Chapter 3 Motion in Two Dimensions (2D) Vectors and Projectile Motion x y.
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Transcript of Chapter 3 Motion in Two Dimensions (2D) Vectors and Projectile Motion x y.
Chapter 3 Motion in Two Dimensions (2D)
Vectors and Projectile Motion
x
y
What is a …?
• Vector– Has Magnitude & Direction
12 m/s Northeast– Examples:
• Drawn as arrows• Angles measured from 0
• Scalar– No direction– Examples:
12 m/s
tip
tail
0°
+
Vector addition• Vectors can be added together 2 ways:
graphically or analytically (with math!)• The answer to a vector addition problem is
called the Resultant, which is also a vector, so it has a magnitude (size) and direction
• The Resultant (R) goes from the starting tail to the ending tip
• Any number of vectors can be added with these methods
• http://illuminations.nctm.org/Activity.aspx?id=6598
v1
v2R
Graphical Addition
• Add using “Tip to Tail” method—tail of the next vector is drawn on the tip of the previous
• Draw to scale, keeping size and direction. Use ruler & protractor.
• Start at an origin!• http://lectureonline.cl.msu.edu/~mmp/kap3/cd052a.htm
• Examples next
Vector Direction: ExamplesWrite the vector angle in terms of 0 to 360°
A. 50m North
B. 100m West
C. 50m South
D. 100m East
0°Now add (sketch) A+D and B+C
AB
C
D
http://phet.colorado.edu/simulations/sims.php?sim=Vector_Addition
Vector addition
• Properties of vectors:– Vectors can be moved parallel to themselves
in a diagram– Vectors can be added in any order– To subtract a vector, add the vector pointing
in the opposite direction (or negative)– Multiplying or dividing vectors by scalars
results in vectors– http://www.nctm.org/standards/content.aspx?id=26779
– http://illuminations.nctm.org/Activity.aspx?id=6598 storm
Analytical AdditionWe’ll only look at vectors which are perpendicular to each other, so Right Triangles!
•Use Pythagorean theorem and/or Trigonometry to find exact values for Resultant SOH CAH TOA
– One axis is x, the other is y—they are independent of each other
– Make a sketch to help—add “tip to tail”, draw “x” first– Always indicate from 0, use sketch to reduce errors
•Carefully determine the correct quadrant for Resultant, R
Example: Perpendicular vectors Little red riding hood walks to grandma’s house 165 m North and 125 m to the East. What is her resultant displacement?
Example in packet
R = 207 m
E
N
R
tan = y/x = 165m/125m
= tan-1 (1.32) = 52.9°
Next…
• When you know the vector, but want the x and y that make it up:
• Breaking the vectors into its pieces is called Resolving the vector
• Use Trig to get the horizontal and vertical parts, or components, of the original vector
Resolving Vectors into Components
sin
siny
yAop
A
p
hyp A
A
cos
cosx
xAad
A
j
hyp A
A
Vector A and its x and y components
Often vector length and angle are known
FORMULAS FOR FROM X AXIS ONLY
Example What are the components that make up the given vector V?
V = 10 m/sec
60°
Vx = Vcos = 10cos60° = 5 m/s
Vy = Vsin = 10sin60° = 8.66 m/s
Examples
cos = Vx / V
sin = Vy / V
What is a Projectile?
Projectile motion• Projectile: Moving object that is only
affected by gravity.• The motion of a projectile is determined
only by the object’s initial velocity and gravity (g). vi has vy and vx components.
• Use x & y dimensions to analyze motion.• Vertical motion (y) of a projected object is
independent of its horizontal motion. • Vertical motion of a projectile is nothing
more than free fall (acceleration).• Horizontal motion (x) is constant
velocity.• The common variable between the
horizontal and vertical motion is time
Path of a Projectile• The path of a projectile is called its
trajectory.• The trajectory of a projectile in free fall is a
parabola. • http://regentsprep.org/Regents/physics/phys06/amotproj/sld005.htm
i
Variables:
∆xvx
∆t∆yviy
vfy
ay
vi and so also:vx= vicosviy= visin
Projectile Motion Problems
• The first type of projectile motion is: Horizontal Launch Projectiles – Path appears to be ½ a parabola
– Projectile has no initial vertical velocity, viy = 0
– Vertical is done like a dropped object (free fall)
– Projectile has no horizontal acceleration, ax = 0
– Horizontal motion portion is done the same as a constant velocity object (vix = vfx = vx)
– Use previous equations with revised subscripts
Question ?
• If a gun is fired horizontally and simultaneously a bullet is dropped from the same height, which bullet hits the ground first?
Variables:
∆xvx
∆t∆yViy = 0
vfy
Ay
Angle = 0
Projectile formulasx-direction y-direction
cos sin
x y
x
v v v v
xv
t
2 2
2
2
1
21
2
fy iy y
fy iy y
fy iy
iy y
v v a t
v v a y
y v v t
y v t a t
** Note:
vi is the vector sum (resultant) of vx and viy
vf is the vector sum of vx and vfy
Constant v
Free fall
Example A ball is launched horizontally at 10 m/s off
of a 1.2 m tall table. Calculate the following:
A) The time in the air.B) How far from the table does the ball land
(x)?C) What is the magnitude of the velocity of
the ball just as it hits the ground?
Examples in packet
Other horizontal examples
• Drop a bomb
• Shoot
• Throw or drive off a cliff…..
• All are initially going horizontally only, the x direction (an angle of 0 or 180)
• True Lies…
• two more questions…
..\My Videos\robbiemadison motorcycle.mp4
Horizontal Projectile
• An airplane flying steadily and horizontally drops a bowling ball. If it is in free fall, where does it hit relative to the plane?
• In front, directly under, or behind?
Horizontal Projectile
• What would happen if you shoot a projectile from a moving vehicle and the projectile has the same speed as the vehicle, but opposite direction?
• https://www.youtube.com/watch?v=BLuI118nhzc
Projectile Motion Problems• The second type of projectile motion is for objects
Launched at an Angle (Vi and )• These objects will follow a full parabola• The up ½ of the motion = the down ½ at
corresponding heights, other than direction (it is symmetric)
• Initial velocity is a vector at an angle which must be resolved into x & y components
• Same equations work as for Horizontal projectiles, but viy is not = 0.
• Use vx = vi·cos, viy = vi·sin..\My Videos\robbiemadison motorcycle.mp4
y=0 Projectile Motion Problems
http://jersey.uoregon.edu/newCannon/nc4.html
http://jersey.uoregon.edu/newCannon/nc1.html
http://www.mhhe.com/physsci/physical/giambattista/proj/projectile.html
Total time in the air (ttotal) = time to go up + time to go down
vi = initial velocity (total), includes both vx and viy
vf = final velocity (total); = magnitude of vi (or vx + vfy )
ymax = maximum height (top). Note: vy = 0
xmax = total horizontal distance, called Range (R)
Range symmetry and maximum
http://phet.colorado.edu/simulations/projectilemotion/projectile.swf
Problem Solving
• Angled launch problems are solved similar to horizontal
• Same procedure except for the start: Resolve velocity vector into x & y components (viy is not zero)
• Use symmetry and Vy = 0 at top when needed
• examples
Example A cannon ball gets fired at a 30 degree angle at 26 m/sec.A) How far away does it land?B) How long is it in the air?C) What is the maximum height?D) How fast is it moving when it hits the ground?E) How fast is the ball moving and how high is the
ball 0.6 sec after launch?
http://jersey.uoregon.edu/newCannon/nc4.html
http://jersey.uoregon.edu/newCannon/nc1.html
More practice:
• Practice Problem 1:• A golf ball is hit with a velocity of 31 m/s at 48
above the horizontal. – (A) Find how long it is in the air. – (B) How high did the ball go? – (C) What was the range of the ball?
• Challenge Problem:• A cannon shoots a pumpkin at an angle of 30
degrees which lands 120 m from the cannon. – A) With what velocity was the projectile fired? – B) What is the pumpkin’s velocity as it strikes the
ground?– https://www.youtube.com/watch?v=b0xmTzSXL0E
Monkey Hunter!!
http://www.phy.ntnu.edu.tw/ntnujava/index.php?PHPSESSID=04e5251cd5a765c0ee22dd61a2040312&topic=144.msg721#msg721
http://www.physics.umn.edu/outreach/pforce/circus/
http://www.mhhe.com/physsci/physical/giambattista/proj/projectile.html http://
www.waowen.screaming.net/revision/force&motion/
mandh.htm