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CHAPTER 3 WORK AND ENERGY
3.1 Work
3.2 Energy
3.3 Conservative and
nonconservative forces
3.4 Power, energy and
momentum conservation
3.5 Linear momentum
3.6 Collisions
3.7 Motion in a gravitational potential
Part 5
Linear momentum
Linear Momentum & Impulse
Definition: For a single particle, the momentum p is defined as:
p ≡ mv
(p is a vector since v is a vector)
So px = mvx and so on (y and z directions)
• Newton’s 2nd Law:

dv d 
 m  (mv)
dt dt
F = ma
 dp
F
dt
• This is the most general statement of
Newton’s 2nd Law
Review: which has more momentum?
• A bullet, mass = 100 grams, speed =
1000m/s
100 grams = 0.1kg
p=mv = 0.1 kg x 1000m/s =
100 kg.m/s
• A car, mass=1000kg, speed = 10m/s
p=mv = 1000 kg x 10m/s
= 10,000 kg.m/s
Momentum Conservation
FEXT 
dP
dt
dP
0
dt
FEXT  0
• Momentum conservation - important
principle.
• A vector expression (Px, Py and Pz) .
dPx
Fx 
dt
dPx
0
dt
Fx  0
Momentum Conservation
• Many problems can be addressed
through momentum conservation
even if other physical quantities (e.g.
mechanical energy) are not
conserved
Learning check
• Consider the bullet & block as a system. After the
bullet is shot, there are no external forces acting on
the system in the x-direction.
Momentum is conserved in the x direction!
–Px, i = Px, f
v
–mv = (M+m)V
initial
M  m

v 
V
 m 
x
V
final
Use momentum conservation to compute relation
Between vi and Vf ???
Force and Impulse in Boxers:
Impulse:
(A variable force applied for a given time)

J reflects momentum transfer



dp

J   F dt   (
)dt   dp
dt
t
t
p
0
0
0
Impulse J = area under
this curve !
(Transfer of momentum !)
F
Impulse has units of
Newton-seconds
t
t
ti
tf
Force and Impulse

Two different collisions can have the same impulse since J depends
only on the momentum transfer, NOT the nature of the collision.
F
same area
F
t
t big, F small
t
t
t small, F big
t
Average Force and Impulse
F
Fav
F
Fav
t
t big, Fav small
t
t
t small, Fav big
t
Check: Force & Impulse
• Two boxes, one heavier than the other, are initially at rest on a
horizontal frictionless surface. The same constant force F acts on
each one for exactly 1 second.
• Which box has the most momentum after the force acts ?
• How about Impulse ??
F
light
A. heavier
B. lighter
C.same
D.can’t tell
F
heavy
A. heavier
B. lighter
C. same
D. can’t tell
Check:Momentum and Impulse
• A stunt man jumps from the roof of a tall building, but no injury
occurs because the person lands on a large, air-filled bag. Which
one of the following statements best describes why no injury
occurs?
(a) The bag provides the necessary force to stop the person.
(b) The bag reduces the impulse to the person.
(c) The bag reduces the change in momentum.
(d) The bag decreases the amount of time during which the
momentum is changing and reduces the average force on the
person.
(e) The bag increases the amount of time during which the
momentum is changing and reduces the average force on the
person.
Collisions
Elastic vs. Inelastic Collisions
• An elastic collision: kinetic energy as well as momentum is
conserved before and after the collision.
•
Kbefore = Kafter
– Carts colliding with a spring in between, billiard balls, etc.
vi
•
An inelastic collision: kinetic energy is not conserved before
and after the collision, but momentum is conserved. Kbefore 
Kafter
– Car crashes, collisions where objects stick together, etc.
Exercise: Inelastic collision in 1-D
• A block of mass M is initially at rest on a frictionless horizontal
surface. A bullet of mass m is fired at the block with a muzzle
velocity (speed) v. The bullet lodges in the block, and the block
ends up with a speed V. In terms of m, M, and V :
• What is the final energy of the system?
– Is kinetic energy conserved?
x
v
V
initial
final
M  m
v  
V
 m 
Solution
• Now consider the kinetic energy of the system before and after:
Before:
1 2 1 Mm 2 1Mm
2


E B  mv  m
V

M

m
V



2
2  m 
2 m 
After:
1
E A  M  m V 2
2
So
M
mM
EB  
 E A  (1  )E A  E A
m
 m 
2
•
•
Kinetic energy is NOT conserved! (friction stopped the bullet)
However, momentum was conserved, and this was useful.
Test: Inelastic Collision in 1-D
M
m
V
v=?
–What is the initial speed of the M car ?
–What is the initial energy of the system?
–What is the final energy of the system?
–Is kinetic energy conserved?
v=0
M+m
ice
(no friction)
Check: Momentum Conservation
• Two balls of equal mass are thrown
horizontally with the same initial velocity.
They hit identical stationary boxes resting on
a frictionless horizontal surface.
• The ball hitting box 1 bounces back, while the
ball hitting box 2 gets stuck.
– Which box ends up moving faster?
(a) Box 1
(b) Box 2
(c) same
x
V1
1
2
V2
Explanation
• Since the total external force in the x-direction is zero,
momentum is conserved along the x-axis.
• In both cases the initial momentum is the same (mv of ball).
• In case 1 the ball has negative momentum after the collision,
hence the box must have more positive momentum if the
total is to be conserved.
• The speed of the box in case 1 is biggest!
x
V1
1
2
V2
Prove
mvinit = MV1 - mvfin
mvinit = (M+m)V2
V1 = (mvinit + mvfin) / M
V2 = mvinit / (M+m)
V1 numerator is bigger and its denominator is smaller
than that of V2.
V1 > V2
x
V1
1
2
V2
Test: Inelastic collision in 2-D
• Consider a collision in 2-D (cars crashing at a
slippery intersection...no friction).
• Finding angle after collision ?
V
v1
m1 + m2
m1
m2
v2
before
after
Explanation
• There are no net external forces acting.
– Use momentum conservation for both
components.
X:
y:
Px ,i  Px ,f
v1
m1v1  m1  m 2 Vx
Vx 
m1
v1
m1  mV2 
Py ,i  Py ,f
m 2v 2  m1  m 2 V y
Vy 
m2
v2
m1  m 2 
m1 + m2
m1
m2
v2
before
after
Solution
• So we know all about the motion after
m1
the collision!
V 
v
x
V = (Vx,Vy)
m1  m 2 
1
Vy
m2
Vy  
v2
m1  m 2 
VV
x
m2v 2 p2
tan  

Vx m1v1 p1
y
Inelastic collision in 2-D...
• We can see the same thing using
vectors:
P
P
p2

p1
p2
p2
tan  
p1
p1
Explosion (inelastic un-collision)
fine the relation bewteen V1 and V2
Before the explosion:
M
After the explosion:
v1
v2
m1
m2
Explanation for Explosion...
• No external forces, so P is
conserved.
M
Initially: P = 0
Finally: P = m1v1 + m2v2 = 0
m1v1 = - m2v2
v1
v2
m1
m2
Learning check
VR= ??
V2= ?? V1
Motion in a gravitational potential

Work dWg done on an object by gravity in a displacement dr is
given by:
.
.
dWg = Fg dr = (-GMm / R2)r (dRr + Rd.)
dWg = (-GMm / R2) dR
.
.
(since r  = 0, r r = 1)
dR
Rd
Fg
d
R
M
dr
m
Gravitational potential

Integrate dWg to find the total work done by gravity in a “big”
displacement:

R2

R2
Wg = dWg = (-GMm / R2) dR = GMm (1/R2 - 1/R1)
R1
R1
Fg(R2)
R2
Fg(R1)
R1
M
m
Gravitational potential

Work done depends only on R1 and R2, not on the path taken.
 1
1 

Wg  GMm

R

 2 R1 
m
R2
R1
M
Gravitational potential

Suppose R1 = RE and R2 = RE + y
 R2  R1 
 RE  y   RE  
 GM 




Wg  GMm
 GMm
 m 2 y
 RR 
 R  y R  
R 
 1 2 


 E 
E
E
 GM 
but we have learned that  2   g
 RE 
So:
m
Wg = -mgy
R2
m
RE+ y
RER1
M
M
Potential Energy
• For any conservative force F we can define a
potential energy function U in the following way:
W =
 F.dr = -U
– The work done by a conservative force is equal
and opposite to the change in the potential
U2
r2
energy function.
• This can be written as:
F.dr
r2
U = U2 - U1 = -W = -
r1
r1
U1
Test : Gravitational Potential Energy
• We have seen that the work done by gravity near
the Earth’s surface when an object of mass m is
lifted a distance y is
Wg = -mg y
• Compute the change in potential energy of this object
from work done??
• Give potential energy at initial point ??
j
•
U = U1 - U2 =-W
m g = mg y
y
Wg = -mg y
Solution
• the change in U near the Earth’s surface is:
U = -Wg = mg y = mg(y2 -y1).
So U = mg y + U0 where U0 is an arbitrary constant.
Having an arbitrary constant U0 is equivalent to saying
that we can choose the y location where U = 0 to be
anywhere we want to.
j
m
y2
y1
Wg = -mg y
REVIEW
Definitions of impulse and momentum
Impulse imparted to object 1 by object 2:
I12 = F12 t
Momentum of an object:
p = mv
Now: return your home work
Home work
Thesis
1. The applications of conservation
laws
2. Human energy and their applications
3. Power and energy transfer in Biology
4. The shooting technique
5. Particle collisions