1 sace stage 2 physics motion in 2 dimensions. 2 motion in 2 - dimensions errors in measurement...
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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
5
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
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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.
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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
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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.
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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
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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,
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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.
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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
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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
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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.