i  m 2 v 2i  m1 v1f  m 2 v 2 f
m1v1i  m 2 v 2i  m1v1f  m 2 v 2f (1)
– Use subscripts for identifying the object, initial and final,
and components
1 1 1 1 m1v1ix  m 2 v 2ix  m1v1fx  m 2 v 2 fx and
m1v12i  m 2 v 22i  m1v12f  m 2 v 22 f (2)
2 2 2 2 m1v1iy  m 2 v 2iy  m1v1fy  m 2 v 2 fy

A Collision in Two Dimensions

Use momentum conservation to determine
the magnitude and direction of the final
velocity of ball 1 after the collision.

m1v f 1x  m 2 v f 2 x  m1v o1x  m 2 v o 2 x

m1v f 1y  m 2 v f 2 y  m1v o1y  m 2 v o 2 y

 Center of Mass
The center of mass is a point that represents the average
location for the total mass of a system.

m1x1  m 2 x 2
x cm 
m1  m 2

Center of Mass Coordinates Velocity of Center of Mass

• The coordinates of the
center of mass are
m x i i
x CM  i

m i
i

m y i i
m1x1  m 2 x 2 x1 x 2
y CM  x cm  m1  m2
x cm
i

m i
Two masses on x-axis m1  m 2  t t
t m1  m 2
m1x1  m 2 x 2
i

x CM  m1v1  m 2 v 2
 m i = M is the total mass v cm  pcm  M vcm  p1  p2
i
of the system m1  m 2 m1  m 2
In an isolated system, the total linear momentum does not change,
therefore the velocity of the center of mass does not change.

Example: Center of Mass Motion Motion of the Center of Mass

vcm = const. In isolated system • The system will move as if an
external force were applied to a
BEFORE single particle of mass M located at
the center of mass
m1v1  m 2 v 2
v cm  0
m1  m 2

AFTER

v cm 
88 kg  1.5 m s   54 kg  2.5 m s   0.002  0
88 kg  54 kg