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Motion in a Plane - Resolution of Vectors

Grade 11CBSEPhysics

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

🔑Concepts

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Resolution of a vector is the process of splitting a single vector into two or more vectors (called components) in such a way that their combined effect is the same as the original vector.

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The most common way to resolve a vector is into its rectangular components, which are perpendicular to each other, usually along the xx and yy axes in a 2D plane.

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If a vector A⃗\vec{A} makes an angle θ\theta with the positive xx-axis, its horizontal component is Ax=Acos⁡θA_x = A \cos \theta and its vertical component is Ay=Asin⁡θA_y = A \sin \theta.

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In terms of unit vectors i^\hat{i} and j^\hat{j}, the vector can be expressed as A⃗=Axi^+Ayj^\vec{A} = A_x \hat{i} + A_y \hat{j}.

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The magnitude of the vector is given by ∣A⃗∣=Ax2+Ay2|\vec{A}| = \sqrt{A_x^2 + A_y^2} and the direction (angle θ\theta) is given by tan⁡θ=AyAx\tan \theta = \frac{A_y}{A_x}.

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Resolution of vectors is essential for solving problems involving motion in a plane, such as projectile motion or forces acting at various angles.

📐Formulae

Ax=Acos⁡θA_x = A \cos \theta

Ay=Asin⁡θA_y = A \sin \theta

A⃗=Axi^+Ayj^\vec{A} = A_x \hat{i} + A_y \hat{j}

∣A⃗∣=Ax2+Ay2|\vec{A}| = \sqrt{A_x^2 + A_y^2}

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

💡Examples

Problem 1:

A force of 50 N50\text{ N} acts on a particle at an angle of 30∘30^\circ with the horizontal. Find the horizontal and vertical components of the force.

Solution:

Fx=Fcos⁡θ=50cos⁡30∘=50×32=253 N≈43.3 NF_x = F \cos \theta = 50 \cos 30^\circ = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3}\text{ N} \approx 43.3\text{ N}. Fy=Fsin⁡θ=50sin⁡30∘=50×12=25 NF_y = F \sin \theta = 50 \sin 30^\circ = 50 \times \frac{1}{2} = 25\text{ N}.

Explanation:

By applying the trigonometric resolution of vectors, we find the effect of the force along the xx-axis (horizontal) and yy-axis (vertical) separately.

Problem 2:

A velocity vector is given by v⃗=(3i^+4j^) m/s\vec{v} = (3\hat{i} + 4\hat{j})\text{ m/s}. Find the magnitude of the velocity and the angle it makes with the xx-axis.

Solution:

Magnitude ∣v⃗∣=32+42=9+16=25=5 m/s|\vec{v}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ m/s}. Direction θ=tan⁡−1(43)≈53.13∘\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ.

Explanation:

We use the Pythagorean theorem for the magnitude and the inverse tangent of the ratio of the yy and xx components to find the direction.

Resolution of Vectors - Revision Notes & Key Formulas | CBSE Class 11 Physics