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Introduction to 2-Dimensional
Motion
Vectors
Two vectors are equal if they have the same
magnitude (how big) and same direction.
How to add vectors:
Line them up head to tail
Draw a vector that connects the tail of the first
arrow to the head of the last arrow
Make sure your calculator is set to DEGREES
Always use UNITS
2-Dimensional Motion
Definition: motion that occurs with
both x and y components.
Example:
Playing pool .
Throwing a ball to another person.
Each dimension of the motion can
obey different equations of motion.
Solving 2-D Problems
Resolve all vectors into components
x-component
Y-component
Work the problem as two one-dimensional
problems.
Each dimension can obey different equations of
motion.
Re-combine the results for the two
components at the end of the problem.
Sample Problem
You run in a straight line at a speed of 5.0 m/s in a
direction that is 40o south of west.
a) How far west have you traveled in 2.5 minutes?
b) How far south have you traveled in 2.5 minutes?
Sample Problem
You run in a straight line at a speed of 5.0 m/s in a
direction that is 40o south of west.
a) How far west have you traveled in 2.5 minutes?
b) How far south have you traveled in 2.5 minutes?
v = 40 m/s
Sample Problem
You run in a straight line at a speed of 5.0 m/s in a
direction that is 40o south of west.
a) How far west have you traveled in 2.5 minutes?
b) How far south have you traveled in 2.5 minutes?
vx v = 5 m/s, Q = 40o,
t = 2.5 min = 150 s
vy vx = v cosQ vy = v sin Q
v = 5 m/s
vx = 5 cos 40 vy = 5 sin 40
vx = vy =
x = vx t y = vyt
x = ( )(150) y = ( )(150)
x= y=
Sample Problem
A particle passes through the origin with a speed of 6.2 m/s
traveling along the y axis. If the particle accelerates in the
negative x direction at 4.4 m/s2.
a) What are the x and y positions at 5.0 seconds?
Sample Problem
A particle passes through the origin with a speed of 6.2 m/s
traveling along the y axis. If the particle accelerates in the
negative x direction at 4.4 m/s2.
a) What are the x and y positions at 5.0 seconds?
Vo,y = 6.2 m/s, Vo,x = 0 m/s, t = 5 s,
ax = -4.4 m/s2, ay = 0 m/s2
x=? y=?
x = vo,x + at y = vo,y + at
x = 0 + (-4.4)(5) y = 6.2 + (0)(5)
x= y=
Sample Problem
A particle passes through the origin with a speed of 6.2 m/s
traveling along the y axis. If the particle accelerates in the
negative x direction at 4.4 m/s2.
b) What are the x and y components of velocity at this time?
Sample Problem
A particle passes through the origin with a speed of 6.2 m/s
traveling along the y axis. If the particle accelerates in the
negative x direction at 4.4 m/s2.
b) What are the x and y components of velocity at this time?
vx = vo,x + axt vy = voy + axt
vx = 0 + (-4.4)(5) vy = 6.2 + 0(5)
vx = vy =
Mechanics Quiz
A mass experiences a force vector
with components 30N to the right,
and 40 N down . ExplainN how to
determine the magnitude and
direction (angle) of the force vectors.
The resultant vector can found
graphically or mathematically.
30N2 + 40N2 = 2500N2 = 50N
Projectiles
Projectile Motion
Something is fired, thrown, shot, or
hurled near the earth’s surface.
Horizontal velocity is constant.
Vertical velocity is accelerated.
Air resistance is ignored.
1-Dimensional Projectile
Definition: A projectile that moves in a
vertical direction only, subject to
acceleration by gravity.
Examples:
Drop something off a cliff.
Throw something straight up and catch it.
You calculate vertical motion only.
The motion has no horizontal component.
2-Dimensional Projectile
Definition: A projectile that moves both
horizontally and vertically, subject to
acceleration by gravity in vertical
direction.
Examples:
Throw a softball to someone else.
Fire a cannon horizontally off a cliff.
Shoot a monkey with a blowgun.
You calculate vertical and horizontal
motion.
Horizontal Component of
Velocity
Is constant
Not accelerated
Not influence by gravity
Horizontal component velocity
vx = v(cosQ)
Follows equation:
x = Vo,xt
Horizontal Component
of Velocity
Vertical Component of
Velocity
Undergoes accelerated motion
Accelerated by gravity (9.8 m/s2
down)
Vertical component of velocity
vy = v(sinQ)
Vy = Vo,y - gt
y = yo + Vo,yt - 1/2gt2
Vy2 = Vo,y2 - 2g(y – yo)
Horizontal and Vertical
Horizontal and Vertical
Zero Launch Angle Projectiles
Launch angle
Definition: The angle at which a
projectile is launched.
The launch angle determines what the
trajectory of the projectile will be.
Launch angles can range from -90o
(throwing something straight down)
to +90o (throwing something straight
up) and everything in between.
Zero Launch angle
vo
A zero launch angle implies a perfectly
horizontal launch.
Sample Problem
The Zambezi River flows over Victoria Falls in Africa. The falls are
approximately 108 m high. If the river is flowing horizontally at 3.6 m/s
just before going over the falls, what is the speed of the water when it
hits the bottom? Assume the water is in freefall as it drops.
Sample Problem
The Zambezi River flows over Victoria Falls in Africa. The falls are
approximately 108 m high. If the river is flowing horizontally at 3.6 m/s
just before going over the falls, what is the speed of the water when it
hits the bottom? Assume the water is in freefall as it drops.
yo = 108 m, y = 0 m, g = -9.8 m/s2, vo,x = 3.6 m/s
v=?
v vx vy
2 2
Sample Problem
The Zambezi River flows over Victoria Falls in Africa. The falls are
approximately 108 m high. If the river is flowing horizontally at 3.6 m/s
just before going over the falls, what is the speed of the water when it
hits the bottom? Assume the water is in freefall as it drops.
yo = 108 m, y = 0 m, g = 9.8 m/s2, vo,x = 3.6 m/s
v=?
Gravity doesn’t change horizontal velocity.
v vx vy
2 2
vo,x = vx = 3.6 m/s
Vy2 = Vo,y2 - 2g(y – yo)
Vy2 = (0)2 – 2(9.8)(0 – 108)
Vy =
Sample Problem
The Zambezi River flows over Victoria Falls in Africa. The falls are
approximately 108 m high. If the river is flowing horizontally at 3.6 m/s
just before going over the falls, what is the speed of the water when it
hits the bottom? Assume the water is in freefall as it drops.
yo = 108 m, y = 0 m, g = 9.8 m/s2, vo,x = 3.6 m/s
v=?
Gravity doesn’t change horizontal velocity.
v vx vy
2 2
vo,x = vx = 3.6 m/s
v (3.6)2 ( )2 Vy2 = Vo,y2 - 2g(y – yo)
v= Vy2 = (0)2 – 2(9.8)(0 – 108)
Vy =
Sample Problem
An astronaut on the planet Zircon tosses a rock horizontally with a
speed of 6.75 m/s. The rock falls a distance of 1.20 m and lands a
horizontal distance of 8.95 m from the astronaut. What is the
acceleration due to gravity on Zircon?
Sample Problem
An astronaut on the planet Zircon tosses a rock horizontally with a
speed of 6.75 m/s. The rock falls a distance of 1.20 m and lands a
horizontal distance of 8.95 m from the astronaut. What is the
acceleration due to gravity on Zircon?
vo,x = 6.75 m/s, x = 8.95 m, y = 0 m, yo = 1.2, Vo,y = 0 m/s
g=?
y = yo + Vo,yt - 1/2gt2 x = vo,xt
g = -2(y - yo - Vo,yt)/t2 t = x/vo,x
t = 8.95/6.75
g = -2[0 – 1.2 – (0)( )]/( ) 2
t=
g=
Sample Problem
Playing shortstop, you throw a ball horizontally to the second
baseman with a speed of 22 m/s. The ball is caught by the second
baseman 0.45 s later.
a) How far were you from the second baseman?
b) What is the distance of the vertical drop?
Should be able to do this on your own!
General Launch Angle
Projectiles
General launch angle
vo
Projectile motion is more complicated when the
launch angle is not straight up or down (90o or –
90o), or perfectly horizontal (0o).
General launch angle
vo
You must begin problems like this by resolving
the velocity vector into its components.
Resolving the velocity
Use speed and the launch angle to find
horizontal and vertical velocity components
Vo Vo,y = Vo sin
Vo,x = Vo cos
Resolving the velocity
Then proceed to work problems just like
you did with the zero launch angle
problems.
Vo Vo,y = Vo sin
Vo,x = Vo cos
Sample problem
A soccer ball is kicked with a speed of 9.50 m/s at an angle
of 25o above the horizontal. If the ball lands at the same
level from which is was kicked, how long was it in the air?
Sample problem
A soccer ball is kicked with a speed of 9.50 m/s at an angle
of 25o above the horizontal. If the ball lands at the same
level from which is was kicked, how long was it in the air?
vo = 9.5 m/s, = 25o, g = -9.8 m/s2, Remember: because it lands
at the same height: Dy = y – yo = 0 m and vy =- vo,y
Find: Vo,y = Vo sin and Vo,x = Vo cos
Vo,y = 9.5 sin 25 Vo,x = Vo cos
Vo,y = Vo,x =
t=?
Vy = Vo,y - gt
t = (Vy - Vo,y )/g
t = [( ) – ( )]/9.8 don’t forget vy =- vo,y
t=
Sample problem
Snowballs are thrown with a speed of 13 m/s from a roof
7.0 m above the ground. Snowball A is thrown straight
downward; snowball B is thrown in a direction 25o above the
horizontal. When the snowballs land, is the speed of A
greater than, less than, or the same speed of B? Verify your
answer by calculation of the landing speed of both
snowballs.
We’ll do this in class.
Projectiles launched over
level ground
These projectiles have highly
symmetric characteristics of motion.
It is handy to know these
characteristics, since a knowledge of
the symmetry can help in working
problems and predicting the motion.
Lets take a look at projectiles
launched over level ground.
Trajectory of a 2-D
Projectile
y
x
Definition: The trajectory is the path
traveled by any projectile. It is plotted
on an x-y graph.
Trajectory of a 2-D
Projectile
y
x
Mathematically, the path is defined by
a parabola.
Trajectory of a 2-D
Projectile
y
x
For a projectile launched over level
ground, the symmetry is apparent.
Range of a 2-D Projectile
y
x
Range
Definition: The RANGE of the projectile is
how far it travels horizontally.
Maximum height of a
projectile
y
Maximum
Height
x
Range
The MAXIMUM HEIGHT of the projectile
occurs when it stops moving upward.
Maximum height of a
projectile
y
Maximum
Height
x
Range
The vertical velocity component is zero at
maximum height.
Maximum height of a
projectile
y
Maximum
Height
x
Range
For a projectile launched over level ground, the
maximum height occurs halfway through the flight
of the projectile.
Acceleration of a projectile
y
g
g g
g g
x
Acceleration points down at 9.8 m/s2 for
the entire trajectory of all projectiles.
Velocity of a projectile
y
v
v
v
vo
vf x
Velocity is tangent to the path for the
entire trajectory.
Velocity of a projectile
y
vx
vy vx
vx vy
vy vx
vx vy x
The velocity can be resolved into
components all along its path.
Velocity of a projectile
y
vx
vy vx
vx vy
vy vx
vx vy x
Notice how the vertical velocity changes
while the horizontal velocity remains
constant.
Velocity of a projectile
y
vx
vy vx
vx vy
vy vx
vx vy x
Maximum speed is attained at the beginning,
and again at the end, of the trajectory if the
projectile is launched over level ground.
Velocity of a projectile
vo
-
vo
Launch angle is symmetric with landing angle
for a projectile launched over level ground.
Time of flight for a
projectile
t
to = 0
The projectile spends half its time
traveling upward…
Time of flight for a
projectile
t
to = 0 2t
… and the other half traveling down.
Position graphs for 2-D
projectiles
y y x
x t t
Velocity graphs for 2-D
projectiles
Vy Vx
t t
Acceleration graphs for 2-D
projectiles
ay ax
t t
Projectile Lab
Projectile Lab
The purpose is to collect data to plot a trajectory for
a projectile launched horizontally, and to calculate the
launch velocity of the projectile. Equipment is
provided, you figure out how to use it.
What you turn in:
1. a table of data
2. a graph of the trajectory
3. a calculation of the launch velocity of the ball
obtained from the data
Hints and tips:
1. The thin paper strip is pressure sensitive.
Striking the paper produces a mark.
2. You might like to hang a sheet of your own
graph paper on the brown board.
More on Projectile Motion
The Range Equation
Derivation is an important part of
physics.
Your book has many more equations
than your formula sheet.
The Range Equation is in your
textbook, but not on your formula
sheet. You can use it if you can
memorize it or derive it!
The Range Equation
R = vo2sin(2)/g.
R: range of projectile fired over level
ground
vo: initial velocity
g: acceleration due to gravity
: launch angle
Deriving the Range Equation
Sample problem
A golfer tees off on level ground, giving the ball an initial
speed of 42.0 m/s and an initial direction of 35o above the
horizontal.
How far from the golfer does the ball land?
Vo = 42m/s, = 35o, g = 9.8 m/s2
R=?
Sample problem
A golfer tees off on level ground, giving the ball an initial
speed of 42.0 m/s and an initial direction of 35o above the
horizontal.
b) The next golfer hits a ball with the same initial speed, but at a
greater angle than 45o. The ball travels the same horizontal
distance. What was the initial direction of motion?
Review Day
Practice Problems
A canoe is paddled due north across a lake at
2.0 m/s relative to still water. The current in
the lake flows toward the east; its speed is 0.5
m/s. Which of the following vectors best
represents the velocity of the canoe relative to
shore?
a. 2.5 m/s b. 2.1 m/s c. 2.5 m/s
d. 1.9 m/s e. 1.9 m/s
The hypotenuse of a right triangle has to be bigger
than either leg, but less than the algebraic sum of the
27.0 N
740
Practice Problems
A Force vector A has magnitude 27.0 N and is
direction 740 below the vertical as shown
above. Which of the following are the
horizontal and vertical components of vector
A?
Vertical HorizontalAnswers B & E are wrong
because the vertical
A. 7.7 N 26.4 N component is bigger than the
B. 26.4 N 7.7 N horizontal component.
C. 8.5 N 28.6 N Answers C & D are wrong
D. 15.0 N 27.0 N because the horizontal
E. 23.3 N 13.5 N component is greater than the
initial vector
Practice Problems
A B C
Force vectors A, B, and C are shown
above. Which of the following forces
gives the sum of zero?
To get a resultant vector of zero both the
horizontal and vertical components of the vector
sum must go to zero.
E = A-B-C
Practice Problem
Which of the following is a scalar
quantity?
a. Electric force
b. Gravitational force
c. Weight
d. Mass
e. friction
Free-Response
An airplane attempts to drop a bomb on a target. When
the bomb is released, the plane is flying upward at an
angle of 300 above the horizontal at a speed of 200 m/s.
At the point of release, the plane’s altitude is 2.0 km.
The bomb hits the target.
a. Determine the magnitude and direction of the
vertical component of the bomb’s velocity at the point of
release.
vy = vi sin Q Since sine of 300 is 0.5 the vertical
component of the velocity is 100 m/s
Free-Response
An airplane attempts to drop a bomb on a target. When
the bomb is released, the plane is flying upward at an
angle of 300 above the horizontal at a speed of 200 m/s.
At the point of release, the plane’s altitude is 2.0 km.
The bomb hits the target.
b. Determine the magnitude and direction of the
horizontal of the bomb’s velocity at the point when the
bomb contacts the target.
vx = vi cos Q Since cosine of 300 is 0.86 the
horizontal component of the velocity is 173m/s to
the right throughout the flight
Free-Response
An airplane attempts to drop a bomb on a target. When
the bomb is released, the plane is flying upward at an
angle of 300 above the horizontal at a speed of 200 m/s.
At the point of release, the plane’s altitude is 2.0 km.
The bomb hits the target.
c. Determine how much time it takes for the bomb to
hit the target after it is released.
vx = vi cos Q Since cosine of 300 is 0.86 the
horizontal component of the velocity is 173m/s to
the right throughout the flight
Mechanics Quiz
Quickly identify as a vector or a scalar:
Acceleration
Velocity
Work
Force
Speed
Mass
Momentum
Displacement
Kinetic energy
A ball is shot at a velocity of 25 m/s
from a cannon pointed at an angle of 300
above the horizontal. How far does it
travel before hitting the ground?
Vx = vi cos = 21.65 m/s t = 2.6 s
Vy = vi sin = 12.5 m/s x = vx t
y= vy t + ½ gt2 x = (21.65 m/s)(2.6 s)
y=0 x = 56.29 m
-2vy/g = t
