1. Paper One – Composition(Narrative & Discussion), Situational Writing 2. Paper Two – Comprehension, Vocabulary, Summary
3. Paper Three- Oral Reading, Conversation, Picture Discussion

Mathematics

1. Lower Secondary Topics- Formulas
2. Secondary Three Topics – Indices, Quadratic Eqns, Linear inequalities, Congruence & Similarity, Functions & Graphs, Properties of Circles, Trigonometry, Applications of trigonometry, Coordinate Geometry, Arc Lengths & Sector Areas, Quartiles & Percentiles 3. Secondary Four Topics- Standard Deviation, Probability, Matrices, Vectors in Two Dimensions, Mathematics in Practical situation, Graphs in Practical Situations 4. Ten Years Series

Additional Mathematics

1. Secondary Three Topics- Simultaneous Eqn, Surds Indices & Log, Quadratic Eqn & inequalities, Polynomials & Partial Fractions, The Modulus Function, Biomial Theorem, Coordinate Geometry, Linear Law, Trigo Functions 2. Secondary Four Topics- Simple Trigo Identities & Eqn, Further Trigo Identities, Differentiation, Rates of Change, Maxima & Minima Problems, Derivatives of Trigo Function, Exponential & Log Functions, Integration, Area Of a region, Kinematics, Curves & Circles, Plane Geometry. 3. Ten Years Series, Redspot Book

Chemistry

1. Secondary Three Topics- Kinetic Particle Theory, Measurement & Experimental Techniques, Purification & Separation, Elements Compounds & Mixtures, Atomic Structure, Ionic Bonding, Covalent & Metallic Bonding, Writing Eqn, The Mole, Chemical Calculations, Acid & Bases 2. Secondary Four Topics- Salts, Oxidation & Reduction, Metals, Electrolysis, The Periodic Table, Energy Changes, Speed of Reaction, Ammonia, The Atmosphere & environment, Organic Chemistry, Alkanes, Alkenes, Alcohols & Carboxylic Acids, Macromolecules

Physics

1. Secondary Three Topics- Measurement, Kinematics, Forces, Mass Weight & Density, Turning Effects of Forces, Energy Work & Power, Pressure, Temperature, Kinetic...

...Part B
Now, suppose that Zak's younger cousin, Greta, sees him sliding and takes off her shoes so that she can slide as well (assume her socks have the same coefficient of kinetic friction as Zak's). Instead of getting a running start, she asks Zak to give her a push. So, Zak pushes her with a force of 125 \rm N over a distance of 1.00 \rm m. If her mass is 20.0 \rm kg, what distance d_2 does she slide after Zak's push ends?
Remember that the frictional force acts on Greta during Zak's push and while she is sliding after the push.
F= Fp-Fr
E= F*Lp= (Fp-Fr)*Lp= Fr*Lr
Lr= Lp*((Fp/Fr)-1)
Lr= 1*((125/(20*9.8*0.25))-1)= 1.6 m
Mark pushes his broken car 150 m down the block to his friend’s house. He has
to exert a 110 N horizontal force to push the car at a constant speed. How much
thermal energy is created in the tires and road during this short trip?
thermal energy is created in the tires and road
= 110 * 150
=16500 J
A 30 kg child slides down a playground slide at a constant speed. The slide has a height of 4.0 m and is 7.0 m long.
Using energy considerations, find the magnitude of the kinetic friction force acting on the child.
The friction force F is parallel to the slope and is constant in magnitude, so its work is
W = - F d
with d = length of the slide.
ΔU = m g Δh
Therefore:
- F d = m g Δh
F = - m g Δh / d = - 30 x 9.8 x (- 4.0) / 7.0 = 168N
A block of weight w = 15.0 N...

...Example problems involving collisions 1) On a horizontal frictionless surface a puck of mass m initially at speed u collides head-on (without rotation) with a stationary puck of mass M. Find the velocities of both puck after the collision if: i) the collision is fully elastic ii) the collision if fully inelastic. i) momentum: kineticenergy: mu = mv+MV (+ve in direction of initial u) 1 /2 m u2 = 1/2 m v2 + 1/2 M V2
2 eqns in 2 unknowns: V = (u - v) m/M substitute in K eqn: u2 = v2 + (M/m) V2 = v2 + (M/m) (u - v)2 (m/M)2 = v2 + (u - v)2 (m/M) let ρ = (m/M) ⇒ v2 (1 + ρ) - 2ρ u v + u2 (ρ - 1) = 0 quadratic eqn: b2-4ac = 4ρ2 u2 - 4 (1 + ρ) (ρ - 1) u2 = 4ρ2 u2 - 4 (ρ2 - 1) u2 = 4u2
v = [2ρ u ± (4 u2)1/2]/{2 (1 + ρ)} = [2ρ u ± 2 u]/{2 (1 + ρ)} = u (ρ ± 1)/(1 + ρ) + ⇒ v = u , and V = 0 (no collision occurs!) - ⇒ v = u (ρ - 1)/(1 + ρ) , and V = ρ (u - v) = 2ρ2 u/(1 + ρ) e.g. ρ = 1 ⇒ v = 0, V = u . as ρ → 0 ⇒ v → - u , V → 0
ii) momentum: let ρ = (m/M) kineticenergy: ratio: then
m u = (m + M) v v = u 1/(1 + 1/ρ)
1
⇒
v = u m/(m + M)
before: K1 =
/2 m u2 ,
after:
K2 =
1
/2 (m + M) v2
K2/K1 = (1 + 1/ρ) v2/u2 = 1/(1 + 1/ρ) ⇒ v = u/2 , K2/K1 = 1/2 . as ρ → 0 ⇒ v → 0 , K2/K1 → 0
e.g. ρ = 1
2) A point mass m swings under gravity from a fixed pivot on a massless cord through an angle θ to collide with a stationary block of mass M. Assuming a fully elastic collision find the distance the...

...Potential and KineticEnergy lab report
Caty Cleary
4th period
Problem statement:
How does the drop height (gravitational potential energy) of a ball affect the bounce height (kineticenergy) of the ball?
Variables:
Independent variable- drop height
Dependent variable- bounce height
Controlled variables (constants) - type of ball, measurement(unit), place bounced, and the materials used for each experiment.
Hypothesis:
If the gravitational potential energy (drop height) of the ball is increased, then the kineticenergy (bounce height) will increase because the ball will pick up speed on its way down which will cause it to apply more force to the ground, making the ball bounce higher.
Materials and Procedure:
Ball(s), meter stick, balance and a flat surface.
Procedure-
1. Tape the meter stick to the side of the table with the 0-cm end at the bottom and the 100-cm end at the top. Be sure that the meter stick is resting flat on the floor and is standing straight up.
2. Choose a ball type and record the ball type in the data table.
3. Use the triple beam balance to determine the mass of the ball and record the ball’s mass in the data table.
4. Calculate the gravitational potential energy (GPE) for the ball at each drop height. Record GPE in the data table.
5. For Trial 1, hold the ball at a height of 40 cm, drop the ball carefully...

...Learning Goals:
• Predict the kinetic and potential energy of objects.
• Examine how kinetic and potential energy interact with each other.
In the space provided, define the following words:
Kineticenergy-is the energy of motion. An object that has motion - whether it is vertical or horizontal motion
Potential energy-is the energy of an object or a system due to the position of the body or the arrangement of the particles of the system
Open Internet Explorer. From the FMS Jump Page. Click the Potential and KineticEnergy Skate Park link. Or type in http://phet.colorado.edu/en/simulation/energy-skate-park
Observations of KE and PE
Start your skater at the top of the track. Draw or write what happens to the skater.
|Position of Skater |Result |Possible reasons why it happened. |
|[pic] |The Skater keeps going back and fourth. |It never stops because it keeps getting KE |
| | |& PE. |
Click on [pic] and run your skater through the track again. Use this tool to help you label the spots on the ramp where there is the greatest KE and...

...KINETICENERGY
Objects have energy because of their motion; this energy is called kineticenergy. Kineticenergy of the objects having mass m and velocity v can be calculated with the formula given below;
K=1/2mv²
Kineticenergy is a scalar quantity; it does not have a direction. Unlike velocity, acceleration, force, and momentum, thekineticenergy of an object is completely described by magnitude alone. Like work and potential energy, the standard metric unit of measurement for kineticenergy is the Joule. As might be implied by the above equation, 1 Joule is equivalent to 1 kg(m/s) 2.
Examples
1. Determine the kineticenergy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s.
Answer:
KE = 0.5mv2
KE = (0.5)(625 kg)(18.3 m/s)2
KE = 1.05 x105 Joules
2. If the roller coaster car in the above problem were moving with twice the speed, then what would be its new kineticenergy?
Answer:
KE = 0.5mv2
KE = 0.5(625 kg)(36.6 m/s)2
KE = 4.19 x 105 Joules
Work-Energy Theorem
Relationship between KE and W: The word done on an object...

...has mass 2.0 x 10 kg. When the sphere is released, it eventually
falls in the liquid with a constant speed of 6.0 cm s .
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(i) For this sphere travelling at constant speed, calculate
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1. its kineticenergy,
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kineticenergy = ...................................... J
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2. its rate of loss of gravitational potential energy.
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rate = ...................................... J s
[5]
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(ii) Suggest why it is possible for the sphere to have constant kineticenergy whilst
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losing potential energy at a steady rate.
...................................................................................................................................
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...KineticEnergy:
Consider a baseball flying through the air. The ball is said to have "kineticenergy" by virtue of the fact that its in motion relative to the ground. You can see that it is has energy because it can do "work" on an object on the ground if it collides with it (either by pushing on it and/or damaging it during the collision).
The formula for Kineticenergy, and for some of the other forms of energy described in this section will, is given in a later section of this primer.
Potential Energy:
Consider a book sitting on a table. The book is said to have "potential energy" because if it is nudged off, gravity will accelerate the book, giving the book kineticenergy. Because the Earth's gravity is necessary to create this kineticenergy, and because this gravity depends on the Earth being present, we say that the "Earth-book system" is what really possesses this potential energy, and that this energy is converted into kineticenergy as the book falls.
Thermal, or heat energy:
Consider a hot cup of coffee. The coffee is said to possess "thermal energy", or "heat energy" which is really the collective, microscopic, kinetic and potential...