CAIE AS Level Mathematics Notes

Scalar and Vector Quantities

Within the study of kinematics, quantities are categorized based on whether they possess direction. Scalar quantities are defined as those that only have magnitude and do not have direction. Examples of scalar quantities include distance, speed, mass, and time, as each of these is fully described by a numerical value. In contrast, vector quantities possess both magnitude and direction. Examples include displacement, velocity, and acceleration. These values can be positive or negative, where the sign serves as an indicator of the direction of the motion or force relative to a defined axis.

Kinematics Equations

Kinematics equations are specific formulas used to describe the motion of objects. However, these formulas can only be utilized under the strict condition that acceleration is a constant value. The primary equations are as follows:

$v = u + at$

$s = ut + \frac{1}{2}at^2$

$s = vt - \frac{1}{2}at^2$

$s = \frac{1}{2}(u + v)t$

$v^2 = u^2 + 2as$

In these equations, $s$ represents displacement, $u$ represents initial velocity, $v$ represents final velocity, $a$ represents acceleration, and $t$ represents time.

Graphical Analysis of Motion

Graphs provide a visual representation of how motion variables change over time. In a Displacement-Time Graph, the gradient of the line at any given point represents the velocity of the object. In a Velocity-Time Graph, the gradient of the line represents the acceleration, while the area under the graph represents the change in displacement, denoted as $\Delta x$.

Consider the example from {S22-P43} Question 3. A particle's motion is analyzed between $t = 5$ and $t = 10$. (a) Since speed is the gradient of a distance-time graph and the line is linear, the speed is constant. The gradient is calculated as $\frac{\Delta y}{\Delta x} = \frac{150 - 50}{10 - 5} = \frac{100}{5} = 20\,ms^{-1}$. (b) To find acceleration between $t = 5$ and $t = 10$, we note the particle starts from rest ($u = 0\,ms^{-1}$ at $t = 5$) and reaches $v = 20\,ms^{-1}$ at $t = 10$. Using $v = u + at$, we rearrange to $a = \frac{v - u}{t} = \frac{20 - 0}{5} = 4\,ms^{-2}$. (c) If the particle passes point R at $t = 15$, which is $200\,m$ from the start, and travels a further $200\,m$ to reach a total of $400\,m$ at $t = 20$, the average speed is $\frac{\text{total distance}}{\text{total time}} = \frac{400}{20} = 20\,ms^{-1}$.

Average and Relative Velocities

For an object moving with constant acceleration over a specific period, several quantities are equal: the average velocity, the mean of the initial and final velocities, and the instantaneous velocity at the exact midpoint of the time interval. When analyzing two particles, A and B, traveling distances $s_A$ and $s_B$, if a collision occurs at a point C, the relationship is expressed as $s_A + s_B = D$, where D is the initial separation. This analysis is applicable to both horizontal and vertical motion.

Newton’s Laws of Motion

Newton's first law of motion states that an object will remain at rest or continue to move with a constant velocity unless an external force is applied to it. The second law of motion is quantitatively expressed by the formula $F = ma$, where force equals mass times acceleration. The third law of motion states that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A.

Vertical Motion and Projectiles

In vertical motion scenarios, weight is always directed vertically downwards, while the normal contact force acts perpendicular to the plane of contact. To find the time taken to reach the maximum height for a projectile, one should set the final velocity to zero ($v = 0$) in the equation $v = u + at$ and solve for $t$. The total time to return to the original position is double this value. To find the maximum height ($H$) above a launch point, one uses $v^2 = u^2 - 2as$ with $v = 0$. To find the time interval during which a particle is above a specific height ($H$), one sets $s = H$ in the displacement equation $s = ut + \frac{1}{2}at^2$, which results in a quadratic equation in $t$. Solving this provides two values of $t$, and the difference between them is the required time interval.

Example {S04-P04} describes particle P1 projected upwards at $30\,ms^{-1}$ from the ground, while P2 is projected at the same instant from a tower of height $25\,m$ at $10\,ms^{-1}$. Solving the quadratic $25 = (30)t - \frac{1}{2}(10)t^{2}$ for P1 gives $5t^{2} - 30t + 25 = 0$, resulting in $t = 1\,s$ and $t = 5\,s$. Thus, P1 is above the tower for $5 - 1 = 4\,seconds$. For part (ii), the displacement relationship is $s_1 = 25 + s_2$. Substituting kinematics into this gives $30t + \frac{1}{2}(-10)t^{2} = 25 + 10t + \frac{1}{2}(-10)t^{2}$, which simplifies to $20t = 25$, so $t = 1.25\,s$. The velocities at this instant are $v_1 = 30 - 10(1.25) = 17.5\,ms^{-1}$ and $v_2 = 10 - 10(1.25) = -2.5\,ms^{-1}$.

Resolving Forces and Lami’s Theorem

If a force $F$ makes an angle $\theta$ with a given direction, its effect in that direction is $F\cos(\theta)$. Other components include $F\cos(90 - \theta) = F\sin(\theta)$ and $F\sin(90 - \theta) = F\cos(\theta)$. When forces are in equilibrium, the resultant force is zero, and if drawn, they form a closed polygon. Methods for solving equilibrium include constructing a force triangle or resolving forces into $x$ and $y$ components where the sum of each equals zero. Lami's Theorem states that for three forces P, Q, and R in equilibrium:

$\frac{P}{\sin(\alpha)} = \frac{Q}{\sin(\beta)} = \frac{R}{\sin(\gamma)}$

Friction and Limiting Equilibrium

Friction always acts in the direction opposite to motion. Limiting equilibrium occurs when an object is on the point of moving or slipping, and the frictional force is at its maximum value. Smooth contact implies friction is negligible. The formula for friction is $F = \mu r$, where $\mu$ is the coefficient of friction and $r$ is the contact force. On a horizontal plane, the contact force equals the weight ($mg$). On an inclined plane, the contact force equals the vertical component of the weight, $mg\cos(\theta)$.

In {W11-P43} Question 6, a ring of mass $2\,kg$ on a rough horizontal rod ($\mu = 0.24$) is in limiting equilibrium. Scenario 1: The ring is about to move up. Resultant = $T\sin(30) - \text{friction} - \text{Weight} = 0$. Contact Force = $T\cos(30)$. Friction = $0.24 \times T\cos(30)$. Solving $0 = T\sin(30) - 0.24T\cos(30) - 20$ yields $T = 68.5\,N$. Scenario 2: The ring is about to move down. Here, friction acts in the opposite direction: $0 = T\sin(30) + 0.24T\cos(30) - 20$, yielding $T = 28.3\,N$.

On a rough plane, the force $P$ required for equilibrium varies. To find the maximum value (particle about to move up), friction acts down the slope: $P = F + mg\sin(\theta)$. To find the minimum value (particle about to slip down), friction acts up the slope: $F + P = mg\sin(\theta)$. In scenario {W12-P43}, with a friction magnitude of $0.36 \times 6\cos(25)$, the maximum $P = 6\sin(25) + \text{friction} = 4.49\,N$ and minimum $P = 6\sin(25) - \text{friction} = 0.578\,N$.

Connected Particles and Pulleys

When particles are connected, such as a train pulling carriages with a force of $2500\,N$ and resistances of $90\,N$, $150\,N$, and $200\,N$, the system is treated as a single object to find acceleration: $2500 - (200 + 150 + 90) = 1900a$, giving $a = 1.08\,ms^{-2}$. Tension in couplings is then found by looking at individual carriages ($T_1$ and $T_2$).

In pulley systems, the tension $T$ is uniform throughout the string if the pulley is smooth. In {W05-P04}, forces are resolved at point A vertically ($W_1\cos(40) + W_2\cos(60) = 5$) and horizontally ($W_1\sin(40) = W_2\sin(60)$). Solving these gives $W_2 = 3.26\,N$ and $W_1 = 4.40\,N$. In {S12-P41}, particles P ($0.6\,kg$) and Q ($0.4\,kg$) on slopes with $\sin(\theta) = 0.8$ yield weight effects of $4.8\,N$ and $3.2\,N$. The equations of motion are $4.8 - T = 0.6a$ and $T - 3.2 = 0.4a$. Solving gives $T = 3.84\,N$ and $a = 1.6\,ms^{-2}$. The total time for P to reach the ground ($v = 2\,ms^{-1}$) and for Q to reach max height is the sum of $t_1 = 1.25\,s$ and $t_2 = 0.25\,s$, totaling $1.5\,s$.

Force Exerted by String on Pulley

There are three primary cases for the force exerted on a pulley:

1. Case 1: Weights hanging vertically. The force on the pulley is $2T$ acting downwards.
2. Case 2: One weight vertical and one horizontal. The force is $T\sqrt{2}$ acting along the line bisecting the angle.
3. Case 3: Strings at an angle $\theta$. The force is $2T\cos(\frac{1}{2}\theta)$ acting inwards along the line that bisects $\theta$.

Work, Energy, and Power

The Principle of Conservation of Energy states that energy cannot be created or destroyed, only transformed. Key formulas include:

• Work Done: $W = Fs$
• Kinetic Energy: $E_k = \frac{1}{2}mv^2$
• Gravitational Potential Energy: $E_p = mgh$
• Power: $P = \frac{W}{t}$ and $P = Fv$

Energy changes are governed by the equation $\epsilon_f - \epsilon_i = (\text{Work})<em>{\text{engine}} - (\text{Work})</em>{\text{friction}}$, where $\epsilon_f$ and $\epsilon_i$ are final and initial kinetic energies. In {S05-P04}, a $1200\,kg$ car with $20\,kW$ power and $500\,N$ resistance moves from $10\,ms^{-1}$ to $25\,ms^{-1}$ in $30.5\,s$. The driving force at A is $F = \frac{P}{v} = \frac{20000}{10} = 2000\,N$, and acceleration is $\frac{2000-500}{1200} = 1.25\,ms^{-2}$. Work done by the engine is $20000 \times 30.5 = 610000\,J$. Change in kinetic energy is $375000 - 60000 = 315000\,J$. Distance $s$ is found via $610000 = 315000 + 500s$, resulting in $s = 590\,m$.

Momentum

Linear momentum is a vector quantity defined as $p = mv$, measured in Newton-seconds ($Ns$). The Principle of Conservation of Linear Momentum states that total momentum remains constant if no external forces act. Formulas vary based on direction:

• Same direction: $m_Au_A + m_Bu_B = m_Av_A + m_Bv_B$
• Towards each other: $m_Au_A - m_Bu_B = m_Av_A + m_Bv_B$
• Sticking together (coalescing): $m_Au_A + m_Bu_B = (m_A + m_B)v$

In the example with spheres A ($4\,kg$), B ($2\,kg$), and C ($3\,kg$), A moves at $6\,ms^{-1}$ toward stationary B and C. Momentum before is $(4 \times 6) = 24\,kgms^{-1}$. After A hits B and slows to $2\,ms^{-1}$, momentum is $(4 \times 2) + 2v = 24$, so B moves at $v = 8.0\,ms^{-1}$. When B subsequently coalesces with C to form D ($5\,kg$), momentum is $16 = 5v$, giving D a speed of $3.2\,ms^{-1}$.

General Motion in a Straight Line

General motion is analyzed using calculus. A particle is at instantaneous rest or at maximum displacement when velocity $v = 0$. Maximum velocity occurs when acceleration $a = 0$. Acceleration is the derivative of velocity ($\frac{dv}{dt}$), and displacement is the integral of velocity ($\int v\,dt$).

In {W10-P42}, velocity is $v = 0.002t^3 - 0.12t^2 + 1.8t + 5$. Acceleration is $\frac{dv}{dt} = 0.006t^2 - 0.24t + 1.8$. Setting $a = 0$ gives $t = 10$ and $t = 30$. The distance $OP$ at $t = 30$ is found by integrating $v$ from $0$ to $30$, resulting in $[0.0005t^4 - 0.04t^3 + 0.9t^2 + 5t]$ evaluated from $0$ to $30$, which equals $285\,m$.

In {S13-P42}, a complex motion over three intervals is analyzed to show the displacement $s = 0.375t^2 - 13t + 202$ for $20 \le t \le 26$. This is done by finding the displacement after the first two intervals ($s_1 = 92\,m$) and adding the displacement for the final interval ($s_2 = 2(t-20) + \frac{1}{2}(0.75)(t-20)^2$).