Lecture 11
Goals:
•
Chapter 9: Momentum & Impulse
 Understand what momentum is and how it relates to
forces
 Employ momentum conservation principles
 In problems with 1D and 2D Collisions
 In problems having an impulse (Force vs. time)
 Chapter 8: Use models with free fall
Assignment:


Read through Chapter 10
MP HW5, due Wednesday 3/3
Physics 207: Lecture 11, Pg 1
Problem 7.34 Hint
Suggested Steps
 Two independent free body diagrams are necessary
 Draw in the forces on the top and bottom blocks
 Top Block
 Forces: 1. normal to bottom block 2. weight 3. rope tension
and 4. friction with bottom block (model with sliding)
 Bottom Block
 Forces:
1. normal to bottom surface
2. normal to top block interface
3. rope tension (to the left)
4. weight (2 kg)
5. friction with top block
6. friction with surface
7. 20 N
Use Newton's 3rd Law to deal with the force pairs
(horizontal & vertical) between the top and bottom block.
Physics 207: Lecture 11, Pg 2
Locomotion: how fast can a biped
walk?
Physics 207: Lecture 11, Pg 3
How fast can a biped walk?
What about weight?
(a) A heavier person of equal
height and proportions
can walk faster than a
lighter person
(b) A lighter person of equal
height and proportions
can walk faster than a
heavier person
(c) To first order, size doesn’t
matter
Physics 207: Lecture 11, Pg 4
How fast can a biped walk?
What about height?
(a) A taller person of equal
weight and proportions
can walk faster than a
shorter person
(b) A shorter person of equal
weight and proportions
can walk faster than a
taller person
(c) To first order, height
doesn’t matter
Physics 207: Lecture 11, Pg 5
How fast can a biped walk?
What can we say about the walker’s
acceleration if there is UCM (a
smooth walker) ?
Acceleration is radial !
So where does it, ar, come from?
(i.e., what external forces are on
the walker?)
1. Weight of walker, downwards
2. Friction with the ground, sideways
Physics 207: Lecture 11, Pg 6
Orbiting satellites vT = (gr)½
Physics 207: Lecture 11, Pg 7
Geostationary orbit
Physics 207: Lecture 11, Pg 8
Geostationary orbit

The radius of the Earth is ~6000 km but at 36000 km you are
~42000 km from the center of the earth.

Fgravity is proportional to r-2 and so little g is now ~10 m/s2 / 50

vT = (0.20 * 42000000)½ m/s = 3000 m/s

At 3000 m/s, period T = 2p r / vT = 2p 42000000 / 3000 sec =
= 90000 sec = 90000 s/ 3600 s/hr = 24 hrs

Orbit affected by the moon and also the Earth’s mass is
inhomogeneous (not perfectly geostationary)

Great for communication satellites
(1st pointed out by Arthur C. Clarke)
Physics 207: Lecture 11, Pg 9
Impulse & Linear Momentum

Transition from forces to conservation laws
Newton’s Laws  Conservation Laws
Conservation Laws  Newton’s Laws
They are different faces of the same physics
NOTE: We have studied “impulse” and “momentum”
but we have not explicitly named them as such
Conservation of momentum is far more general than
conservation of mechanical energy
Physics 207: Lecture 11, Pg 10
Collisions are a fact of life
Physics 207: Lecture 11, Pg 11
Forces vs time (and space, Ch. 10)
Underlying any “new” concept in Chapter 9 is
(1) A net force changes velocity (either magnitude or
direction)
(2) For any action there is an equal and opposite
reaction


If we emphasize Newton’s 3rd Law and look at the
changes with time then this leads to the
Conservation of Momentum Principle
Physics 207: Lecture 11, Pg 12
-
10
F (N)

F
Example 1
+
0
time (s)
2
A 2 kg block, initially at rest on frictionless horizontal surface, is
acted on by a 10 N horizontal force for 2 seconds (in 1D).
What is the final velocity?


F is to the positive & F = ma thus a = F/m = 5 m/s2
v = v0 + a Dt = 0 m/s + 2 x 5 m/s = 10 m/s (+ direction)
Notice: v - v0 = a Dt  m (v - v0) = ma Dt  m Dv = F Dt
If the mass had been 4 kg … now what final velocity?
Physics 207: Lecture 11, Pg 13

Same force

Same time
Before
10
F (N)

F
Twice the mass
0

Half the acceleration (a = F / m’)

Half the velocity ! ( 5 m/s )
2
Time (sec)
Physics 207: Lecture 11, Pg 14
Example 1

Notice that the final velocity in this case is inversely
proportional to the mass (i.e., if thrice the
mass….one-third the velocity).

Here, mass times the velocity always gives the same
value. (Always 20 kg m/s.)
10
F (N)
Area under curve is still the
same !
0
2
Time (sec)
Force x change in time =
mass x change in velocity
Physics 207: Lecture 11, Pg 15
Example 1
There many situations in which the sum of the
products “mass times velocity” is constant over time
 To each product we assign the name, “momentum”
and associate it with a conservation law.
(Units: kg m/s or N s)
 A force applied for a certain period of time can be
graphed and the area under the curve is the “impulse”

F (N)
10
0
Area under curve : “impulse”
With: m Dv = Favg Dt
2
Time (sec)
Physics 207: Lecture 11, Pg 16
Force curves are usually a bit different in the
real world
Physics 207: Lecture 11, Pg 17
Example 1 with Action-Reaction
Now the 10 N force from before is applied by person
A on person B while standing on a frictionless surface
 For the force of A on B there is an equal and opposite
force of B on A
10

F (N)
A on B
0
B on A
-10
0
2
Time (sec)
MA x DVA = Area of top curve
MB x DVB = Area of bottom curve
Area (top) + Area (bottom) = 0
Physics 207: Lecture 11, Pg 18
Example 1 with Action-Reaction
MA DVA + MB DVB = 0
MA [VA(final) - VA(initial)] + MB [VB(final) - VB(initial)] = 0
Rearranging terms
MAVA(final) +MB VB(final) =
MAVA(initial) +MB VB(initial)
which is constant regardless of M or DV
(Remember: frictionless surface)
Physics 207: Lecture 11, Pg 19
Example 1 with Action-Reaction
MAVA(final) +MB VB(final) = MAVA(initial) +MB VB(initial)
which is constant regardless of M or DV
Define MV to be the “momentum” and this
is conserved in a system if and only if the
system is not acted on by a net external
force (choosing the system is key)
Conservation of momentum is a special
case of applying Newton’s Laws
Physics 207: Lecture 11, Pg 20
Applications of Momentum Conservation
Radioactive decay:
238U
Alpha
Decay
234Th
v2
4He
v1
Explosions
Collisions
Physics 207: Lecture 11, Pg 21
Impulse & Linear Momentum
Definition: For a single particle, the momentum p is defined as:

p ≡ mv
(p is a vector since v is a vector)
So px = mvx and so on (y and z directions)

Newton’s 2nd Law:

dv d

 m  (mv)
dt dt

F = ma

 dp
F
dt
This is the most general statement of Newton’s 2nd Law
Physics 207: Lecture 11, Pg 22
Momentum Conservation
FEXT
dP

dt
dP
0
dt
FEXT  0

Momentum conservation (recasts Newton’s 2nd Law when
F = 0) is an important principle

It is a vector expression (Px, Py and Pz) .
 And applies to any situation in which there is NO net
external force applied (in terms of the x, y & z axes).
Physics 207: Lecture 11, Pg 23
Momentum Conservation
Many problems can be addressed through
momentum conservation even if other physical
quantities (e.g. mechanical energy) are not
conserved
 Momentum is a vector quantity and we can
independently assess its conservation in the x, y and
z directions
(e.g., net forces in the z direction do not affect the
momentum of the x & y directions)

Physics 207: Lecture 11, Pg 24
Exercise 1
Momentum is a Vector (!) quantity


A.
B.
C.
D.
A block slides down a frictionless ramp and then falls and
lands in a cart which then rolls horizontally without friction
In regards to the block landing in the cart is momentum
conserved?
Yes
No
Yes & No
Too little information given
Physics 207: Lecture 11, Pg 25
Exercise 1
Momentum is a Vector (!) quantity
x-direction: No net force so Px is conserved.
 y-direction: Net force, interaction with the ground so
depending on the system (i.e., do you include the Earth?)
Py is not conserved (system is block and cart only)

2 kg
5.0 m
Let a 2 kg block start at rest on
a 30° incline and slide
vertically a distance 5.0 m
and fall a distance 7.5 m into
the 10 kg cart
What is the final velocity of the
cart?
30°
10 kg
7.5 m
Physics 207: Lecture 11, Pg 26
Inelastic collision in 1-D: Example 2

A block of mass M is initially at rest on a frictionless
horizontal surface. A bullet of mass m is fired at the block
with a muzzle velocity (speed) v. The bullet lodges in the
block, and the block ends up with a speed V. In terms of m,
M, and V :
What is the momentum of the bullet with speed v ?
x
v
V
before
after
Physics 207: Lecture 11, Pg 27
Inelastic collision in 1-D: Example 2
What is the momentum of the bullet with speed v ?

mv
 Key question: Is x-momentum conserved ?
Before
After
aaaa
mv  M 0  (m  M )V
v
V
after
x
before
Physics 207: Lecture 11, Pg 28
Example 2
Inelastic Collision in 1-D with numbers
Do not try this at home!
ice
(no friction)
Before: A 4000 kg bus, twice the mass of the car, moving
at 30 m/s impacts the car at rest.
What is the final speed after impact if they move together?
Physics 207: Lecture 11, Pg 29
Exercise 2
Momentum Conservation

Two balls of equal mass are thrown horizontally with the
same initial velocity. They hit identical stationary boxes
resting on a frictionless horizontal surface.

The ball hitting box 1 bounces elastically back, while the ball
hitting box 2 sticks.
 Which box ends up moving fastest ?
A.
B.
C.
Box 1
Box 2
same
1
2
Physics 207: Lecture 11, Pg 30