This document provides information about momentum and related concepts in physics. It defines momentum as p = mv, where p is momentum, m is mass, and v is velocity. It discusses that momentum is a vector quantity and is conserved during interactions between objects. The document also introduces impulse as the product of force and time (J=Ft), and proves the impulse-momentum theorem. It then explains how to use conservation of linear momentum to solve 1D and 2D collision problems through vector diagrams or component equations. Finally, it introduces angular momentum as dependent on linear momentum and distance from a point.

1. MOMENTUM!
Momentum
Impulse
Conservation of Momentum in
1 Dimension
Conservation of Momentum in
2 Dimensions
Angular Momentum
Torque
Moment of Inertia

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

3. 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.

4. 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
9 km/s
p = 45 kg·m/s
at 26º N of E
26º

5. 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

6. 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.

7. Impulse Defined
Impulse is defined as the product force acting on an
object and the time during which the force acts. The
symbol for impulse is J. So, by definition:
J = F t
Example: A 50 N force is applied to a 100 kg boulder
for 3 s. The impulse of this force is J = (50 N) (3 s)
= 150 N·s.
Note that we didn’t need to know the mass of the
object in the above example.

8. Impulse Units
J = Ft shows why the SI unit for impulse is the Newton ·second.
There is no special name for this unit, but it is equivalent to a kg·m/s.
proof: 1 N·s = 1 (kg·m/s2
) (s) = 1 kg·m/s
{
Fnet = ma shows this is
equivalent to a newton.
Therefore, impulse and momentum have the same units, which leads
to a useful theorem.

9. 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 = mat = m(∆v/ t)t = m∆v = ∆p

10. 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. A pole vaulter can fall
from a great height without getting hurt because the mat applies a
smaller force over a longer period of time than the ground alone would.
Stopping Time
Ft = F t

11. Impulse - Momentum Example
A 1.3 kg ball is coming straight at a 75 kg soccer player at 13 m/s who
kicks it in the exact opposite direction at 22 m/s with an average force
of 1200 N. How long are his foot and the ball in contact?
answer: We’ll use Fnet t = ∆p. Since the ball
changes direction, ∆p = m∆v = m(vf - v0)
= 1.3 [22 - (-13)] = (1.3 kg) (35 m/s)
= 45.5 kg·m/s. Thus, t = 45.5 / 1200
= 0.0379 s, which is just under 40 ms.
During this contact time the ball compresses substantially and then
decompresses. This happens too quickly for us to see, though. This
compression occurs in many cases, such as hitting a baseball or golf
ball.

12. Fnet vs. t graph
Fnet (N)
t (s)
6
A variable strength net force acts on an object in the positive direction
for 6 s, thereafter in the opposite direction. Since impulse is Fnet t, the
area under the curve is equal to the impulse, which is the change in
momentum. The net change in momentum is the area above the curve
minus the area below the curve. This is just like a v vs. t graph, in
which net displacement is given area under the curve.
Net area = ∆p

13. Conservation of Momentum in 1-D
Whenever two objects collide (or when they exert forces on each
other without colliding, such as gravity) momentum of the system
(both objects together) is conserved. This mean the total momentum
of the objects is the same before and after the collision.
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
m1 v1 - m2 v2 = - m1 va + m2 vb

14. 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.
m1 v1 - m2 v2 = - m1 va + m2 vb
m1 m2
v1
v2
m1 m2
va vb

15. Sample 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
continued on next slide

16. 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.

17. Sample Problem 2
7 kg
v = 0
700 m/s
35 g
Same as the last problem except this time it’s a block of wood rather than
butter, and the bullet does not pass all the way through it. How fast do
they move together after impact?
v
7. 035 kg
(0.035)(700) = 7.035v v = 3.48 m/s
Note: Once again we’re assuming a frictionless surface, otherwise there
would be a frictional force on the wood in addition to that of the bullet,
and the “system” would have to include the table as well.

18. Proof of Conservation of Momentum
The proof is based on Newton’s 3rd
Law. Whenever two objects collide
(or exert forces on each other from a distance), the forces involved are an
action-reaction pair, equal in strength, opposite in direction. This means
the net force on the system (the two objects together) is zero, since these
forces cancel out.
M m
FF
force on m due to Mforce on M due to m
For each object, F = (mass)(a) = (mass)(∆v/ t) = (mass ∆v)/t = ∆p/ t.
Since the force applied and the contact time is the same for each mass,
they each undergo the same change in momentum, but in opposite
directions. The result is that even though the momenta of the individual
objects changes, ∆p for the system is zero. The momentum that one
mass gains, the other loses. Hence, the momentum of the system before
equals the momentum of the system after.

19. 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!

20. 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 = M v

21. 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.

22. Conservation of Momentum in 2-D
m1
m2
v1
v2
m1 m2
va
vb
θ1
θ2
θa
θb
To handle a collision in 2-D, we conserve momentum in each
dimension separately. Choosing down & right as positive:
before:
px = m1 v1cosθ1 - m2 v2 cosθ2
py = m1 v1sinθ1 + m2 v2sinθ2
after:
px = -m1 vacosθa + m2 vbcosθb
py = m1 vasinθa + m2 vbsinθb
Conservation of momentum equations:
m1 v1cosθ1 - m2 v2cosθ2 = -m1 vacosθa + m2 vbcosθb
m1 v1sinθ1 + m2 v2sinθ2 = m1 vasinθa + m2 vbsinθb

23. Conserving Momentum w/ Vectors
m1
m2
p1
p2
m1 m2
pa pb
θ1
θ2
θa
θb
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

24. Exploding Bomb
Acme
before
after
A
c
m
e
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.
continued on next slide

25. Exploding Bomb (cont.)
Since the momentum of the bomb was zero before the
explosion, it must be zero after it as well. Each piece does
have momentum, but the total momentum of the exploded
bomb must be zero afterwards. This means that it must be
possible to place the momentum vectors tip to tail and form a
closed polygon, which means the vector sum is zero.
If the original momentum of
the bomb were not zero,
these vectors would add up
to the original momentum
vector.

26. 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
θ
v
152(34)sin 40° = 452v sinθ
152(34)cos 40° - 300(5) = 452v cosθ
Working in grams and taking left & down as + :
Dividing equations : 1.35097 = tanθ
θ = 53.4908°
Substituting into either of the first two
equations : v = 9.14 m/s

27. 4132.9736
Alternate Solution
5168
40°
1500
β
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 β
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°
β = 13.4908°
Angle w/ resp. to horiz. = 40° + 13. 4908 ° ≈ 53.49°

28. 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.

29. Angular Momentum
r
v
m
Angular momentum depends on linear momentum and the distance
from a particular point. It is a vector quantity with symbol L. If r
and v are ⊥ then the magnitude of angular momentum w/ resp. to
point Q is given by L = rp = mvr. In this case L points out of the
page. If the mass were moving in the opposite direction, L would
point into the page.
The SI unit for angular momentum
is the kg⋅m2
/s. (It has no special
name.) Angular momentum is a
conserved quantity. A torque is
needed to change L, just a force is
needed to change p. Anything
spinning has angular has angular
momentum. The more it has, the
harder it is to stop it from spinning.
Q

30. Angular Momentum: General Definition
r
v
m
If r and v are not ⊥ then the angle between these two vectors must
be taken into account. The general definition of angular momentum is
given by a vector cross product:
L = r × p
This formula works regardless of the angle. As you know from our
study of cross products, the magnitude of the angular momentum
of m relative to point Q is: L = r psinφ = m v r. In this case, by
the right-hand rule, L points out of the page. If the mass were
moving in the opposite direction, L would point into the page.
Q
φ

31. Moment of Inertia
Any moving body has inertia. (It wants to keep moving at constant
v.) The more inertia a body has, the harder it is to change its linear
motion. Rotating bodies possess a rotational inertial called the
moment of inertial, I. The more rotational inertia a body has, the
harder it is change its rotation. For a single point-like mass w/ respect
to a given point Q, I = mr2
.
I = mr2
m
r
For a system, I = the sum of each mass
times its respective distance from the
point of interest.
r2
m2
r1
m1
I = Σmi ri
2
= m1 r1
2
+ m2r2
2
Q
Q

32. Moment of Inertia Example
Two merry-go-rounds have the same mass and are spinning with the
same angular velocity. One is solid wood (a disc), and the other is a
metal ring. Which has a bigger moment of inertia relative to its
center of mass?
ω ω
m
m
r r
answer: I is independent of the angular speed. Since their masses and
radii are the same, the ring has a greater moment of inertia. This is
because more of its mass is farther from the axis of rotation. Since I
is bigger for the ring, it would more difficult to increase or decrease its
angular speed.

33. Angular Acceleration
As you know, acceleration is when an object speeds up, slows down,
or changes directions. Angular acceleration occurs when a spinning
object spins faster or slower. Its symbol is α, and it’s defined as:
α = ∆ω/t
Note how this is very similar to a = ∆v/t for linear acceleration.
Ex: If a wind turbine spinning at 21 rpm speeds up to 30 rpm over
10 s due to a gust of wind, its average angular acceleration is
9 rpm/10 s. This means every second it’s spinning 9 revolutions per
minute faster than the second before. Let’s convert the units:
9 rpm
10 s
=
9 rev/min
10 s
=
9 rev
min ⋅ 10 s
=
9 ⋅ (2π rad)
(60 s) ⋅ 10 s
= 0.094 rad/s2
Since a radian is really dimensionless (a length divided by a length),
the SI unit for angular acceleration is the “per second squared” (s-2
).

34. Torque & Angular Acceleration
Newton’s 2nd
Law, as you know, is Fnet = ma
The 2nd
Law has a rotational analog: τnet = Iα
A force is required for a body to undergo acceleration. A
“turning force” (a torque) is required for a body to undergo
angular acceleration.
The bigger a body’s mass, the more force is required to
accelerate it. Similarly, the bigger a body’s rotational inertia,
the more torque is required to accelerate it angularly.
Both m and I are measures of a body’s inertia
(resistance to change in motion).

35. Linear Momentum & Angular Momentum
If a net force acts on an object, it must accelerate, which means its
momentum must change. Similarly, if a net torque acts on a body, it
undergoes angular acceleration, which means its angular momentum
changes. Recall, angular momentum’s magnitude is given by
L = mvr
r
v
m
So, if a net torque is applied, angular velocity
must change, which changes angular momentum.
proof: τnet = r Fnet = r ma
= r m∆v/t = ∆L/t
So net torque is the rate of change of angular momentum, just as net
force is the rate of change of linear momentum. continued on next slide
(if v and r are perpendicular)

36. Linear & Angular Momentum (cont.)
Here is yet another pair of similar equations, one linear,
one rotational. From the formula v = r ω, we get
L = mvr = mr(r ω) = m r2
ω = I ω
This is very much like p = mv, and this is one reason I is
defined the way it is.
In terms of magnitudes, linear momentum is inertia times
speed, and angular momentum is rotational inertia times
angular speed.
L = I ω
p = mv

37. Spinning Ice Skater
Why does a spinning ice skater speed up when she pulls her arms in?
Suppose Mr. Stickman is sitting on a stool that swivels holding a pair of
dumbbells. His axis of rotation is vertical. With the weights far from
that axis, his moment of inertia is large. When he pulls his arms in as
he’s spinning, the weights are closer to
the axis, so his moment of inertia gets
much smaller. Since L = Iω and L is
conserved, the product of I and ω is a
constant. So, when he pulls his arms in,
I goes down, ω goes up, and he starts
spinning much faster.
Iω = L = Iω

38. Comparison: Linear & Angular Momentum
Linear Momentum, p
• Tendency for a mass to continue
moving in a straight line.
• Parallel to v.
• A conserved, vector quantity.
• Magnitude is inertia (mass)
times speed.
• Net force required to change it.
• The greater the mass, the greater
the force needed to change
momentum.
Angular Momentum, L
• Tendency for a mass to continue
rotating.
• Perpendicular to both v and r.
• A conserved, vector quantity.
• Magnitude is rotational inertia
times angular speed.
• Net torque required to change it.
• The greater the moment of
inertia, the greater the torque
needed to change angular
momentum.