Momentum

What are the 3 conservation laws?
 In physics, a conservation law states that
a particular measurable property of an
isolated physical system does not change
as the system evolves over time.
 Exact conservation
laws include conservation of
energy, conservation of linear
momentum, conservation of angular
momentum, and conservation of electric
charge.

What is meant by an isolated system?
• An isolated system is a thermodynamic system that
cannot exchange either energy or matter outside the
boundaries of the system.
• An isolated system differs from a closed system by
the transfer of energy.
• Closed systems are only closed to matter, energy can
be exchanged across the system's boundaries

Is an isolated system possible?
• An isolated system does not exchange energy or
matter with its surroundings.
• In fact, there are a few, if any, systems that exist
in this world that are completely isolated
systems.
• The universe of course is an another
approximation of an isolated system

MOMENTUM!
Momentum
Impulse
Conservation of Momentum in
1 Dimension
Conservation of Momentum in
2 Dimensions

Momentum Defined
p = mv
p = momentum vector
m = mass
v = velocity vector

Momentum Facts
• p = mv
• Momentum is a vector quantity!
•Velocity and momentum vectors point in the same direction.
•SI unit for momentum: kg·m/s (no special name).
•Momentum is a conserved quantity (this will be proven later).
•A net force is required to change a body’s momentum.
•Momentum is directly proportional to both mass and speed.
•Something big and slow could have the same momentum as
something small and fast.

Can momentum be conserved for a system if
there are external forces acting on the system?
• During projectile motion and where air resistance is
negligible, momentum is conserved in the
horizontal direction because horizontal forces are
zero.
• Conservation of momentum applies only when the
net external force is zero.
• The conservation of momentum principle is
valid when considering systems of particles.

How do you know if momentum is
conserved?
• Momentum is not conserved if there is friction,
gravity, or net force (net force just means the total
amount of force).
• What it means is that if you act on an object, its
momentum will change.
• This should be obvious, since you are adding to or
taking away from the object's velocity and therefore
changing its momentum.

Momentum Principle
In any real situations, the momentum ( and a velocity)
of a moving object are continuously changing due to interactions
of the objects with its surrounding
The momentum principle makes a quantitative connection
between the amount of interaction and the change in momentum
The momentum principle is the first of the three fundamental
principles of mechanics that together make it possible to predict
and explain a very broad range of real-world phenomena
The momentum principle is also known as Newton’s second law
∆𝑃 = 𝐹𝑛𝑒𝑡∆𝑡
The concept of force was used to quantify interactions between two
objects
𝐹𝑛𝑒𝑡 =
∆𝑃
∆𝑡

It is fundamental because..
• 𝐼𝑡 𝑖𝑠 𝑎𝑝𝑝𝑙𝑖𝑐𝑎𝑏𝑙𝑒 𝑡𝑜 𝑎𝑛𝑦 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
• 𝑛𝑜 𝑚𝑎𝑡𝑡𝑒𝑟 ℎ𝑜𝑤 𝑙𝑎𝑟𝑔𝑒 𝑜𝑟 𝑠𝑚𝑎𝑙𝑙.
• No matter how fast or slow and is true for every kind of interactions
• It relates an effect(change in momentum) to a cause (interaction)
𝑃 = 𝑃0 + 𝐹𝑛𝑒𝑡∆𝑡
𝑃𝑓𝑢𝑡𝑢𝑟𝑒 = 𝑃𝑛𝑜𝑤 + 𝐹𝑛𝑒𝑡∆𝑡
𝑫𝒖𝒓𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝑰𝒏𝒕𝒆𝒓𝒂𝒄𝒕𝒊𝒐𝒏:
The length of time during which the force acts on the object
affects the change of momentum of the object

𝑃𝑓𝑢𝑡𝑢𝑟𝑒 − 𝑃𝑛𝑜𝑤 = 𝐹𝑛𝑒𝑡∆𝑡
The quantity on the right of the equation was named an impulse
Hence the change in the momentum of the system is equal to
the net impulse applied to the system
𝑨𝒔𝒔𝒖𝒎𝒑𝒕𝒊𝒐𝒏𝒔:
𝑇ℎ𝑒 𝑛𝑒𝑡 𝑓𝑜𝑟𝑐𝑒 𝑖𝑠 𝑛𝑒𝑎𝑟𝑙𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑑𝑢𝑟𝑖𝑛𝑔 𝑎
𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛
The amount of interaction affecting the object include both
the strength of interaction and the duration of the interaction
Either a bigger force or applying the force for a longer time
will cause the change in momentum
Impulse is defined as the product net force acting on an object
and the time during which the force acts.

Impulse - Momentum Theorem
The impulse due to all forces acting on an object (the net force) is
equal to the change in momentum of the object:
Fnet t = p
We know the units on both sides of the equation are the same
(last slide), but let’s prove the theorem formally:
Fnet t = mat
= m(v/t) t
= mv
= p

Momentum Examples
10 kg
3 m/s
10 kg
30 kg·m/s
Note: The momentum vector does not have to be drawn 10 times
longer than the velocity vector, since only vectors of the same
quantity can be compared in this way.
5 g
p = 45 kg·m/s
at 26º above +
x axis
26º

Equivalent Momenta
Bus: m = 9000 kg; v = 16 m/s
p = 1.44·105 kg·m/s
Train: m = 3.6·104 kg; v = 4 m/s
p = 1.44·105 kg·m/s
Car: m = 1800 kg; v = 80 m/s
p = 1.44·105 kg·m/s
continued on next slide

Equivalent Momenta (cont.)
The train, bus, and car all have different masses and
speeds, but their momenta are the same in magnitude.
The massive train has a slow speed; the low-mass car has a
great speed; and the bus has moderate mass and speed.
Note: We can only say that the magnitudes of their
momenta are equal since they’re aren’t moving in the same
direction.
The difficulty in bringing each vehicle to rest--in terms of a
combination of the force and time required--would be the
same, since they each have the same momentum.

Imagine a car hitting a wall and coming to rest.
The force on the car due to the wall is large (big F), but that force only
acts for a small amount of time (little t).
Now imagine the same car moving at the same speed but this time
hitting a giant haystack and coming to rest.
The force on the car is much smaller now (little F), but it acts for a
much longer time (big t).
In each case the impulse involved is the same since the change in
momentum of the car is the same.
Any net force, no matter how small, can bring an object to rest if it
has enough time.
Stopping Time
F t = F  t

Conservation of Momentum in 2-D
m1
m2
v1
v2
m1 m2
va
vb
1
2
a
b
To handle a collision in 2-D, we conserve momentum in each
dimension separately. Choosing Cartesian plane
directions: before:
𝑃𝑥 = m1 v1cos1 - m2 v2 cos2
𝑃𝑦 = - m1 v1sin1 - m2 v2sin2
after:
𝑃𝑥 = -m1 vacosa + m2 vbcos
𝑃𝑦 = - m1 vasina -m2 vbsinb
Conservation of momentum equations:
m1v1cos1 - m2 v2cos2 = -m1vacosa + m2vbcosb
-m1v1sin1-m2 v2sin2 = -m1 vasina - m2 vbsinb

Home work Questions
Chapter 7 problems no 25, 34, 39, 40 and 45
• No net external force acts on the plate
parallel to the floor; therefore, the
component of the momentum of the plate
that is parallel to the floor is conserved as
the plate breaks and flies apart.
• The drawing in the text shows the pieces in the plane parallel to
the floor just after the collision.
• Initially, the total momentum parallel to
the floor is zero. After the collision with
the floor, the component of the total
momentum parallel to the floor must
remain zero.

• Clearly, the linear momentum in the plane
parallel to the floor has two components;
therefore the linear momentum of the plate
must be conserved in each of these two
mutually perpendicular directions.
• Using the drawing in the text, we choose the
following direction:
• with the positive directions taken to be up
and to the right,
x direction   m v m v1 1 2 2
(sin 25.0 ) + (cos 45.0 ) = 0
y direction m v m v m v1 1 2 2 3 3
(cos 25.0 ) + (sin 45.0 ) – = 0 
m1
 1.00 kg
m2
 1.00 kg
Subtracting (2) from (1), and noting that cos 45.0º = sin 45.0º, gives
. Substituting this value into either (1) or (2) then yields
.

Conserving Momentum w/ Vectors
m1
m2
p1
p2
m1 m2
pa pb
1
2
a
b
B
E
F
O
R
E
A
F
T
E
R
This diagram shows momentum vectors, which are parallel to
their respective velocity vectors. Note p1 + p2 = pa + pb and
𝑃before = 𝑃after as conservation of momentum demands.
p1
p2
pbefore
pa
pb
pafter

Exploding Bomb
Acme
before
after
A bomb, which was originally at rest, explodes and shrapnel flies
every which way, each piece with a different mass and speed. The
momentum vectors are shown in the after picture.

Conservation of Momentum
 Momentum must be conserved.
 This fact means that the momentum in a specific direction
before a collision must be equal to the momentum in that same
direction after the collision.
 Consider the two identical pucks to the right.
 In the first collision, momentum is conserved because the first
puck completely stops and the second puck departs with the
same initial momentum as the first puck in the same direction
as the first puck.
 In the second collision, momentum is conserved because both
pucks move off with one half the speed of the first puck before
the collision and in the same direction as the first puck.

Momentum – Elastic Collisions
 Elastic Collision – a collision in which the colliding bodies do not
stick together
 Both momentum and kinetic energy is conserved.
 The equation used for elastic collisions is as follows.
FFII vmvmvmvm 22112211 

Momentum – Inelastic Collisions
 Inelastic Collision – a collision in which the colliding bodies stick
together.
 Only momentum is conserved, kinetic energy is not
 The equation used for elastic collisions is as follows.
1 1 2 2 Fm v m v MV 

Momentum
 Identify the number and types of collisions in the animation below.
FFII vmvmvmvm 22112211  FII MVVmvm  2211

Proving if collision is elastic or not
• Calculate the total kinetic energies before
• Calculate the total kinetic energies after
• If total kinetic energies before is not equal
to total kinetic energies after
• Hence the collision is INELASTIC

Intersection Collision Problems
 Two cars approach an intersection
with a malfunctioning stop light.
 The red car (m = 2000.00 kg)
approaches the intersection from the
North.
 The blue car (m = 2250.00 kg)
approaches the intersection from the
West at the speed limit (40.0 km/hr)
on both roads.
 Cars are designed to collide in-
elastically in order to minimize
injury to passengers.
 After the collision, the cars move at
a velocity of 31.7 km/hr @ 48.01
below the + x-axis.
 If you were the police officer
investigating the accident, then
would you write one of the drivers a
citation? Explain your answer.

Conservation of Momentum in 1-D
The momentum of an isolated system is conserved in magnitude and
direction.
before: 𝑃 = m1 v1 - m2 v2
after: 𝑃 = - m1 va + m2 vb
m1 m2
v1
v2
(Choosing right as the +
direction, m2 has - momentum.)
m1 m2
va vb
m1v1 - m2v2 = - m1va + m2 vb

Directions after a collision
On the last slide the boxes were drawn going in the opposite direction
after colliding. This isn’t always the case. For example, when a bat hits
a ball, the ball changes direction, but the bat doesn’t. It doesn’t really
matter, though, which way we draw the velocity vectors in “after”
picture. If we solved the conservation of momentum equation (red box)
for vb and got a negative answer, it would mean that m2 was still moving
to the left after the collision. As long as we interpret our answers
correctly, it matters not how the velocity vectors are drawn.
m1v1 - m2v2 = - m1va + m2 vb
m1 m2
v1
v2
m1 m2
va vb

Sample Homework Problem 1
7 kg
v = 0
700 m/s
A rifle fires a bullet into a giant slab of butter on a frictionless surface.
The bullet penetrates the butter, but while passing through it, the bullet
pushes the butter to the left, and the butter pushes the bullet just as hard
to the right, slowing the bullet down. If the butter skids off at 4 cm/s
after the bullet passes through it, what is the final speed of the bullet?
(The mass of the rifle matters not.)
35 g
7 kg
v = ?
35 g
4 cm/s

Sample Problem 1 (cont.)
7 kg
v = 0
700 m/s
35 g
7 kg
v = ?
35 g
4 cm/s
𝑃before = 7(0) + (0.035)(700)
= 24.5 kg·m/s
Let’s choose left to be the + direction & use conservation of
momentum, converting all units to meters and kilograms.
𝑃after = 7(0.04) + 0.035v
= 0.28 + 0.035v
𝑃before = 𝑃after 24.5 = 0.28 + 0.035v v = 692 m/s
v came out positive. This means we chose the correct
direction of the bullet in the “after” picture.

Conservation of Momentum applies only
in the absence of external forces!
In the first two sample problems, we dealt with a frictionless surface.
We couldn’t simply conserve momentum if friction had been present
because, as the proof on the last slide shows, there would be another
force (friction) in addition to the contact forces. Friction wouldn’t
cancel out, and it would be a net force on the system.
The only way to conserve momentum with an external force like
friction is to make it internal by including the tabletop, floor, or the
entire Earth as part of the system. For example, if a rubber ball hits a
brick wall, p for the ball is not conserved, neither is p for the ball-
wall system, since the wall is connected to the ground and subject to
force by it. However, p for the ball-Earth system is conserved!

Sample Problem 3
Earth
M
apple
An apple is originally at rest and then dropped. After falling a short
time, it’s moving pretty fast, say at a speed V. Obviously, momentum
is not conserved for the apple, since it didn’t have any at first. How can
this be?
m
F
F
answer:
• Gravity is an external force on the apple, so
momentum for it alone is not conserved.
• To make gravity “internal,” we must define a
system that includes the other object responsible
for the gravitational force--Earth.
• The net force on the apple-Earth system is zero,
and momentum is conserved for it. During the
fall the Earth attains a very small speed v. So,
by conservation of momentum:
V
v
mV = Mv

Sample Problem 4
before
after
3 kg 15 kg
10 m/s 6 m/s
3 kg 15 kg
4.5 m/s v
A crate of raspberry donut filling collides with a tub of lime Kool Aid
on a frictionless surface. Which way on how fast does the Kool Aid
rebound? answer: Let’s draw v to the right in the after picture.
3(10) - 6(15) = -3(4.5) + 15v v = -3.1 m/s
Since v came out negative, we guessed wrong in drawing v to the
right, but that’s OK as long as we interpret our answer correctly.
After the collision the lime Kool Aid is moving 3.1 m/s to the left.

2-D Sample Problem
0.3 kg
34 m/s
40
152 g
A mean, old dart strikes an innocent
mango that was just passing by
minding its own business. Which
way and how fast do they move off
together?
before
after
5 m/s
452 g

v
152(34)sin 40 = 452v sin (1)
152(34)cos 40 - 300(5) = 452v cos (2)
Working in grams and taking left & down as + :
Dividing equations : 1.35097 = tan
 = 53.4908
Substituting into either of the first two equations :
v = 9.14 m/s
Vertically
Horizontally
Below the negative - axes

Is your system isolated?
• The net external force acting
on the two-puck system is
zero (the weight of each ball
is balanced by an upward
normal force,
• and we are ignoring friction
due to the layer of air on the
hockey table)..
   A 0A A fA B fBcos65 cos37m v m v m v    (1)
while momentum conservation in the y direction yields
Solving equation (2) for vfB, we find that
 
 
A fA
fB
B
sin65
sin37
m v
v
m



Substituting equation (3) into Equation (1) leads to

Solving for vfA gives
From equation (3), we find that

Proving the system is
isolated:
Considering the boat and the stone
as a single system, there is no net
external horizontal force acting on
the stone-boat system, and thus the
horizontal component of the
system’s linear momentum is
conserved:
1 f1 2 f2 1 01 2 02x x x xm v m v m v m v  
Because the boat is initially at rest, v02x = 0 m/s,
1 f1 2 f2 1 01x x xm v m v m v 
 1 01 f1
2 f2 1 01 1 f1 2
f2
or x x
x x x
x
m v v
m v m v m v m
v

  
As noted above, the boat’s final velocity is horizontal, so vf2x = vf2.
The horizontal components of the stone’s initial and final velocities
are, respectively, v01x = v01 cos 15° and vf1x = vf1 cos 12°
       1 01 f1
2
f2
0.072 kg 13 m/s cos15 11 m/s cos12
0.062 kg
2.1 m/s
x xm v v
m
v
     
o o

Proving and isolated system:
The system in this problem consists
of the coal and the car.
There is no net external force acting
on this system as far as the
horizontal direction is concerned;
the gravitational force acts in the
vertical (not the horizontal)
direction and the weight of the
car (and its contents) is balanced
by the normal force supplied by
the track.
Thus, linear momentum in the
horizontal direction is conserved.
coal car f coal coal car car
Total momentum in Total momentum in
horizontal direction horizontal direction
after collision before collision
( ) cos25.0m m v m v m v   
1 4 4 2 4 4 3 1 4 4 4 44 2 4 4 4 4 43
coal coal car car
f
coal car
cos 25.0
(150 kg)(0.80 m/s)cos 25.0 (400 kg)(0.50 m/s)
0.56 m/s
150 kg + 400 kg
m v m v
v
m m
 


 
 

The two balls constitute the system.
• The tension in the wire is the only
nonconservative force that acts on
the ball.
• The tension does no work since it is
perpendicular to the displacement of
the ball. Since Wnc = 0 J,
• the principle of conservation of
mechanical energy holds and can be
used to find the speed of the 1.50-kg
ball just before the collision.
• Momentum is conserved during the collision, so the principle of
conservation of momentum can be used to find the velocities of
both balls just after the collision.
• Once the collision has occurred, energy conservation can be
used to determine how high each ball rises.
Applying the principle of energy conservation to the 1.50-kg ball, we have
1
2
2 1
2
2
0
mv mgh
E
mv mgh
E
f f
f
0 0
  
       
If we measure the heights from the lowest point in the
swing, hf = 0 m, and the expression above simplifies to
1
2
2 1
2
2
mv mv mghf 0 0
  v v ghf 0 0
2 2
( m/s) 2(9.80 m/s )(0.300 m) 5.56 m/s    2
2 5 00.

If we assume that the collision is elastic, then the velocities of
both balls just after the collision can be obtained from
Equations 7.8a and 7.8b:
You are expected to derive these two equations
v
m m
m m
vf1



F
HG
I
KJ1 2
1 2
01
v
m
m m
vf2


F
HG
I
KJ
2 1
1 2
01
Since v01 corresponds to the speed of the 1.50-kg ball just before the collision, it is
equal to the quantity vf calculated in part (a). With the given values of
m1
 1.50 kg and m2
kg 4 60. , and the value of v01
= 5.56 m / s obtained in
part (a), Equations 7.8a and 7.8b yield the following values:
v vf1 f2
= –2.83 m/ s and = +2.73 m/ s
The minus sign in vf1 indicates that the first ball reverses its
direction as a result of the collision.

c. If we apply the conservation of mechanical energy to either
ball after the collision we
have 1
2
2 1
2
2
0
mv mgh
E
mv mgh
E
f f
f
0 0
  
       
where v0 is the speed of the ball just after the collision, and hf is the final height to which the ball rises. For either
ball, h0 = 0 m, and when either ball has reached its maximum height, vf = 0 m/s. Therefore, the expression of
energy conservation reduces to
Thus, the heights to which each ball
rises after the collision are
1.50- kg ball
(2.83 m/ s)
(9.80 m/ s
0.409 mf
0
2
2
h
v
g
  
2
2 2 )

4132.9736
Alternate Solution
5168
40
1500

p
Shown are momentum vectors (in g m/s).
The black vector is the total momentum
before the collision. Because of
conservation of momentum, it is also the
total momentum after the collisions. We
can use trig to find its magnitude and
direction.
Law of Sines :
40
sin 
1500
Law of Cosines : p2 = 51682 + 15002 - 25168 1500 cos40
p = 4132.9736 g m/s
Dividing by total mass : v = (4132.9736 g m/s) / (452 g) = 9.14 m/s
=
sin 40
 = 13.4908
Angle w/ resp. to horiz. = 40 + 13. 4908   53.49

Comments on Alternate Method
• Note that the alternate method gave us the exact
same solution.
• This method can only be used when two objects
collide and stick, or when one object breaks into
two. Otherwise, we’d be dealing with a polygon
with more sides than a triangle.
• In using the Law of Sines (last step), the angle
involved (ß) is the angle inside the triangle. A little
geometry gives us the angle with respect to the
horizontal.

Momentum

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