Download Topic 2 Mechanics Part 3 and 4 projectile, friction,10

NOTES AND SOLUTIONS TOPIC 2 PART 3 AND 4
Topic 2 New- Projectile Motion,
Friction
SL and HL
Topic 10 ( NEW)
Gravitational Field HL ONLY ONLY
Text Tsokos pp. 132-157
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Topic 2 New- Projectile Motion
1. Components of Motion
Objects launched at an angle are called projectiles and have two components:
vx and vy . vx is always constant in magnitude and direction throughout the flight. vy
decreases going up due to gravity pushing down. vy increases going down due to gravity.
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2. Projectiles at an Angle
Objects launched at an angle are called projectiles and have two components:
vx and vy . Vx is always constant in magnitude and direction throughout the flight. V y
decreases going up due to gravity pushing down. Vy increases going down due to gravity.
vy at the top = 0. vx is constant throughout the flight and NOT zero at the top.
vy0 = initial velocityy Do not confuse vy0 with vy at the top = 0.
Time in the air make sure you understand how to calculate:
Time in the air (t) = tup + t down . Remember from free fall tup = tdown. So total time = 2 x tup
From free fall formula ( see below) v = u + gt :
tup = u = initial vy
g
9.8
= vy0
9.8
Formulas Projectiles
vy = v at the top = 0 so:
height of object: y = vy tup + ½ g (tup )2
Data Booklet Formulas – Free Fall
v = u + gt
( g = - 9.8 ms-2 )
s = ut + ½ g t2
height
s= (v+ u )t
2
v2 = u2 + 2gs
x = vx t total = Range
Don’t confuse total time ( ttotal) with tup
tup = vy0
g
Range (x) = horizontal displacement
Range x uses total time: ttotal = 2 x tup
x = vx t total
*height of object:
y = vy tup + ½ g (tup )2
Height of object ( y) only depends on time up:
g = -9.8ms-2
y = vy tup + ½ g (tup )2
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Fig. 3.21 – Range (x) = horizontal displacement from point of origin = Vx t
Range is longest at 45 degrees.
Example Range : Two identical balls are shot with an initial velocity of 20.0 ms-1.
One ball is shot from an angle of 15.00 and another is shot from an angle of 60.00.
Find the range ( x) and height (y) of each. See solution next page.
Ball shot at 15.00
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Ball shot at 60.00
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3. Horizontal Projectile Motion
Formulas
Vx doesn´t change. Vy changes due to gravity.
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Example . 3.8 – determining how long the ball is in the air
See solution next page……
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Ex.3 - Solution
Thinking It Through. In looking at the components of motion, we find that part
(a) involves the time it takes the ball to fall vertically, analogous to a ball dropped
from that height. Thls time is also the time the ball travels in the horizontal
direction .The horizontal speed is constant ,so we can find the horizontal
distance, requested in part (b).
Solution. Writing the data with the origin chosen as the point from which the
ball is thrown and downward taken as the negative direction, we have:
y = -25.0m
vx = 8.25 ms-1
a)
b)
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General Questions and Problems : Solutions next page …..-->
1. On Cartesian axes, the x component of a vector is generally associated with :
a) cosine b) sine c) tangent d) none of the above
2. Can an x component of a vector be greater than the magnitude of the vector? Explain.
3. A golf ball is hit with an initial velocity of 35ms-1 at an angle less than 450 above
the horizontal. The horizontal component of the velocity is :
a) greater than b) equal to or c) less than the vertical component
Explain the answer above:
4. If the initial velocity of the ball is 35 ms-1 and the ball is hit at an angle of 370 , what are
the horizontal and vertical components of the velocity?
5. If the magnitude of a velocity vector is 7.0 ms-1 and the x component is 3.0 ms-1, what is
the y component?
6. The x component of the velocity vector has an angle of 370 to the x axis and a
magnitude of 4.8 ms-1 .
a) what is the magnitude of the velocity vector?
b) what is the magnitude of the y component ?
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7.
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General Questions and Problems : Solutions
1. On Cartesian axes, the x component of a vector is generally associated with :
a) cosine b) sine c) tangent d) none of the above
Answer - a
2. Can an x component of a vector be greater than the magnitude of the vector? Explain.
Answer – No. Both components , X and Y, add together to get the resultant.
3. A goofball is hit with an initial velocity of 35ms-1 at an angle less than 450 above the
horizontal. The horizontal component of the velocity :
a) greater than b) equal to or c) less than the vertical component
Answer – a)
Explain the answer above:
Range (x) = horizontal displacement from point of origin = Vx t
Conceptual Note : Projectiles shot at angles less than 45o will have a larger Vx than
Vy . Basically the horizontal component of velocity increases as the angle gets
smaller or “ closer to the ground” . This can be proven mathematically because cos
of angles ( θ) less than 45 are numerically
greater than sin of angles (θ) less
than 45
Fig. 3.21 – range is longest at 45 degrees
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4. If the initial velocity of the ball is 35 ms-1 and the ball is hit at an angle of 370 , what are
the horizontal and vertical components of the velocity?
Answer : Vx = V cos θ = 28ms-1
Vy = V sin θ = 21ms-1
5. If the magnitude of a velocity vector is 7.0 ms-1 and the x component is 3.0 ms-1, what is
the y component?
Answer:
V2 = Vx2 + Vy2
Vy = √V2 – Vy2
Vy = 6.3ms-1
6. The x component of the velocity vector has an angle of 37 0 to the + x axis and a
magnitude of 4.8 ms-1 .
a) what is the magnitude of the velocity vector?
Vx = V cos θ
V = Vx
=
cos θ
6.0 ms-1
b) what is the magnitude of the y component ?
Vy = V sin θ = 3.6ms-1
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7.
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More Example Questions: see solutions next page …..-->
1.
2.
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More examples - Solutions
1. D
2.
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Topic10 New (HL ONLY) - Gravitational Field
Tsokos pp. 142-146
This is a complicated topic. The book goes into too much detail so focus on the following
.
concepts , formulas and example problems . These formulas are given in the data booklet
Work being done to move
masses is stored as
gravitational potential
energy.
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NOTE : Do not confuse gravitational potential energy formula ( Ep ) with
the gravitational force formula ( Fg ) .
The gravitational potential energy
formula is derived form the gravitational force formula ( Fg ) as follows:
Fg = GMm
R2
given in data booklet
Energy = Work done = Force x distance
Distance = R
Work = Ep = Fg x R
Work = Ep = GMm X R
R2
Work = Ep = GMm
R
in data booklet
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Gravitational Potential (V)
V=W
m
V=
GMm
r .
m
=
GM
r
Example: The gravitational field close to the surface of a planet is uniform. The gravitational
potential difference between the surface of the planet and a point 30.0m above is 15 Jkg -1.
a) What is the work done to raise a mass of 1.0 kg from the surface of the planet to 30.0 m
above the surface?
b) What is the work done to raise a mass of 4.0 kg from the surface of the planet to 30.0 m
above the surface?
c) What is the work done to raise a mass of 4.0 kg from the surface of the planet to 10.0 m
above the surface?
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M is the planet
The escape velocity is derived by the following way:
Change in gravitational potential energy is equal to the change in kinetic energy (½ m v 2) .
The mass of the object does not matter and velocity is solved in the following way:
Ep = GMm = ½ m v2
R
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Examples - see solutions next page
Example 1 – Escape Velocity
Example 2 - Gravitational Potential Energy
A satellite in orbit about Earth moves to another orbit that is closer to the surface of Earth.
When the satellite moves into the orbit closer to earth, which of the following correctly
describes the change in speed of the satellite and the change in ts gravitational potential
energy ?
Choice
Speed
A
B
C
D
decreases
decreases
increases
increases
Gravitational potential
energy
decreases
increases
increases
decreases
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Example 3
Example 4
Example 5
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Equipotential surfaces and fields
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Example 6
2020
Gravity vs Electricity
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SOLUTIONS
Example 2 - Gravitational Potential Energy
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Example 5
Example 6
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TOPIC 2 MECHANICS PART 4 NEW - FRICTION
Basic Friction
Basically 4 forces acting on the box:
W = weight of the box = N
N = normal force of table pushing up on the box = W
Fa = applied force to the right
fs = force of static friction between the box and the surface of the table. This opposes the
applied force.
f s = Fa
Text Tsokos p. 74 : 12 - Worked Example
Answer : F = 1343 N
Frictional force = fs = horizontal component Fx = Fa = 1163
F = pulling force ( resultant force of Fx and Fy )
Since Fx = F cos 300
F=
Fx
= 1343 N
Cos 30
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Friction in Detail : Static and Kinetic
Formulas
fs = µs N
fk = µk N
N=w=mg
F=ma
Types of Friction:

Static friction – when the frictional force is sufficient to prevent relative motion
between surfaces.

 Sliding/ kinetic friction - static friction is overcome and there is relative motion
between surfaces in contact.

Rolling friction – rotating motion with no sliding or slipping.

Use fs = µs N

µs = coefficient of static friction and depends on type of surfaces involved, for
example glass on glass is .94 and dry rubber on dry concrete is 1.2 .

Once there is motion use fk = µk N .

µk is less than µs due to object in motion and varies with the slow speed of surfaces
involved. At fast speeds ( 10 m / s or more ) µk is fairly constant.
when there is no motion
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Examples - check answers next page
Example 1
A box of mass m = 20.0 kg and the coefficients of static and kinetic friction between the box
and the table are 0.800 and 0.500 respectively. A string is attached to the box and used to
pull the box directly horizontal at no angle
Draw a free body diagram. Label the normal force ( N) the weight ( W) the force due to static
friction ( fs) and the applied force in the string ( Fa):
A) what is the minimum tension in the string to start the box in motion.
B) after the box is moving and the tension is kept constant, what is the acceleration of the
box?
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Example 2
A box of mass m = 40.0 kg and the coefficients of static and kinetic friction between the box
and the table are 0.650 and 0.500 respectively. A string is attached to the box and used to
pull the box directly horizontal at no angle.
A) what is the minimum tension in the string to start the box in motion.
B) after the box is moving and the tension is kept constant, what is the acceleration of the
box?
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Answers to examples 1 and 2 – Friction
Example 1
a) Fa = fs = s N
N = mg = (40.0)(9.8) = 392 N
Fa = (392) (0.8) = 314 N
b) Fnet = Fa - fk
From a) above need Fa of 314 N to get crate moving
fk = k N = (0.5)(314) = 157 N
Fnet = Fa - fk = 314 N – 157 N = 157 N
a = Fnet
m
=
= 3.93 ms-2
157 N
40.0
Example 2
a) Fa = fs = s N
N = mg = (40.0)(9.8) = 392 N
Fa = (392) (0.65) = 255 N
b) Fnet = Fa - fk
From a) above need Fa of 255 N to get crate moving
fk = k N = (0.5)(392) = 196 N
Fnet = Fa - fk = 255 N – 196 N = 59 N
a = Fnet
m
=
59 N
40.0
= 1.48 ms-2
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MISCELLANEOUS PROBLEMS Wilson Buffa pp. 137 – 139 : 74, 75, 78, 82 – 85 , 96
Solutions at end of packet
78. Is it easier to push a lawn mower at an angle ? Draw a free body diagram and explain.
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SOLUTIONS
MISCELLANEOUS PROBLEMS Wilson Buffa pp. 137 – 139 : 74, 75, 78, 82 – 85 , 96
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