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Chapter 9: Linear Momentum
9.1: Linear Momentum and Its Conservation
9.5: The Center of Mass
9.2: Impulse and Momentum
9.6: Motion of a System of Particles: velocity
of the center of mass, total momentum of a
system of particles, acceleration of the center
of mass, Newton's second law for a system of
particles
9.3: Collisions in One Dimension: collision, elastic
collision, inelastic collision
9.4: Two-Dimensional Collisions
9.7: Rocket Propulsion
9.1: Linear Momentum and Its Conservation
Our book states that in isolated systems momentum is conserved.
Another way to state this is that we can account for all the parts involved in the problem.
ΣF
=
ma
=0

ΣF
ΣF
ΣF
ΣF
=
=
=
=
m dv/dt + m2 dv2 / dt = 0
d (mv
+ m2v2) / dt = 0
d (mv) /dt
dp/ dt

pbefore = pafter
+
m2a2


If two particles strike each with equal and opposite
force then
And since sum is constant...the total momentum in a
system must also be constant
Where a is in the opposite direction of a2 and a = dv/dt
So, what are the units of momentum
p = mv has the units of kg m/s...

Demo: Bi-Ball: ME-N-BB
A couple of interesting notes:
The derivative with respect to time is zero...
thus the sum must be constant or conserved
9.2: Impulse and Momentum
Let's play with units...
kg m/s
kg m/s (s/s)
kg m/s (s/s)
F
F
F dt
∫ F dt
FΔt
FΔt
FΔt
=
=m a
= m dv/dt
= m dv
= ∫m dv
= m Δv
= Δmv
= Δp
Multiply both sides by seconds / seconds
Newton second
So this gives us a hint we need to integrate with respect to time.
Let’s assume F is independent of time
Impulse (F dt) = change of momentum (Δp = Δmv)
Demo: Half-Liter Paper Rocket: ME-N-PR
Estimate the distance of the deformed tip of the rocket. Use d = ½at2 to determine time of impact.
Calculate Force exerted on Rocket during this time.
9.3: Collisions in One Dimension
BIG SUV…small car (BUG)
Smash small car on grill of big SUV
pbefore
= pafter
when   
If we account for all parts (objects) then
momentum is conserved during the collision (this
is a closed system)
Let’s say a 7000 kg SUV (totally loaded) is traveling North at 90 mph (about 40 m/s) and a little car,
let’s call it a bug, and is traveling to the south at 65 mph (about 30 m /s). These are both controlled
remotely. We ram them together. What happens?
pbefore = msuvvsuv + mbugvbug;
pbefore
msuvvsuv + mbugvbug
7000(40) + 1000(-30)
250000
vf = 33 1/3 m/s
pafter = msuvvsuv-f + mbugvbug-f
= pafter
= (msuv + mbug) vf but bug is squashed on grill, thus final velocity is identical
= (7000 + 1000) vf
= 8000 vf
The final velocity is 33 1/3 m/s.
The bug is traveling in the opposite
The Massive SUV goes from 40 to 33 1/3 m/s for direction…thus a negative.
a change of 6 2/3 m/s.
So the bug starts at -30 m/s and end up at 33 1/3
That’s like hitting a concrete wall doing about 14
m/s…or a change of 63 1/3 m/s or about 140 mph
to 15 mph. The SUV is wrecked badly.
directly in a concrete wall.
Like we said earlier…the bug is squash on the grill of the SUV.
Demo: Newton’s Cradle: ME-N-NC
9.4: Two-Dimensional Collisions
First off, this is an artificial situation. 3D is a must; we're
looking at this as a 2D equation for simplicity.
Σpxi = constant
Σpxf = constant
KEi = KEf
A 2D Look at a Neutron Moderator
Initial
Final
px
= (m) 3.5 x 105 + 0
px = (m) v1x
px = (m) v1 cos 37°
+ (m) v2x
+ (m) v2x cos φ
py
=0
py = (m) v1y
py = (m) v1 sin 37°
+ (m) v2y
+ (m) v2 sin φ
Kfinal = ½(m) v12
+ ½(m) v22
Kinitial = ½(m) (3.5 x 105)2 + 0
Kinetic energy isn't broke into components (v2, a scaler).
Kinetic energy is conserved in elastic collisions only; where Ktotalbefore = Ktotalafter
Eq (1)
px
px
 (m) 3.5 x 105

3.5 x 105
Eq (2)
py

Eq (3)
K

0
(3.5 x 105)2
v1f = 2.80 x 105 m/s
v2f = 2.11 x 105 m/s
φ = 53.0°
= (m) v1 cos 37°
=
v1f cos 37°
+ (m) v2x cos φ
+
v2f cos φ
=
+
v2f sin φ
+
v2f2
=
v1f sin 37°
v1f2
Solve these 3 equations simultaneously
9.5: The Center of Mass
Demo: Leaning Tower: ME-J-LT
If 3D then
CMx =
CMx =
Σmi*xi / mTotal
1 /mT ∫x dm
CMy =
CMy =
Σmi*yi / mT
1 /mT ∫y dm
CMz =
CMz =
Σmi*zi / mT
1 /mT ∫z dm
Where is the center of mass in a rod is
λ = mass / length
How do we find the center of mass of a
homogeneous symmetric object
(Like solid 2D square
or the figure to the right)?
IT's AT THE CENTER
Just as obvious is the center of mass of the above bar
bell set, where the center of mass is marked by the “X”.
This time…let’s write the expression for the center of
mass.   
Now we just extend this principle to many more objects
for the x, y, and z axis.
xCM =
m1x1 + m2x2
——————
m1 + m2
And rCM = xCM i + yCM j + yCM k
Σi mixi
xCM = ————
Σi mi
yCM =
Σi miyi
————
Σi mi
Σi mizi
zCM = ————
Σi mi
rCM =
Σi miri
————
Σi mi
Example 9.14: Where is the center of mass of a right triangle?
We need mass / unit area of the triangle (total mass / total area)
dm = mass / unit area
dm = ( M / ½ a b)
*
dm = 2M/ab
dm = 2M/ab
dm = 2M/ab
dm = 2M/ab
dm = 2M/a² (
* area of green strip
y dx
*
y dx
*
y dx
* (x*b/a) dx
* (b/a) xdx
x dx)
xCM = 1/M ∫ x dm
xCM = 1/M ∫ x (2M/a² * xdx)
combine like terms
xCM = 2/a² ∫ x² dx
integrate from 0 to a
xCM = 2/3 a
Expression for equivalent
proportions
x
/y = a/b 
y = x*b/a
Example 3: Center of Mass for varying linear density
λ is mass per unit length (g / m)
λ = 2 + 40x²
What is its mass if it's 30 cm long?
m = ∫ λ dx =
m = ∫ (2 + 40x²)dx =
m = 2x + 40x³/3 from 0 to 0.3 m
m = 0.6 g + 0.36 g
m = 0.96 grams
Where is its center of mass from x = 0?
xCM = ∫ x dm / mT
xCM = 1 / mT ∫ x λ
dx
xCM = 1 / mT ∫ x (2 + 40x²) dx
xCM = 1 / mT ∫ 2xdx + 40∫ x³dx
xCM = 1 / mT
(x² + 10x4)
from 0 to 0.3
xCM = 1/0.96 (0.09 + 0.081)
xCM = 0.178 m past 0 m
Demo: Walking the Spool: ME-K-WS
Demo: Center of Mass O.C Ruler: ME-J-CE
Demo: Center of Mass, Irregular Object: ME-J-CI
9.6: Motion of a System of Particles
A rocket is fired vertically upward. At the instant it reaches an altitude of 1000 meters and a speed
of 300 m/s it explodes into three fragments. The 1st fragment has twice the mass of either the 2nd
or 3rd fragments and continues to move upward with a speed of 200 m/s. The second fragment has a
speed of 240 m/s and is moving east right after the explosion. What is the velocity of the 3
fragment?
Whole = 1st frag + 2nd frag + 3rd frag
4m
= 2m + m
+ m
Vertical components
Horizontal components
pbefore
= pafter
4m(300) = 2m(v1) + m(v3y)
pbefore = pafter
0
= 2m(v2) + m(v3x)
4m(300) = 2m(v1) + m(v3y)
1200
= 2(200) + (v3y)
v3y = 800 m/s
0
= m(v2) + m(v3x)
0
= (240) + (v3x)
v3x = -240 m/s
(v1 is all y-comp)
v3rd
v3rd
= (2402 + 8002)½
= 835.2 m/s @ 106.7
(v2 is all x-comp)
3rd = tan-1(800/240)
3rd = 73.3
We want 180 - 3rd
9.7: Rocket Propulsion
pbefore
procket&fuel
(M+m)v
Mv +mv
0
Mv
Mdv
Mdv
dv
v
=
pafter
= procket
= M(v + v)
= Mv + Mv
= Mv
= m vexh
= vexh dm
+
pfuel
+ m(v - vexhaust)
+ mv - mvexh
- mvexh
(the velocity of the
exhaust is a constant)
(limit of m  0)
= -vexh dM
= -vexh dM / M
= vexh ln (Mi / Mf)
FThrust = m
a
FThrust = m dv/dt
FThrust = vexh dM/dt
M is the mass of the rocket (which may also
containing some remaining fuel)
We know downward momentum of the exhaust
opposes the upward momentum of the rocket…
thus vexhaust dm = -v dM
We know that the dm = -dM
dM is the mass of the rocket with the remaining fuel
and dm is the spent fuel. As more and more fuel
(+dm) exits the rocket (-dM, mass of rocket) is
reduced (-) accordingly