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CHAPTER 3 Motion in Two Dimensions;

Vectors

http://www.physicsclassroom.com/Class/vectors/U3L3a.html

• Vectors and Scalars • Addition of Vectors – Graphical Methods • Subtraction of Vectors, and Multiplication of a Vector by a Scalar • Adding Vectors by Components • Projectile Motion • Solving Problems Involving Projectile Motion • Projectile Motion Is Parabolic • Relative Velocity

Vector vs. Scalar Review

• All physical quantities encountered in this text will be either a scalar or a vector • A vector quantity has both magnitude (size) and direction • A scalar is completely specified by only a magnitude (size)

Vectors and Scalars

Vectors Examples of Scalars and vectors:

A vector has magnitude as well as direction. Some vector quantities: displacement, velocity, force, momentum A scalar has only a magnitude. Some scalar quantities: mass, time, temperature

Properties of Vectors • Equality of Two Vectors

– Two vectors are equal if they have the same magnitude and the same direction • Movement of vectors in a diagram

– Any vector can be moved parallel to itself without being affected • Negative Vectors

– Two vectors are negative if they have the same magnitude but are 180° apart (opposite directions)

• Resultant Vector

– The resultant vector is the sum of a given set of vectors Addition of Vectors – Graphical Method

Scalars Distance Mass Speed Time Work

Vectors Displacement Weight Velocity Acceleration Momentum

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For vectors in one dimension, simple addition and subtraction are all that is needed. You do need to be careful about the signs, as the figure indicates.

If the motion is in two dimensions, the situation is somewhat more complicated.

Adding the vectors in the opposite order gives the same result:

Even if the vectors are not at right angles, they can be added graphically by using the “tail-to-tip” method.

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The parallelogram method may also be used; here again the vectors must be “tail-to-tip.”

Subtraction of Vectors, and Multiplication of a Vector by a Scalar

In order to subtract vectors, we define the negative of a vector, which has the same magnitude but points in the opposite direction.

Adding Vectors by Components

Adding vectors: 1. Draw a diagram; add the vectors graphically. 2. Choose x and y axes. 3. Resolve each vector into x and y components. 4. Calculate each component using sine and cosine. Add the components in each direction. 5. To find the length and direction of the vector, use:

Example 1: A mail carrier leaves the post office and drives 22.0 km in a northerly direction. She then drives in a direction 60 degrees south of east for 47.0 km. What is her displacement from the post office?

Resolve each displacement vector.

Resultant vector D

1D 1 10 22.0x yD D km

2D2

2

(47.0 )(cos60 ) (47.0 )(0.500) 23.5

(47.0 )(sin 60 ) (47.0 )(0.866) 40.7

o

x

o

y

D km km km

D km km km

1 2

1 2

0 23.5 23.5

22.0 ( 40.7 ) 18.7

x x x

y y y

D D D km km km

D D D km km km

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Combine the vectors Use the theorem of Pythagoras

The angle of displacement

Example 2: An airplane travels east for 620 km: then travels southeast for 440 km; and finally 53 degrees south of west for 550 km. What is the plane’s total displacement?

Combine the vectors

1 2 3

1 2 3

620 311 331 600

0 311 439 750

x x x x

y y y y

D D D D km km km km

D D D D km km km km

Displacement is: Angle of displacement is:

23.5 18.7x yD km D km

2 2 2 2(23.5 ) ( 18.7 )x yD D D km km 30.0km

18.7tan 0.796 38.5

23.5

y o

x

D km

D km

1 1 1 1

1 1

2 2 2

2 2

3 3 3

3 3

: cos0 620

sin 0 0

: cos 45 (440 )(0.707) 311

sin 45 (440 )(0.707) 311

: cos53 (550 )(0.602) 331

sin 53 (550 )(0.

o

x

o

y

o

x

o

y

o

x

o

y

D D D D km

D D km

D D D km km

D D km km

D D D km km

D D km

799) 439km

2 2 2 2(600 ) ( 750 ) 960R x yD D D km km km

750tan 1.25 = 51

600

y o

x

D km

D km

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Projectile Motion

A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

It can be understood by analyzing the horizontal and vertical motions separately.

• An object may move in both the x and y directions simultaneously

– It moves in two dimensions • The form of two dimensional motion we will

deal with is called projectile motion

Assumptions of Projectile Motion

• We may ignore air friction • We may ignore the rotation of the earth • With these assumptions, an object in projectile motion will follow a parabolic path

Rules of Projectile Motion

• The x- and y-directions of motion are completely independent of each other • The x-direction is uniform motion

– ax = 0

• The y-direction is free fall – ay = g (for objects falling downward and –g for all other directions)

• The initial velocity can be broken down into its x- and y-components _ cos sinOx O O Oy O Ov v v v

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The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g. This photograph shows two balls that start to fall at the same time. The one on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.

If an object is launched at an initial angle of θo with the horizontal, the

analysis is similar except that the initial velocity has a vertical component.

Projectile Motion at Various Initial Angles

• Complementary values of the initial angle result in the same range

– The heights will be different • The maximum range occurs at a projection angle

of 45o

Some Details about the Rules

• x-direction – ax = 0

– – x = vxot

• This is the only operative equation in the x-direction since there is uniform velocity in that direction

cos constantxo o o xv v v

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• y-direction – – free fall problem

• a = – g – take the positive direction as upward – uniformly accelerated motion, so the motion equations all hold

Velocity of the Projectile • The velocity of the projectile at any point of its motion is the vector sum of its x and y

components at that point

– Remember to be careful about the angle’s quadrant

Solving Problems Involving Projectile Motion

1. Read the problem carefully, and choose the object(s) you are going to analyze. 2. Draw a diagram. 3. Choose an origin and a coordinate system. 4. Decide on the time interval; this is the same in both directions, and includes only the time

the object is moving with constant acceleration g. 5. Examine the x and y motions separately. 6. List known and unknown quantities. Remember that vx never changes, and that vy = 0 at

the highest point. 7. Plan how you will proceed. Use the appropriate equations; you may have to combine some

of them.

Some Variations of Projectile Motion

• An object may be fired horizontally • The initial velocity is all in the x-direction

– vo = vx and vy = 0

• All the general rules of projectile motion apply

Example 3: A plane drops a package of rations to a hiker. The plane is traveling horizontally at 40.0 m/s at

a height of 21.00 10x m above the ground. (a) Where does the package strike the ground

relative to the point at which it was released?

2

2 2 2

1

2

1.00 10 (4.90 / )

4.52

o oyy y y v t gt

x m m s t

t s

sinyo o ov v

2 2 1tany

x y

x

vv v v and

v

(40.0 / )(4.52 ) 181

o oxx x x v t

x m s s m

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(b) What are the horizontal and vertical components of the velocity of the package just before it hits the ground?

Example 4: A jet traveling horizontally at 21.0 10 /x m s drops a rocket from a high altitude. The rocket immediately

fires its engines, accelerating at 220.0 /m s in the x-direction while falling under the influence of gravity in the y-direction. When the rocket has fallen 1.00 km, find: (a) its velocity In the y-direction.

(b) Its velocity in the x-direction.

(c) the magnitude and direction of its velocity. 2 2 2 2 2(386 / ) ( 1.40 10 / ) 411 /x yv v v m s x m s m s

Non-Symmetrical Projectile Motion

• Follow the general rules for projectile motion • Break the y-direction into parts

– up and down – symmetrical back to initial height and then the

rest of the height

Example 5: A stone is thrown upward from the top of a building at

an angle of 30.0o to the horizontal and with an initial speed of 20.0 m/s. The point of release is 45.0 m above the ground. (a) How long does it take for the stone to hit the ground?

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cos (40.0 / )cos0 40.0 /

sin 0 (9.80 / )(4.52 ) 44.3 /

o

x o

y o

v v m s m s

v v gt m s s m s

2 2 2y oyv v g y 2 2 30 2( 9.8 / )( 1.00 10 )yv m s x m

21.40 10 /yv x m s

y oy yv v a t 2 21.40 10 / 0 (9.80 / ) 14.3x m s m s t t s 2 21.00 10 / (20.0 / )(14.3 )x ox xV v a t x m s m s s 386 /m s

21 1 1.40 10 /

tan tan 19.9386 /

y o

x

v x m s

v m s

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cos (20.0 / )(cos30.0 ) 17.3 /

sin (20.0 / )(sin30.0 ) 10.0 /

o

ox o o

o

oy o o

v v m s m s

v v m s m s

2 245 (10.0 / ) (4.90 / )m m s t m s t

(b) Find the stone’s speed at impact.

(c) What is the horizontal range of the stone?

Relative Velocity

• Relative velocity is about relating the measurements of two different observers • It may be useful to use a moving frame of reference instead of a stationary one • It is important to specify the frame of reference, since the motion may be different in

different frames of reference • There are no specific equations to learn to solve relative velocity problems

Relative Velocity Notation

• The pattern of subscripts can be useful in solving relative velocity problems • Assume the following notation:

– E is an observer, stationary with respect to the earth – A and B are two moving cars

Example 6: A motorcycle speeds horizontally off a 50.0m high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0m from the base of the cliff? given find formula

21

2o oyy y y v t gt 4.22t s

210 / (9.80 / )(4.22 )y oyv v gt m s m s s 31.4 /m s

2 2 2 2(17.3 / ) ( 31.4 / )x yv v v m s m s 35.9 /m s

( cos ) (20.0 / )( 30 )(4.22 )o

o ox x x v t m s cos s 73.1m

2

0

90.0

50.0

0

9.80 /

0

o o

x

y

yo

x y

x m

y m

a

a g m s

v

xov

t

21

2

2

y gt

yt

g

21

2o xo x

xo

xo

x x v t a t

x v t

xv

t

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Example 7: A football is kicked at an angle 37.0o

o with a velocity of 20.0m/s.

Calculate: (a) The maximum height:

During this time interval, the acceleration is g downward. The time to reach maximum height is:

This time is needed to calculate height.

Or

(b) The time of travel before the football hits the ground: (c) How far away it hits the ground:

2

2 2( 50.0 )3.19

9.80 /

y mt s

g m s

90.0

28.2 /3.19

xo

x mv m s

t s

cos37.0 (20.0 / )(0.799) 16.0 /

sin37.0 (20.0 / )(0.602) 12.0 /

o

xo o

o

yo o

v v m s m s

v v m s m s

2

(12.0 / )1.22

(9.80 / )

yov m st s

g m s

21

2yoy v t gt 2 21

(12.0 / )(1.22 ) (9.80 / )(1.22 ) 7.352

m s s m s s m

2 2 2 2

2

(12.0 / ) (0 / )7.35

2 2(9.80 / )

yo yv v m s m sy m

g m s

2

2 2

2

2

10 0 (12.0 / ) (9.80 / )

2

1[ (9.80 / ) 12.0 / ] 02

2(12.0 / )2.45

(9.80 / )

o yoy y v t gt

m s t m s t

m s t m s t

m st s

m s

(16.0 / )(2.45 ) 39.2xox v t m s s m

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(d) The velocity vector at the maximum height: At the highest point, there is no vertical component to the velocity. There is only the horizontal component.

(e) The acceleration vector at maximum height is the same throughout, 29.80 /m s down. Example 8: A cannon has a muzzle velocity of 60 m/s. At what angle should it have been aimed to strike a target 320 m away? The use of the range equation can be used as long as the cannon is on level ground and the projectile lands on the same level ground.

Thus, 12 60.6 (0.871sin )o is a solution, but 2 180 60.6 119.4o o o

o is a solution. These

solutions are 2 so divide each by two:

Relative Position Equations

• AEr is the position of car A as measured by E

• BEr is the position of car B as measured by E

• ABr is the position of car A as measured by car B

• AB AE EBr r r

Relative Position

• The position of car A relative to car B is given by the vector subtraction equation

The rate of change of the displacements gives the relationship for the velocities

cos37 16.0 /o

xo ov v v m s

2 sin 2o ovR

g

2

2 2

(320 )(9.80 / )sin 2 0.871

(60.0 / )o

o

Rg m m s

v m s

30.3 and 59.7o o

o

AB AE EBv v v

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Problem-Solving Strategy: Relative Velocity

• Label all the objects with a descriptive letter • Look for phrases such as “velocity of A relative to B”

– Write the velocity variables with appropriate notation – If there is something not explicitly noted as being relative to something else, it is

probably relative to the earth • Take the velocities and put them into an equation

– Keep the subscripts in an order analogous to the standard equation • Solve for the unknown(s)

Example 9: A boat’s speed in still water is 1.85 /BWv m s . If the boat is to travel directly

across a river whose current has speed

1.20 /WSv m s , at what upstream angle must the

boat head?

Example 10: The same boat 1.85 /BWv m s

now heads directly across the river whose current is still 1.20 m/s. (a) What is the velocity (magnitude and direction) of the boat relative to the shore?

(b) If the river is 110 m wide, how long will it take to cross and how far downstream will the boat be then?

1.20 /sin 0.6486

1.85 /

WS

BW

v m s

v m s

40.4o

BS BW WSv v v

2 2 2 2(1.85 / ) (1.20 / )BS BW WSv v v m s m s 2.21 /m s

tan / (1.20 / ) /(1.85 / ) 0.6486WS BWv v m s m s 33.0o

/ 110 /1.85 / 60BWv D t t m m s s

(1.2 / )(60 ) 72WSd v t m s s m

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CHAPTER 3

MOTION IN TWO DIMENSIONS CONCEPTS

1. A river flows due east at 1.5 m/s. A ship leaves the north shore of the river and heads due south at 2.0 m/s, as shown to the right. The vector that best represents the resultant velocity of the ship relative to the riverbank is A.

2. The vector that represents the resultant of the concurrent vectors A and B in the diagram to the right is B.

3. The vector that best represents the resultant of forces 1F and 2F acting

concurrently on point P is B.

4. A ball rolls down a curved ramp from a table top. The dotted line that best represents the path of the ball after leaving the ramp is B.

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5. An arrow, A is shot horizontally from a bow at a speed of 50 m/s as shown to the right. A second arrow, B, is dropped from the same height and at the same as A is fired. Neglecting air friction, compared to the amount of time A takes to strike the ground, the amount of time B takes to strike the ground is the same. 6. A ball is fired vertically upward at 5.0 m/s from a cart moving horizontally to the right at 2.0 m/s. The vector that best represents the resultant velocity of the ball when fired is B. 7. The diagram to the right shows a ball projected horizontally with an initial velocity of 20.0 m/s east, off a cliff 100-m high. During the flight of the ball, the direction of its acceleration is westward. 8.A cannon elevated at an angle of 35o to the horizontal fires a cannonball, which travels the path shown in the diagram to the right. If the angle of elevation of the cannon is decreased from 35o to 30o, the vertical component of the ball’s initial velocity will increase and its horizontal component will increase. 9. A projectile is launched at an angle 60o above the horizontal. Compared to the vertical component of the initial velocity of the projectile, the vertical component of the projectile’s velocity when it has reached its maximum height is greater. 10. An object is thrown into the air and follows the path in the diagram to the right. The vector that best represents the acceleration of the object at point A is B. 11. A vector cannot have a zero magnitude if one of its components is not zero. 12. An operation that will not change a vector is when it is translated parallel to itself. 13. If the acceleration of an object always directed perpendicular to its velocity, the object is turning. 14. If three forces, each having a magnitude of 30N, pull on an object in directions that are 120o apart from each other, the resultant force is zero.

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15. If several projectiles are fired at the same time, the one that will be in the air the longest is the one with the highest maximum elevation. 16. A water-gun should be aimed at 45o in order for the water to land with the greatest horizontal range. 17. The acceleration of gravity on the moon is only one-sixth of that on earth. If you hit a baseball on the moon with the same effort (and at the same speed and angle) that you would on earth, the ball would land 6 times as far. 18. Ignoring air resistance, the vertical component of a projectile’s velocity continuously decreases. 19. You hit a tennis ball over the net. When the ball reaches its maximum height, its speed is less than its initial speed. 20. A pilot drops a bomb from a plane flying horizontally. When the bomb hits the ground, the horizontal location of the plane will be over the bomb. 21. A stone is thrown horizontally from the top of a tower at the same instant that a ball is dropped vertically. The stone is traveling faster when it hits the level ground below. 22. An Olympic athlete throws a javelin at four different angles above the horizontal, each with the same speed: 30o, 40o, 60o, 80o. The 30o and 60o throws cause the javelin to land at the same place. 23. If three vectors add to zero, they must be in the same plane. 24. Ignoring air resistance, the horizontal component of a projectile’s velocity remains constant. 25. Ignoring air resistance, the horizontal component of a projectile’s acceleration is zero. 26. A projectile is launched at an θ to the horizontal with an initial velocity v. In the absence of any air resistance the horizontal velocity remains the same.

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27. A projectiles acceleration remains constant when in flight near the earth. 28. For a projectile launched at an angle of 45o to the horizontal, the vertical speed decreases on the way up and increases on the way down. 29. Two forces act on an object concurrently. The resultant will be greatest when the angle between the forces is 0o. 30. A single force can be resolved into an unlimited number of components.