1

SACE Stage 2 Physics

Motion in 2 Dimensions

2

Motion in 2 - Dimensions

Errors in Measurement

Suppose we want to find the area of a piece of paper (A4)

Length = 297 ± 0.5 mm

Width = 210 ± 0.5 mm

Areamax = 62623.75 mm2

Areamin = 62116.75 mm2

Area = 62370 ± 253.5 mm2

3

Motion in 2 - Dimensions

Significant Figures

When calculating data, the accuracy of the answer is only as accurate as the information that is least accurate.

307.63 – 5 significant figures

0.00673 – 3 significant figures

12000 – can be 2,3,4, or 5 significant figures depending on whether the zeros are just place holders for the decimal point.

12.45 x 1012 – has 4 significant figures

4

Motion in 2 - Dimensions

Scientific Notation

The diameter of the solar system is 5 946 000 000 000 metres.

Can write this as 5.946 x 1012m.

The decimal place has moved 12 places to the left.

Calculations

baba

baba

baba

10)10(

101010

101010

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Motion in 2 - Dimensions

Scientific Notation

Example

Evaluate where,

k = 9.00 x 109,

q1 = 1.60 x 10-19,

q2 = 3.20 x 10-19,

r = 6.273 x 10-11

221

r

qkq

6

Motion in 2 - Dimensions

Scientific Notation

221

r

qkq

Example

Evaluate where,

k = 9.00 x 109,

q1 = 1.60 x 10-19,

q2 = 3.20 x 10-19,

r = 6.273 x 10-11

7

22

29

211

19199

2

211

19199

221

1017.1

10

10171.1

)10(

101010

273.6

2.36.19

)10273.6(

)102.3()1060.1()109(

r

qkq

Answer given to three significant figures as the least accurate piece of data was given to three sig. figs.

7

Motion in 2 - Dimensions

Equations of Motion

Average Velocity

Average Acceleration

221 vv

t

sv

t

νν

t

va

12

8

Motion in 2 - Dimensions

Equations of MotionUsing average velocity and average acceleration to derive two other equations.

(a) Assuming velocity and acceleration remain constant,

t

vva and

vv

t

s

1221

2Become,

t

vva and v

t

sv

1212

2

9

Motion in 2 - Dimensions

Equations of MotionCombining,

21 1

s

v v at

t

tavt

s

122

2122 tatvs

21 2

1tatvs

10

Motion in 2 - Dimensions

Equations of Motion

12 2

21 tvv

s2

vv

t

s

1

(b)

21212 a

vvt

t

vva

equation (1) = equation (2)

a

vv

vv

s 12

21

2

11

Motion in 2 - Dimensions

Equations of Motion

21122 vvvvsa Hence,

Ie, savv 221

22

Note:(1) the acceleration is constant,(2) the directions for velocity and acceleration are used

correctly

12

Motion in 2 - Dimensions

Uniform Gravitational Field

1. Gravity acts vertically downwards.

2. A mass can only accelerate in the direction of gravity in the absence of all other forces (including air resistance).

3. Gravity g = 9.8 ms-2 vertically down.

13

Motion in 2 - Dimensions

Uniform Gravitational Field – vector diagram

vH

vH

vH

vH

v2

vv

vv

vv

vH

vH

vH

vH

v1

vv

vv

vv

a = g = 9.8 m s-2

a = g = 9.8 m.s-2

14

Motion in 2 - Dimensions

Uniform Gravitational Field – multi-image photograph

1. Vertical separation the same for both balls at the same time interval.

2. Horizontal separation constant.

3. Vertical and horizontal components are independent of each other.

15

Motion in 2 - Dimensions

Vector Resolution

A vector can be resolved into components at right angles to each other.

v

vh = v cos

vv = v sin

16

Motion in 2 - Dimensions

Example 1 – Known vector

30o

v = 40 m s-1

vvertical

vhorizontal

vvertical = 40 sin 30o

= 20 m s-1

vhorizontal = 40 cos 30o

= 34.6 m s-1

Trigonometric ratios,

17

Motion in 2 - Dimensions

Example 2 – Unknown vector

v = ?vv= 20m s-1

vh= 50m s-1

Pythagoras’ Theory,

0

1

1

22

22

222

8.21

5020tan

5020tan

9.53

2050

msv

v

vvv

vvv

vh

vh

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Motion in 2 - Dimensions

Time of Flight

Note:

1. Acceleration present is from gravity and remains constant.

2. Horizontal velocity remains constant (Ignore air resistance)

3. Vertical motion is independent of horizontal motion.

4. The launch height is the same as the impact height.

We can now determine the time of flight by only considering the vertical motion of the projectile.

19

Motion in 2 - Dimensions

Time of Flight

savv tatvs

t

vva

vv

t

sv

2)4()3(

)2(2

)1(

21

22

21

1221

2

1

Can use the following equations for the vertical motion, (a = -g = 9.8ms-2)

Can use the following equation for the horizontal velocity,

221 vv

t

sv

20

Motion in 2 - Dimensions

Time of Flight

We assume the launch point has position s1 = 0. The projectile is launched with some initial horizontal velocity (vh1) and some initial vertical velocity (vv1). The only acceleration is due to gravity acting vertically downwards. It reaches a maximum height at the time tmax, when,

a gv v

gv

v

g

v v

v

v

2 1

1

1

t

t

t

max

max

max

(take a =-g assuming acceleration down & vv1 up - ie.

up is a positive direction)

vh1

vv1

a = 9.8ms-2 down

21

Motion in 2 - Dimensions

Time of Flight

2111 )(

2

1)(

g

vg

g

vvheight vv

v

At the time the maximum height is reached,

21

1

2

1tatvs into

g

vt v

max

gives,

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Motion in 2 - Dimensions

Time of Flight

Time of impact occurs when S = 0. ie, 01

212 v gv t t

This equation has two solutions, at t = 0 and

t 2 1v

gv equation for the time of flight

Comparing the two equations, and

The time of flight is exactly twice the time taken to reach the maximum height.

t 2 1v

gv tmax

v

gv1

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Motion in 2 - Dimensions

Range

The range is simply the horizontal distance attained at the time t = tflight.

s vv v

grange hh v 1

1 12t flight

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Motion in 2 - Dimensions

ExampleA rugby player kicks a football from ground level with a speed of 35 ms-1 at an angle of elevation of 250 to the horizontal ground surface. Ignoring air resistance determine; (a) the time the ball is in the air, (b) the horizontal distance travelled by the ball before hitting the ground (c) the maximum height reached by the ball.

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Motion in 2 - Dimensions

Example(a) the time the ball is in the air,

vH m s-1

vv m s-1

25o

35 m s-1

vH = v cos

= 35cos(25) = 31.72 m s-1 vv = v sin(25)

= 35(sin25) = 14.79 m s-1

Using vertical components to determine time to reach maximum height

vv = vo + at t = 14.79/9.8 = 1.509 = 1.5 s

Hence time in the air = 2(1.509) = 3.02 s

26

Motion in 2 - Dimensions

Example

(b) the horizontal distance travelled by the ball before hitting the ground sH = vHt

= (31.72)(2(1.5)) = 2(47.8766) = 2(47.9) = 96 m

(c) the maximum height reached by the ball.

2o t

2

1t avs

s = (14.79)(1.5) + (0.5)(-9.8)(1.5)2 = 11.16 = 11.2 m

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Motion in 2 - Dimensions

Launch Angle and Range

The following diagram shows the trajectories of projectiles as a function of elevation angle. Note that the range is maximum for q = 45o and that angles that are equal amounts above or below 45o yield the same range, eg, 30o and 60o.

Projectile ranges for various angles of launch

0

50

100

150

200

250

300

350

400

450

500

0 200 400 600 800 1000 1200

range

heig

ht

Ignoring air resistance

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Motion in 2 - Dimensions

Air Resistance

1. Affects all moving through air.

2. The force due to air resistance always acts in the opposite direction to the velocity of the object.

3. Air resistance is proportional to the speed of the object squared.

4. As speed changes, the air resistance must also change.

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Motion in 2 - Dimensions

Air Resistance

Projectile ranges with / without air resistance

0

50

100

150

200

250

300

0 200 400 600 800 1000 1200

range

hei

gh

t

no air resistance

with air resistance

1. Horizontal velocity always decreasing.

2. No vertical air resistance at max height as vv = 0.

3. Time of Flight is reduced.

4. Range also reduced.

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Motion in 2 - Dimensions

Application: Projectiles in Sport

1. Launch height affects the range of the football.

2. Maximum distance achieved for elevation angle of 45o.

3. Air resistance will depend on the type of projectile, ie, basketball, football, ball of paper.