krit.club logo

Motion, Forces and Energy - Physical quantities and measurement techniques

Grade 12IGCSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Physical quantities consist of a numerical magnitude and a unit. Standard International (SI) base units include the meter (mm), kilogram (kgkg), second (ss), ampere (AA), and kelvin (KK).

•

Scalar quantities have magnitude only (e.g., distance, speed, time, mass, energy). Vector quantities have both magnitude and direction (e.g., displacement, velocity, acceleration, force, momentum).

•

Vectors can be added using the parallelogram law or the head-to-tail method. A vector VV at an angle θ\theta to the horizontal can be resolved into components: Vx=Vcos⁡(θ)V_x = V \cos(\theta) and Vy=Vsin⁡(θ)V_y = V \sin(\theta).

•

Precision refers to how close a series of measurements are to one another, while accuracy refers to how close a measurement is to the true value.

•

Systematic errors (e.g., zero error in a micrometer) affect accuracy and cannot be reduced by averaging. Random errors (e.g., parallax error, reaction time) affect precision and can be reduced by taking multiple readings and calculating a mean.

•

Measurement instruments: A ruler has a precision of 1 mm1\text{ mm}, a vernier caliper 0.1 mm0.1\text{ mm}, and a micrometer screw gauge 0.01 mm0.01\text{ mm}.

📐Formulae

Density(ρ)=mV\text{Density} (\rho) = \frac{m}{V}

Percentage Uncertainty=UncertaintyMeasured Value×100%\text{Percentage Uncertainty} = \frac{\text{Uncertainty}}{\text{Measured Value}} \times 100\%

R=Fx2+Fy2R = \sqrt{F_x^2 + F_y^2}

θ=tan⁡−1(FyFx)\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)

💡Examples

Problem 1:

A student measures the diameter of a wire using a micrometer screw gauge. The readings are 1.22 mm1.22\text{ mm}, 1.23 mm1.23\text{ mm}, and 1.21 mm1.21\text{ mm}. The micrometer has a zero error of +0.02 mm+0.02\text{ mm}. Calculate the corrected average diameter.

Solution:

Mean observed diameter=1.22+1.23+1.213=1.22 mm\text{Mean observed diameter} = \frac{1.22 + 1.23 + 1.21}{3} = 1.22\text{ mm} Corrected diameter=Observed value−Zero error\text{Corrected diameter} = \text{Observed value} - \text{Zero error} Corrected diameter=1.22 mm−0.02 mm=1.20 mm\text{Corrected diameter} = 1.22\text{ mm} - 0.02\text{ mm} = 1.20\text{ mm}

Explanation:

To find the true value, the mean of the observed readings must be calculated first, and then the systematic zero error must be subtracted from that mean.

Problem 2:

Calculate the resultant force of two perpendicular forces: F1=3.0 NF_1 = 3.0\text{ N} acting North and F2=4.0 NF_2 = 4.0\text{ N} acting East.

Solution:

R=F12+F22=3.02+4.02=9+16=5.0 NR = \sqrt{F_1^2 + F_2^2} = \sqrt{3.0^2 + 4.0^2} = \sqrt{9 + 16} = 5.0\text{ N} θ=tan⁡−1(3.04.0)=36.9∘ North of East\theta = \tan^{-1}\left(\frac{3.0}{4.0}\right) = 36.9^\circ \text{ North of East}

Explanation:

Since the forces are perpendicular, Pythagoras' theorem is used to find the magnitude of the resultant vector, and trigonometry (SOH CAH TOA) is used to find the direction.

Physical quantities and measurement techniques Revision - Grade 12 Physics IGCSE