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Questions- A.1 Kinematics SL Paper 1

 A block of mass \(2.0 \mathrm{~kg}\) accelerates uniformly at a rate of \(1.0 \mathrm{~m} \mathrm{~s}^{-2}\) when a force of \(4.0 \mathrm{~N}\) acts on it. The force is doubled while resistive forces stay the same. What is the block’s acceleration?

A. \(4.0 \mathrm{~ms}^{-2}\)

B. \(3.0 \mathrm{~m} \mathrm{~s}^{-2}\)

C. \(2.0 \mathrm{~m} \mathrm{~s}^{-2}\)

D. \( 1.0 \mathrm{~ms}^{-2}\)

We can use Newton’s second law to find the initial resistive force (\(R_1\)) and the final resistive force (\(R_2\)):

1. For the initial case (\(F_1\)), the net force (\(F_{\text{net}_1}\)) is responsible for the initial acceleration (\(a_1\)):

\[F_{\text{net}_1} = F_1 – R_1 = m \cdot a_1\]

Solving for \(R_1\):

\[R_1 = F_1 – m \cdot a_1\] \[R_1 = 4.0 \, \mathrm{N} – (2.0 \, \mathrm{kg} \cdot 1.0 \, \mathrm{m/s^2}) = 4.0 \, \mathrm{N} – 2.0 \, \mathrm{N} = 2.0 \, \mathrm{N}\]

So, in the initial case, the resistive force (\(R_1\)) is \(2.0 \, \mathrm{N}\).

2. For the final case (\(F_2\)), the net force (\(F_{\text{net}_2}\)) is responsible for the final acceleration (\(a_2\)):

\[F_{\text{net}_2} \Rightarrow F_2 – R_1 = m \cdot a_2\]

Solving for \(a_2\):

\[2.0 \, \mathrm{kg}\cdot a_2 = 8.0 \, \mathrm{N} – 2.0 \, \mathrm{kg} \]

\[6.0 \, \mathrm{N} = 2.0 \, \mathrm{kg} \cdot a_2\]

\[a_2 = 3.0 \, \mathrm{m/s^2}\]

[qdeck ” ]

[h] IB Mathematics AA HL Flashcards- Standard Form

[q] Standard form

A number in the format , \( \pm a \times10^k \) , where \(1\leqslant a\leqslant 10 \) and \(k\in Z\) (k is an integer)

Number in the form  \(a \times 10^ k\)  such as \(4.7 \times 10^5\)  instead of 475,000 may be called Specific notation or Standard form  or not given any specific name.

CONDITIONS // The number 470,000 could theoretically be written: \( 4.7 \times 10^5 \), \( 47 \times 10^4 \), \( 0.47 \times 10^6 \), and so on. However, to simplify things, we say that ‘a’ & ‘k’ must have the following:

– \( 1 \leq |a| < 10 \) – \( k \in \mathbb{Z} \)

Meaning: ‘a’ is between 1 and 10 (+ve or -ve), and ‘k’ is an integer.

E.G.1 // \( 5 \times 10^{3.2} \): NOT in the correct form (k is not an integer)

E.G.2 // \( 0.89 \times 10^{-4} \): NOT, because \( a < 1 \). Should be: \( 8.9 \times 10^{-5} \)

E.G.3 // \( 2.3 \times 10^8 \): ✓, this means 230,000,000

GENERAL USAGE // This format of writing numbers is frequently used in science, where we want to save time by avoiding writing very large or very small numbers in full. For example, the width of an atom is ~\( 3 \times 10^{-10} \) m.

I.B. MATH USAGE // It is extremely unlikely that you will ever see a standalone question about this. Far more likely is to see this used as a small element of a question on some other topic.

E.G.4 // Find \( x \) (in the form \( a \times 10^k \), \( 1 \leq a < 10 \), \( k \in \mathbb{Z} \))

Using sine rule: \[ \frac{x}{\sin 21^\circ} = \frac{1.6 \times 10^7}{\sin 57^\circ} \] Solving for \( x \): \[ x = 0.834 \times 10^7 \] Change to correct form: \[ x = 8.34 \times 10^6 \]

[q] Notes on GDC

[q] Standard Form – Solved Examples 1

The diameter of a spherical planet is 6 × 10 4 km . (a) Write down the radius of the planet. [1] The volume of the planet can be expressed in the form π(a × 10 k ) km 3 where 1 ≤ a <10 and k ∈  \(\mathbb{Z}\) (b) Find the value of a and the value of k .

[a] Answer – Solved Examples 1

(a)  \(3 \times 10^4\) (b) \(\frac{4}{3}\pi(3 \times 10^4)^3\) \(=\frac{4}{3}\pi\times 27 \times 10^{12}\) \(=\pi(3.6 \times 10^{13})(km)^3\) Hence \(a= 3.6 \) ,  \(k =13\)

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E. Water: The Versatile Molecule

➢ In water, electrons are not shared equally in the bonds between hydrogen and oxygen

•  Hydrogen atoms have a partial positive charge while oxygen atoms has a partial negative charge

 ■ Water is polar

➢ Hydrogen bonds

• Weak attractions that result of water’s polarity

              ■ Positive end of another polar molecule attracted to oxygen negative charge, and vice versa with the hydrogen end               ■ Hydrogen atom covalently bonded to one electronegative atom is also attracted to another electronegative atom               ■ Weak Individually, but strong on a larger scale

• Lends watermany special properties ■ Cohesion     ● Tendency for water to stick to water     ● Important during transpiration
•  Water evaporates, pulls other water molecules with it, pulling all the way down from leaves to roots ■ Adhesion   ● Tendency of water to stick to other substances   ● $\text{Cohesion + Adhesion =}$ capillary action
•  Allows water to flow up roots/trunks/branches of trees in thin vessels ■ Surface tension ● Results from cohesion of water molecules ● Ex. water striders can sit on top of water without sinking ■ High heat capacity         ● Heat Capacity=ability of a substance to resist temperature changes         ● Keeps ocean temperatures stable         ● Allows organisms to keep constant body temperature, since most life forms are mostly made up of water        ● Heat is absorbed when hydrogen bonds break, released when hydrogen bonds form ■ High heat of vaporization        ● Heat a liquid must absorb for 1g to be converted to gas        ● Evaporative cooling
•  As a liquid evaporates, its remaining surface cools ■ How sweat works to cool body down ■ Expansion on freezing       ● Lattice structure of ice causes water to expand on freezing       ● Allows ice to float on top of lakes in winter
•  Animal life can live beneath ice ■ Versatility as a solvent       ● Solution is a liquid that is a homogenous mix of substances       ● Solvent is the dissolving agent of a solution       ● Solute is the substance that is dissolved       ● Aqueous solution is one where water is the solvent       ● Polarity of water allows it to be a versatile solvent
•  Can form hydrogen bonds easily ● Hydrophobic substances do not dissolve in water, but hydrophilic ones will

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