lOMoARcPSD|29059196
Fsk 116 fsk exam proofs
Physics for Engineers (University of Pretoria)
Scan to open on Studocu
Studocu is not sponsored or endorsed by any college or university
Downloaded by PBS Sibanda (pbssibanda@gmail.com)
lOMoARcPSD|29059196
FSK Exam Proofs
Proof 1: Kinematic Equations (by substitution)
∆𝑣
𝑎=
∆𝑡
𝑣𝑓 − 𝑣𝑖
𝑎=
𝑡
𝑎. 𝑡 = 𝑣𝑓 − 𝑣𝑖
(𝟏) 𝒗𝒇 = 𝒗𝒊 + 𝒂. 𝒕
𝒗𝒇 + 𝒗𝒊
𝟐
𝑥𝑓 − 𝑥𝑖
𝑣𝑎𝑣𝑔 =
𝑡
∴ 𝑣𝑎𝑣𝑔 . 𝑡 = 𝑥𝑓 − 𝑥𝑖
𝑥𝑓 = 𝑥𝑖 + 𝑣𝑎𝑣𝑔 . 𝑡
𝟏
(𝟑) 𝒙𝒇 = 𝒙𝒊 + (𝒗𝒇 + 𝒗𝒊 ) 𝒕
𝟐
(𝟐) 𝒗𝒂𝒗𝒈 =
1
𝑥𝑓 = 𝑥𝑖 + (𝑣𝑖 + 𝑣𝑖 + 𝑎. 𝑡) 𝑡
2
𝟏
(𝟒) 𝒙𝒇 = 𝒙𝒊 + 𝒗𝒊 . 𝒕 + (𝒂. 𝒕𝟐 )
𝟐
𝑣𝑓 = 𝑣𝑖 + 𝑎. 𝑡
𝑣𝑓 −𝑣𝑖
=𝑡
𝑎
𝑣𝑓 −𝑣𝑖
1
)
𝑥𝑓 = 𝑥𝑖 + (𝑣𝑓 + 𝑣𝑖 ) (
𝑎
2
1
𝑥𝑓 = 𝑥𝑖 +
(𝑣 2 − 𝑣𝑖 2 )
2𝑎 𝑓
(𝟓) 𝒗𝒇 𝟐 = 𝒗𝒊 𝟐 + 𝟐𝒂(𝒙𝒇 − 𝒙𝒊 )
Downloaded by PBS Sibanda (pbssibanda@gmail.com)
lOMoARcPSD|29059196
𝟏
𝟐
Deriving 𝒙𝒇 = 𝒙𝒊 + 𝒗𝒊 . 𝒕 + 𝒂. 𝒕𝟐 , using acceleration and velocity formulas.
𝒅𝒗
𝒂 = 𝒅𝒕
(Move ‘dt’ to the other side)
𝒂. 𝒅𝒕 = 𝒅𝒗
(Since ‘d’ is on both sides, you can now integrate both sides)
𝒕
𝒗
∫𝒐 𝒂. 𝒅𝒕 = ∫𝒗 𝒇 𝒅𝒗
𝒊
(Always integrate from initial to final (technically the time should be ti-tf but they just assumed ti =0)
𝒂. 𝒕 = 𝒗𝒇 − 𝒗𝒊
(Acceleration is treated as constant since the kinematic equations make the constant acceleration
assumption. Since there is no function in the integral on the right, you treat it as ‘1’ and then after
the integration it becomes final-initial; hence, vf-vi)
𝒗𝒇 = 𝒗𝒊 + 𝒂. 𝒕
(The equation is just rearranged to make ‘vf’ the subject)
------------------------------------------------------------------------------------------------------------------------------------𝒅𝒙
𝒗 = 𝒅𝒕
(This is a new calculation in the proof where the same method is done but with ‘v’ instead of ‘a’ as
done above)
𝒕
𝒙𝒇
∫𝟎 𝒗. 𝒅𝒕 = ∫𝒙𝒊 𝒅𝒙
(Before integrating, ‘vf’ is substituted into ‘v’ as shown in the step below.)
𝒕
∫𝟎 (𝒗𝒊 + 𝒂𝒕)𝒅𝒕 = 𝒙𝒇 − 𝒙𝒊
𝟏
𝒗𝒊 . 𝒕 + 𝟐 𝒂. 𝒕𝟐 = 𝒙𝒇 − 𝒙𝒊
(The terms are now integrated where ‘vi’ and ‘a’ are treated as constants)
𝟏
𝒙𝒇 = 𝒙𝒊 + 𝒗𝒊 . 𝒕 + 𝟐 𝒂. 𝒕𝟐
(The equation is just rearranged to make ‘xf’ the subject)
Downloaded by PBS Sibanda (pbssibanda@gmail.com)
lOMoARcPSD|29059196
Proof 2: Range and Maximum Height
• Deriving equations for the maximum height and horizontal range of a projectile
1. To determine h (maximum height)
1
2
𝑣𝑦 = 0
∴ 𝑣𝑦 = 𝑣𝑦𝑖 − 𝑔𝑡 = 0
∴ 0 = 𝑣𝑖 sin(𝜃) − 𝑔𝑡
𝑣𝑖 sin(𝜃)
𝑡ℎ =
𝑔
Substitute this into: 𝑦 = 𝑦𝑖 + 𝑣𝑦𝑖 𝑡 − 𝑎𝑡 2
2
𝑣𝑖 sin(𝜃)
1
𝑣𝑖 sin(𝜃)
)− 𝑔(
)
𝑔
2
𝑔
𝑣𝑖 2 sin2(𝜃)
ℎ=
2𝑔
2. To determine R (maximum range) – only apples to where the starting and finishing
horizontal plane are on the same level
ℎ = 𝑣𝑖 sin(𝜃) (
2𝑡ℎ = 𝑡𝑅
Using: 𝑥 = 𝑣𝑥𝑖 𝑡
𝑅 = 𝑣𝑖 cos(𝜃) 2𝑡ℎ
𝑣𝑖 sin(𝜃)
𝑅 = 𝑣𝑖 cos(𝜃) 2 (
)
𝑔
𝑣𝑖 2 cos(𝜃) sin(𝜃)
𝑅 = 2(
)
𝑔
𝑣𝑖 2 sin(2𝜃)
𝑅=
𝑔
Proof 3: Uniform Circular Motion
𝑣̅𝑖
𝑟̅𝑖
̅̅̅
∆𝑟
∆𝜃
𝑟̅𝑓
̅̅̅ = 𝑟̅𝑓 − 𝑟̅𝑖
∆𝑟
𝑣̅𝑖
∴
|∆𝑣|
𝑣
=
|∆𝑟|
𝑟
|𝑟̅𝑓 | = |𝑟̅𝑓 | = 𝑟
∆𝑣 = 𝑣𝑓 − 𝑣𝑖
𝑣𝑓 = 𝑣𝑖 = 𝑣
Because ∆𝑣 is tangential to the path, a is as
well
∆𝑟|||∆𝑣
|∆𝑣|
𝑡
𝑣|∆𝑟|
|𝑎𝑎𝑣𝑔 | =
𝑟∆𝑡
|𝑎𝑎𝑣𝑔 | =
Find instantaneous velocity:
𝑎=
𝑣 𝑑|∆𝑟|
.
𝑟 𝑑𝑡
𝑎𝑐 =
𝑣2
– Always directed perpendicular to the path towards the centre of the circle
𝑟
Downloaded by PBS Sibanda (pbssibanda@gmail.com)
lOMoARcPSD|29059196
Proof 4: Work done by a spring
•
If we compress the spring to x = -xmax and release it, we can calculate the work done by the
spring from –xmax to 0
•
Similarly if the spring is moved to x = xmax then
o
•
•
−𝑘
2
𝑤 = 2 𝑥𝑚𝑎𝑥
Thus the net work from –xmax to xmax will equal zero
In general
o
o
𝑥𝑓
𝑤 = ∫𝑥𝑖 −𝑘𝑥. 𝑑𝑥
𝑘
𝑘
𝑤 = 2 𝑥𝑖2 − 2 𝑥𝑓2
Proof 5: Work Energy Theorem
Downloaded by PBS Sibanda (pbssibanda@gmail.com)
lOMoARcPSD|29059196
Proof 6: Conservation of Energy in an Isolated System
Thus
o
o
𝐸𝑚𝑒𝑐ℎ,𝑖 = 𝐸𝑚𝑒𝑐ℎ,𝑓
𝐾𝑖 + 𝑈𝑖 = 𝐾𝑓 + 𝑈𝑓
Proof 7: Momentum Conservation
In an isolated system (no external forces act on the system) of two particles that interact with one
another.
Downloaded by PBS Sibanda (pbssibanda@gmail.com)
lOMoARcPSD|29059196
Proof 8: Impulse
Proof 9: Rotation under constant acceleration
Same as part 2 of Proof 1
Proof 10: Rigid Object under Torque
Downloaded by PBS Sibanda (pbssibanda@gmail.com)
lOMoARcPSD|29059196
Proof 11: Acceleration due to gravity
Proof 12: Gravitational Potential Energy
Downloaded by PBS Sibanda (pbssibanda@gmail.com)
lOMoARcPSD|29059196
Proof 13: SMH formula
• If an object is attached to a spring on a frictionless surface
o Hooke’s Law: 𝐹𝑠 = −𝑘𝑥
•
Using Newton’s second law:
•
Determining an equation for x(t) – using differential equations
Downloaded by PBS Sibanda (pbssibanda@gmail.com)
lOMoARcPSD|29059196
Proof 14: Mechanical Energy for SHO
• If the system is frictionless
o Mechanical energy is conserved
o 𝐸𝑚𝑒𝑐ℎ = 𝐾 + 𝑈
•
1
1
1
𝑘𝐴2 = 2 𝑘𝑥 2 + 2 𝑚𝑣 2
2
o
Can be simplified to: 𝑣 = ±𝜔√𝐴2 − 𝑥 2
Downloaded by PBS Sibanda (pbssibanda@gmail.com)
lOMoARcPSD|29059196
Proof 15: Superposition
• Superposition of sinusoidal waves
o For two waves traveling in the positive, x direction, with the same frequency,
wavelength and amplitude:
•
If two waves travel in opposite directions with the same frequency, wavelength and
amplitude
Downloaded by PBS Sibanda (pbssibanda@gmail.com)