Transcript f - Images

Physics Chapter 6
Momentum and Its
Conservation
Linear Momentum
The
velocity and mass of an
object determine what is needed
to change its motion.
Linear Momentum (ρ) is the
product of mass and velocity
ρ =mv
Unit is
kgm/s
Example 1
Example 1. Find the magnitude of the momentum of
each cart shown below, before the collision.
Example 1. Find the magnitude of the momentum of
each cart shown below, before the collision.
Given:
mA = 450 kg mB = 550 kg
nA = 4.50 m/s nB = 3.70 m/s
Find: ρ A and ρ B
Solution:
ρ A = mAnA = (450 kg)(4.50 m/s) = 2025 = 2.0 x 103 kg . m/s
ρ B = mBnB = (550 kg)(3.70 m/s) = 2035 = 2.0 x 103 kg . m/s
Example 2. The man in this picture has a mass of
85.0 kg. If he strikes the ground at 7.7 m/s, what is
his change in momentum (Dρ) Assume his mass
remains the same after the collision.
Given: m = 85.0 kg nf = 0.0 m/s
ni = -7.7 m/s (negative for down)
Find: D ρ
Solution: D ρ = m(nf - ni)
= (85.0 kg)((0.0 m/s – (-7.7 m/s)) = 650 kg . m/s
If
the velocity is changed by
an outside force, then the
momentum is also changed.
F=ma = mDv/Dt = (D ρ /Dt)
FDt = mD v = mvf-mvi= D ρ
FDt = D ρ = “impulse” in N

Dt
s
is the time during which the
force is applied.
This
is called the “ImpulseMomentum Theorem”.
Example 3. A player exerts a force
(assume constant) of 12.5 N on a ball
over a period of 0.35 s. Find the total
impulse .
Given: F = 12.5 N
Dt = 0.35 s
Find: Impulse
Solution: Impulse = FDt = (12.5 N)(0.35 s) = 4.4 N . s
Example 4:
Water leaves a hose at a rate of 1.5
kg/s with a speed of 20 m/s and is
aimed at the side of a car, which
stops it without splashing it back
(kind of a fake problem, but that’s
OK). What is the force exerted by
the water on the car each second.
m = 1.5 kg
vi=20m/s vf=0m/s Dt=1s
FC means
force on
FC Dt =D ρ
water by
FC=D ρ /Dt
car (Fw is
FC = (ρ f – ρi) / Dt
force on car
by water
FC = (mvf – mvi) / Dt
which is
equal and
FC = m (vf – vi) / Dt
opposite)
FC = 1.5 kg (0 m/s – 20 m/s) / 1 s
FCWC,x
= -30
N
30=N30 N
= -30
NF
WF=
CW,x
Stopping Distance
Stopping
distance depends on
the Impulse-Momentum
Theorem.
A
heavy object takes longer to
stop…
Also,
A change in momentum
over a longer time requires less
force.
Ex:
nets, air mattresses,
padding (see fig. 6-5)
Calculating time and
distance
Stopping
time can be
calculated using FDt = D p, and
solving for Dt.
Stopping distance can be
calculated using Dx = ½(vi+vf)Dt
Example
A
720.0 kg beryllium sphere
traveling to the west slows down
uniformly from 230.0 m/s to
12.00 m/s. How long does it
take the sphere to decelerate if
the force on the sphere is 840 N
to the east? How far does the
sphere travel during the
deceleration?
First find change in time
m=720.0kg
vi=-230.0 m/s vf=-12.00m/s
F= +840N
FDt=Dp
Dt=Dp/F=(mvf-mvi)/F
=[(720kgX-12.00m/s)(720.0kgX-230.0m/s)]/840N=186.9s
Then find change in x
Dx=1/2
(vi+vf)Dt=
1/2(-230.0m/s-12.00m/s)(186.9s)
=-22614.9m
= 22610 m to the West
Newton’s
rd
3
Law
As
two objects collide, the
force that one exerts on the
other multiplied by the time
(FDt) is the impulse for each.
Newton’s 3rd law tells us that
the force on mass1 is equal and
opposite to the force on mass2
when they collide, which leads
to conservation of momentum.
Forces
in real collisions are not
constant (see fig. 6-9, pg. 220)
Conservation of
Momentum in Collisions
When two objects collide, the
momentum of the individual
objects change, but the total
momentum remains the same.
p1,i + p2,i = p1,f + p2,f
m1v1,i + m2v2,i = m1v1,f +
Friction is disregarded
m2v2,f
Example 1
A
76 kg person, initially at
rest in a stationary 45 kg boat,
steps out of the boat and onto
the dock. If the person moves
out of the boat with a velocity
of 2.5 m/s to the right, what
is the final velocity of the
boat?
List Knowns and Unknowns
and formula…
vi,boat=0
vi,person=0
vf,boat=?
vf,person=2.5m/s
mperson=76kg
mboat=45kg
mpvp,i + mbvb,i = mpvp,f + mbvb,f
0 + 0 = (76kg)(2.5m/s) + (45kg)(vb,f)
Vbf= - (76X2.5)/45 = -4.2m/s
(negative means the boat is moving in
the opposite direction)
Example 2. A proton with an initial velocity of
235 m/s strikes a stationary alpha
particle(He) and rebounds off with a velocity
of –188 m/s. Find the final velocity of the
alpha particle.
Mass of proton= 1.67 x 10 -27 kg
Mass of alpha particle (He)= 6.64 x 10 -27 kg
-27 kg
-27 kg
m
=
1.67
x
10
m
=
6.64
x
10
Given: p
He
np,i = 235 m/s nHe,i = 0.0 m/s np,f = -188 m/s
Find: nHe,f
Original Formula:
mp ν p,i  mHe ν He,i  mp ν p,f  mHe ν He,f
Now, the initial velocity of He (alpha particle) is zero,
so cross it out!
-27 kg
-27 kg
m
=
1.67
x
10
m
=
6.64
x
10
Given: p
He
np,i = 235 m/s nHe,i = 0.0 m/s np,f = -188 m/s
Find: nHe,f
Original Formula:
mp ν p,i  mp ν p,f  mHe ν He,f
Solve for n
He,f
Working Formula:
ν He,f 
m p ν p,i - m p νp, f
m He
-27 kg
-27 kg
m
=
1.67
x
10
m
=
6.64
x
10
Given: p
He
np = 235 m/s nHe = 0.0 m/s np,f = -188 m/s
Find: nHe,f
Working Formula: The mass of the proton doesn’t change, so
ν He,f 
m p (ν p,i - ν p,f )
m He
ν He,f 
1.67x10
 27
 106m / s
kg (235 m/s - (-188 m/s))
- 27
6.64 x 10 kg
Perfectly Inelastic
Collisions
A
collision is perfectly
inelastic when they collide
and move together as one
mass with a common
velocity.
m1v1,i
+ m2v2,i = (m1 + m2)vf
Example 1. A moving railroad
car, mass=M, speed=3m/s,
collides with an identical car at
rest. The cars lock together as
a result of the collision. What is
their common speed
afterward?
Vi1
Vi2=0
M
M
before
Vf?
M
after
M
x
Vi1
Vi2=0
M
M
x
Vf?
M M
before
after
p1i = p2f
M(+Vi1) + 0= (M+M)(Vf)
Vf = MVi1 / 2M = Vi1 / 2
Vf = (3m/s) / 2 = 1.5m/s
Kinetic Energy and Perfectly
Inelastic Collisions
During
perfectly inelastic collisions,
some kinetic energy is lost because the
objects are deformed.
Once mass and velocities are known
(using conservation of momentum),
solve for initial and final kinetic
energies (1/2mv2) using the formula
from Ch. 5 to determine energy lost.
KEi = KE1,i + KE2,i
KEf = KE1,f + KE2,f
DKE = KEf - KEi
Example 1. A red clay ball with a mass of 1.5 kg and an
initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black
clay ball at rest. Find the final velocity of the combined
mass.
Given: mr = 1.5 kg
Find: nf
mb = 2.8 kg
nr,i = 4.5 m/s
nb,i = 0 m/s
Original Formula:
m r ν r,i  m b ν b,i  (m r  m b ) νf
Example 1. A red clay ball with a mass of 1.5 kg and an
initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black
clay ball at rest. Find the final velocity of the combined
mass.
Given: mr = 1.5 kg
Find: nf
mb = 2.8 kg
nr,i = 4.5 m/s
nb,i = 0 m/s
Original Formula: The black ball starts at rest, so
m r ν r,i  m b ν b,i  (m r  m b ) νf
Example 1. A red clay ball with a mass of 1.5 kg and an
initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black
clay ball at rest. Find the final velocity of the combined
mass.
Given: mr = 1.5 kg
Find: nf
mb = 2.8 kg
nr,i = 4.5 m/s
Original Formula: The black ball starts at rest, so
m r ν r,i  (m r  m b ) νf
nb,i = 0 m/s
Example 1. A red clay ball with a mass of 1.5 kg and an
initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black
clay ball at rest. Find the final velocity of the combined
mass.
Given: mr = 1.5 kg
Find: nf
mb = 2.8 kg
nr = 4.5 m/s
Working Formula:
mr νr
νf 
(m r  m b )
nb = 0 m/s
Example 1. A red clay ball with a mass of 1.5 kg and an
initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black
clay ball at rest. Find the final velocity of the combined
mass.
Given: mr = 1.5 kg
Find: nf
mb = 2.8 kg
nr,i = 4.5 m/s
nb,i = 0 m/s
Solution:
(1.5 k g)(4.5 m/s)
νf 

 1.569  1.6 m/s
(m r  m b ) (1.5 k g  2.8 k g)
m r ν r,i
Recall the formula for kinetic energy.
KE 
1
2
mv
2
Was the kinetic energy conserved in
example 1?
KEred 
1
mr vr  2 (1.5 kg)(4.5 m/s)  15.1875 J
2
2
2
1
KEblack  2 mr vr  12 (2.8 kg)(0.0 m/s) 2  0 J
1
2
KEbefore = 15 J
KEafter  2 mr b vf  2 (4.3 kg)(1.6 m/s)  5.3 J
1
2
1
2
KEbefore = 15 J
KEafter = 5.3 J
DKE = 5.3 J-15 J = -9.7 J
Where does the missing kinetic energy go?
Some energy is converted to internal energy,
some is converted into sound.
Two
Elastic Collisions
objects collide and return
to their original shape with no
loss of Kinetic Energy.
The two objects move
separately after the collision.
Total momentum and KE remain
constant. See page 230.
Most
collisions lose energy as
sound, internal elastic potential
energy and friction…we assume
perfect collisions.
Formula for elastic
collisions
m1v1,i
½
+ m2v2,i = m1v1,f + m2v2,f
m1v1,i2 + ½ m2v2,i2 = ½ m1v1,f2 + ½ m2v2,f2
Example 1. A 0.025 kg blue marble sliding to the
right on a frictionless surface with a velocity of 17 m/s
makes an elastic head-on collision with 0.025 kg red
marble moving to the left with a velocity of 15 m/s.
After the collision, the
blue marble moves to the left
with a velocity of 15.0 m/s. Find
the velocity of the red marble.
Example 1.
Given: mb = 0.025 kg
nr,i, = -15 m/s
mr = 0.025 kg
nb,f = -15 m/s
nb,i = 17 m/s
Find: nr,f
Original Formula:
m b ν b,i  m r ν r,i  m b ν b,f  m r ν r,f
ν r,f 
m b ν b,i  m r ν r,i - m b ν b,f
mr
ν r,f 
m b ν b,i  m r ν r,i - m b ν b,f
mr
(0.025 kg)(17 m/s)  (0.025 kg)(-15 m/s) - (0.025 kg)(-15 m/s)

0.025 kg
=17m/s
Now…you
can plug the numbers into
the conservation of kinetic energy
formula to check your work which they
will ask you to do!
(Remember – there is NO LOSS OF
KINETIC ENERGY with elastic
collisions!!)
½
m1v1,i2 + ½ m2v2,i2 = ½ m1v1,f2 + ½ m2v2,f2
½ m1v1,i2 + ½ m2v2,i2 = ½ m1v1,f2 + ½ m2v2,f2
½ (.025)(17)2+ ½ (.025)(-15)2 = ½ (.025)(-15)2+ ½ (.025)(17)2
6.425J = 6.425J