Match each physical scenario involving scalar and vector quantities on the left with its c… — JAMB UTME

Match each physical scenario involving scalar and vector quantities on the left with its corresponding resultant value or component magnitude on the right.

A particle undergoes successive horizontal displacements of $10\text{ m}$ East, $12\text{ m}$ North, and $5\text{ m}$ West. The magnitude of its net displacement. $13\text{ m}$
Two equal coplanar forces, each of magnitude $F$ , act at an angle of $60^\circ$ to each other. The magnitude of their resultant force. $F\sqrt{3}$
A force vector of magnitude $40\text{ N}$ is inclined at an angle of $60^\circ$ to the vertical axis. The magnitude of its vertical component. $20\text{ N}$
Two concurrent forces of magnitudes $8\text{ N}$ and $15\text{ N}$ act at an angle of $90^\circ$ to one another. The magnitude of their resultant force. $17\text{ N}$

Answer

The correct pairings are: (1) The particle's net displacement corresponds to 13 m; (2) The resultant of two equal forces of magnitude F at 60 degrees corresponds to F√3; (3) The vertical component of a 40 N force inclined at 60 degrees to the vertical corresponds to 20 N; (4) The resultant of perpendicular forces of 8 N and 15 N corresponds to 17 N.

Each scenario correctly applies vector algebra: 2D displacement resolution yields a 5-12-13 right triangle; the parallelogram rule for equal forces at 60 degrees produces F√3; resolving a force adjacent to the vertical axis uses cos(60°) to give 20 N; and perpendicular 8 N and 15 N forces synthesize to a 17 N resultant using the Pythagorean theorem.

Step-by-Step Solution

Calculate net displacement for Item 1

Net x-component:

10\text{ m} - 5\text{ m} = 5\text{ m}

East. Net y-component:

12\text{ m}

North. Magnitude

R = \sqrt{5^2 + 12^2} = 13\text{ m}

Displacements along parallel lines subtract scalar-wise, and perpendicular components combine via the Pythagorean theorem.

Determine the resultant of two equal forces at 60 degrees for Item 2

R = \sqrt{F^2 + F^2 + 2(F)(F)\cos(60^\circ)} = \sqrt{2F^2 + 2F^2(0.5)} = \sqrt{3F^2} = F\sqrt{3}

Applying the parallelogram law of vector addition.

Resolve the force vector along the vertical direction for Item 3

F_{\text{vertical}} = F \cos(\theta_{\text{vertical}}) = 40 \cos(60^\circ) = 40 \times 0.5 = 20\text{ N}

The component adjacent to the reference angle uses the cosine function.

Compute resultant magnitude of orthogonal forces for Item 4

R = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\text{ N}

Vectors at right angles sum directly using Pythagorean synthesis.

Key Concept

Vector resolution, component synthesis, and parallelogram law of vector addition

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