NOTES AND SOLUTIONS TOPIC 2 PART 3
Topic 2 New- Projectile Motion – SL and HL
Topic 10 ( NEW) - Gravitational Field HL 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. V y
decreases going up due to gravity pushing down. Vy increases going down due to gravity.
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2. Projectiles at an Angle
Formulas
x = vx t
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 force 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
.
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Fig. 3.21 – range is longest at 45 degrees
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. 3-8 - Solution
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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 golfbal 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
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 37 0 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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Topic 2 New- Projectile Motion - Solutions
Problems:
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 force 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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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 (HLONLY) - 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 not given in the data
booklet
.
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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
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The escape velocity is derived by the following way:
Change in gravitational potential energy is equal to the change in kinetic energy (½ m v2) .
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 at end of packet
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
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Gravity vs Electricity
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SOLUTIONS
Example 1 - Escape Velocity
Example 2 - Gravitational Potential Energy
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Example 3
Example 4
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Example 5
Example 6
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