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Physics 3A: Linear Momentum
Shoup – 166
Recall that there was more to motion than just speed
A more complete description of motion is the conceptof linear momentum: Being a product of a scalar (m) and a vector (v), momentumis a vector: Newton's 2
nd
law tells us how forces cause acceleration.
Accelerations are changes in motion; thus forces must causechanges in momentum as well:
p
m
v
p
x
m v
x
p
y
m v
y
p
z
m v
z
F
m
a
m
d
v
d t
d
m
v
d t
d
p
d t
F
d
p
d t
Shoup – 167
How can we use this concept of momentum to analyze
situations? We will find (as in Chapter's 6 & 7) that another conservationlaw holds which we can employ
Consider two interacting particles:
Assume they form an isolated system (F
ext
From figure: From Newton's 3
rd
law:
So:
m
1
m
2
p
1
= m
1
v
1
p
2
= m
2
v
2
F
12 F
21
Physics 3A: Linear Momentum
F
21
d
p
1
d t
F
12
d
p
2
d t F
12
F
21
F
12
F
21
d
p
1
d t
d
p
2
d t
d
d t
p
1
p
2
Shoup – 168
If the time derivative of a quantity is zero, then the quantity is
constant over time, so: This is true for any number of particles in the system: We can also break (8.5) into components as well: This is the concept of conservation of linear momentum:
or
The total linear momentum of an isolated system is constant
Physics 3A: Linear Momentum
p
tot
constant
p
1 i
p
2 i
p
1 f
p
2 f
system
p
ix
system
p
fx
system
p
iy
system
p
fy
system
p
iz
system
p
fz
system
p
i
system
p
f
Shoup – 169
Now lets look at case where net force isn't zero
from (8.3): We (you!) can integrate this: We define the integral of the force over the time it is applied as
the
Impulse (I)
m
p
i
= m v
i
Σ
F
acts for
∆
t = t
f
-t
i
(before)
m
p
f
= m v
f
p
i
∆
p
(after)
Physics 3A: Linear Momentum
d
p
F dt
p
i
p
f
dp
t
i
t
f
F dt
p
f
p
i
p
t
i
t
f
F dt
I
t
i
t
f
F dt
p
Shoup – 170
This is also true for systems of particles where an external force
acts on the system:
If the force is constant, then the integral is simply F
t, so:
We can also compute the "constant" force that would result in the
same impulse:
So that:
F
x
t
i
t
f
F
x
t
i
t
f
areas equal
Physics 3A: Linear Momentum
I
t
i
t
f
F
ext
dt
p
tot
I
F
t
F
t
t
i
t
f
F dt
I
F
t
F
Shoup – 171
m
1
m
2
F
12
F
21
Now lets see how to use this very powerful tool of conservation
of momentum for isolated systems
Collisions are a large class of problems which can be solved using
it First, some concepts & assumptions about collisions:
Basically, they are interactions between two or more objectsin the given system Sometimes they involve "physical contact", sometimes not
+
F
21
++
F
12
p
4
He
Physics 3A: Linear Momentum
Shoup – 172
Concepts & assumptions about collisions (continued):
We can model some of them as
perfectly inelastic
collisions:
particles "stick" together momentum is conserved kinetic energy is not conserved maximum amount of K istransformed if collision is head-on(one dimensional) then:
m
1
m
2
v
1i
v
2i
m
1
2
v
f
Before collision
After collision
Physics 3A: Linear Momentum
m
1
v
1 i
m
2
v
2 i
m
1
m
2
v
f
v
f
m
1
v
1 i
m
2
v
2 i
m
1
m
2
Shoup – 173
Concepts & assumptions about collisions (continued):
We can model some of them as elastic
collisions:
particles don't deform in collision momentum is conserved kinetic energy is conserved if collision is head-on(one dimensional), then:
With these two equations we can solve problems with twounknowns
m
1
m
2
v
1i
v
2i
v
2f
Before collision
After collision
v
1f
momentum conserved
kinetic energy
conserved
Physics 3A: Linear Momentum
m
1
v
1 i
m
2
v
2 i
m
1
v
1 f
m
2
v
2 f
m
1
v
2 1 i
m
2
v
2 2 i
m
1
v
2 1 f
m
2
v
2 2 f
Shoup – 178
We have conservation of momentum (always for isolated systems):
Before
After
Physics 3A: Linear Momentum
p
i x
m
1
v
1 i x
p
i y
p
f x
m
1
v
1 f x
m
2
v
2 f x
p
f x
m
1
v
1 f
cos
m
2
v
2 f
cos
p
f y
m
1
v
1 f y
m
2
v
2 f y
p
f y
m
1
v
1 f
sin
m
2
v
2 f
sin
Shoup – 178
We have conservation of momentum (always for isolated systems):
Before
After
X
Physics 3A: Linear Momentum
p
i x
m
1
v
1 i x
p
i y
p
f x
m
1
v
1 f x
m
2
v
2 f x
p
f x
m
1
v
1 f
cos
m
2
v
2 f
cos
p
f y
m
1
v
1 f y
m
2
v
2 f y
p
f y
m
1
v
1 f
sin
m
2
v
2 f
sin
p
i x
p
f x
m
1
v
1 i x
m
1
v
1 f
cos
m
2
v
2 f
cos
Shoup – 178
We have conservation of momentum (always for isolated systems):
Before
After
p
i x
p
f x
X
Y
Physics 3A: Linear Momentum
p
i x
m
1
v
1 i x
p
i y
p
f x
m
1
v
1 f x
m
2
v
2 f x
p
f x
m
1
v
1 f
cos
m
2
v
2 f
cos
p
f y
m
1
v
1 f y
m
2
v
2 f y
p
f y
m
1
v
1 f
sin
m
2
v
2 f
sin
m
1
v
1 i x
m
1
v
1 f
cos
m
2
v
2 f
cos
p
i y
p
f y
m
1
v
1 f
sin
m
2
v
2 f
sin
m
1
v
1 f
sin
m
2
v
2 f
sin
Shoup – 179
If an elastic collision, we also have: Lets expand our horizons and consider "extended" objects
Approach is to consider a system of discrete point particles, thenexpand this to a "continuous medium", i.e. solid objects So consider the following system of two particles attached bya rigid rod:
conservation of kinetic
energy
rotates
"center of mass"
Physics 3A: Linear Momentum
m
1
v
1 i
2
m
1
v
2 1 f
m
2
v
2 2 f
Shoup – 180
If we push at CM, then system does
not rotate, but moves as a"point-particle": so we can simplify if we use CM Center of mass for two particles is:("weighted average") For a larger system of particles:
y
x
x
1
x
2
x
cm
CM
m
1
m
2
Physics 3A: Linear Momentum
x
CM
m
1
x
1
m
2
x
2
m
1
m
2
x
CM
m
1
x
1
m
2
x
2
m
n
x
n
m
1
m
2
... m
n
i
m
i
x
i
i
m
i
i
m
i
x
i
M
Shoup – 181
X
CM
is the x coordinate of the center of mass
We can use similar equations for y & z, so we have: These are the coordinates of the Center of Mass. The position
vector is:
Physics 3A: Linear Momentum
x
CM
i
m
i
x
i
M
y
C M
i
m
i
y
i
M
z
CM
i
m
i
z
i
M
r
CM
x
CM
i
y
CM
j
z
CM
k
r
CM
i
m
i
x
i
i
i
m
i
y
i
j
i
m
i
z
i
k
M
r
CM
i
m
i
x
i
i
y
i
j
z
i
k
M
r
C M
i
m
i
r
i
M
Shoup – 182
So, to compute the CM of a collection of particles, sum over all
particle position vectors:
Now extend this to a continuous body (extended body):
sum over all small mass elements: take limit:
r
1
r
2
r
3
r
4
y
x
z
m
1
m
2
m
3
m
4
CM
(fig 8.14)
same for y & z
Physics 3A: Linear Momentum
r
C M
i
m
i
r
i
M
x
Cm
i
x
i
m
i
M
x
C m
lim
m
i
0
i
x
i
m
i
M
M
x d m
Shoup – 183
As before, can put into position vector form: If the following bodies are homogeneous, where are there
centers of mass?
sphere
rod
doughnut
cone
CM
CM
CM
CM
Physics 3A: Linear Momentum
r
CM
M
r dm
