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Revision on VPM, Electric Circuits, Electrochemistry
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This document adressess challenges in VPM, Electric Circuits, Electrochemistry. For Grade 12 Learners preparing for their final examination.
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PHYSICAL SCIENCESCAMP MATERIAL
OCTOBER 2022
, 2
TABLE OF CONTENTS
Page
Vertical Projectile Motion 3
• Key concepts to be learnt 3
• Using equations of motion to solve vertical projectile motion problems 3
• Graphs of motion 8
• Exercises and questions from past examination papers 9
Electric circuits 18
• Ohm’s Law 19
• Total Potential Difference in a Circuit 20
• Power 21
• Calculations 22
• Internal Resistance and Series and Parallel Networks 28
• Questions from past examination papers 31
Electrochemistry 44
• Oxidation Numbers 44
• Voltaic / Galvanic cells 49
• Electrolytic cells 52
• Questions from past examination papers 63
Preparing for and answering multiple choice questions 67
• Multiple choice questions from past examination papers 69
Additional questions from past examination papers 75
Easy to score marks 84
Definitions, Principles and Laws from the examination guidelines 98
, 3
VERTICAL PROJECTILE MOTION
Key Concepts to be learnt:
1. Using equations of motion to determine position, velocity and displacement of a projectile.
2. Using graphs of position versus time, velocity versus time and acceleration versus time to
determine the position, displacement, velocity and acceleration of a projectile.
3. Describing the motion of a projectile when given a graph of position versus time, velocity
versus time or acceleration versus time
4. The types of projectiles to be studied are:
• An object dropped form a certain height.
• An object thrown vertically upwards and dropping to the same level of projection.
• An object thrown vertically upwards from a certain height and dropping lower than
the level of projection.
• An object thrown vertically downwards.
• Involving two objects thrown at different time intervals or at the same instant (same
direction or opposite directions).
• Bouncing objects and rockets.
Definitions:
• Vertical projectile motion is the motion of an object that moves vertically upwards or
downwards through the air while gravity is the only force acting on it.
• A projectile is an object upon which the ONLY force acting on it is the force of gravity.
• Free fall is the motion of an object under the influence of the gravitational force ONLY.
USING EQUATIONS OF MOTION TO SOLVE
VERTICAL PROJECTILE MOTION PROBLEMS
• When the projectile is in free-fall the acceleration of the projectile is always 9,8 m·s−2
downwards.
THE FOLLOWING EQUATIONS ARE USED TO SOLVE PROJECTILE MOTIONS
PROBLEMS:
v f = v i + a t Δx = v i Δt + 21 at 2 or/of Δy = v i Δt + 21 at 2
v + vf v + vf
v f = v i + 2ax or/of v f = v i + 2ay
2 2 2 2
Δx = i Δt or/of Δy = i Δt
2 2
Where:
• vi = initial velocity (measured in m·s−1)
• vf = final velocity (measured in m·s−1)
• Δx/ Δy = displacement (measured in m)
• Δt = time interval / change in time (measured in s)
• a = acceleration (measured in m·s−2)
, 4
WORKED EXAMPLE 1:
Question: A ball is thrown upwards with an initial velocity of 10 m·s−1.
1.1. Determine the maximum height reached above the thrower’s hand.
1.2. Determine the time it takes the ball to reach its maximum height.
Solutions:
Question 1.1
Step 1: Determine how to approach the problem
• Decide whether you want the downward direction to be positive or negative. We will choose
down to be positive.
Step 2: Identify what is required and what is given
• We are required to determine the maximum height reached by the ball and how long it takes
to reach this height.
• We are given the initial velocity vi = 10 m·s−1
Step 3: Identify other information that is known
• The acceleration due to gravity is not given in the problem but we know that it is ‘g’ = 9,8
m·s−2.
• We also know that at the maximum height the velocity of the ball is 0 m·s −1.
• We therefore have the following values:
Δy Vi Vf a=g
? -10 m·s−1 0 m·s−1 9,8 m·s−2
Step 4: Identify the appropriate equation to determine the height.
We can use the equation:
v f = vi + 2ax
2 2
to solve for the height.
Step 5: Substitute the values in and find the height.
v f = v i + 2ay
2 2
(0)2 = (−10)2 + (2) (9, 8) (Δy)
−100 = 19,6Δy
Δy = - 5,102 m
The value is negative
Why is the answer negative, do we leave it like that in the final answer?
The value for the displacement is negative because the object is displaced in
an upwards direction and we have chosen the downward direction to be
positive The answer has to be written as a positive value.
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