krit.club logo

Measurements and Uncertainties - Vector and Scalar Quantities

Grade 11IBPhysics

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

🔑Concepts

•

Scalar quantities are physical quantities that have magnitude (size) only. Examples include distance dd, speed vv, mass mm, time tt, and energy EE.

•

Vector quantities have both magnitude and direction. Examples include displacement s⃗\vec{s}, velocity v⃗\vec{v}, acceleration a⃗\vec{a}, force F⃗\vec{F}, and momentum p⃗\vec{p}.

•

Vectors are represented graphically by arrows where the length indicates the magnitude and the arrow points in the direction of the quantity.

•

Resultant vectors can be found using the 'tip-to-tail' method or the parallelogram law of vector addition.

•

Vector resolution is the process of breaking a single vector into two perpendicular components, usually horizontal (AxA_x) and vertical (AyA_y).

•

The magnitude of a resultant vector RR derived from two perpendicular components AxA_x and AyA_y is found using the Pythagorean theorem: R=Ax2+Ay2R = \sqrt{A_x^2 + A_y^2}.

•

The direction θ\theta of a resultant vector relative to the horizontal is given by θ=tan⁡−1(AyAx)\theta = \tan^{-1}\left(\frac{A_y}{A_x}\right).

•

Vector subtraction is defined as adding the negative of a vector: A⃗−B⃗=A⃗+(−B⃗)\vec{A} - \vec{B} = \vec{A} + (-\vec{B}).

📐Formulae

Ax=Acos⁡θA_x = A \cos \theta

Ay=Asin⁡θA_y = A \sin \theta

R=Ax2+Ay2R = \sqrt{A_x^2 + A_y^2}

θ=tan⁡−1(AyAx)\theta = \tan^{-1} \left( \frac{A_y}{A_x} \right)

R⃗=A⃗+B⃗\vec{R} = \vec{A} + \vec{B}

💡Examples

Problem 1:

A hiker walks 5.0 km5.0\text{ km} due North and then 12.0 km12.0\text{ km} due East. Determine the magnitude and direction of the hiker's total displacement.

Solution:

Magnitude: s=5.02+12.02=25+144=169=13.0 kms = \sqrt{5.0^2 + 12.0^2} = \sqrt{25 + 144} = \sqrt{169} = 13.0\text{ km}. Direction: θ=tan⁡−1(12.05.0)=67.4∘\theta = \tan^{-1}\left(\frac{12.0}{5.0}\right) = 67.4^\circ East of North.

Explanation:

Since the movements are perpendicular, the Pythagorean theorem is used to find the magnitude of the resultant displacement. The angle is found using the inverse tangent of the ratio of the Eastward component to the Northward component.

Problem 2:

A force of 10.0 N10.0\text{ N} acts at an angle of 30.0∘30.0^\circ to the horizontal. Calculate the horizontal and vertical components of this force.

Solution:

Fx=10.0cos⁡(30.0∘)=10.0×0.866=8.66 NF_x = 10.0 \cos(30.0^\circ) = 10.0 \times 0.866 = 8.66\text{ N}. Fy=10.0sin⁡(30.0∘)=10.0×0.500=5.00 NF_y = 10.0 \sin(30.0^\circ) = 10.0 \times 0.500 = 5.00\text{ N}.

Explanation:

Vector resolution uses trigonometric functions cos⁡\cos for the adjacent side (horizontal) and sin⁡\sin for the opposite side (vertical) relative to the given angle.

Problem 3:

A car travels at a constant speed of 20 m s−120\text{ m s}^{-1} around a circular track. Is the velocity constant? Explain.

Solution:

No, the velocity is not constant because velocity is a vector quantity.

Explanation:

Even though the speed (scalar) is constant at 20 m s−120\text{ m s}^{-1}, the direction of travel is continuously changing as the car moves around the circle. Since velocity depends on both magnitude and direction, a change in direction results in a change in velocity.

Vector and Scalar Quantities - Revision Notes & Key Formulas | IB Grade 11 Physics