# MOMENTUM_

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```					M   OMENTUM!
Momentum
Impulse
Conservation of Momentum in
1 Dimension
Conservation of Momentum in
2 Dimensions
Angular Momentum
Torque
Moment of Inertia
Momentum Defined

p = mv
p = momentum
m = mass
v = velocity
Momentum Facts
p = mv
•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.
Calculating Momentum

Car: m = 1800 kg; v = 60 m /s

Bus: m = 9000 kg; v = 100 m /s

Train: m = 3.6 ·104 kg; v = 150 m /s
Momentum of the Train
Bus: m = 9000 kg; v = 100 m /s

Calculate the momentum of the bus.
What is known?    m = 9000 kg v = 100 m/s
What is missing? momentum  p = ?
Equation          p = m·v
Solve             p = (9000 kg)(100 m/s) = 900,000 kg·m/s
Momentum of the Car

Car: m = 1800 kg; v = 60 m /s

Calculate the momentum of the car.
What is known?     m = 1800 kg v = 60 m/s
What is missing? momentum  p = ?
Equation           p = m·v
Solve              p = (1800 kg)(60 m/s) = 108,000 kg·m/s
Momentum of the Train

Calculate the momentum of the train.
What is known?    m = 3.6 ·104 kg      v = 150 m/s
What is missing? momentum  p = ?
Equation          p = m·v
Solve             p = (3.6 x 104 kg)(150 m/s) = 5,400,000 kg·m/s

Train: m = 3.6 ·104 kg; v = 150 m /s
Comparing Momentum

Car: m = 1800 kg; v = 60 m /s

Bus: m = 9000 kg; v = 100 m /s
p = 108,000 kg·m/s

p = 900,000 kg·m/s

Train: m = 3.6 ·104 kg; v = 150 m /s

p = 5,400,000 kg·m/s
Calculating Velocity from Momentum

Bus: m = 9000 kg;
Car: p = 108,000 kg·m/s

What velocity does the bus need to travel at in
order to have the same momentum as the car?
Known:      m = 9000 kg           p = 108,000 kg·m/s
Missing:    velocity = v = ?
Equation:   v = p/m =
Solve:      v = (108,000 kg·m/s) / (9000 kg) = 12 m/s
Calculating Velocity from Momentum

Car: m = 1800 kg;
Train: p = 5,400,000 kg·m/s

What velocity does the car need to travel at in
order to have the same momentum as the train?
Known:      m = 1800 kg           p = 5,400,000 kg·m/s
Missing:    velocity = v = ?
Equation:   v = p/m =
Solve:      v = (5,400,000 kg·m/s) / (1800 kg) = 3000 m/s
Calculating Mass
Car #1: p = 108,000 kg·m/s;
Car #2: v = 50 m/s

Another car has the same momentum as the
first car, but a larger mass. If it is traveling at
50 m/s, what is its mass?
Known:      v = 50 m/s           p = 108,000 kg·m/s
Missing:    mass = m = ?
Equation:   m = p/v =
Solve:      m = (108,000 kg·m/s) / (50 m/s) = 2160 kg
Conservation of Momentum

• Momentum in a system is ALWAYS
conserved.
• The total momentum two or more objects
have prior to a collision is equal to the total
momentum after the collision.
Conservation of Momentum in 1-D
The total momentum of the objects is the same before and after the
collision.
Positive is defined as the right direction.
before: pT = m1 v1 - m2 v2

v1                  v2
m1                                    m2

m1 v1 - m2 v2 = - m1 va + m2 vb
after: pT = - m1 va + m2 vb
va                               vb
m1              m2
Directions after a collision
Some collisions cause the objects to go in opposite directions.
Other collisions cause the objects to go in similar directions.

v1
m1                                v2
m2

va
m1                          vb
m2
Elastic vs. Inelastic
Elastic collisions occur when the object “bounce” off each other and no
energy is lost to changes in shape.
Inelastic collision occur when the objects get “stuck” to each other, or
energy is lost to changes in shape.

v1
m1                                v2
m2

m1           vb
m2
Sample Problem 1
35 g
7 kg
700 m/s
v=0
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=?                  4 cm/s
continued on next slide
Sample Problem 1                 (cont.)
Let’s choose left to be the + direction & use conservation of
momentum, converting all units to meters and kilograms.
35 g
p before = 7 (0) + (0.035) (700)          7 kg
700 m/s
= 24.5 kg · m /s                   v=0

35 g                                    p after = 7 (0.04) + 0.035 v
7 kg
v=?                        4 cm/s                   = 0.28 + 0.035 v

p before = p after        24.5 = 0.28 + 0.035 v        v = 692 m/s

v came out positive. This means we chose the correct
direction of the bullet in the “after” picture.
Sample Problem 2
35 g
7 kg                  700 m/s
v=0
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) = 7.035 v                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.
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.
F                   F
M m
force on M due to m                            force 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.
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
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?                     answer: Gravity is an external force on the
apple, so momentum for it alone is not
apple       m            conserved. To make gravity “internal,” we
V                    must define a system that includes the
F             other object responsible for the
gravitational force--Earth. The net force
v             on the apple-Earth system is zero, and
momentum is conserved for it. During the
fall the Earth attains a very small speed v.
Earth                    So, by conservation of momentum:
M         F
mV = M v
Sample Problem 4
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) + 15 v                 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.

before
10 m/s                 6 m/s
3 kg                                  15 kg

after
4.5 m/s                                v
3 kg              15 kg
Conservation of Momentum in 2-D
To handle a collision in 2-D, we conserve momentum in each
dimension separately.               Choosing down & right as positive:
m2        before:
m1                                   px = m1 v1 cos1 - m2 v2 cos2
1            2 v
2
v1                                py = m1 v1 sin1 + m2 v2 sin2
after:
m1       m2             px = -m1 va cosa + m2 vb cos b
a                      b
va            vb
py = m1 va sina + m2 vb sin b
Conservation of momentum equations:

m1 v1 cos1 - m2 v2 cos2 = -m1 va cosa + m2 vb cos b
m1 v1 sin1 + m2 v2 sin 2 = m1 va sina + m2 vb sin b
Conserving Momentum w/ Vectors
B                         m2                            p1
E m1
1              2
F
p2                 p before
O
R    p1
p2
E

pa
A           m1    m2
F      a                 b                        p after
T
E       pa                pb
pb
R
This diagram shows momentum vectors, which are parallel to
their respective velocity vectors. Note p1 + p 2 = p a + p b and
p before = p after as conservation of momentum demands.
Exploding Bomb

Acme

after
before

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
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,
to the original momentum
vector.
2-D Sample Problem
152 g
A mean, old dart strikes an innocent
before           40                    mango that was just passing by
34 m/s           minding its own business. Which
way and how fast do they move off
together?
5 m/s
0.3 kg                 Working in grams and taking left & down as + :

152 (34) sin 40 = 452 v sin
152(34) cos 40 - 300 (5) = 452 v cos
after
Dividing equations : 1.35097 = tan
452 g
 = 53.4908
                   Substituting into either of the first two
v                  equations :      v = 9.14 m/s
40                        Alternate Solution
Shown are momentum vectors (in g m/s).
5168                    The black vector is the total momentum
before the collision. Because of
p            conservation of momentum, it is also the
total momentum after the collisions. We
40                        can use trig to find its magnitude and
1500                       direction.

Law of Cosines : p2 = 5168 2 + 1500 2 - 2 5168  1500 cos 40
p = 4132.9736 g m/s
Dividing by total mass : v = (4132.9736 g m/s) / (452 g) = 9.14 m/s
sin        sin 40
Law of Sines :               =                      = 13.4908
1500      4132.9736
Angle w/ resp. to horiz. = 40 + 13. 4908   53.49
• 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.
Angular Momentum
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
v name.) Angular momentum is a
conserved quantity. A torque is
needed to change L, just a force is
r                 m needed to change p. Anything
spinning has angular has angular
Q                                   momentum. The more it has, the
harder it is to stop it from spinning.
Angular Momentum: General Definition
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 p sin = 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.

v


r               m

Q
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 = m r 2. For a system, I = the sum of each mass
m                   times its respective distance from the
point of interest.
r
m2
Q                              m1
r1
I=   mr2                                                           r2

I=    mi ri 2               Q

= m1 r12 + m2 r22
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?

r                                 r

                                 
m
m

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.
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 , and it’s defined as:
 =   /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   9 rev/min    9 rev        9  (2 rad)
=           =            =               = 0.094 rad/s2
10 s      10 s     min  10 s   (60 s)  10 s

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).
Torque & Angular Acceleration
Newton’s 2nd Law, as you know, is Fnet = m a

The 2nd Law has a rotational analog:         net = I 
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).
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          (if v and r are perpendicular)

v     So, if a net torque is applied, angular velocity must
r       m     change, which changes angular momentum.

proof:    net = r Fnet = r m a
= 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
Linear & Angular Momentum            (cont.)

Here is yet another pair of similar equations, one linear,
one rotational. From the formula v = r , we get

L = mvr = m r (r ) = m r 2  = I 
This is very much like p = m v, 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
p = mv
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 = I  and L is
conserved, the product of I and  is a
constant. So, when he pulls his arms in,
I goes down,  goes up, and he starts
spinning much faster.

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

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