Download Momentum

Momentum and
Momentum Conservation
• Momentum
• Impulse
• Conservation
of Momentum
• Collision in 1-D
• Collision in 2-D
So What’s Momentum ?
• Momentum = mass x velocity
• This can be abbreviated to :
.
momentum = mv
• Or, if direction is not an important factor : .
.
momentum = mass x speed
• So, A really slow moving truck and an
extremely fast roller skate can have the
same momentum.
Question :
• Under what circumstances would the roller
skate and the truck have the same
momentum ?
• The roller skate and truck can have the same
momentum if the ratio of the speed of the
skate to the speed of the truck is the same as
the ratio of the mass of the truck to the mass
of the skate.
• A 1000 kg truck moving at 0.01 m/sec has the
same momentum as a 1 kg skate moving at
10 m/sec. Both have a momentum of 10 kg
m/sec. ( 1000 x .01 = 1 x 10 = 10 )
Linear Momentum
• A new fundamental quantity, like force, energy
• The linear momentum p of an object of mass
m moving with a velocity v is defined to be the
product of the mass and velocity:


p  mv
– The terms momentum and linear momentum will be
used interchangeably in the text
– Momentum depend on an object’s mass and velocity
Newton’s Law and Momentum
• Newton’s Second Law can be used to relate the
momentum of an object to the resultant force
acting on it




v (mv )
Fnet  ma  m

t
t
• The change in an object’s momentum divided by
the elapsed time equals the constant net force
acting on the object


p change in momentum

 Fnet
t
time interval
Impulse and Momentum
• If momentum changes, it’s because mass or
velocity change.
• Most often mass doesn’t change so
velocity changes and that is acceleration.
• And mass x acceleration = force
• Applying a force over a time interval to an
object changes the momentum
• Force x time interval = Impulse
• Impulse = F t or Ft
Ft == mv
mv
Impulse-Momentum Theorem
• The theorem states
that the impulse
acting on a system is
equal to the change in
momentum of the
system

 
p  Fnet t  I




I  p  mv f  mvi
MOMENTUM
• An object at rest has no momentum, why?
• Because anything times zero is zero
• (the velocity component is zero for an object at rest)
FORCE
• To INCREASE MOMENTUM,
apply the greatest force possible for as long
as possible.
• Examples :
• pulling a sling shot
•
•
•
drawing an arrow in a bow all the way back
a long cannon for maximum range
hitting a golf ball or a baseball
. (follow through is important for these !)
TIME
MOMENTUM
• SOME VOCABULARY :
• impulse : impact force X time (newton.sec)
.
• impact :
.
Ft = impulse
the force acting on an object (N)
usually when it hits something.
• impact forces : average force of impact
MOMENTUM
• Decreasing Momentum
• Which would it be more safe to hit in a car ?
mv
mv
Ft
Ft
• Knowing the physics helps us understand why
hitting a soft object is better than hitting a hard one.
MOMENTUM
• In each case, the momentum is decreased by the same
amount or impulse (force x time)
• Hitting the haystack extends the impact time (the time in
which the momentum is brought to zero).
• The longer impact time reduces the force of impact and
decreases the deceleration.
• Whenever it is desired to decrease the force of impact,
extend the time of impact !
DECREASING MOMENTUM
• If the time of impact is increased by 100 times (say from .01
sec to 1 sec), then the force of impact is reduced by 100
times (say to something survivable).
•
•
•
•
EXAMPLES :
Padded dashboards on cars
Airbags in cars
or
safety nets in circuses
Moving your hand backward as you catch a fast-moving ball
with your bare hand
or
a boxer moving with a punch.
• Flexing your knees when jumping from a higher place to the
ground.
or
elastic cords for bungee jumping
• Using wrestling mats instead of hardwood floors.
• Dropping a glass dish onto a carpet instead of a sidewalk.
•
EXAMPLES OF DECREASING
MOMENTUM
Bruiser Bruno on boxing …
F = change in
t
momentum
t = change in
F
momentum
• Increased impact time reduces force of impact
Ft = Δmv applies here.
• Barbie on bungee Jumping …
mv = the momentum gained before the cord
begins to stretch that we wish to change.
Ft = the impulse the cord supplies to
reduce the momentum to zero.
Because the rubber cord stretches for
a long time the average force on the
jumper is small.
Questions :
• When a dish falls, will the impulse be less if it lands on a
carpet than if it lands on a hard ceramic tile floor ?
• The impulse would be the same for either surface
because there is the same momentum change for each.
It is the force that is less for the impulse on the carpet
because of the greater time of momentum change.
There is a difference between impulse and impact.
• If a boxer is able to increase the impact time by 5 times
by “riding” with a punch, by how much will the force of
impact be reduced?
• Since the time of impact increases by 5 times, the force
of impact will be reduced by 5 times.
BOUNCING
• IMPULSES ARE GREATER WHEN AN OBJECT
BOUNCES
• The impulse required to bring an object to a stop and
then to throw it back upward again is greater than
the impulse required to merely bring the object to a
stop.
• When a martial artist breaks boards,
• does their hand bounce?
• Is impulse or momentum greater ?
• Example :
• The Pelton Wheel.
Calculating the Change of Momentum
p  pafter  pbefore
 mvafter  mvbefore
 m(vafter  vbefore )
For the teddy bear
p  m 0  (v)  mv
For the bouncing ball
p  m v  (v)  2mv
How Good Are the Bumpers?
In a crash test, a car of mass 1.5103 kg collides with a wall and
rebounds as in figure. The initial and final velocities of the car are vi=-15
m/s and vf = 2.6 m/s, respectively. If the collision lasts for 0.15 s, find
(a) the impulse delivered to the car due to the collision
(b) the size and direction of the average force exerted on the car

How Good Are the Bumpers?
In a crash test, a car of mass 1.5103 kg collides with a wall and
rebounds as in figure. The initial and final velocities of the car are vi=-15
m/s and vf = 2.6 m/s, respectively. If the collision lasts for 0.15 s, find
(a) the impulse delivered to the car due to the collision
(b) the size and direction of the average force exerted on the car

pi  mvi  (1.5 103 kg)( 15m / s)  2.25 10 4 kg  m / s
p f  mv f  (1.5 103 kg)(2.6m / s)  0.39 104 kg  m / s
I  p f  pi  mv f  mvi
 (0.39 10 4 kg  m / s )  (2.25  10 4 kg  m / s )
 2.64 10 4 kg  m / s
p I
2.64 104 kg  m / s
Fav 


 1.76 105 N
t t
0.15s
CONSERVATION OF MOMENTUM
• To accelerate an object, a force must be applied.
• The force or impulse on the object must come
from outside the object.
• EXAMPLES : The air in a basketball,
sitting in a car and pushing on the dashboard
or sitting in a boat and blowing on the sail
don’t create movement.
• Internal forces like these are balanced and cancel
each other.
• If no outside force is present, no change in
momentum is possible.
The Law of Conservation of Momentum
• In the absence of an external force, the momentum of
a system remains unchanged.
• This means that, when all of the forces are internal
(for EXAMPLE: the nucleus of an atom undergoing
.
radioactive decay,
.
cars colliding, or
.
stars exploding
the net momentum of the system before and after the
event is the same.
Conservation of Momentum
• In an isolated and closed
system, the total momentum of
the system remains constant in
time.
– Isolated system: no external forces
– Closed system: no mass enters or
leaves
– The linear momentum of each
colliding body may change
– The total momentum P of the
system cannot change.
Conservation of Momentum
• Start from impulse-momentum
theorem



F21t  m1v1 f  m1v1i



F12t  m2v2 f  m2v2i
• Since


F21t   F12t




• Then m1v1 f  m1v1i  (m2v2 f  m2 v2i )
• So




m1v1i  m2 v2i  m1v1 f  m2 v2 f
QUESTIONS
• 1. Newton’s second law states that if no net force is
exerted on a system, no acceleration occurs. Does it
follow that no change in momentum occurs?
• No acceleration means that no change occurs in
velocity of momentum.
• 2. Newton’s 3rd law states that the forces exerted on
a cannon and cannonball are equal and opposite.
Does it follow that the impulse exerted on the
cannon and cannonball are also equal and opposite?
• Since the time interval and forces are equal and
opposite, the impulses (F x t) are also equal and
opposite.
COLLISIONS
• ELASTIC COLLISIONS
Momentum transfer from one
Object to another .
Is a Newton’s cradle like the one
Pictured here, an example of an
elastic or inelastic collision?
• INELASTIC COLLISIONS
Problem Solving #1
• A 6 kg fish swimming at 1 m/sec swallows a 2 kg fish
that is at rest. Find the velocity of the fish immediately
after “lunch”.
•
net momentum before = net momentum after
•
(net mv)before = (net mv)after
• (6 kg)(1 m/sec) + (2 kg)(0 m/sec) = (6 kg + 2 kg)(vafter)
•
6 kg.m/sec = (8 kg)(vafter)
•
vafter = 6 kg.m/sec / 8 kg
vafter =
•
8 kg
•
vafter = ¾ m/sec
Problem Solving #2
• Now the 6 kg fish swimming at 1 m/sec swallows a 2
kg fish that is swimming towards it at 2 m/sec. Find
the velocity of the fish immediately after “lunch”.
•
net momentum before = net momentum after
•
(net mv)before = (net mv)after
• (6 kg)(1 m/sec) + (2 kg)(-2 m/sec) = (6 kg + 2 kg)(vafter)
• 6 kg.m/sec + -4 kg.m/sec = (8 kg)(vafter)
•
vafter = 2 kg.m/sec / 8 kg
vafter =
•
8 kg
•
vafter = ¼ m/sec
Problem Solving #3 & #4
• Now the 6 kg fish swimming at 1 m/sec swallows a 2
kg fish that is swimming towards it at 3 m/sec.
•
(net mv)before = (net mv)after
• (6 kg)(1 m/sec) + (2 kg)(-3 m/sec) = (6 kg + 2 kg)(vafter)
• 6 kg.m/sec + -6 kg.m/sec = (8 kg)(vafter)
•
vafter = 0 m/sec
• Now the 6 kg fish swimming at 1 m/sec swallows a 2
kg fish that is swimming towards it at 4 m/sec.
•
(net mv)before = (net mv)after
• (6 kg)(1 m/sec) + (2 kg)(-4 m/sec) = (6 kg + 2 kg)(vafter)
• 6 kg.m/sec + -8 kg.m/sec = (8 kg)(vafter)
•
vafter = -1/4 m/sec
The Archer
An archer stands at rest on frictionless ice and fires a 0.5-kg arrow
horizontally at 50.0 m/s. The combined mass of the archer and bow is
60.0 kg. With what velocity does the archer move across the ice after
firing the arrow?

pi  p f
m1v1i  m2 v2i  m1v1 f  m2 v2 f
m1  60.0kg, m2  0.5kg, v1i  v2i  0, v2 f  50m / s, v1 f  ?
0  m1v1 f  m2 v2 f
m2
0.5kg
v1 f  
v2 f  
(50.0m / s )  0.417 m / s
m1
60.0kg
Types of Collisions
• Momentum is conserved in any collision
• Inelastic collisions: rubber ball and hard ball
– Kinetic energy is not conserved
– Perfectly inelastic collisions occur when the objects
stick together
• Elastic collisions: billiard ball
– both momentum and kinetic energy are conserved
• Actual collisions
– Most collisions fall between elastic and perfectly
inelastic collisions
Simple Examples of Head-On Collisions
(Energy and Momentum are Both Conserved)
Collision between two objects of the same mass. One mass is at rest.
Collision between two objects. One at rest initially has twice the mass.
Collision between two objects. One not at rest initially has twice the mass.


p  mv
Example of Non-Head-On Collisions
(Energy and Momentum are Both Conserved)
Collision between two objects of the same mass. One mass is at rest.
If you vector add the total momentum after collision,
you get the total momentum before collision.


p  mv
Collisions Summary
• In an elastic collision, both momentum and kinetic
energy are conserved
• In an inelastic collision, momentum is conserved but
kinetic energy is not. Moreover, the objects do not stick
together
• In a perfectly inelastic collision, momentum is conserved,
kinetic energy is not, and the two objects stick together
after the collision, so their final velocities are the same
• Elastic and perfectly inelastic collisions are limiting
cases, most actual collisions fall in between these two
types
• Momentum is conserved in all collisions
More about Perfectly Inelastic
Collisions
• When two objects stick together
after the collision, they have
undergone a perfectly inelastic
collision
• Conservation of momentum
m1v1i  m2 v2i  ( m1  m2 )v f
m1v1i  m2 v2i
vf 
m1  m2
• Kinetic energy is NOT conserved
An SUV Versus a Compact

An SUV with mass 1.80103 kg is travelling eastbound at +15.0 m/s, while
a compact car with mass 9.00102 kg is travelling westbound at -15.0 m/s.
The cars collide head-on, becoming entangled.
Find the speed of the entangled cars after
the collision.
(b) Find the change in the velocity of each
car.
(c) Find the change in the kinetic energy of
the system consisting of both cars.
(a)
An SUV Versus a Compact
(a)
Find the speed of the entangled cars after
the collision.
m1  1.80 103 kg, v1i  15m / s
m2  9.00 102 kg, v2i  15m / s
pi  p f
m1v1i  m2 v2i  (m1  m2 )v f
m1v1i  m2 v2i
vf 
m1  m2
v f  5.00m / s
An SUV Versus a Compact
(b)
Find the change in the velocity of each car.
v f  5.00m / s
m1  1.80 103 kg, v1i  15m / s
m2  9.00 102 kg, v2i  15m / s
v1  v f  v1i  10.0m / s
v2  v f  v2i  20.0m / s
m1v1  m1 (v f  v1i )  1.8 104 kg  m / s
m2 v2  m2 (v f  v2i )  1.8 104 kg  m / s
m1v1  m2v2  0
An SUV Versus a Compact
(c)
Find the change in the kinetic energy of the
system consisting of both cars.
m1  1.80 103 kg, v1i  15m / s
m2  9.00 102 kg, v2i  15m / s
v f  5.00m / s
1
1
2
KEi  m1v1i  m2 v22i  3.04 105 J
2
2
1
1
2
KE f  m1v1 f  m2 v22 f  3.38 10 4 J
2
2
KE  KE f  KEi  2.70 105 J
More About Elastic Collisions
• Both momentum and kinetic energy
are conserved
m1v1i  m2 v2i  m1v1 f  m2 v2 f
1
1
1
1
2
2
2
m1v1i  m2 v2i  m1v1 f  m2 v22 f
2
2
2
2
• Typically have two unknowns
• Momentum is a vector quantity
– Direction is important
– Be sure to have the correct signs
• Solve the equations simultaneously
Elastic Collisions
• A simpler equation can be used in place of the KE
equation
1
1
1
1
2
2
2
m1v1i  m2 v2i  m1v1 f  m2 v22 f
2
2
2
2
m1 (v12i  v12f )  m2 (v22 f  v22i )
m1 (v1i  v1 f )( v1i  v1 f )  m2 (v2 f  v2i )( v2 f  v2i )
m1v1i  m2 v2i  m1v1 f  m2 v2 f
m1 (v1i  v1 f )  m2 (v2 f  v2i )
v1i  v1 f  v2 f  v2i
m1v1i  m2 v2i  m1v1 f  m2 v2 f
Summary of Types of Collisions
• In an elastic collision, both momentum and kinetic
energy are conserved
v1i  v1 f  v2 f  v2i
m1v1i  m2 v2i  m1v1 f  m2 v2 f
• In an inelastic collision, momentum is conserved but
kinetic energy is not
m1v1i  m2 v2i  m1v1 f  m2 v2 f
• In a perfectly inelastic collision, momentum is conserved,
kinetic energy is not, and the two objects stick together
after the collision, so their final velocities are the same
m1v1i  m2 v2i  ( m1  m2 )v f
Problem Solving for 1D Collisions, 1
• Coordinates: Set up a
coordinate axis and define
the velocities with respect
to this axis
– It is convenient to make your
axis coincide with one of the
initial velocities
• Diagram: In your sketch,
draw all the velocity
vectors and label the
velocities and the masses
Problem Solving for 1D Collisions, 2
• Conservation of
Momentum: Write a
general expression for the
total momentum of the
system before and after
the collision
– Equate the two total
momentum expressions
– Fill in the known values
m1v1i  m2 v2i  m1v1 f  m2 v2 f
Problem Solving for 1D Collisions, 3
• Conservation of Energy:
If the collision is elastic,
write a second equation for
conservation of KE, or the
alternative equation
– This only applies to perfectly
elastic collisions
v1i  v1 f  v2 f  v2i
• Solve: the resulting
equations simultaneously
One-Dimension vs Two-Dimension
Two-Dimensional Collisions
• For a general collision of two objects in twodimensional space, the conservation of momentum
principle implies that the total momentum of the
system in each direction is conserved
m1v1ix  m2v2ix  m1v1 fx  m2v2 fx
m1v1iy  m2v2iy  m1v1 fy  m2v2 fy
Two-Dimensional Collisions
• The momentum is conserved in all directions
m1v1ix  m2v2ix  m1v1 fx  m2v2 fx
• Use subscripts for
– Identifying the object
m1v1iy  m2v2iy  m1v1 fy  m2v2 fy
– Indicating initial or final values
– The velocity components
• If the collision is elastic, use conservation of
kinetic energy as a second equation
– Remember, the simpler equation can only be used
for one-dimensional situations
v1i  v1 f  v2 f  v2i
Glancing Collisions
• The “after” velocities have x and y components
• Momentum is conserved in the x direction and in the
y direction
• Apply conservation of momentum separately to each
direction
mv m v mv m v
1 1ix
2 2 ix
1 1 fx
2 2 fx
m1v1iy  m2v2iy  m1v1 fy  m2v2 fy
Problem Solving for TwoDimensional Collisions, 3
• Conservation of Energy: If the collision is
elastic, write an expression for the total
energy before and after the collision
– Equate the two expressions
– Fill in the known values
– Solve the quadratic equations
• Can’t be simplified
Problem Solving for TwoDimensional Collisions, 4
• Solve for the unknown quantities
– Solve the equations simultaneously
– There will be two equations for inelastic
collisions
– There will be three equations for elastic
collisions
• Check to see if your answers are
consistent with the mental and pictorial
representations. Check to be sure your
results are realistic
Collision at an Intersection
A car with mass 1.5×103 kg traveling
east at a speed of 25 m/s collides at
an intersection with a 2.5×103 kg van
traveling north at a speed of 20 m/s.
Find the magnitude and direction of
the velocity of the wreckage after the
collision, assuming that the vehicles
undergo a perfectly inelastic collision
and assuming that friction between the
vehicles and the road can be
neglected.

mc  1.5 103 kg, mv  2.5 103 kg
vcix  25m / s, vviy  20m / s, v f  ?  ?
Collision at an Intersection
mc  1.5 103 kg, mv  2.5 103 kg
vcix  25m / s, vviy  20m / s, v f  ?  ?
p
p
xi
 mc vcix  mv vvix  mc vcix  3.75 104 kg  m / s
xf
 mc vcfx  mv vvfx  (mc  mv )v f cos
3.75 104 kg  m / s  (4.00 103 kg)v f cos
p
p
yi
yf
 mc vciy  mv vviy  mv vviy  5.00 104 kg  m / s
 mc vcfy  mv vvfy  (mc  mv )v f sin 
5.00 104 kg  m / s  (4.00 103 kg)v f sin 
Collision at an Intersection
mc  1.5 103 kg, mv  2.5 103 kg
vcix  25m / s, vviy  20m / s, v f  ?  ?
5.00 104 kg  m / s  (4.00 103 kg)v f sin 
3.75 104 kg  m / s  (4.00 103 kg)v f cos
5.00 104 kg  m / s
tan  
 1.33
4
3.75 10 kg  m / s
  tan 1 (1.33)  53.1
5.00 104 kg  m / s
vf 
 15.6m / s
3

(4.00 10 kg) sin 53.1
MOMENTUM VECTORS
• Momentum can be analyzed by using vectors
• The momentum of a car accident is equal to the
vector sum of the momentum of each car A & B
before the collision.
A
B
MOMENTUM VECTORS (Continued)
• When a firecracker bursts, the vector sum of the momenta
of its fragments add up to the momentum of the firecracker
just before it exploded.
• The same goes for subatomic elementary particles. The
tracks they leave help to determine their relative mass and
type.