SPH 4U(G) GRADE TWELVE PHYSICS MR. MOORS

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Rodney.Moors@ocdsb.ca general booklet

Physics is the original science that studies the physical world. It used to be known as natural philosophy. Today it is mostly concerned with the behaviour and structure of matter.The study of physics often leads to mathematical relationships. These relation-ships are useful because they describe the nature of matter and they allow the accurate prediction of future events.

PHYSICS IS MATHEMATICAL.

PHYSICS CAN BE ABSTRACT, YOU MUST THINK.

PHYSICS DEMANDS HARD WORK.

SAY THE F-WORD OR ABOVE??????

(I decide what is above or below, e.g. the s-word is below)

THREE CHOICES

1. DO TEN PUSH-UPS

2. DO TWENTY CRUNCHES

3. SING I’M A LITTLE TEAPOT (with proper actions) IN FRONT OF CLASS

(this could also be foisted upon you in the halls!!)

PROPORTIONALITY STATEMENTS

The expression of how one quantity varies in relation to another is called a proportionality statement. In science and many other areas of thought people are looking for patterns that exist between different types of quantities. If there is a pattern to the data it is important to be able to deduce this relationshipLet’s start by graphing some given data. Graph this data on one quarter of a standard piece of graph paper. Place distance on the y-axis and time on the x-axis. Label your graph properly.

time (s) 1 2 3 4 5

distance (m) 0.2 1.6 5.4 12.8 25.0

The goal of this exercise is to find the pattern that exists (if one exists at all) between distance and time. The general proportionality statement is, y xn, while for this set of data it is d tn. In the first graph d vs. t1 was plotted. How does one decide if distance is proportional to time? Two quantities are proportional to each other if they produce a linear relationship when graphed. A linear relationship means that the ratio of any two pairs of data are equal or proportional. After graphing this data you should see that distance is not proportional to time.Plot d vs.t2 on another quarter of your graph paper. Is distance proportional to the square of time? Plot d vs. t3 on another quarter of your graph paper. Is distance proportional to the cube of time?

Since this graph is linear the proportionality statement is d t3. A couple of items to note, firstly we will only be studying data that includes the origin as a data point even though it will not be listed. Secondly, if we had graphed time on the y-axis and distance on the x-axis we would have obtained t d1/3 , after a graphical analysis.A proportionality statement can be analyzed more completely by finding the constant of proportionality, k.

y xn y = kxn (proportionality equation)

Therefore for our example d = kt3

We can determine the value of k by calculating the slope of the line in our final graph. This means the proportionality equation for the data given is d = 0.2t3 .

It is important to see the visual relationship in each of the graphs but there is a faster way to determine the

proportionality equation without graphing the data.

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

loglog

)(

x

xn

y

y

x

x

y

y

x

x

y

y

kx

kx

y

y

n

n

n

n

n

Any two sets of data points could be used.

n = 3

Express all exponents in integer or rational form.

Use any set of data to calculate k.

k = 0.232.0 td

This equation can be rearranged to another form of the equation.

dt

td

5

53

3

Write n values as fractions or whole numbers at all times in this unit. Write k values as decimals at all times.

Up to this point we have dealt with artificial data where the proportionality expression is perfectly upheld by all of the data points. In experiments that will be performed later the data obtained will not behave the same way and we will return to our graphing techniques to discover the proportionality expression. Graphing will determine the exponent while calculation of the slope of the best-fit line will provide the proportionality constant.

3

1

710.1 dt

Of course there are many mathematical relationships between variables that have already been discovered such as the following:circumference = 2r surface area of sphere = 4r2 F = ma F = Gm1m2/r2

If one was to plot circumference vs. radius one would expect to find the slope to be 2Plotting SA vs. r2 should produce a straight line with slope 4Plotting F vs. a should produce a straight line with a slope of whatever the constant mass is. Plotting F vs. 1/r2 should produce a straight line with a slope of Gm1m2 (these are held constant).Homework: Booklet Proportion 1-4

don’t do graphing for any

Using Proportionality Equations

A cylindrical water tank holds 10 000 L of water. The volume of the tank is given by the following equation.

hrV 2a) If the radius of the tank is 9 m. Then what is the height of the tank?b) What is the constant of proportionality for proportion between V and r?c) If the volume of the tank needs to be increased by a factor of 3 then what must the new radius be (assume h remains the same)?d) If the radius is increased by a factor of 5 then by what factor does the volume of the tank increase?

mhr

Vh

hrV

mr

mV

LV

03930.0

9

10

10000

2

2

3

a) b)

2

2

2

1235.0 rV

hrV

hrV

mh 1235.0Constants of proportionality can have units!!

mr

r

rV

mV

LV

59.15

1235.030

1235.0

30

30000

2

2

3

c)

d)

25

)5

(

)(

1235.0

1235.0

1

2

2

1

1

1

2

2

1

2

1

2

21

22

1

2

V

V

r

r

V

V

r

r

V

V

r

r

V

V

VECTORSVector quantities have magnitude and direction. A vector is displayed with an arrow.

[E 32o N]

[32o N of E]

22 m/s [58o]

[32o above the horizon]The product of a vector and a scalar is a vector with the same direction but with a different magnitude (unless multiplied by one). Multiplying by a negative scalar results in a vector in the opposite direction.

v

v3

v2

The component of a vector is the projection of a vector along an axis of a rectangular coordinate system.

Rectangular components are always perpendicular to each other

and are said to be orthoganal components.

To add vector quantities the arrows representing the vectors are first joined head to tail. Next the resultant vector is drawn from the tail of the first vector to the head of the last vector.

The result of adding vectors is called the vector addition, resultant vector, resultant, net vector or vector sum.

Remember that it does not matter which order vectors are added or how they are grouped.

Subtraction of vectors can be thought of in two equal ways.

A

B

AB

AB

B

A

tail to tail

Solving vector problems involves using trigonometry. Two general methods are used. The first involves the cosine law and the sine law. Remember . . .

cos2222 BABAR

sinsinsin

RBA

The second method utilizes vector components. This method is especially good for adding more than two vectors.

All answers should include:-a ruled vector drawing, properly labeled-a vector equation-proper vector notation**Remember sin-1(0.5) = 30o or 150o

DISPLACEMENT and VELOCITY AS VECTORS

a) Determine the resultant displacement of a dog that runs 12.6 m [N 21o W] then 18.7 m [S 57o W].

21

ddd R

1

d]21[6.121 WNmd o

2

d

]57[7.182 WSmd o

Rd

o57

o21

oR ddddd 78cos2 21

2

2

2

1

2

oRd 78cos)7.18)(6.12(27.186.12 22

2

mdR 26.20

78sinsin

2

Rdd

o78sin

26.20

sin

7.18

ooor 5.11553.64

o53.64

]53.85[26.20 WNmd oR

by looking at the diagram one can see the angle is acute

look at the diagram to see how this direction was calculated

a)

b) Mr. Anderson walked 25 m [S 32o E] then 41.2 m [N 51o W]. What was his resultant displacement?

c) A person walks with an average velocity of 6.25 km/hr [N 38o E] for 90 minutes. The trip consisted of two stages. The first stage was a trip of 5.8 km [E]. What was the displacement of the second stage of the trip?

]38[375.9

)5.1])(38[25.6(

ENkmd

hrENhr

kmd

tvd

t

dv

o

o

Rd ]38[375.9 ENkmd oR

21 dddR

2d

1d

][8.51 Ekmd

o52

oRR ddddd 52cos2 1

22

1

2

2

]5956.0[388.72 WNkmd o

c)

Use cos law twice because angle is too close to 90. This angle is off a bit ffrom cos law calc

COMPONENT METHOD

a) Determine the average velocity of a dog that runs 14.0 m [S 21.5o E] then 10.5 m [E] then 25.6 m [S 71.1o W] in 35 seconds.

1

d

]5.21[0.141 ESmd o

2

d

][5.102 Emd

3

d

]9.18[6.253 SWmd o

Rd

321

dddd R

Rxd

Ryd

yd1

xd1

xd3

yd3

21.5o

71.1o

Let left and down be positive.

xxxRx dddd 321

ooRx dddd 1.71sin5.21sin 321

ooRxd 1.71sin6.255.105.21sin0.14

mdRx 589.8

yyyRy dddd 321

ooRy ddd 1.71cos5.21cos 31

ooRyd 1.71cos6.255.21cos0.14

mdRy 32.21

Rdyd

xd

222

yxR ddd

222

32.21589.8

Rd

md R 99.22

y

x

d

d

tan32.21

589.8tan o94.21

]94.21[99.22 WSmd oR

t

dv

]94.21[6569.0 WSs

mv o

st 35

In the preceding question you do not have to draw in any right angled triangles, you do not need to show me Pythagoras theorem work or trigonometric ratio work. You need not show me the velocity calculation. All other work must be shown for full marks.

b) Determine the average speed and velocity of a car that travels 12 km [SE], 20 km [S 10o W] and 10 km [S 80o W] in a total of 45 minutes.

Rd

1

d

2

d

3

d

kmdRx 836.4

kmdRy 92.29

]181.9[41.40

]181.9[31.30

WShr

kmv

WSkmd

o

oR

left and down are positive

c) A plane flies 265 km [N 52o W], then it flies 189 km [N 88o W] followed by an unknown stage. If the resultant displace-ment was 512 km [S 82o W] then what was the displacement of the third leg?

2d

1d

3d

Rd

321 ddddR

]52[2651 WNkmd o

]88[1892 WNkmd o

]82[512 WSkmd oR

down and left are positive

213

dddd R

xxRxx dddd 213 ooo

Rx dddd 88sin52sin82sin 213

rearranging negatives direction positives

kmd x 3.1093

yyRyy dddd 213

oooRy dddd 88cos52cos82cos 213

kmd y 0.2413

]34.24[6.2643 WSkmd o

VELOCITY AND ACCELERATION AS VECTORS

tavv

12

vvv 12

a) A car is moving 60 km/hr [W], soon after it is moving 45 km/hr [SSW]. What was the car’s change in velocity?

1v

][601 Whr

kmv

v

12 vvv

2v][452 SSW

hr

kmv

67.5o

These are vector equations!!!

]82.45[65.59 EShr

kmv o

b) A car is travelling 50 km/hr [E] when it accelerates at 30 km/hr2 [NE] for 30 minutes. What is the car’s final velocity?

tavv

12

][501 Ehr

kmv

][15

5.0][302

NEhr

kmta

hrNEhr

kmta

1

v

2

v

ta

135o

otavtavv 135cos2 1

22

1

2

2

]07.80[53.612 ENhr

kmv o

c) A car is travelling 80 km/hr [SE] after it accelerated at 40 km/hr2 [S20oE] for 45 minutes. What was the car’s initial velocity?

tavv

21

][802 SEhr

kmv

]20[30

75.0]20[402

EShr

kmta

hrEShr

kmta

o

o

2v

1v

ta

otavtavv 25cos2 2

22

2

2

1

]50.58[31.541 EShr

kmv o

25o

obtuse

CONSTANT ACCELERATION KINEMATIC EQUATIONS

- one-dimensional motion thus vector notation is not required but you still must set up a direction system

- if the force of gravity is the net unbalanced force then the object is accelerating at g

g = 9.81 m/s2 [down]

davv 221

22

tavv 12 tvv

d

)2

( 12

21 2

ta

tvd

22 2

ta

tvd

-these problems are a review of grade 11 kinematics

- two of the equations are quadratic equations of elapsed time and you might be required to use the quadratic formula to solve for elapsed time (one of the answers is extraneous)

-when an object is thrown upwards its maximum height occurs when its velocity is zero

A ball is thrown upwards from a height of 20 m with a velocity of 20 m/s [up].

a) What height does the ball reach?

b) What time does the ball reach max. height?

c) How long does the ball take to reach the ground?

d) What velocity does the ball hit the ground with?

20 m

At maximum height v2 is 0

For the whole flight the ball accelerates at 9.81 m/s2 [down]

The initial velocity of the ball is 20 m/s [up]

a)

m 40.39 ofheight a reached ball the

39.20 81.9

2

vv 0

2v v20

up is

2

21

22

2

21

221

mds

ma

adv

das

mv

sta

vvt

t

vva

039.2

12

12

b)

c)

sorst

tt

tt

tatvd

md

8308.0908.4

02020905.4

2

81.92020

2

20

2

2

2

1

d)

s

mv

tavvt

vva

15.282

12

12

PROJECTILE MOTION

An object that moves through the air without a propulsion system is called a projectile and it experiences projectile motion.

Projectiles follow a parabolic curved path called a trajectory. The shape of the parabola depends on the initial velocity given to the projectile.

When analyzing projectile motion the projectile’s velocity must be resolved into its horizontal and vertical components.

v

yv

xv

cos

vvx

sin

vvy

Analysis of projectile motion allows two conclusions to be drawn

•The horizontal component of a projectile’s velocity is constant.

•The vertical component of a projectile’s velocity experiences acceleration due to gravity.

A ball is thrown off an 85 m cliff such that it’s initial velocity is 65 m/s [28o above the horizon].

a) What are the horizontal and vertical components of the initial velocity?b) What is the maximum height of the ball?c) What is the horizontal range of the ball?d) What is the final velocity of the ball when it hits the ground?

discuss what happens when an object is launched horizontally

85 m

horizontal range

maximum height

1v

081.9)

52.3039.57

sincos

2865

positive is up a)

22

1

111

1

y

yx

yx

o

vs

mab

s

mv

s

mv

vvvv

aths

mv

ground. theabove m 132.48 of

height maximum a reaches ball the

48.47

)81.9(2

)52.30(0

2

2

2

21

22

21

22

md

d

a

vvd

davv

yy

yy

The horizontal range is the horizontal displacement of the projectile when the projectile lands on the ground.

To calculate horizontal range one needs the horizontal velocity and the amount of time the projectile is in the air.

storst

tt

tt

tat

m

086.2308.8

08552.30905.4

2

81.952.3085

2vd

85d c)

2

2

2

1yy

y

md

d

tvd

x

x

xx

8.476

)308.8)(39.57(

The final velocity of the ball (just before it hits the ground) is the sum of the projectile’s horizontal and vertical components.

The horizontal component velocity remains constant but the vertical component velocity changes with time.

s

mv

v

tavvt

vvad

y

y

yy

yy

98.50

)308.8)(81.9(52.30

)

2

2

12

12

vx

v2y

2v

]61.41[76.76v2 bths

m o

SPECIAL HORIZONTAL RANGE EQUATION

xd

1

v

0 yd

cos

motionx

1

v

dt

v

dt

t

dv

x

x

x

xx

2)(sin0

20

2d

motiony

2

1

2

1

2

1y

tatv

tatv

tatv

y

y

02

sin

0

)2

sin(0

1

1

tav

ort

tavt

dy is zero at this time.

0cos2

sin

02

)cos

(

sin

1

1

1

1

v

dav

v

da

v

x

x

cos1

v

dt

remember

x

a

vd

vda

vv

da

x

x

x

2sin

cossin2

sincos2

2

1

2

1

1

1

in this final equation substitute g for a and get rid of all + and – since they mean nothing to the x motion

g

vd x

2sin2

1