Vectors

Unit I Part I

Notes

• In physics, there are two types of measurable quantities: vectors and scalars

• Scalars have magnitude (size) only

• Examples include: time, mass, volume, area, distance, speed, work, energy, power

Notes

• Examples of scalars:– Distance (d, m) is the length of a path travelled.

For example, running north for 300m and then running south for 200m means you ran a total distance of 500m

– Speed (v, m/s) is the rate of distance travelled (also known as distance travelled in a time interval)

Notes

• Vectors have both magnitude and direction

• Examples include: displacement, velocity, acceleration, force, momentum, impulse, field strength– Displacement (d, m) is the change in position of

an object.

– For example, running north for 300m and then running south for 200m means your total displacement is 100m north from your starting point

Notes

– Velocity (v, m/s) is the rate of change in position (also known as net displacement in a time interval)

– Acceleration (a, m/s2) is the rate of change in velocity (also known as the change in velocity divided by time)

Notes

• With vectors, you must include direction in your calculations

• Types of Directions:

( - )

Example

• A man walks 20m east, then 30m west. What is the distance travelled by the man? What is his displacement?

Example

• A car drives 10km north in 20 minutes. What is the speed of the car in m/s? What is the velocity of the car?

Notes

• To show a vector graphically, draw an arrow in the direction of the vector and scale the length to reflect the vector’s magnitude

3 km east

5 km east

3 km north

Scale 1cm:1km

Notes

• To show a vector that is not N, S, E or W, use a grid

• The direction of this vector can be described in 2 ways:– 25o north of due east, or 25oN of E– 65o east of due north, or 65oE of N

E

N

25o65o

Practice

• Draw three vectors on a sheet of paper. Pass the piece of paper to your partner and have him/her describe each vector in terms of magnitude and direction

• Example:

80o

75o

Scale 1cm:1km

Practice

• Once you’ve had a chance to answer your partners’ diagrams, hand it back to them for them to evaluate. Discuss any disagreements.

• Hand in each paper when you and your partner have completed the activity

• Collect a worksheet after you’ve handed in your paper

Notes

• Vectors can be added together in two ways:– Tip-to-tail Method

– Component Method

Notes• In the tip-to-tail method, draw the vectors one

after another. The resultant (vector sum) will extend from the tail of the first vector to the tip of the second.

• In this method, it is crucial that you draw vectors to scale

• You can then measure the length and angle of the resultant

+ = =V1

V2

V1 V1

V2 V2R

θ

Example

• A man walks 250m due east, then 250m 60o north of east. Determine the magnitude and direction of the resultant displacement

Example

• Add the displacements D1 + D

2 + D

3 = R where

D1 = 6km north, D

2 = 3km east, and D

3 = 4km

at 45oS of E

Notes

• Subtracting vectors (A – B) is the same as adding a negative vector (A + (-B) )

• To make a vector negative, change its direction

A B+ =R

A + -B = R

Example

• Given the vectors shown, draw diagrams and find R for:1. A + B = R

2. A – B = R

18o

A = 6.3cm

B =5.0cm

53o

Notes• You can also use trigonometry when solving

vector addition and subtraction problems– Draw a diagram using the tail to tip method

(should be close, but doesn’t need to be to scale)

– Use the pythagorean theorem to find the magnitude of the resulting vector

– Use trigonometry to find the direction

Example

• Given these vectors: A=15m/s N B=25m/s E

• Find the magnitude and direction of the following:1. A + B = R

2. A – B = R

Notes

• Vectors are important when dealing with relative velocity

• Imagine a car moving 20m/s north on a highway. Another car moves south at 25m/s. These are observations you make if you are stationary observer outside the frame of reference– If you are in car A, car B appears to be moving 45m/s

south– If you are in car B, car A appears to be moving 45m/s

north

Example

• A plane heads due north at 145m/s. What is the plane’s velocity relative to the ground if there is a 25m/s wind blowing due west?

Notes

• Navigators of boats and planes must correct for wind and water currents

• Using the last example, what direction would the pilot have to aim the plane in order end up straight north?

Example

• There is a 15.0m/s wind blowing due east as you ride your bike north at 9.0m/s. What is the velocity of the wind in your face?

Example

• A boat is capable of 12m/s in still water. If a river flows at 7.0m/s due east and is 500m wide:– What is the velocity of the boat relative to the

shore if the boat heads south, perpendicular to the current?

– How long would it take to cross the river?

Notes

• The other method of adding vectors is completed by breaking down the vector into its components (this is called resolving a vector)

• In general, components are either:– Horizontal and vertical

– Parallel and perpendicular

Notes

• To resolve a vector:1. Place the vector on an x-y grid

2. Drop perpendicular lines to each axis

3. Draw your components on each axis OR draw them tail to tip

y

x

y-comp.

x-comp.

y-comp.

Notes

• Once you’ve drawn the vector components, use trigonometry to find the magnitude of each component

• In the diagram below,

θx

yR

therefore

therefore x = R cosθ

Example

• A cannon is shot at a muzzle velocity of 1500m/s at an angle of 60o to the horizontal. What are the vertical and horizontal components of the velocity?

Example

• A boy pulls a wagon with a force of 100N at 40 degrees to the horizontal. Find the pulling force (F

x) and the lifting force (F

y)

Notes

• Components are not always drawn horizontally and vertically

• Sometimes it’s easier to show components that are parallel and perpendicular to the surface on which the object is on

Example

• A ball rests on a slope. What component of its weight presses down onto the surface and what component of its weight acts to pull the ball down the plane? The ball weighs 600N and the slope is 40o to the horizontal

Notes

• When adding vectors that have been resolved into their components, simply add the components that are in the same direction together

• This usually means adding all the x-components together and then adding all the y-components together

Example

• Add the following vectors:– 8N (N)

– 5N (45o N of W)

– 4N (30o S of E)

– 6N (20o N of E)

– 6N (S)

• Always start by drawing a diagram!

Projectiles

Unit I Part II

Notes• These equations should be familiar from Physics

11

• v=final velocity(m/s) vo

=initial velocity (m/s)

• a=acceleration (m/s2 ) d=displacement (m)

• t=time (s)

Notes

• In Physics 12, we will be looking at objects that are launched into the air

• In this case, the only acceleration considered is gravity which will always act vertically downwards regardless of how the object is travelling

Example

• A rocket is projected upwards at an initial velocity of 750m/s. If there is no air friction, – How long does the rocket take to reach its highest

point?

– How high does it go?

– If it lands at the same level as it was launched, how long is the rocket in the air?

Example

• Using a spring loaded launcher, a cart is quickly accelerated from rest over a distance of 1.62m in a time of 0.86s. It then continues at a constant speed along a horizontal table until it reaches a ramp that is sloped at 11.5o . Assuming negligible friction, how far up the ramp does the cart go before stopping?

Notes

• Let’s look at a distance-time graph of a baseball that is thrown straight upwards from ground level •Remember that

the slope of a d-t graph is equal to velocity•On curved graphs, slope is found using tangent lines

vo

v1/2

=0

vf = - v

o

d

t

Notes

• Now, here’s a velocity-time graph of the baseball’s motion

vo

-vo

0

•Remember that the slope of a v-t graph is equal to acceleration•The area under the line equals the net displacement of the ball

t

Notes• The first triangle is a positive area and

represents the upward displacement of the baseball

vo

-vo

0

•The second triangle is a negative area and shows the downward displacement from maximum height back to the ground

t

Example

• A stone is thrown vertically upward with an initial velocity of 11m/s. If the stone returns to its initial position, calculate the time the stone is in the air

Notes

• We also need to consider scenarios where the final position might be different from the initial

• Type 1: Landing above a launch position– v

o is positive

– a = - 9.8m/s2 – There will be two times (t)when displacement (d) is reached

Notes

• Type 2: Landing below a launch position– v

o is positive

– a = - 9.8m/s2

– There will be one time (t) when displacement (d) is reached

Example

• Wile E. Coyote wants to jump onto a cliff 30.0m high so he jumps straight upwards at 25.0m/s. He rises slightly higher than the edge of the cliff and then safely lands on the cliff edge. – How long is he in the air?

– How high did he actually jump?

Notes

• Now we will be looking at projectile motion problems that take place in two-dimensions

• These problems can be broken down into two types:– Horizontal Projectiles

– Projectiles at an angle

Notes

– In this diagram, the ball has both vertical and horizontal velocity components, even though initially the ball was thrown horizontally

• Horizontal Projectiles:

Notes• With any motion, it is important to note that:– Horizontal motion is unrelated to vertical motion

– Gravity only acts in the vertical direction

– Because gravity only acts in the vertical direction, horizontal velocity is constant (no acceleration)

– Time is the same for both directions

Vertical displacement

Horizontal displacement (range)

Hor. Motion Vert. Motion

V=constant Vo = 0

a = 0 a = -9.8m/s2

Notes• The horizontal motion is uniform motion

(when we neglect air friction)– Use uniform motion equation

• The vertical motion is uniformly accelerated motion – Use uniformly accelerated motion equations

Example

• A car going 22.0m/s runs off a 50.0m high cliff– How long does the car take to hit the ground?

– What is the range of the car?

– With what speed and direction does it hit?

Notes

• To find the final velocity, use the horizontal and vertical components of the velocity to create a triangle and solve for the resultant

Notes

• Projectiles at an angle:– In this diagram, the

initial velocity is in both the horizontal and vertical directions

– First, figure out the ‘x’ and ‘y’ components of the velocity

– Then, solve problems the same way as you would horizontal projectiles

Example

• An artillery shell is fired over level ground, at 400 m/s and at an angle of 50o to the horizontal. Find:– The total time in the air

– How high the shell rises

– The height of the shell after 25.0 s

– The range of the shell

Example

• A mass is shot at 25o above the horizontal at a velocity of 120m/s from a height of 65.0m. Find:– The time in the air

– The highest height of the projectile above the ground

– The range

– The final velocity