Transcript Unit B Changes in Motion

Unit B Changes in Motion
Chapter 1
Describing Motion
Science 20 Unit 2
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Average Speed
• Average speed is equal to the total distance
traveled divided by the total time.
• Average speed = total distance
•
elapsed time
• v= d
•
t
• Pg 169 # 1-3
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Uniform Motion
• Uniform motion is motion in a straight line
at a constant speed.
• Uniform motion is rare
• Non-uniform motion is when there is a
change in speed (speeding up or slowing
down) or a change in direction.
• Instantaneous speed is the speed at any one
point in time
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Scalar Quantity
• Scalar quantities consist of magnitude only
and no indication of direction.
• Speed, time and volume are scalar
quantities
• Pg 172 #6-8
• Pg 173 # 2-4
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Velocity
• Position is a vector quantity describing the
location of a point relative to a reference
point
• Vector quantity is a quantity consisting of
magnitude and direction
• Sign convention – north is positive, to the
right is positive – south is negative and to
the left is negative
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• Displacement is a vector quantity
describing the length and direction in a
straight line from the starting position to the
final position.
• Average velocity is a vector quantity
describing the change in position over a
specified time
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Example
• A high school athlete runs 100 m south in
12.20 s. What is the velocity in m/s and
km/hr?
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Scale Diagrams
• Resultant displacement is the vector sum of
individual displacements.
• Head-to-tail method: a method where the
tail of a succeeding vector arrow begins at
the head of the preceding vector arrow.
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• Draw a vector diagram to represent the
following. A person walks 300 m south and
then turns around and walks 150 m north.
What is the persons displacement?
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• Pg 181 #15
• Pg 184 # 17
• Pg 185 # 2-5
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Lab Activity
• Using a ticker tape timer pull a dynamics
cart at constant speed.
• Mark ‘0’time when uniform motion starts –
dots are equally spaced.
• Count every five dots and mark ticker tape
• Measure and chart data from the ticker tape
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• Graph a Position vs Time Graph, calculate
the slope of the line.
• Graph a Distance vs Time Graph
• Graph a Velocity vs Time Graph, calculate
displacement at time = to 5 tocks
• Hand in Chart and three graphs
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Work
• Pg 193 #3&4
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Acceleration
• Acceleration is a change in velocity during
a time interval(speeding up or slowing
down)
• Acceleration is a vector quantity
• A force is required to change motion in
some way
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Acceleration cont’d
• Units for acceleration – m/s2
 v
• Formula is a  t
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• v = vf – vi
• vf = final speed (m/s)
• vi = initial speed (m/s)
 v
a
t

a


v f  vi 
t
f
 ti 
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Example
• A car traveling a 50 m/s speeds up to 95 m/s
over 6 s. What is the car’s acceleration?
• vi = 50 m/s
• vf = 95 m/s
• t=6s
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Examples
• The velocity of a car increases from 2 m/s at
1.0s to 16 m/s at 4.5 s. What is the car’s
average acceleration?
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Rearrange Formula
• vf = vi + at
• Use this formula to find final speed when an
object is accelerating
• Example – If a car with a velocity of 2.0
m/s at time zero, accelerates at a rate of
+4.0 m/s for 2.5 s, what is its velocity at the
end of its acceleration?
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Work
• Pg 200 #25
• Pg 203 #26 & 28
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Acceleration Lab
• Using the ticker tape timer drop an object
from the top of the stair well.
• Chart data from ticker tape
–
–
–
–
1. Time
2. Position
3. Velocity
4. Acceleration
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Lab Continued
• Graphs to be completed
– 1. Position vs Time graph
– 2. Velocity vs Time graph – calculate slope
– 3. Acceleration vs Time graph
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Displacement Equation



1
d  2 (v f  vi )t
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Acceleration Due to Gravity
• Acceleration due to gravity is 9.81 m/s2
• Gravitational acceleration will act on any
object moving up or down in the
atmosphere
• Example jumping, throwing a ball up,
falling off a building etc
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Example
• A rock is thrown straight up in the air. It reaches a height
of 18.6 m in 2.1 s. Calculate the initial velocity
t = 2.1 s
d = 18.6 m
a = - 9.81
Vf = 0
Vi = ?
Vi = vf - at
= 0 – (-9.81)(2.1)
Vi = 20.6 m/s
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Work
•
•
•
•
Pg 199 # 24
Pg 203 # 27
Pg 208 # 32
Pg 209 # 33
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Another Distance
(Displacement) Equation
• When final velocity is not given in the
original data and acceleration is given use
the following formula
 

2
1
d  vi t  2 at
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Example 1
• A boy leaves the surface of a trampoline
with an initial velocity of 11.8 m/s, straight
up. Determine the displacement after 0.8 s.
• vi = 11.8 m/s
• t = 0.8 s
• a = -9.81 m/s2
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Example 2
• A diver steps off the ledge of a platform and
enters the water 5.0 m below. If the initial
velocity of the diver was zero, determine
the time it took the diver to reach the water.
• d = -5.0 m
• a = -9.81 m/s2
• vi = 0
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Reaction Distance
• Reaction time is critical in the stopping of a
vehicle.
• Includes – the time it takes the drivers brain
to recognize there is a need to stop and the
time it takes the driver’s foot to move from
the gas pedal to the brake pedal.
• Reaction time varies from person to person
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• Reaction distance is the distance the
vehicle travels while the driver is reacting.
• Braking distance is the distance a vehicle
travels from the moment the brakes are first
applied to the time the vehicle stops.
• Stopping Distance = Reaction Distance +
Braking Distance
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Example
• The typical reaction time for most drivers is
considered to be about 1.50 s. This includes the
time required to identify the danger (0.75 s) and
the time required to react to the danger (0.75 s)
The ability of vehicles to decelerate varies greatly
however. Traffic safety engineers often use a
deceleration value of 5.85m/s2 to calculate the
minimum stopping distance for a vehicle on
smooth, dry pavement.
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• Determine the distance traveled while
reacting, the distance traveled while braking
and the minimum stopping distance of a
vehicle traveling 110 km/h.
• While reacting
d = vt = (31)(1.50) = 46.5 m
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•
•
•
•
While stopping (braking):
Vi = 31 m/s
Vf = 0
a = - 5.85 m/s
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Work
•
•
•
•
P. 213 #3, 4
Pg 216 # 38
Pg 218 # 39-40
Pg 220 #1 & 4
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Braking
• Force of friction is contact between two surfaces
that acts to oppose the motion of one surface past
the other.
• Friction is a force
• All forces are a push or pull.
• Forces are measured in Newtons – N.
• Brakes – particularly brake pads and rotors – are
designed to produce additional friction between
the rotating wheels and the fixed frame of the
vehicle
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• Net force – is the vector sum of all forces
acting on an object.
• In the case of braking the net force includes
– 1. Force of air resistance
– 2. Force of road resistance
– 3. Force applied by the braking system
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• Another factor that affects the rate of
deceleration is the mass of the vehicle.
• Larger trucks require a much greater
stopping distance
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Newton’s Second Law of
Motion
• Newton’s Second Law of Motion states that
an object will accelerate in the direction of
the net force applied.
Fnet = ma
Units are N = kg m/s2
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Example
• A vehicle with a mass of 1250 kg is
traveling 45 km/h east, when the driver
engages the brakes to stop at an
intersection.If the net force on the vehicle is
7000 N west, determine the magnitude and
direction of the deceleration of the vehicle
while the net force is applied.
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• Determine the length of time the net force
must be applied to stop the vehicle.
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Work
• Pg. 222 #43
• Pg 226 # 45
• Pg 227 # 1-4 (copy questions)
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Speeding Up
• Newton’s second law also explains what
happens when a vehicle increases its
velocity.
• The additional force required to make the
vehicle move faster is called the applied
force.
• The net force results from the vector sum of
the applied force and the force of friction
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Newton’s First Law of Motion
(Inertia)
• This law states that in the absence of a net
force, an object in motion will tend to
maintain its velocity, and an object at rest
will tend to remain at rest.
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Example - One
• The engine of a motorcycle supplies an
applied force of 1880 N, west, to overcome
frictional forces of 520 N, east. The
motorcycle and rider have a combined mass
of 245 kg. Determine the acceleration.
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Example - Two
• A car with a mass of 1075 kg is traveling on
a highway, The engine of the supplies an
applied force of 4800 N, west, to overcome
frictional forces of 4800 N, east. Determine
the acceleration.
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Inertia
• Inertia is the property of an object to resist
changes in its state of motion.
• Example an object at rest will remain at rest
or an object in motion will remain in motion
• The amount of inertia an object has depends
upon its mass. The greater the mass the
greater the inertia.
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• Try to push a small car stopped on the road
or a transport truck.
• Work Pg 233 # 47-49
• Work Pg 235 # 1-13, 16-20
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Momentum (p)
• Momentum is the product of an objects
mass and its velocity.
p = mv
p – momentum(kg m/s)
m – mass (kg)
v – velocity (m/s)
• Momentum is a vector quantity.
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• Rearranged formulas:
m=p
v
v=p
m
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Example 1
• Determine the momentum of a vehicle with
a mass of 2100 kg moving at a velocity of
22 m/s (E).
m = 2100 kg
v = 22 m/s
p=?
p = mv = (2100)(22)
p = 46,200 kg m/s
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Example 2
• An airplane has a momentum of 8.7 x 107 kg m/s .
If the airplane is flying at a velocity of 990 km/h,
determine its mass.
p = 8.7 x 107 kg
v = 990 km/h = 275 m/s
m=?
m = p/v
m = (8.7 x 107)/(275)
m = 3.2 x 105 kg
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Work
• Read Pg 244 – 245
• P. 244 #1-3
• Pg 245 # 2 – 6 (copy questions)
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Newton vs Momentum
• Newton’s second law can show that a net
force will cause a mass to accelerate in the
direction of the applied force.
F = ma
a = vf – vi
t
Therefore
F = m( vf – vi)
t
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F = change in momentum
Or
time
Ft = m( vf – vi)
Ft = Impulse
m( vf – vi) = Change in momentum
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Example
• A 2.1 kg barn owl flying at a velocity of 15
m/s (E) strikes head-on with the windshield
of a car traveling 30 km/h (W).
• If the time interval for the impact was 6.7 x
10-3 s, determine the force that acted on the
owl.
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m = 2.1 kg
vi = 15 m/s
vf = - 30 km/h = - 8.3 m/s
t = 6.7 x 10-3 s
F=?
F = m(vf – vi)
t
F = 2.1(-8.3 – 15)
6.7 x 10-3
F = - 7303 N
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•
•
•
•
Read Pg 249 – 250
Discuss results on Pg 250
Do questions Pg 247 # 4 –6
Pg 251 # 2 - 7
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Impulse
• Impulse is the product of the net force applied to
an object and the time interval during which the
force is applied.
• Impulse does not have its own symbol.
• It is represented by FΔt and has the units kgm/s
• When it comes to roadside safety some types of
barriers are less damaging to vehicles and their
occupants.
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• Remember that FΔt = Δp
• Example
• A raw egg drops to the floor. If the floor
exerts a force of 9.0 N over a time interval
of 0.030 s, determine the impulse required
to change the egg’s momentum.
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Example 2
• A raw egg with a mass of 0.065 kg falls to
the floor. At the moment the egg strikes the
floor, it is travelling 4.2 m/s. Assuming that
the final velocity of the egg is zero after
impact, determine the impulse required to
change the momentum of the egg.
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Work
•
•
•
•
Pg 254 # 9-12
Read Pg 254
Pg 255 # 13-14
Pg 256 # 1,3,5,7,8,10-12
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Collisions
• There are 3 classes of collisions
• 1. Primary collision – the vehicle colliding with
another object, such as another vehicle
• 2. Secondary collision – the occupant colliding
with the interior of the vehicle
• 3. Tertiary collision – the occupant’s internal
organs colliding within the occupant’s body
Pg 257
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Newton’s Third Law
• Newton’s Third Law states that whenever one
object exerts a force on a second object, the
second object exerts an equal but opposite force
on the first object.
• F1on 2 = F 2on1
• Forces occur in pairs- action vs reaction
• Forces are the same magnitude but push in
opposite directions
• Forces push on different objects.
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Computer Work
• Pg 260 – 261
• Pg 262 # 1-5
• Computer Lab Pg 265-267
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Momentum is Conserved
• Collisions observed in the activity were one
of three kinds: hit and stick, hit and
rebound, or explosion.
• In all cases the momentum of one object
was transferred to another such that the total
momentum always remained the same.
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• Law of Conservation of momentum – if
the net force acting on a system is zero, the
sum of the momentum before an interaction
equals the sum of the momentum after the
interaction.
• Σ p before = Σ p after
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Example 1
• A 10 000 kg freight car traveling west with
a velocity of 1.5 m/s collides with a
• 20 000 kg freight car at rest. After the
collision, the freight cars stick together.
Determine the velocity of the fright cars
after the collision.
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Example 2
• A 3.0 kg ball rolling east with a velocity of
1.5 m/s collides with another 6.0 kg ball at
rest. After the collision, the first ball
rebounds and is traveling at a velocity of
0.50 m/s west.
• A) Determine the velocity of the second ball
after the collision
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• B) Determine the momentum values of the
balls before and after the collision
• C) Use scale diagrams of momentum
vectors to demonstrate that momentum is
conserved in this collision.
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Example 3
• A 0.020 kg firecracker at rest explodes into
two pieces. If a 0.015 kg piece flies off to
the right at a velocity of 3.00 m/s, determine
the velocity of the other 0.0050 kg piece.
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Work
• Pg 270 # 17-19
• Pg 271 # 20
• Pg 271 # 1-8
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Designing a Helmet
• You are to design a helmet for an egg. An
egg is much like your head, it has a thin
hard shell protecting a soft inside. Your
skull is a hard shell covering soft brain
tissue.
• Your design must address three
characteristics of helmets
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• 1. The egg test dummy must be protected
from a frontal collision against a rigid
barrier
• 2. The egg test dummy must have no
covering over its eyes and ears.
• 3. The egg test dummy’s helmet must
continue to provide protection after a
number of severe frontal impacts.
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• Canadian Standards Association (CSA) tests all
helmets to ensure that they will remain on the
head during an impact and will provide sufficient
protection.
• To test your helmet the egg test dummy with its
helmet on will be placed in a plastic bag and
suspended at the end of a long string and then
swung till it hits a solid barrier.
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• When the egg is pulled back prior to release
the work done to change the position of the
egg is equal to the gravitational potential
energy of the egg.
• Ep(grav) = W
•
= FΔd
•
= mgh
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• Then according to the Law of Conservation
of Energy, the gravitational potential energy
should be converted to kinetic energy as it
swings toward the barrier
• Ek = ½ mv2
• v=
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Example
• The mass of the egg, plastic bag, helmet,
and paper clip is 0.085 kg. If they are pulled
back such that they are now 0.40 m higher
than they were at rest and then released to
swing forward and hit a solid barrier.
• A) Calculate the gravitational potential
energy of the egg when it is 0.40 m above
the resting point.
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• Eg = mgh
•
= (0.085 kg)(9.81m/s2)(0.40 m)
•
= 0.33 J
• B) Determine the kinetic energy of the egg
just before it hits the barrier
• Ek = Eg
•
= 0.33 J
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• C) Calculate the speed of the egg just before
impact.
• Ek = ½ mv2
• v =
•
= 2(0.33 J)
•
0.085kg
•
= 2.8 m/s
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• D) Calculate the magnitude and direction of
the momentum of the egg just before
impact.
• The forward motion will be positive
• p = mv
• = (0.085 kg)(2.8 m/s)
• = 0.24 kgm/s
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• E) Assuming the egg stops immediately
upon impact, calculate the change in
momentum upon impact.
• Δp = pf – pi
(pf is zero)
•
= - pi
•
= - 0.24 kgm/s
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• F) Calculate the impulse required to stop the
egg during impact.
• impulse = FΔt
•
= Δp
•
= - 0.24 kgm/s
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• G) If the impact lasts for 0.040 s, determine
the force that acted upon the egg during the
collision using the equation for impulse.
• FΔt = Δp
•
F = Δp
Δt
= - 0.24 kgm/s
= - 6.0 N
0.040 s
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• H) Determine the acceleration of the egg
over the 0.040 s
• a = vf – vi
Δt
•
= 2.8 m/s
0.040 s
•
= - 70 m/s2
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• I) Use Newton’s second law to confirm your
answer to question g by calculating the
force that acted on the egg.
• F = ma
• = (0.085 kg)(- 70 m/s2)
• = - 6.0 N
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Practice
• Pg 277 # 23
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Do Lab
• Pg 277-279
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Work
• Pg 281 # 2-6
• Pg 282-3 # 1,2,5,7,9,11,13,15,19,20
• Pg 285-291 # 1,3 – 7, 17, 19
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Name
Symbol
Unit
Formula
Momentum
p
kgm/s
p=mv
Δmomentum Δp
kgm/s
p1 - p2
Impulse
Impulse
Ns
FΔt = pf-pi
Newtons
F1on2 =F2on1 N
F1on2 = F2on1
Conservatio Σp = Σp
kgm/s
Σpbefore = Σpafter
Work
J
W = Fd
J
E = 1/2mv2
J
E = mgh
W
Kinetic
KE
Energy
Gravitational Eg
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