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Transcript
PHYS 172: Modern Mechanics
Lecture 3 - The Momentum Principle
Summer 2012
Read 2.1-2.6
The Momentum Principle
An object moves in a straight line and at constant speed
except to the extent that it interacts with other objects
The Momentum Principle
Dp = Fnet Dt
Change of momentum is equal to the net force
acting on an object times the duration of the
interaction
F units: N  m  kg/s2
• Fnet constant during t!
What is “force” F? • measure of interaction.
• defined by the momentum principle.
The principle of superposition
Fpush
Dp = Fnet Dt
Fearth
Net force
The Superposition Principle:
The net force on an object is the vector sum of all the
individual forces exerted on it by all other objects
Each individual interaction is unaffected by the
presence of other interacting objects
Definition of net force:
Fnet = F1 + F2 + ...
Misconception: need constant force to maintain motion
Fgravity
Neglected
friction!
Impulse
The Momentum Principle
Definition of impulse
Dp = Fnet Dt
Impulse º Fnet Dt
Note: small t
Fnet ~ const
Momentum principle:
The change of the momentum of an object is equal to the net impulse applied to it
Measuring force
Fspring = kS s
L0
L
1. Use knowledge of specific
Interaction: Hooke’s Law,
Gravitational Law, etc.
Fgrav on 2by1  G
m2 m1
r21
2
rˆ2 1
The Momentum Principle
2. Use the momentum principle
Dp = Fnet Dt
Predictions using the Momentum Principle
The Momentum Principle
Dp = Fnet Dt
p f - pi = Fnet Dt
p f = pi + Fnet Dt
p fx , p fy , p fz  pix , piy , piz  Fnet , x , Fnet , y , Fnet , z t
For components:
p fx  pix  Fnet , x t
p fy  piy  Fnet , y t
p fz  piz  Fnet , z t
Short enough,
F~const
Example
Dp = Fnet Dt
p f = pi + Fnet Dt
Force: provided by a spring stretched by L=4 cm
Find momentum pf if pi=<0,0,0> kg.m/s 1 second later
1. Force:
Fspring = kS DL
Fspring = 500(N/m)0.04(m)=20 N
x
NB: force must not
change during t
Fspring = 20,0,0 N
2. Momentum:
p f = pi + Fnet Dt
( )
p f =< 0,0,0 > kg × m/s + 20,0,0 N × 1 s
p f =< 20,0,0 > kg × m/s
N.s = kg.m/s2.s
= kg.m/s
System and surroundings
p
system
System: an object for which we calculate some property (force, momentum, etc)
a system can consist of several objects
Surroundings: objects which interact with system (earth, man, air…)
p f = pi + Fnet Dt
Only external forces matter !
Internal forces cancel
Applying the Momentum Principle to a system:
predicting motion
1. Choose system and surroundings
2. Make a list of objects in surroundings that exert significant forces on system
3. Apply the Momentum Principle
p f = pi + Fnet Dt
4. Apply the position update formula if needed
5. Check for reasonableness (units, etc.)
rf = ri + vavg Dt
Example: a hockey puck
A hockey puck with a mass of 0.16 kg is initially at rest. A player hits it
applying force F  400, 400,0 N during t = 4 ms. Where would the puck be
2 seconds after it loses contact with hockey stick?
Solution:
1. Choose a system and
surroundings:
Example: a hockey puck
A hockey puck with a mass of 0.16 kg is initially at rest. A player hits it
applying force F  400, 400,0 N during t = 4 ms. Where would the puck be
2 seconds after it loses contact with hockey stick?
Solution:
1. Choose a system and
surroundings:
2. Make a list of objects in surroundings that exert significant forces on system
3. Apply the Momentum Principle
p f = pi + Fnet Dt
p f = 0,0,0
( m kg/s) +
p f = 1.6,1.6,0 m× kg/s
Hockey stick
Earth
Ice (floor) (normal force, ~friction)
(
400,400,0 N × 4×10-3s
)
Example: a hockey puck
A hockey puck with a mass of 0.16 kg is initially at rest. A player hits it
applying force F  400, 400,0 N during t = 4 ms. Where would the puck be
2 seconds after it loses contact with hockey stick?
Solution:
y
3. Momentum
pf
p f = 1.6,1.6,0 m× kg/s
x
4. The position update formula
rf = ri + vavg Dt
p » mv
*
Choose coordinate system origin:
initial position of puck
p
v » = 10,10,0 m/s
m
( )
rf = 0,0,0 m + 10,10,0 m/s× 2 s
rf = 20,20,0 m
rf
Constant Gravitational Field
Clicker question #2:
g  9.8 N/kg (or m/s2 )
Due to the gravity, which components
of the velocity will change?
A) x, y and z
B) only x and y
C)only x
D)only y
E) only z
y
F  0, mg ,0
x
z
p fx  pix  Fnet , x t
p fy  piy  Fnet , y t
p fz  piz  Fnet , z t
Shoot the monkey
lecture demo
Clicker question #5:
The hunter aims right at the monkey and shoots. As the bullet
leaves the rifle, the monkey sees the flash and releases the
branch entering freefall. Will the bullet:
A) Hit the monkey
B) Undershoot the monkey
C) Overshoot the monkey
(ignore air resistance)
Note: the trajectory is a parabola
(if air friction is ignored)