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2D MotionHomework Solutions
ANSWERS:1. The Wooley Bear2.
3. 26.8 sec4. 12. 4 sec
Vwest = .235 m/sVNorth = .348 m/s
MMMMM Dinner!
.42 m/s
34o
N
S
EW
1.4 cm/s
*Done by applying trig to distances given.
Trig...
Kinematics...
1.4 cm/s1.1 cm/s
*Done by applying trig to velocity vector.
Trig...
Kinematics...
*note it's the same answer and the same amount of work using either method.
Conversion:
Given
1.59 cm/s2
2.4 cm
/s2
Trig applied to acceleration vector:
4ft = 121.9 cm ay = asinθ = (2.4 cm/s2)sin41o =1.59 cm/s2
*Note: Like #3, there are two ways to solve this one.You can apply trig to the distance information given or to the acceleration vector.
Mo4 Motion Deluxe#1 Can the x‐component of a vector be greater than the magnitude of the vector?
How about the y‐component? Explain...
θ
V
Vx
Vy
#2 A golf ball is hit with an initial speed of 35 m/s at an angle less than 45o above the horizontal.
The horizontal velocity component is...
(a) greater than , (b) equal to, (c) less than the vertical velocity component. Why?
35 m/s
#3 If the ball is hit at an angle of 37o, what are the initial horizontal and vertical velocity components?
#4 The x‐ and y‐components of an acceleration vector are 3.0 m/s2 and 4.0 m/s2, respectively. The magnitude of the acceleration vector is...
#5 What are the magnitude and direction of the "resultant" acceleration vector shown in #4?
3 m/s2
4 m/s
2 a
θ
(a) less than 3.0 m/s2
(b) betwen 3.0 m/s2 and 4.0 m/s2
(c) between 4.0 m/s2 and 7.0 m/s2
(d) equal to 7.0 m/s2
#6 If the magnitude of a velocity vector is 7.0 m/s and the x‐component is 3.0 m/s, what is the y‐component?
3 m/s
? m/s
7 m/s
θ
#7 The x‐component of a velocity vector that has an angle of 37o to the x‐axis has a magnitude of 4.8 m/s.
(a) What is the magnitude of the velocity?(b) What is the magnitude of the y‐component of the velocity?
4.8 m/s
? m/s
? m/s
37o
θ = 37o
Vx= (35m/s)cos37o = 28 m/s
Vy = (35m/s)sin37o = 21 m/s
Vx
Vy35 m
/s
a2 + b2 = c2... PYTHAGORIZE IT!
PYTHAGORIZE IT!
Use any trig function to get θ....
tanθ = (4/3)
θ = tan1(4/3) = 53o
a) V = 6 m/s
b) Vy = 3.6 m/s
P.M.P.M.D.D.Processed Meat Projectile Motion Demo Deluxe µs Timer
P.M.P.M.D.D.
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SPAM
2D Mo Intro1.
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Answer = 733 m
http://www.potatorock.com/
2.
VerticalHorizontal
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Answer = 2.3 m
3.
VerticalHorizontal
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Answer = 1.6 m/s2
Team Name #1 #2 #3 #4 #5 #6 #7 #8
The It's
All Ball
Bearing
s Nowad
ays
2D Kin
ematics
Relay De
luxeThe
It's All
Ball
Bearing
s Nowad
ays
2D Kin
ematics
Relay De
luxe
Touch and DragThe BallBearing
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#1
* No horizontal calculation needed.
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#2*Find both Δx and Δy.
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#3
* No horizontal calculation needed.
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#4*Note: Its the same cliff fromproblem #3 so Δt = 1.59 sec
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#5* Find Δt using vertical clues,then sub into horizontal.
#6Vx = 3 m/s
Vy = 4 m/s V =?θ =?
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#7* Find Δt using horizontal clues,then sub into vertical.
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#8 Vx = 6 m/s
Vy = ?
V = ?
θ = ?
*Find Vy using the vertical kinematics clues, then do trig on the vectors.
Vx = 6 m/s
Vy = 68.6 m/s
V = ?
θ = ?
*any of these work
Vx
.4 m
*Note that this is the same scenario as the Flick the ant problem from
Mo Mo Mo Motion worksheet, or the Bas A$$ bonus problem on Mo4.
*This can be seen in the saw problem on Mo Mo Mo Motion worksheet.
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*note that this one is the same exact scenario as problem #7 from the relay race.
Aww come on guys....maybe you need a refresher course.
100 m/s
Vyfinal
