Transcript Impulse and Momentum

Students are responsible for ALL of the material in this presentation . Use this
as an initial guide through the material. Please consult your text for clarification
and ask questions in class.
 Linear Momentum  a vector quantity describing
the product of an object’s mass and velocity.
 Momentum is represented by a lowercase “p”
 Mathematically, momentum is represented:
p = mv
Momentum
(kgm/s)
=
(Ns)
Mass
(kg)
Don’t believe me? Check for yourself… Ns = kgm/s
recall that N = kgm/s/s
so if you multiply N x s
you will get (kgm/s/s) x s = kgm/s 
Velocity at a
given moment
(m/s)
Question:
 Consider applying a net force (F)
to an object as shown. If this
force is applied for a time frame
of Δt, then the box will accelerate
to some resulting (final) velocity.
F
 If the time frame is doubled, what
happens to the resulting velocity?
Obviously, the resulting velocity will increase (because it experiences an
acceleration for a longer period of time.)
 Impulse is a vector quantity describing the product of
the (average) force (applied to an object) and the time
over which that force is applied.
 Impulse is represented by a capital “J”
 Mathematically, impulse is represented:
J = FΔt
Impulse
(Ns)
Force
applied
(N)
Time during
which the force
is applied
(s)
 Objects respond to an impulse (J) by changing their
momentum (p).
* A large impulse will produce a large response (i.e. a baseball
will move faster if subjected to a larger impulse) to a given
mass….BUT…the same impulse will have a different effect on
different masses (i.e. That same impulse applied to a bowling
ball will not result in the same speed as the baseball had).
J
J
 We know that F = ma
and
So
vf  vo
a
t
F = m ( v f  vo )
t
Ft = m(vf – vo)
We already stated that the quantity Ft is impulse
(J). We can also rewrite the right side of the
equation giving us:
J = mvf - mvo
The quantity mass x velocity is known as
momentum. This gives us:
J = pf – po = Δp  Impulse-Momentum Theory
This is a HUGE DEAL! Think about it….
Consider that you get a job as a stuntman (or stunt
woman as it might be) in the next Marvel Comic
movie. You will be filming a scene in which you crash
a speeding car into a wall. The director asks you,
“would you rather perform the stunt by actually
crashing into a brick wall or would you rather crash
into a Styrofoam replica placed in front of a bunch of
hay bales?”
…obviously you would choose the Styrofoam replica.
You intuitively know that the Styrofoam/hay
combination will provide a “softer” stop than the
actual brick wall. But, did you ever stop to think why?
The “why” is because of the impulse-momentum
theorem.
Let’s assign some values. Let us say that the 1100-kg car is
initially traveling east at 30 m/s. No matter which object
the car hits, it will stop in both cases. So the change in
momentum is the same for both cases….
vo =30 m/s
vf = 0 m/s
m = 1100-kg
Δp = pf – po
Δp = mvf – mvo = -33000 kgm/s
FOR BOTH THE BRICK
WALL AND THE STYROFOAM
WALL.
The change in momentum will be -33000 kgm/s for
either case. But HOW they stop will indicate whether a
large or small force was responsible.
The brick wall would stop the car VERY QUICKLY…let’s
say it brings the car to rest in 2.1-seconds.
The Styrofoam replica would stop the car SLOWLY….let’s
say it brings the car to rest in 5.8-seconds.
Because the TIME of IMPACT is different, the FORCE
applied will be different as well.
Hitting the Brick Wall….
Δp = -33000 kgm/s
t = 2.1 s
F=?
Hitting the Styrofoam….
Δp = -33000 kgm/s
t = 5.8 s
F=?
Δp = Ft
-33000 = F(2.1)
F = -15714.3 N
NOTE – the negative
signs on the forces
indicate the
direction of the
force.
Δp = Ft
-33000 = F(5.8)
F = -5689.7 N
The Shorter the time frame, the greater the force.
The Longer the time frame, the weaker the force.
THIS IS A BIG DEAL!
Work on the following practice problems BEFORE you
continue to the next topic.
Page 199: 13
Page 201: 1 4
Page: 203: 13
You can find solutions to these problems on the class
webpage. You should (as always) use the solutions to check your
work…not as a substitute for trying it on your own.
Before we get into the law, let’s look at a few definitions:
 Internal Forces  forces that act within the system
 objects within the system exert
forces on each other
 External Forces  forces exerted on the objects (of
the system) by external agents.
System  the collection of objects being studied.
 an isolated system is one for which all
external forces are neglected.
 Consider an isolated
system of two masses as
shown.
 When they collide, the
objects will exert equal and
opposite forces on each
other (F12 = -F21) for a
period of time, t. Thus,
there is an impulse on each
object which will change
their momentum.
Since F12 = F21
J = Δp
J1 + J2 = Δp1 + Δp2
F12t + F21t = (p1f – p1o) + (p2f – p2o)
0 = p1f – p1o + p2f – p2o
p1o + p2o = p1f + p2f This is the conservation
of momentum.
For an isolated system,
the total initial
momentum of the
system MUST BE
CONSERVED. That is
the total initial
momentum (p1o + p2o )
must equal the total final
momentum (p1f + p2f).
…these forces do NOT cancel each other out on in
individual object b/c they are acting on different
objects. HOWEVER, when dealing with a system we
are looking at the combination of the objects (as if
they were one object) so F12 + F21 = F12 -F21 = 0
The net force on the SYSTEM
is zero even though individual
forces exist!
 A collision is an interaction between masses in which
there is a transfer of momentum and energy.
 For ALL collisions, (total) momentum is ALWAYS
conserved!
 There are two MAIN categories of collisions (but three
types if you consider a sub-group).
 Elastic Collisions
 Inelastic Collisions
 Perfectly inelastic collisions
Collisions are classified according to changes in the kinetic energy.
1) elastic collisions
 total kinetic energy IS CONSERVED!
This is the defining characteristic.
 the objects will move independently after they collide
(this, however, does not make it elastic)
2) inelastic collisions
 total kinetic energy is NOT conserved! This is the defining characteristic.
 the objects may move independently after they
collide (this, however, does not make it inelastic)
3) Completely (or perfectly) inelastic collisions
 Because it is an inelastic collision, KE is not conserved. What
makes it special is that the objects will stick together and move as
one object after colliding. (We will be able to represent this
mathematically.)
Hang on…what is kinetic energy?
Energy is the capacity to do work.
 Energy is the ability for an object to produce a change in itself or
in the environment.
 Mechanical Energy (ME) – the forms of energy due to an
object’s position, orientation and or motion.
Kinetic Energy (KE) – the energy associated with the
motion of a mass
KE = ½ mv2
Kinetic energy is a scalar quantity (no direction) and has
units of kgm2/s2 = Joule
This is what we are covering in class right now. 
A cue ball (mass mA = 0.400kg) moving with speed vA =
1.80m/s strikes ball B, initially at rest, of mass mB =
0.500kg. As a result of the collision, the cue ball travels
(in the same direction) with a speed of 0.20 m/s.
A) What is the velocity of ball B?
B) Is this collision elastic or inelastic?
Cue Ball (A)
mA = 0.40-kg
VAo = 1.80 m/s
VAf = 0.20 m/s
Eight Ball (B)
mB = 0.50-kg
VBo = 0 m/s
VBf = ?
Momentum is conserved…0
mAvAo + mBvBo = mAvAf + mBvBf
mAvAo - mAvAf = mBvBf
(mAvAo – mAvAf ) / mB = vBf
1.28 m/s= vBf
Cue Ball (A)
mA = 0.40-kg
VAo = 1.80 m/s
VAf = 0.20 m/s
Eight Ball (B)
mB = 0.50-kg
VBo = 0 m/s
VBf = 1.28 m/s
If the collision is elastic then KEo = KEf….check it….
0
½ mAvAo2 + ½ mBvBo2 = ½ mAvAf2 + ½ mBvBf2
½ mAvAo2 = ½ mAvAf2 + ½ mBvBf2
0.648 J = 0.418 J
These are NOT equal….kinetic energy was lost.
Therefore, this collision is inelastic.
A cue ball (mass mA = 0.400kg) moving with speed vA =
1.80m/s strikes ball B, initially at rest, of mass mB =
0.500kg. As a result of the collision, ball B is deflected
off at an angle of 30.0o with a speed vB = 1.10m/s.
What is the velocity of ball A?
30
???
The motion occurs in TWO DIMENSIONS….thus we
must (mathematically) analyze BOTH DIMENSIONS.
Cue Ball (A)
mA = 0.40-kg
VAox = 1.80 m/s
VAoy = 0 m/s
VAfx = ?
VAfy = ?
We need These
to findare
thethe
x and
y
components
of
components
for ball A.
the velocities
only in the xdirection (East-West)
Eight Ball (B)
mB = 0.50-kg
VBox = 0 m/s
VBoy = 0 m/s
VBfx = 1.10cos30
VBfy = 1.10sin30
These are the x and y
These are the components of
components of the
the velocities only in the yresulting velocities.
direction (North-South)
Cue Ball (A)
mA = 0.40-kg
VAox = 1.80 m/s
VAoy = 0 m/s
VAfx = ?
VAfy = ?
Eight Ball (B)
mB = 0.50-kg
VBox = 0 m/s
VBoy = 0 m/s
VBfx = 1.10cos30
VBfy = 1.10sin30
Analyze the motion in the x-direction (East-West)
mAvAox + mBvBox = mAvAfx + mBvBfx
mAvAox – mBvBfx = mAvAfx
(mAvAox –mBvBfx) / mA = vAfx
0.609 m/s= vAfx
Cue Ball (A)
mA = 0.40-kg
VAox = 1.80 m/s
VAoy = 0 m/s
VAfx = ?
VAfy = ?
Eight Ball (B)
mB = 0.50-kg
VBox = 0 m/s
VBoy = 0 m/s
VBfx = 1.10cos30
VBfy = 1.10sin30
Analyze the motion in the y-direction (North-South)
mAvAoy + mBvBoy = mAvAfy + mBvBfy
– mBvBfy = mAvAfy
(–mBvBy) / mA = vAfy
-0.688 m/s= vAfy
We now know that the resulting motion has two dimensions. Now
we just need to combine these motions.
0.608 m/s
a2 + b2 = c2
(0.608)2 + (0.688)2 = c2
c = 0.92m/s = resulting speed
0.688=m/s
Direction
 = tan-1(0.688/0.608)
= 48.5
So, the velocity is 0.92 m/s 48.5 SE
A Ford F-150 (4690 – Lbs) is stopped at a red light when
it is rear-ended by a Honda Civic (2600-Lbs) traveling
north at 20.0 m/s. The Honda becomes wedged
beneath the rear bumper of the Ford, causing the two
vehicles to travel together after the crash.
A) What is the resulting velocity of the wreckage?
B) Verify (mathematically) that the collision is
inelastic.
Honda Civic (A)
mA = 1181.82-kg
VAo = 20 m/s
Ford F150 (B)
mB = 2131.82-kg
VBo = 0 m/s
Vf = ?
mAvAo + mBvBo = mAvAf + mBvBf
The vehicles get STUCK TOGETHER, so they will have the SAME FINAL VELOCITY.
mAvAo = (mA + mB) vBf
7.13 m/s= vf
Work on the following practice problems BEFORE you
continue to the mixed review. All of these problems
deal with the conservation of momentum and
collisions.
Page 214: 1, 3, 5
Page 216: 1, 3
Page 219: 1,3
You can find solutions to these problems on the class webpage.
should (as always) use the solutions to check your work…not as a substitute for
trying it on your own.
You
Work on the following practice problems BEFORE you
continue to the next topic. This is a mixed review of all
of the problems covered in this section.
Page 223 – Theory: 2, 3, 4, 6, 8, 24, 25,
Page 223 – Problems: 11, 12, 14, 22, 23, 28, 30, 40
Worksheet
You can find solutions to these problems on the class
webpage. You should (as always) use the solutions to check your
work…not as a substitute for trying it on your own.
 http://www.youtube.com/watch?v=T9lehHxv-C8
The clip of the “Rube Goldberg Device” featuring the
Honda Accord is from an actual commercial. To view
the commercial in its entirety, click here.
 Don’t think it’s real??? Read this
 Do you like Rube Goldberg Machines??? Me too!
Check this one out!