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The Conservation of Mechanical Energy
Example: A Daredevil Motorcyclist
A motorcyclist is trying to leap across the canyon by driving
horizontally off a cliff 38.0 m/s. Ignoring air resistance, find
the speed with which the cycle strikes the ground on the other
side.
The Conservation of Mechanical Energy
Ef = Eo
2
f
2
o
mgh f + mv = mgho + mv
1
2
1
2
gh f + 12 v 2f = gho + 12 vo2
The Conservation of Mechanical Energy
gh f + 12 v 2f = gho + 12 vo2
v f = 2g ( ho − h f ) + vo2
v f = 2 ( 9.8m s
2
) (35.0m) + (38.0 m s)
2
= 46.2 m s
The Conservation of Mechanical Energy
Conceptual Example: The Favorite Swimming Hole
The person starts from rest, with the rope
held in the horizontal position,
swings downward, and then lets
go of the rope. Three forces
act on him: his weight, the
tension in the rope, and the
force of air resistance.
Can the principle of
conservation of energy
be used to calculate his
final speed?
The Conservation of Mechanical Energy
Example: Vertical spring with a mass
A 0.20-kg ball is attached to a vertical spring. The spring constant
is 28 N/m. When released from rest, how far does the ball fall
before being brought to a momentary stop by the spring?
The Conservation of Mechanical Energy
E f = E0
1
2
mv 2f + mgh f + 12 k(Δy f )2 = 12 mv02 + mgh0 + 12 k(Δy0 )2
v0 = v f = 0
Δy0 = 0 , since spring initially unstretched
Δy f = h0
hf = 0
∴ 12 kh02 = mgh0
2mg
h0 =
k
2 ( 0.20 kg) ( 9.8m s2 )
=
= 0.14 m
28N m
Nonconservative Forces and the Work-Energy Theorem
Example of a nonconservative force problem: Fireworks
A 0.20 kg rocket in a fireworks display is launched from rest
and follows an erratic flight path to reach the point P, as in the
figure. P is 29 m above the starting point. In the process, 425 J
of work is done on the rocket by the nonconservative force
generated by the burning propellant. Ignoring air
resistance and the mass loss due to the burning
propellant, find the speed vf of the rocket at point P.
THE WORK-ENERGY THEOREM
Wnc = E f − E o
Wnc = ( mgh f + 12 mv 2f ) −
(mgh
o
2
o
+ mv
1
2
)
Nonconservative Forces and the Work-Energy Theorem
2
f
2
o
Wnc = mgh f − mgho + mv − mv
1
2
Wnc = mg ( h f − ho ) + mv
1
2
1
2
2
f
425 J = ( 0.20 kg) ( 9.80 m s2 ) ( 29.0 m )
+ 12 ( 0.20 kg) v 2f
v f = 61 m s
Power
DEFINITION OF AVERAGE POWER
Average power is the rate at which work is done, and it
is obtained by dividing the work by the time required to
perform the work.
Work W
P=
=
Time
t
joule s = watt (W)
Power
From the work-energy theorem, doing work results in a
change in energy for the system, so alternatively:
Change in energy
P=
Time
1 horsepower = 550 foot ⋅ pounds second = 745.7 watts
Another way to express the power generated by a force
acting on an object moving at some average speed,
P = Fv
Chapter 7
Linear Momentum
The Impulse-Momentum Theorem
There are many situations when the
force on an object is not constant.
The Impulse-Momentum Theorem
DEFINITION OF IMPULSE
The impulse of a force is the product of the average
force and the time interval during which the force acts:
 
J = F Δt
Impulse is a vector quantity and has the same direction
as the average force.
newton ⋅ seconds (N ⋅ s)
The Impulse-Momentum Theorem
 
J = F Δt
The Impulse-Momentum Theorem
DEFINITION OF LINEAR MOMENTUM
The linear momentum of an object is the product
of the object’s mass times its velocity:


p = mv
Linear momentum is a vector quantity and has the same
direction as the velocity.
kilogram ⋅ meter/second (kg ⋅ m/s)
The Impulse-Momentum Theorem
 
 vf − vo
a=
Δt


∑ F = ma
 mv f − mv o
∑ F = Δt
!
!
! ! !
∑ F Δt = mv f − mv o = pf − p0
( )
The Impulse-Momentum Theorem
IMPULSE-MOMENTUM THEOREM
When a net force acts on an object, the impulse of
this force is equal to the change in the momentum
of the object
impulse
!
J=
!
! !
∑ F Δt = pf − p0
( )
final momentum
initial momentum
The Impulse-Momentum Theorem
!
! !
!
∑ F Δt = pf − p0 = Δp
( )
à Divide both sides of the Impulse-Momentum Theorem by Δt
! Δp!
∴∑ F =
Δt
ç  More general version of Newton’s 2nd Law
allows for mass changing with time (e.g. rocket)
!
!
!
&
Δ
m
v
!)
Δp
Δv
(
)
=
=m
= ma +
( for mass constant in time →
Δt
Δt
Δt
(
+
!
!
( and then recover our familiar version
+
F
=
m
a
∑
'
*
The Impulse-Momentum Theorem
Example: Hitting a pitched baseball. A baseball of mass
0.14 kg is pitched at a batter with an initial velocity of
-38 m/s (negative is towards the bat). The bat applies an
average force that is much greater than the weight of the
ball, and the ball departs from the bat with a final velocity
of +58 m/s. Assuming that the time of contact with the bat
is 1.6 x 10-3 s, find the average force exerted on the ball
by the bat.



J = F Δt = mv f − mv o
= (0.14)(58) - (0.14)(-38) = +13.4 kg m/s

 J
13.4
F= =
= +8400 N
−3
Δt 1.6 ×10
The Impulse-Momentum Theorem
Example: A Rain Storm
Rain comes down with a velocity of -15 m/s and hits the
roof of a car. The mass of rain per second that strikes
the roof of the car is 0.060 kg/s. Assuming that rain comes
to rest upon striking the car, find the average force
exerted by the rain on the roof.



∑ F Δt = mv f − mv o
( )
The Impulse-Momentum Theorem
Neglecting the weight of
the raindrops, the net force
on a raindrop is simply the
force on the raindrop due to
the roof.



F Δt = mv f − mv o

& m #
F = −$ ! v o
% Δt "

F = −(0.060 kg s )(− 15 m s ) = +0.90 N
The Impulse-Momentum Theorem
Conceptual Example: Hailstones Versus Raindrops
Instead of rain, suppose hail is falling. Unlike rain, hail usually
bounces off the roof of the car.
If hail fell instead of rain, would the force be smaller than, equal
to, or greater than that calculated in the previous Example?
The Principle of Conservation of Linear Momentum
WORK-ENERGY THEOREM óCONSERVATION OF ENERGY
IMPULSE-MOMENTUM THEOREM ó???
Apply the impulse-momentum theorem to the midair collision
between two objects…..
The Principle of Conservation of Linear Momentum
The midair collision between two
objects.
Internal forces – Forces that objects within
the system exert on each other.
External forces – Forces exerted on objects
by agents external to the system.
The Principle of Conservation of Linear Momentum



∑ F Δt = mv f − mv o
( )
OBJECT 1
(
 


W1 + F12 Δt = m1v f1 − m1v o1
)
External
forces
(gravity)
(
Internal
forces
OBJECT 2




W2 + F21 Δt = m2 v f 2 − m2 v o2
)
The Principle of Conservation of Linear Momentum
(
 


W1 + F12 Δt = m1v f1 − m1v o1
)
+
(
(




W2 + F21 Δt = m2 v f 2 − m2 v o2
)








W1 + W2 + F12 + F21 Δt = ( m1v f1 + m2 v f 2 ) − ( m1v o1 + m2 v o2 )
)


F12 = − F21
Internal forces cancel
from Newton’s 3rd law

Pf
Total final
momentum

Po
Total initial
momentum
The Principle of Conservation of Linear Momentum
The internal forces cancel out.
(


 
W1 + W2 Δt = Pf − Po
)
More generally,
 
(sum of average external forces) Δt = Pf − Po
(

∑F
EXT
)
 
Δt = Pf − P0
The Principle of Conservation of Linear Momentum
(

∑F
EXT
)
 
Δt = Pf − P0
If the sum of the external forces is zero, then
 
0 = Pf − Po


Pf = Po
PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM
The total linear momentum of an isolated system is constant
(conserved). An isolated system is one for which the sum of
the average external forces acting on the system is zero.
The Principle of Conservation of Linear Momentum
Conceptual Example: Is the Total Momentum Conserved?
Imagine two balls colliding on a billiard
table that is friction-free. Use the momentum
conservation principle in answering the
following questions. (a) Is the total momentum
of the two-ball system the same before
and after the collision? (b) Answer
part (a) for a system that contains only
one of the two colliding
balls.
The Principle of Conservation of Linear Momentum
PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM
The total linear momentum of an isolated system is constant
(conserved). An isolated system is one for which the sum of
the average external forces acting on the system is zero.
In the top picture the net external force on the
system is zero.
In the bottom picture the net external force on the
system is not zero.
The Principle of Conservation of Linear Momentum
Applying the Principle of Conservation of Linear Momentum
1. Decide which objects are included in the system.
2. Relative to the system, identify the internal and external forces.
3. Verify that the system is isolated.
4. Set the final momentum of the system equal to its initial momentum.
Remember that momentum is a vector.
The Principle of Conservation of Linear Momentum
Example: Ice Skaters
Starting from rest, two skaters
push off against each other on
ice where friction is negligible.
One is a 54 kg woman and
one is a 88 kg man. The woman
moves away with a velocity of
+2.5 m/s. Find the recoil velocity
of the man.
The Principle of Conservation of Linear Momentum


Pf = Po
m1v f 1 + m2 v f 2 = 0
vf 2 = −
vf 2
m1v f 1
m2
(
54 kg )(+ 2.5 m s )
=−
= −1.5 m s
88 kg