Mathematics (9709) AS Compact cheat sheet

    Mathematics (9709) · CAIE · AS

    Compact
    3 pages
    20 formulas, 16 concepts
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    All Mathematics (9709) AS cheat sheets

    M1 - Kinematics of Motion in a Straight Line

    • SUVAT 1: Velocity-Time
      v=u+atv = u + at
      Use when you know/need to find u, v, a, t, but displacement (s) is not involved.
    • SUVAT 2: Displacement-Time (without v)
      s=ut+12at2s = ut + \frac{1}{2}at^2
      Use when you know/need to find s, u, t, a, but final velocity (v) is not involved.
    • SUVAT 3: Velocity-Displacement
      v2=u2+2asv^2 = u^2 + 2as
      Use when you know/need to find v, u, a, s, but time (t) is not involved. This is the only non-time-dependent equation.
    • SUVAT 4: Displacement-Time (without a)
      s=12(u+v)ts = \frac{1}{2}(u+v)t
      Use when you know/need to find s, u, v, t, but acceleration (a) is not involved. Represents displacement as average velocity times time.
    • Calculus: Velocity from Displacement
      v=dsdtv = \frac{ds}{dt}
      For variable acceleration, velocity (v) is the rate of change of displacement (s) with respect to time (t). Differentiate s to find v.
    • Calculus: Acceleration from Velocity
      a=dvdt=d2sdt2a = \frac{dv}{dt} = \frac{d^2s}{dt^2}
      For variable acceleration, acceleration (a) is the rate of change of velocity (v) with respect to time (t). Differentiate v to find a.
    • Calculus: Integration in Kinematics
      s=∫v dtandv=∫a dts = \int v \,dt \quad \text{and} \quad v = \int a \,dt
      For variable acceleration, integrate acceleration (a) to find velocity (v), and integrate velocity (v) to find displacement (s). Remember the constant of integration.

    Key concepts: **Constant vs. Variable Acceleration**: If acceleration is a constant value (e.g., 5 m/s², or g), use the SUVAT equations. If acceleration is given as a function of time (e.g., a = 3t - 2), you must use calculus (differentiation/integration)., **Velocity-Time Graphs**: The gradient of a velocity-time graph represents acceleration (a = Δv/Δt), and the area under the graph represents displacement (s)., **Displacement-Time Graphs**: The gradient of a displacement-time graph represents velocity (v = Δs/Δt). A straight line indicates constant velocity, while a curve indicates changing velocity (acceleration)., **SUVAT Variables**: s = displacement, u = initial velocity, v = final velocity, a = constant acceleration, t = time. These are vector quantities (except time), so direction is critical.

    Exam tips

    • Always start by defining a positive direction (e.g., 'up is positive'). Any vector quantity (s, u, v, a) pointing in the opposite direction must be given a negative sign. For vertical motion, acceleration due to gravity (g) is typically -9.8 or -10 m/s² if upwards is positive.
    • A very common mistake is applying SUVAT equations to a problem where acceleration is a function of time. If you see 'a = 2t' or similar, you must use calculus. If acceleration is constant, you can use either graphs or SUVAT.

    M1 - Newton's Laws of Motion

    • Newton's Second Law
      Fnet=maF_{net} = ma
      The fundamental equation of dynamics. Use this to find the acceleration of a particle or system when the net force is known, or to find the net force required to produce a certain acceleration. 'F_net' is the resultant or vector sum of all forces acting on the object.
    • Limiting Friction
      Fmax=μRF_{max} = \mu R
      Calculates the maximum possible frictional force between two surfaces. Use when an object is on the point of moving or is moving (in which case friction is constant at this value). 'μ' is the coefficient of friction and 'R' is the normal reaction force.
    • Weight
      W=mgW = mg
      Calculates the force due to gravity on an object of mass 'm'. 'g' is the acceleration due to gravity, typically 9.8 m/s² or 10 m/s² as specified in the question.

    Key concepts: **Newton's Second Law**: The acceleration of an object is directly proportional to the resultant force acting on it and inversely proportional to its mass. The acceleration occurs in the same direction as the resultant force., **Normal Reaction Force (R)**: A contact force exerted by a surface on an object, acting perpendicular to the surface and away from it. It is not always equal to the weight (e.g., on an inclined plane or when a vertical force is applied)., **Frictional Force (F)**: A resistive force that opposes motion or attempted motion between surfaces in contact. It acts parallel to the surface and its magnitude cannot exceed the limiting value F_max = μR.

    Exam tips

    • Before writing any equations, draw a large, clear diagram for each particle or system. Label all forces: Weight (W=mg), Normal Reaction (R), Tension (T), Friction (F), and any applied forces. This is the single most important step to prevent errors.

    M1 - Forces and Equilibrium

    • Resolving Forces into Components
      Fx=Fcos⁡θ,Fy=Fsin⁡θF_x = F \cos \theta, \quad F_y = F \sin \theta
      Used to find the perpendicular components of a force F acting at an angle θ to the x-axis. Essential for almost all force problems.
    • Condition for Equilibrium
      ∑Fx=0,∑Fy=0\sum F_x = 0, \quad \sum F_y = 0
      For a particle to be in equilibrium, the sum of the force components in any two perpendicular directions must be zero. This means all forces are balanced.
    • Friction in Limiting Equilibrium
      Fmax=μRF_{max} = \mu R
      Used ONLY when an object is in 'limiting equilibrium', meaning it is on the point of sliding. F_max is the maximum friction, μ is the coefficient of friction, and R is the normal reaction force.

    Key concepts: **Equilibrium of a Particle**: A particle is in equilibrium if the vector sum of all forces acting on it is zero. This means the particle is either at rest or moving with a constant velocity (zero acceleration)., **Resolving Forces**: The process of splitting a single force vector into two or more components, typically at right angles to each other (e.g., horizontal and vertical, or parallel and perpendicular to a slope)., **Normal Reaction (R)**: The contact force exerted by a surface on an object, which always acts perpendicular to the surface and away from it.

    Exam tips

    • Before any calculation, draw a large, clear diagram showing the particle and ALL forces acting on it as arrows originating from the particle. Label every force (e.g., W, R, T, F, P).

    M1 - Energy, Work and Power

    • Work Done by a Force
      W=Fdcos⁡θW = Fd \cos\theta
      Calculates the work done (W) by a constant force (F) that moves an object a distance (d). Here, θ is the angle between the force vector and the direction of displacement.
    • Kinetic Energy (KE)
      KE=12mv2KE = \frac{1}{2}mv^2
      Calculates the energy an object possesses due to its motion. It depends on the object's mass (m) and its speed (v).
    • Gravitational Potential Energy (GPE)
      GPE=mghGPE = mgh
      Calculates the energy an object possesses due to its position in a gravitational field. It depends on the object's mass (m), the acceleration due to gravity (g), and its vertical height (h) above a reference point.
    • Power, Force and Velocity
      P=FvP = Fv
      Calculates the instantaneous power (P) developed by a driving force (F) acting on an object moving at velocity (v). F must be the driving force in the direction of motion.
    • Work-Energy Principle
      Initial Energy+WD=Final Energy+WR\text{Initial Energy} + W_{D} = \text{Final Energy} + W_{R}
      The core principle relating energy changes. Initial Energy (KE + GPE) plus Work Done by Driving Forces equals Final Energy (KE + GPE) plus Work Done against Resistances.

    Key concepts: **Work-Energy Principle**: The net work done on an object by external forces (driving forces minus resistive forces) is equal to the change in its total mechanical energy (Kinetic Energy + Gravitational Potential Energy)., **Work Done**: Work is done when a force causes displacement, representing a transfer of energy from one form to another. It is a scalar quantity measured in Joules (J)., **Power**: Power is the rate at which work is done or energy is transferred. For a vehicle, it is the rate at which the engine is doing work, measured in Watts (W).

    Exam tips

    • Always use standard SI units before calculation: mass in kg, distance in m, velocity in m/s, time in s. Energy will be in Joules (J) and Power in Watts (W). A common trap is being given mass in tonnes or speed in km/h.

    M1 - Momentum

    • Linear Momentum
      p=mvp = mv
      Calculates the momentum (p) of an object, where 'm' is the mass and 'v' is the velocity. Momentum is measured in Ns or kg m/s.
    • Conservation of Linear Momentum
      m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2
      The total momentum before a direct collision equals the total momentum after. u₁ and u₂ are initial velocities, v₁ and v₂ are final velocities.

    Key concepts: **Linear Momentum**: Linear momentum is the product of an object's mass and its velocity, quantifying its motion. It is a vector quantity, meaning it has both magnitude and direction., **Principle of Conservation of Momentum**: In a closed system with no external forces acting, the total linear momentum before a collision is equal to the total linear momentum after the collision., **Vector Nature of Momentum**: Since velocity is a vector, momentum is also a vector. The direction of motion is critical and must be handled using positive and negative signs in calculations.

    Exam tips

    • Always start by defining a positive direction (e.g., 'to the right is positive'). Any velocity in the opposite direction must be given a negative sign. This is the most common source of errors.