Transcript Document

Momentum and
Impulse
Section 9.1 Objectives
• Define the momentum of an object.
• Determine the impulse given to an object.
• Calculate the momentum and impulse of
an object.
Let’s start with everyday language
What do you say when a sports
team is on a roll?
They may not have the lead but
they may have ___________
MOMENTUM
A team that has momentum is
hard to stop.
What is Momentum?
An object with a lot of momentum is also
hard to stop
Momentum = p = mv
Units: kg∙m/s
m=mass
v=velocity
Momentum is also a vector (it has direction)
Let’s practice
• A 1200 kg car drives west at 25 m/s for 3
hours. What is the car’s momentum?
• Identify the variables:
– 1200 kg = mass
– 25m/s, west = velocity
– 3 hours = time
P = mv = 1200 x 25 = 30000 kg m/s, west
Equivalent Momenta
Car: m = 1800 kg; v = 80 m /s
p = 1.44 ·105 kg · m /s
Bus: m = 9000 kg; v = 16 m /s
p = 1.44 ·105 kg · m /s
Train: m = 3.6·104 kg; v = 4 m /s
p = 1.44 ·105 kg · m /s
How hard is it to stop a moving
object?
To stop an object, we have to apply a force
over a period of time.
This is called Impulse
Impulse = FΔt
Units: N∙s
F = force (N)
Δt = time elapsed (s)
How hard is it to stop a moving
object?
• Using Newton’s 2nd Law we get
FΔt= mΔv
Which means
Impulse = change in momentum
Impulse - Momentum
Theorem
The impulse due to all forces acting on an object (the net force) is
equal to the change in momentum of the object:
Fnet t =  p
We know the units on both sides of the equation are the same
(last slide), but let’s prove the theorem formally:
Fnet t = m a t = m( v / t)t = m v =  p
Why does an egg break or not
break?
• An egg dropped on a tile floor breaks, but an
egg dropped on a pillow does not. Why?
FΔt= mΔv
In both cases, m and Δv are the same.
If Δt goes up, what happens to F, the force?
Right! Force goes down. When dropped on a
pillow, the egg starts to slow down as soon as it
touches it. A pillow increases the time the egg
takes to stops.
Practice Problem
A 57 gram tennis ball falls on a tile floor. The ball
changes velocity from -1.2 m/s to +1.2 m/s in
0.02 s. What is the average force on the ball?
Identify the variables:
Mass = 57 g = 0.057 kg
Δvelocity = +1.2 – (-1.2) = 2.4 m/s
Time = 0.02 s
using FΔt= mΔv
F x (0.02 s) = (0.057 kg)(2.4 m/s)
F= 6.8 N
Car Crash
Would you rather be in a
head on collision with an
identical car, traveling at
the same speed as you, or
a brick wall?
Assume in both situations you
come to a complete stop in the
same amount of time.
Take a guess
http://techdigestuk.typepad.com/photos/uncategorized/car_crash.JPG
Car Crash (cont.)
Everyone should vote now
Raise one finger if you think
it is better to hit another
car, two if it’s better to hit a
wall and three if it doesn’t
matter.
And the answer is…..
Car Crash (cont.)
The answer is…
It Does Not Matter!
Look at FΔt= mΔv
In both situations, Δt, m, and Δv
are the same! The time it
takes you to stop depends on
your car, m is the mass of
your car, and Δv depends on
how fast you were initially
traveling.
Egg Drop connection
• How are you going to use this in your egg
drop?
Which of these variables can you control?
FΔt= mΔv
Which of them do you want to maximize,
which do you want to minimize
(note: we are looking at the force on the
egg. Therefore, m represents the egg
mass, not the entire mass of the project)
Section 9.2 Objectives
• Relate Newton’s third law to conservation
of momentum in collisions.
• Identify conditions that momentum is
conserved.
• Solve conservation of momentum
problems.
Law of Conservation of Momentum
•pi = pf
• The initial momentum = the final
momentum
• pi,A + pi,B = pf,A + pf.B
• The sum of initial momentums = the sum
of final momentums
Conditions for Conservation of
Momentum
• Closed system
• Objects don’t enter or leave
• Isolated system
• No external forces act on the system
Conservation of Momentum
Conservation of Momentum in 1-D
Whenever two objects collide (or when they exert forces on each
other without colliding, such as gravity) momentum of the system
(both objects together) is conserved. This mean the total momentum
of the objects is the same before and after the collision.
(Choosing right as the +
before: p = m1 v1 - m2 v2
v2
v1
m1
direction, m2 has - momentum.)
m2
m1 v1 - m2 v2 = - m1 va + m2 vb
after: p = - m1 va + m2 vb
va
m1
m2
vb
Directions after a collision
On the last slide the boxes were drawn going in the opposite direction
after colliding. This isn’t always the case. For example, when a bat hits
a ball, the ball changes direction, but the bat doesn’t. It doesn’t really
matter, though, which way we draw the velocity vectors in “after”
picture. If we solved the conservation of momentum equation (red box)
for vb and got a negative answer, it would mean that m2 was still moving
to the left after the collision. As long as we interpret our answers
correctly, it matters not how the velocity vectors are drawn.
v2
v1
m1
m2
m1 v1 - m2 v2 = - m1 va + m2 vb
va
m1
m2
vb
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 doesn’t matter)
35 g
v=?
4 cm/s
7 kg
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
= 24.5 kg · m /s
v=0
35 g
4 cm/s
v=?
p before = p after
7 kg
700 m/s
p after = 7 (0.04) + 0.035 v
= 0.28 + 0.035 v
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
p before = 7 (0) + (0.035) (700)
= 24.5 kg· m /s
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?
(0.035kg)+(7kg) = 7.035kg
v
7. 035 kg
p before = p after
24.5kg.m/s = (7.035kg) 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.
Sample Problem 3
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
3 kg
10 m/s
6 m/s
15 kg
after
4.5 m/s
3 kg
15 kg
v
Section 9.2 Objectives
• Explain how the law of conservation of
momentum can be applied to collisions in
two dimensions
• Solve two dimensional collision problems
using vector addition
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:
m1
v1
m2
2 v
2
1
a
m1
va
m2
vb
b
before:
px = m1 v1 cos1 - m2 v2 cos2
py = m1 v1 sin1 + m2 v2 sin2
after:
px = -m1 va cosa + m2 vb cos b
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
E m1
1
F
O
p1
R
E
A
F
T
E
R
a
m1
pa
2
p1
m2
p before
p2
p2
m2
pa
b
pb
p after
pb
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,
these vectors would add up
to the original momentum
vector.
2-D Sample Problem
152 g
before
40
34 m/s
0.3 kg
5 m/s
A mean, old dart strikes an innocent
mango that was just passing by
minding its own business. Which
way and how fast do they move off
together?
Working in grams and taking left & down as + :
152 (34) sin 40 = 452v sin
152 (34) cos 40 - 300 (5) = 452 v cos
after
452 g

v
Dividing equations : 1.35097 = tan
 = 53.4908
Substituting into either of the first two
equations :
v = 9.14 m/s
From a State Standards Test
Copyright © 2004 California Department of Education.
From a State Standards Test
Copyright © 2004 California Department of Education.
From the State Standards Test
Copyright © 2004 California Department of Education.