Chapter 3: Kinematics in 2 dimensions Flashcards
1
Q
What is the horizontal motion equation?
A
Δx = Vxt
2
Q
What does Vx represent in the horizontal equation?
A
The velocity on the x axis (horizontal)
3
Q
What is the equation for verticle motion?
A
Δy = (Vyt)
4
Q
What does Vy represent in the vertical motion equation?
A
The velocity on the y axis (verticle motion)
5
Q
T/F: Verticle motion has constant acceleration
A
True, meaning we need to know the 4 equations of constant acceleration
6
Q
Adam drops a ball from rest from the top floor of a building, at the same time Bob throws a ball horizontally from the same location.
Which ball hits the level ground first?
A)They both hit the ground at the same time
B) Adam’s ball
C) Bob’s ball
D) It depends on how fast Bob throws the ball
E) It depends on how fast the ball falls when Adam drops it
A
A
A) They both hit the ground at the same time
7
Q
Define analytical method
A
the method of determining the magnitude and direction of a resultant vector using the Pythagorean theorem and trigonometric identities
8
Q
Define classical relativity
A
the study of relative velocities in situations where speeds are less than about 1% of the speed of light—that is, less than 3000 km/s
9
Q
Define commutative
A
refers to the interchangeability of order in a function; vector addition is commutative because the order in which vectors are added together does not affect the final sum
10
Q
Define component (of a 2-d vector)
A
A piece of a vector that points in either the vertical or the horizontal direction; every 2-d vector can be expressed as a sum of two vertical and horizontal vector components
11
Q
Define direction (of a vector)
A
the orientation of a vector in space
12
Q
Define head (of a vector)
A
the end point of a vector; the location of the tip of the vector’s arrowhead; also referred to as the “tip”
13
Q
Define head-to-tail method
A
a method of adding vectors in which the tail of each vector is placed at the head of the previous vector
14
Q
Define kinematics
A
the study of motion without regard to mass or force
15
Q
Define magnitude (of a vector)
A
the study of motion without regard to mass or force
16
Q
Define motion
A
displacement of an object as a function of time
17
Q
Define projectile
A
an object that travels through the air and experiences only acceleration due to gravity
18
Q
Define projectile motion
A
the motion of an object that is subject only to the acceleration of gravity
19
Q
Define range
A
the maximum horizontal distance that a projectile travels
20
Q
Define relative velocity
A
the velocity of an object as observed from a particular reference frame
21
Q
Define relativity
A
the study of how different observers moving relative to each other measure the same phenomenon
22
Q
Define resultant
A
the sum of two or more vectors
23
Q
Define resultant vector
A
the vector sum of two or more vectors
24
Q
Define scalar
A
a quantity with magnitude but no direction
25
Define tail
the start point of a vector; opposite to the head or tip of the arrow
26
Define trajectory
the path of a projectile through the air
27
Define vector
a quantity that has both magnitude and direction; an arrow used to represent quantities with both magnitude and direction
28
Define vector addition
the rules that apply to adding vectors together
29
Define velocity
speed in a given direction
30
Graphical method of adding vectors (explanation- not question, just important to know)
Adding vectors A and B involves drawing vectors on a graph and adding them using the head-to-tail method. The resultant vector R is defined such that A + B = R.
○ The magnitude and direction of R are then determined wih a rule and a proteactor
31
Graphical method of subtracting vectors (explanation- not question, just important to know)
Subtracting vector B from A invloves adding the opposite of vector B, which is defined a -B. In this case, A-B = A + (-B) = R. Then the head-to-tail method of addition is dollowed in the usual way to obtain the resultant vector R.
32
Are vectors commutative? What does this mean?
Yes, this means vectors A and B can do the following: A + B = B + A
33
Steps to add vectors A and B using the analytical method (Explanation, not question)
1. Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the equations
i. Ax = Acosꝋ
ii. Bx = Bcosꝋ
AND
iii. Ay = Asinꝋ
iv. By = Bsinꝋ
2. Add the horizontal and vertical components of each vector to determine the components Rx and Ry of the resultant vector, R:
i. Rx = Ax + Bx
AND
ii. Ry = Ay + By
3. Use the Pythagorean theorem to determine the magnitude, R, of the resultant vector R:
i. √(〖R_^2〗_x+〖R_^2〗_y )
4. Use a trigonometric identity to determine the direction, ꝋ or R:
i. ꝋ = tan-1 (Ry / Rx)
34
How to solve projectile motion problems (LONG explanation)
Determine the coordinate system, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position s are given by the quantities x and y and the components of the velocity v are given by vx = v cosꝋ and vy = v sinꝋ, where v is the magnitude of the velocity and ꝋ is its direction
Analyze the motion of the projectile in the horizontal direction using the following equations:
Horizontal motion (ax = 0)
□ X = xo + vxt
□ Vx = vox = vx = velocity is a constant
Analyze the motion of the projectile in the verticle direction using the following equations:
Verticle motion (assuming positive is up ay = -g = -9.81 m/s2)
□ Y = yo + 1/2 (voy+vy)t
□ Vy = voy - gt
□ Y = yo + voyt - 1/2 gt2
□ V2y = v2oy - 2g(y-yo)
Recombine the horizontal and verticle components of location and/or velocity using the following equations:
□ s=√(x^2+y^2 )
□ ꝋ = tan-1 (y/x)
□ v=√(〖v_^2〗_x+〖v_^2〗_y )
□ ꝋv = tan-1(vy/vx)
The maximum height h of a pojectile launched with initial vertical velocity voy is given by:
□ h=〖v_^2〗_oy/2g
The maximum horizontal distance traveled by a projectile is called the range. The range R of a projectile on level ground launed at an angle ꝋo above the horizontal with initial speed vo is given by:
□ R = (〖v_^2〗_o sin〖2ꝋ_o 〗)/g
Velocities in two dimensions are added using the same analytical vector techniques, which are written as:
□ Vx = v cosꝋ
□ Vy = v sin ꝋ
□ V = √(〖v_^2〗_0+〖v_^2〗_o )
□ ꝋ = tan-1 (vy / vx)
35
A chipmunk scampers about collecting in its cheeks safflower seeds that the birds dropped from the feeder hanging overhead. Initially, the little creature is at position vectors r1x = 2.57m and r1y= -1.19m. After filling up, it runs to the hole at position vector r2x =- 1.09m and r2y = 4.21 m that leads to its underground nest.
Find component Δrx (NOT Δry) of the chipmunk's displacmeent vector for this expedition
Δrx = -3.66m
Explanation:
The displacement vector is the difference between the final and initial position vectors: Δ𝑟⃗ =𝑟⃗ 2−𝑟⃗
Where Δ𝑟⃗ is the displacement vector and 𝑟⃗1 and 𝑟⃗ 2 are the initial and final position vectors.
Express this in terms of components.
Δ𝑟𝑥=𝑟2𝑥−𝑟1𝑥
Now, substitute the given values for the position vectors' components to obtain the numerical values for the components of the displacement vector.
Δ𝑟𝑥=(−1.09 m)−(2.57 m)=−3.66 m
36
A chipmunk scampers about collecting in its cheeks safflower seeds that the birds dropped from the feeder hanging overhead. Initially, the little creature is at position vectors r1x = 2.57m and r1y= -1.19m. After filling up, it runs to the hole at position vector r2x =- 1.09m and r2y = 4.21 m that leads to its underground nest.
Find component Δry (NOT Δrx) of the chipmunk's displacement vector for this expedition
Δry = 5.40 m
Explanation:
The displacement vector is the difference between the final and initial position vectors: Δ𝑟⃗ =𝑟⃗ 2−𝑟⃗
Where Δ𝑟⃗ is the displacement vector and 𝑟⃗1 and 𝑟⃗ 2 are the initial and final position vectors.
Express this in terms of components.
Δ𝑟𝑦=𝑟2𝑦−𝑟1𝑦
Now, substitute the given values for the position vectors' components to obtain the numerical values for the components of the displacement vector.
Δ𝑟𝑦=(4.21 m)−(−1.19 m)=5.40 m
37
A ship leaves a harbor and sails at 19.5o to the south of due east. After traveling 395 km, how far EAST is the ship from the harbor?
Distance east? _____ km
Distance east: 372 km
Explanation:
Between south and east, east is the x-component
□ Rx = r cos (𝜃)
□ 𝑟east=𝑟cos(𝜃)
□ 𝑟east=(395 km)cos(19.5∘)=372 km
38
A ship leaves a harbor and sails at 19.5o to the south of due east. After traveling 395 km, how far SOUTH is the ship from the harbor at this point?
Distance south? _____km
Distance south: 132 km
Explanation:
Between south and east, south is the y-component
□ Ry = r sin (𝜃)
□ 𝑟east=𝑟 sin(𝜃)
□ 𝑟east=(395 km)sin(19.5∘)=132 km
39
Adam drops a ball from rest from the top floor of a building at the same time Bob throws a ball horizontally from the same location. Which ball hits the level ground first? (Neglect any affects due to air resistance)
a. They both hit the ground at the same time
b. Adam's ball
c. Bob's ball
d. It depends on how fast Bob throws the ball
e. It depends on how fast the ball falls when Adam drops it
a. They both hit the ground at the same time
Explanation:
The vertical motion of both balls is governed by the same force: gravity. The only way the vertical component of each ball's motion could be different is if they were acted upon by different forces.
40
A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 180 m above the ground below. A cannonball is fired horizontally with an initial speed of 800 m/s.
Assuming air resistance can be neglected, approximately how long will the cannonball be in flight before it hits the ground?
a. 3s
b. 60s
c. 72s
d. 6s
e. 36s
d. 6s
Explanation:
Recall that an object starting from rest and accelerating at a constant rate 𝑎 for a period of time 𝑡 covers a total distance 𝑑 given by:
𝑑=(1/2)𝑎𝑡2
In this case, the vertical velocity of the ball starts at zero and increases at a rate of about 10 m/s2. After one second, then, the ball will have fallen a vertical distance of
𝑑1=(1/2)×(10 m/s2)×(1 s)2=5 m
In the same way, after a total time of 2 s, you find that the ball will have fallen a total vertical distance of about 20 m. After 3 s, it will have fallen a total vertical distance of 45 m, and so on. In general, if you know the total vertical distance fallen, you can find the time taken to fall that distance by solving the free‑fall distance equation for time.
t=√(2d/a)
In this problem, you are told that the cannonball starts out 180 m above the ground, so you can conclude it will take about 6s for the ball to reach the ground.
41
A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 180 m above the ground below. A cannonball is fired horizontally with an initial speed of 800 m/s.
Q2: About how far from the base of the tower will the cannonball land?
a. 5000m
b. 60,000m
c. 60m
d. 800m
e. 360m
a. 5000m
Explanation:
Because the horizontal speed of the ball remains constant at 800 m/s800 m/s for a total time of 6 s,6 s, the total horizontal distance covered is 5000 m,5000 m, obtained by multiplying the constant rate of travel (800 m/s)(800 m/s) by the total travel time (6 s).
42
A jetliner is moving at a speed of 245 m/s. The vertical component of the plane's velocity is 40.6 m/s. Determine the magnitude of the horizontal component of the plane's velocity.
Answer: Vx = 242m/s
Explanation:
Since the plane has a vertical component, it is either climbing, or losing altitude. The answer will be the same for both.
Draw an image. Should be a triangle with vx on the bottom, vy = 40 m/s and the hypotenuse being 245m/s
Pythagoras: v2 = vx2 + vy2
Solve for vx
Vx = √(v^2−v_y^2 )
Plug in:
Vx = √(245m/s^2 −40m/s^2 )
Vx = 242m/s
43
The punter on a football team tris to kick a football so that it stays in the air for a long "hang time." If the ball is kicked with an initial velocity of 25.0 m/s at an angle of 60.0o above the ground, what is the "hang time"?
Answer: T = 4.42 seconds
Explanation:
y= 0 (since initial and final heights are 0)
Ay = -9.8 m/s
Voy = +21.7 m/s
Found using vy = v sin 60o (25 * sin60)
T = ?
Y = yoyt + (1/2)ayt2
0m = (21.7m/s) t - (1/2)(9.81m/s2)t2
Divide both sides by t
0m = (21.7m/s) - (4.9m/s2)t
T = (−21.7(m/s))/(−4.9(m/s^2 ))
T = 4.42 seconds
44
During a baseball game, a fly ball is hit to center field and is caught 115 m from home plate. Just when the ball is caught, a runner on third base takes off for home, and the center fielder throws the ball to the catcher standing on home plate. The runner takes 3.50 seconds to reach home, while the baseball is thrown with a velocity whose horizontal component is 41 m/s. Which reaches home first, the runner or the ball and by how much time?
Answer: The ball by 0.70 seconds
Explanation:
The main idea is to calculate how long the throw from the center fielder takes to reach the home plate
The horizontal ocmponent, Vx, of the ball's velocity is constant throughout the motion. It will determine when the ball is above home plate. So it's a constant velocity problem.
Solve for t:
x=vxt
t= x/vx
Substitute:
T = 115m / 41m/s = 2.80 seconds
Since the runner takes 3.50 seconds to reach home, the ball gets there first by (3.50 - 2.80) seconds = 0.70 seconds
45
A car drives straight off the edge of a cliff that is 54 meters high. The police at the scene of the accident note that the point of impact is 130 m from the base of the cliff. How fast was the car traveling when it went over the cliff?
Answer: 39.2 m/s (or 88mph)
Explanation:
Known variables
Y = 54 m
Ay = 9.81 m/s2
Voy = no initial vertixle component
T = find
Equation: yoyt + (1/2)ayt2
So 54m = 0 + (1/2)(9.81 m/s2)t2
T^2 = 54m/((1/2)9.81(m/s^2 ))=11.02 seconds^2
During the 3.32 second of free fall, the horizontal motion of the car was at constant speed. The horizontal extent was x = 130m so
X = vxt -> vx = x/t = 130m/3.32s = 39.2m/s
This is the speed at which the car drove off the cliff (about 88mph)
46
A soccer player kicks the ball toward a goal that is 29.00m in front of him. The ball leaves his foot at a speed of 19.0 m/s and an angle of 32.0o above the ground. Find the speed of the ball when the goalie catches it in front of the net. (Note: The answer is not 19.0m/s)
A: 17.8m/s
Explanation:
Obtain horiztonal and verticle components of velocity
Vx = v cosꝋ = (19m/s)cos(32) = 16 m/s
Vy = v sinꝋ = (19m/s) sin(32) = 10.1 m/s
From horizontal components find time for ball to reach goalie
Vx = Δx/Δt
Δt = Δx/v = 29m / 16.1m/s = 1.80 sec
Use this to find vy the final y-component of velocity
Ay = -9.81m/s2
Vy = ?
Voy = 10.1 m/s (calculated above)
T = 1.80 seconds
Vy = voy + ayt
10.1m/s - (9.81 m/s2)(1.80s) = -7.55m/s
So ball has a downward component. Use pythagoros to find the magnitude of the final velocity
V = √((16.1m/s)^2 (7.55m/s)^2 )
V = 17.8m/s
47
Freddy starts out skiing Northwards at 8.0 m/s and ends up moving Westward at 6.0 m/s. Using the definition aav = Δv / Δt, what is the direction of his average acceleration?
a = 53o west OR 37o West of South
48
Claudine throws a rock off a cliff at an angle of 33o above the horizontal with speed 24.5 m/s. It takes 3.80 seconds to hit the water below. How high above the water is the rock released?
Answer: 20.3 m above the water
Explanation:
Variables:
Δy = find
A = -9.81 m/s2
T = 3.80sec
Voy = +13.3 m/s
Obtained through drawing a triangle, and using vy = v sin(33)
Vy = (24.5m/s)sin33
Vy = 13.3m/s
Chose equation: Δy = voyt + (1/2)at2 (Note: The 24.5 m/s not used drection in the equation)
Δy = (13.3m/s)(3.80s) - (1/2)(9.81m/s2)(3.80s2)
Δy = -20.3 meters
So the rock is released 20.3 meters above the water
