Chapter 3: Kinematics in 2 dimensions

Question 1

What is the horizontal motion equation?

Answer 1

Δx = Vxt

Question 2

What does Vx represent in the horizontal equation?

Answer 2

The velocity on the x axis (horizontal)

Question 3

What is the equation for verticle motion?

Answer 3

Δy = (Vyt)

Question 4

What does Vy represent in the vertical motion equation?

Answer 4

The velocity on the y axis (verticle motion)

Question 5

T/F: Verticle motion has constant acceleration

Answer 5

True, meaning we need to know the 4 equations of constant acceleration

Question 6

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

Answer 6

A

A) They both hit the ground at the same time

Question 7

Define analytical method

Answer 7

the method of determining the magnitude and direction of a resultant vector using the Pythagorean theorem and trigonometric identities

Question 8

Define classical relativity

Answer 8

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

Question 9

Define commutative

Answer 9

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

Question 10

Define component (of a 2-d vector)

Answer 10

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

Question 11

Define direction (of a vector)

Answer 11

the orientation of a vector in space

Question 12

Define head (of a vector)

Answer 12

the end point of a vector; the location of the tip of the vector’s arrowhead; also referred to as the “tip”

Question 13

Define head-to-tail method

Answer 13

a method of adding vectors in which the tail of each vector is placed at the head of the previous vector

Question 14

Define kinematics

Answer 14

the study of motion without regard to mass or force

Question 15

Define magnitude (of a vector)

Answer 15

the study of motion without regard to mass or force

Question 16

Define motion

Answer 16

displacement of an object as a function of time

Question 17

Define projectile

Answer 17

an object that travels through the air and experiences only acceleration due to gravity

Question 18

Define projectile motion

Answer 18

the motion of an object that is subject only to the acceleration of gravity

Question 19

Define range

Answer 19

the maximum horizontal distance that a projectile travels

Question 20

Define relative velocity

Answer 20

the velocity of an object as observed from a particular reference frame

Question 21

Define relativity

Answer 21

the study of how different observers moving relative to each other measure the same phenomenon

Question 22

Define resultant

Answer 22

the sum of two or more vectors

Question 23

Define resultant vector

Answer 23

the vector sum of two or more vectors

Question 24

Define scalar

Answer 24

a quantity with magnitude but no direction

Question 25

Define tail

Answer 25

the start point of a vector; opposite to the head or tip of the arrow

Question 26

Define trajectory

Answer 26

the path of a projectile through the air

Question 27

Define vector

Answer 27

a quantity that has both magnitude and direction; an arrow used to represent quantities with both magnitude and direction

Question 28

Define vector addition

Answer 28

the rules that apply to adding vectors together

Question 29

Define velocity

Answer 29

speed in a given direction

Question 30

Graphical method of adding vectors (explanation- not question, just important to know)

Answer 30

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

Question 31

Graphical method of subtracting vectors (explanation- not question, just important to know)

Answer 31

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.

Question 32

Are vectors commutative? What does this mean?

Answer 32

Yes, this means vectors A and B can do the following: A + B = B + A

Question 33

Steps to add vectors A and B using the analytical method (Explanation, not question)

Answer 33

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)

Question 34

How to solve projectile motion problems (LONG explanation)

Answer 34

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)

Question 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

Answer 35

Δrx = -3.66m

Explanation:

The displacement vector is the difference between the final and initial positionvectors: Δ𝑟⃗=𝑟⃗2−𝑟⃗

WhereΔ𝑟⃗is the displacement vector and𝑟⃗1and𝑟⃗2are the initial and final positionvectors.

Express this in terms ofcomponents.
Δ𝑟𝑥=𝑟2𝑥−𝑟1𝑥

Now, substitute the given values for the position vectors’ components to obtain the numerical values for the components of the displacementvector.

Δ𝑟𝑥=(−1.09m)−(2.57m)=−3.66m

Question 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

Answer 36

Δry = 5.40 m

Explanation:

The displacement vector is the difference between the final and initial positionvectors: Δ𝑟⃗=𝑟⃗2−𝑟⃗

WhereΔ𝑟⃗is the displacement vector and𝑟⃗1and𝑟⃗2are the initial and final positionvectors.

Express this in terms ofcomponents.
Δ𝑟𝑦=𝑟2𝑦−𝑟1𝑦

Now, substitute the given values for the position vectors’ components to obtain the numerical values for the components of the displacementvector.

Δ𝑟𝑦=(4.21m)−(−1.19m)=5.40m

Question 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

Answer 37

Distance east: 372 km

Explanation:

Between south and east, east is the x-component
□ Rx = r cos (𝜃)
□ 𝑟east=𝑟cos(𝜃)
□ 𝑟east=(395km)cos(19.5∘)=372km

Question 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

Answer 38

Distance south: 132 km

Explanation:

Between south and east, south is the y-component
□ Ry = r sin (𝜃)
□ 𝑟east=𝑟 sin(𝜃)
□ 𝑟east=(395km)sin(19.5∘)=132km

Question 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

Answer 39

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 differentforces.

Question 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

Answer 40

d. 6s

Explanation:

Recall that an object starting from rest and accelerating at a constant rate𝑎for a period of time𝑡covers a total distance𝑑givenby:

𝑑=(1/2)𝑎𝑡2

In this case, the vertical velocity of the ball starts at zero and increases at a rate of about 10m/s2. After one second, then, the ball will have fallen a vertical distanceof

𝑑1=(1/2)×(10m/s2)×(1s)2=5m

In the same way, after a total time of 2s, you find that the ball will have fallen a total vertical distance of about 20m. After 3s, it will have fallen a total vertical distance of 45m, 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 fortime.

t=√(2d/a)

In this problem, you are told that the cannonball starts out180mabove the ground, so you can conclude it will take about6sfor the ball to reach theground.

Question 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

Answer 41

a. 5000m

Explanation:

Because the horizontal speed of the ball remains constant at800m/s800m/sfor a total time of6s,6s,the total horizontal distance covered is5000m,5000m,obtained by multiplying the constant rate of travel(800m/s)(800m/s)by the total travel time(6s).

Question 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 42

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

Question 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 43

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

Question 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 44

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

Question 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 45

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)

Question 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)

Answer 46

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

Question 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?

Answer 47

a = 53o west OR 37o West of South

Question 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 48

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