Scalars and Vectors Practice Questions — JAMB UTME

Question 1Question →

Match each physical quantity on the left with its correct physical definition and scalar or vector classification on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Work Done matches with Scalar quantity defined as the product of force and displacement in the direction of the force; Acceleration matches with Vector quantity defined as the rate of change of velocity with time; Electric Potential matches with Scalar quantity defined as the work done per unit positive charge in bringing it from infinity; Force matches with Vector quantity defined as the rate of change of linear momentum with time.

Work Done matches the scalar definition involving force and displacement. Acceleration matches the vector definition representing rate of change of velocity. Electric Potential matches the scalar definition of work per unit charge. Force matches the vector definition representing rate of change of momentum.

Step-by-Step Solution

1

Classify each physical quantity as a scalar (has magnitude only) or a vector (has both magnitude and direction).

Work Done and Electric Potential are scalar quantities. Acceleration and Force are vector quantities.

Scalars require only numerical value and unit, whereas vectors require direction to be fully specified.

2

Match each physical quantity with its precise physical definition.

Work Done corresponds to force multiplied by displacement in the line of action; Acceleration corresponds to velocity change per unit time; Electric Potential corresponds to work done per unit charge; Force corresponds to rate of change of momentum.

Each definition uniquely identifies the fundamental physical relationship for that quantity.

Key Concept

Classification of physical quantities into scalars and vectors based on their directional properties and definitions

Question 2Question →

Match each vector scenario on the left with its correct resultant magnitude or value on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Match perpendicular forces of 6 N and 8 N to 10 N; minimum resultant of 7 N and 12 N forces to 5 N; two 15 N forces at 120 degrees to 15 N; and 3D displacement vector to 13 m.

Each vector calculation correctly applies the geometric or algebraic properties of vectors: orthogonal vector resolution, opposite-direction subtraction, law of cosines for equal magnitudes at 120 degrees, and 3D component magnitude synthesis.

Step-by-Step Solution

1

Calculate the magnitude of perpendicular vectors

R = \sqrt{6^2 + 8^2} = 10\text{ N}

Perpendicular vectors form a right-angled triangle, so the Pythagorean theorem applies.

2

Find the minimum resultant magnitude of two vectors

R_{\text{min}} = 12\text{ N} - 7\text{ N} = 5\text{ N}

Minimum resultant occurs when vectors act collinear in opposite directions.

3

Determine the resultant of two equal vectors at

120^\circ

R = 15\text{ N}

Using the cosine rule

R = \sqrt{F^2 + F^2 + 2F^2\cos 120^\circ}

, since

\cos 120^\circ = -0.5

,

R = F

.

4

Compute the magnitude of the 3D displacement vector

|\mathbf{r}| = \sqrt{3^2 + 4^2 + 12^2} = 13\text{ m}

3D magnitude is calculated using the square root of the sum of squared orthogonal components.

Key Concept

Vector Addition, Resolution, and Magnitude Evaluation

Question 3Question →

An aircraft flies due East at a velocity of $120\text{ m/s}$ while a crosswind blows due North at $50\text{ m/s}$ . What is the magnitude of the resultant velocity of the aircraft?

130\text{ m/s}

170\text{ m/s}

70\text{ m/s}

110\text{ m/s}

Show answer & explanation

Answer: $130\text{ m/s}$

Answer

The magnitude of the resultant velocity of the aircraft is

130\text{ m/s}

.

Because the aircraft's velocity and the wind's velocity are perpendicular to each other, their vector sum is found using the Pythagorean theorem

R = \sqrt{v_1^2 + v_2^2}

. Substituting

120\text{ m/s}

and

50\text{ m/s}

gives

R = \sqrt{120^2 + 50^2} = \sqrt{16900} = 130\text{ m/s}

.

Step-by-Step Solution

1

Identify the given velocity vectors and their relative direction.

Velocity due East

\vec{v}_e = 120\text{ m/s}

and velocity due North

\vec{v}_n = 50\text{ m/s}

are perpendicular to each other (

90^\circ

angle).

East and North directions are orthogonal to each other.

2

Apply the Pythagorean theorem to calculate the magnitude of the resultant vector

R

.

R = \sqrt{v_e^2 + v_n^2} = \sqrt{120^2 + 50^2} = \sqrt{14400 + 2500} = \sqrt{16900} = 130\text{ m/s}

.

The resultant of two perpendicular vectors forms the hypotenuse of a right-angled triangle.

Key Concept

Vector Addition of Perpendicular Vectors

Estimated Time:1m 0s

Question 4Question →

Match each displacement vector combination on the left with its corresponding resultant displacement magnitude or vector on the right.

Click a left item, then click its matching right item

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Answer

The correct pairs correspond as follows:

3\text{ m}

East and

4\text{ m}

North matches a resultant magnitude of

5\text{ m}

;

5\text{ m}

East and

12\text{ m}

South matches a resultant magnitude of

13\text{ m}

;

8\text{ m}

East and

6\text{ m}

West matches a resultant displacement of

2\text{ m}

East; and

9\text{ m}

North and

12\text{ m}

East matches a resultant magnitude of

15\text{ m}

.

Each vector combination is resolved according to its directional alignment: perpendicular displacements require the Pythagorean theorem (

R = \sqrt{A^2 + B^2}

), whereas anti-parallel collinear displacements require vector subtraction.

Step-by-Step Solution

1

Identify orthogonal vector scenarios

Perpendicular displacement vectors form right-angled triangles.

Directions such as East-North, East-South, and North-East are at

90^\circ

relative to one another.

2

Calculate magnitudes for orthogonal pairs using the Pythagorean theorem

For

3\text{ m}

and

4\text{ m}

:

\sqrt{3^2 + 4^2} = 5\text{ m}

. For

5\text{ m}

and

12\text{ m}

:

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

. For

9\text{ m}

and

12\text{ m}

:

\sqrt{9^2 + 12^2} = 15\text{ m}

.

The magnitude of two perpendicular vectors

\vec{A}

and

\vec{B}

is given by

R = \sqrt{A^2 + B^2}

.

3

Calculate net displacement for opposite collinear vectors

For

8\text{ m}

East and

6\text{ m}

West:

8 - 6 = 2\text{ m}

East.

Vectors pointing in opposite directions along the same axis subtract algebraically, retaining the direction of the vector with the greater magnitude.

Key Concept

Addition of Perpendicular and Collinear Displacement Vectors

Estimated Time:1m 30s

Question 5Question →

A boat moves due north across a river at a speed of $4.0\text{ m s}^{-1}$ relative to the water. If the river current flows due east at a speed of $3.0\text{ m s}^{-1}$ , what is the magnitude of the resultant velocity of the boat relative to the riverbank?

1.0\text{ m s}^{-1}

5.0\text{ m s}^{-1}

7.0\text{ m s}^{-1}

25.0\text{ m s}^{-1}

Show answer & explanation

Answer: $5.0\text{ m s}^{-1}$

Answer

The magnitude of the resultant velocity of the boat relative to the riverbank is

5.0\text{ m s}^{-1}

.

The resultant velocity magnitude is found by applying vector addition for perpendicular components:

v_{resultant} = \sqrt{(4.0)^2 + (3.0)^2} = \sqrt{25.0} = 5.0\text{ m s}^{-1}

.

Step-by-Step Solution

1

Identify the given velocity components and their directions

Northward velocity component

v_y = 4.0\text{ m s}^{-1}

, Eastward velocity component

v_x = 3.0\text{ m s}^{-1}

.

The motion of the boat and the movement of the river current act in perpendicular directions (

90^\circ

to each other).

2

Calculate the resultant magnitude using vector addition (Pythagorean theorem)

v = \sqrt{v_x^2 + v_y^2} = \sqrt{(3.0)^2 + (4.0)^2} = \sqrt{9.0 + 16.0} = \sqrt{25.0} = 5.0\text{ m s}^{-1}

.

For two orthogonal vectors, the magnitude of the vector sum equals the hypotenuse of the right triangle formed by the vector components.

Key Concept

Vector Addition of Perpendicular Quantities

Estimated Time:1m 0s

Question 6Question →

A crate on a smooth surface is simultaneously pulled by two horizontal forces. One force of magnitude $9.0\text{ N}$ acts towards the East, and another force of magnitude $12.0\text{ N}$ acts towards the North. What is the magnitude of the resultant force acting on the crate in newtons?

Show answer & explanation

Answer: 15

Answer

The magnitude of the resultant force acting on the crate is

15.0\text{ N}

.

Because the two pulling forces are perpendicular, their resultant magnitude is calculated using vector addition via the Pythagorean theorem:

R = \sqrt{9.0^2 + 12.0^2} = 15.0\text{ N}

.

Step-by-Step Solution

1

Determine the angle between the two given vectors

The forces are perpendicular (

90^\circ

).

East and North cardinal directions are orthogonal to each other.

2

Compute the resultant magnitude using vector synthesis

R = \sqrt{9.0^2 + 12.0^2} = \sqrt{81 + 144} = \sqrt{225} = 15.0\text{ N}

For perpendicular force vectors, the resultant magnitude is the hypotenuse of the right triangle formed by the vector components.

Key Concept

Vector addition of perpendicular forces using the Pythagorean theorem

Estimated Time:1m 15s

Question 7Question →

A car moves $80\text{ km}$ due East along a straight highway and then turns to travel $60\text{ km}$ due North. What is the magnitude of the displacement of the car from its starting position?

100\text{ km}

140\text{ km}

20\text{ km}

70\text{ km}

Show answer & explanation

Answer: $100\text{ km}$

Answer

The magnitude of the displacement is

100\text{ km}

.

Because displacement is a vector quantity, perpendicular components must be combined vectorially using the Pythagorean theorem rather than scalar addition. The calculation

\sqrt{80^2 + 60^2} = 100\text{ km}

correctly yields the magnitude of the net displacement vector.

Step-by-Step Solution

1

Identify the vector components and their directions.

Eastward displacement

x = 80\text{ km}

, Northward displacement

y = 60\text{ km}

. The directions are mutually perpendicular (

90^\circ

to each other).

Displacement is a vector quantity, so direction must be accounted for when combining components.

2

Apply the Pythagorean theorem to calculate the resultant vector magnitude.

R = \sqrt{x^2 + y^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100\text{ km}

.

When two vector components act at right angles (

90^\circ

), the magnitude of their resultant is given by the hypotenuse of the right-angled triangle formed by the vectors.

Key Concept

Vector Addition of Perpendicular Components

Question 8Question →

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

Click a left item, then click its matching right item

Show answer & explanation

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

1

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.

2

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.

3

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.

4

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

Question 9Question →

Three coplanar forces act simultaneously at a point $O$ . The first force of magnitude $10\text{ N}$ acts due East, and the second force of magnitude $10\text{ N}$ acts at an angle of $60^\circ$ North of East. If a third force keeps the system in static equilibrium, what is the magnitude of this third force?

10\sqrt{3}\text{ N}

20\text{ N}

10\sqrt{2}\text{ N}

10\text{ N}

Show answer & explanation

Answer: $10\sqrt{3}\text{ N}$

Answer

The magnitude of the third force required for equilibrium is

10\sqrt{3}\text{ N}

.

Resolving the forces along the horizontal (East) and vertical (North) axes gives components of

15\text{ N}

and

5\sqrt{3}\text{ N}

respectively. The magnitude of the resultant force is

\sqrt{15^2 + (5\sqrt{3})^2} = \sqrt{300} = 10\sqrt{3}\text{ N}

. Since the third force balances the system, its magnitude must equal that of the resultant, which is

10\sqrt{3}\text{ N}

.

Step-by-Step Solution

1

Resolve the two given forces into horizontal (

x

) and vertical (

y

) Cartesian components.

F_{1x} = 10\text{ N}

,

F_{1y} = 0\text{ N}

;

F_{2x} = 10 \cos(60^\circ) = 5\text{ N}

,

F_{2y} = 10 \sin(60^\circ) = 5\sqrt{3}\text{ N}

.

Vector addition requires breaking non-orthogonal vectors into perpendicular component directions.

2

Calculate the total horizontal (

R_x

) and vertical (

R_y

) components of the resultant of the first two forces.

R_x = 10 + 5 = 15\text{ N}

,

R_y = 0 + 5\sqrt{3} = 5\sqrt{3}\text{ N}

.

Adding aligned components yields the net component along each axis.

3

Compute the magnitude of the resultant force

R

.

R = \sqrt{R_x^2 + R_y^2} = \sqrt{15^2 + (5\sqrt{3})^2} = \sqrt{225 + 75} = \sqrt{300} = 10\sqrt{3}\text{ N}

.

Applying Pythagoras' theorem to orthogonal components determines the net magnitude.

4

Determine the magnitude of the equilibrant (third force).

Equibrant magnitude

= R = 10\sqrt{3}\text{ N}

.

For static equilibrium, the third force must be equal in magnitude and opposite in direction to the resultant of the first two forces.

Key Concept

Vector Addition and Equilibrium of Forces

Question 10Question →

An autonomous drone navigating an obstacle course experiences three mutually perpendicular velocity vectors simultaneously: a horizontal forward velocity of $12\text{ m s}^{-1}$ due East, a horizontal crosswind drift velocity of $9\text{ m s}^{-1}$ due North, and a vertical downdraft velocity of $8\text{ m s}^{-1}$ directed straight downward. What is the magnitude of the resultant velocity vector of the drone in $\text{m s}^{-1}$ ?

Show answer & explanation

Answer: 17

Answer

17 m s⁻¹

Because the three velocity components are mutually perpendicular, the magnitude of the overall resultant velocity is calculated using the 3D Pythagorean theorem:

\sqrt{12^2 + 9^2 + 8^2} = \sqrt{144 + 81 + 64} = \sqrt{289} = 17\text{ m s}^{-1}

.

Step-by-Step Solution

1

Identify the perpendicular vector components

v_x = 12\text{ m s}^{-1}

,

v_y = 9\text{ m s}^{-1}

,

v_z = 8\text{ m s}^{-1}

The three given velocity vectors act along mutually orthogonal spatial axes (East, North, and Downward).

2

Apply the 3D vector resultant magnitude formula

v_r = \sqrt{v_x^2 + v_y^2 + v_z^2}

Since the vector components are perpendicular to one another, the magnitude of their resultant is given by the extension of the Pythagorean theorem to three dimensions.

3

Substitute values and evaluate

v_r = \sqrt{144 + 81 + 64} = \sqrt{289} = 17\text{ m s}^{-1}

Squaring each component, adding them, and taking the principal square root yields the total magnitude of the velocity vector.

Key Concept

Magnitude of three mutually perpendicular vector components using the 3D Pythagorean theorem

Question 11Question →

A particle is subjected to two mutually perpendicular horizontal forces of magnitudes $15\text{ N}$ acting due East and $20\text{ N}$ acting due South. What is the magnitude of the resultant force acting on the particle?

25\text{ N}

35\text{ N}

5\text{ N}

13.2\text{ N}

Show answer & explanation

Answer: $25\text{ N}$

Answer

The magnitude of the resultant force acting on the particle is

25\text{ N}

.

Because the two forces act at right angles (

90^\circ

) relative to each other, vector addition requires applying the Pythagorean theorem. Squaring both component magnitudes (

15^2 = 225

and

20^2 = 400

), summing them to obtain

625

, and taking the square root gives the resultant magnitude of

25\text{ N}

.

Step-by-Step Solution

1

Identify the force components and their geometric orientation

The East force component

F_x = 15\text{ N}

and the South force component

F_y = 20\text{ N}

meet at an angle of

90^\circ

.

Perpendicular vectors form the adjacent and opposite sides of a right-angled vector triangle.

2

Apply the Pythagorean theorem to find the hypotenuse representing the resultant force

R = \sqrt{F_x^2 + F_y^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ N}

.

The resultant of two orthogonal vector quantities equals the square root of the sum of their individual squares.

Key Concept

Vector Addition of Perpendicular Forces

Question 12Question →

Match each physical quantity or scenario on the left with its correct scalar or vector classification and physical property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: (1) Displacement on a closed track matches Vector quantity depending on initial and final positions (evaluating to zero); (2) Work done matches Scalar quantity defined as the dot product of force and displacement; (3) Impulse matches Vector quantity equal to change in momentum with spatial direction; (4) Hydrostatic pressure matches Scalar quantity defined as normal force per unit area acting equally in all directions.

Each physical quantity is categorized accurately based on vector/scalar definition: Displacement is a vector that evaluates to zero over a closed loop. Work done is a scalar derived from the vector dot product. Impulse is a vector defined by momentum change direction. Hydrostatic pressure is a scalar acting equally in all directions at a fluid depth.

Step-by-Step Solution

1

Analyze item 1: Displacement of a particle moving in a complete circular track of radius

r

Displacement is a vector pointing from start to end position. Since the path returns to the start, the magnitude is

0\text{ m}

.

Vectors require direction and position change; returning to origin yields zero vector displacement.

2

Analyze item 2: Work done on a wooden box pushed horizontally

Work is calculated using

W = \vec{F} \cdot \vec{s} = F s \cos\theta

, which yields a scalar value measured in Joules.

The dot product of two vectors produces a scalar quantity.

3

Analyze item 3: Impulse imparted by a vertical wall to a bouncing rubber ball

Impulse is defined as

\vec{J} = \vec{F}_{avg} \Delta t = \Delta \vec{p}

. It is a vector pointing away from the wall.

Impulse has the direction of the applied average force or change in momentum vector.

4

Analyze item 4: Hydrostatic pressure exerted by a fluid at depth

h

Pressure is calculated as

P = \rho g h

. It acts omnidirectionally at any point in fluid statics.

Because pressure at a point has no preferred direction, it is classified as a scalar physical quantity.

Key Concept

Classification of Physical Quantities into Scalars and Vectors based on Directional Properties

Question 13Question →

A hiker walks $12\text{ km}$ due East, then $9\text{ km}$ due South, and finally $4\text{ km}$ due North. What is the magnitude of the hiker's total displacement from the starting point?

13\text{ km}

25\text{ km}

15\text{ km}

7\text{ km}

Show answer & explanation

Answer: $13\text{ km}$

Answer

The magnitude of the hiker's total displacement is

13\text{ km}

.

Displacement is a vector quantity requiring directional vector addition. The horizontal Eastward component is

12\text{ km}

. The net vertical component along the North-South line is

9\text{ km \text{ South}} - 4\text{ km \text{ North}} = 5\text{ km \text{ South}}

. Because the East and South directions are perpendicular (

90^\circ

), the resultant displacement magnitude is

\sqrt{12^2 + 5^2} = \sqrt{169} = 13\text{ km}

.

Step-by-Step Solution

1

Set up a coordinate system for horizontal (East-West) and vertical (North-South) components.

East is

+x

direction and North is

+y

direction.

Resolving vectors into orthogonal components allows simple independent summation along each axis.

2

Calculate net displacement along the x-axis and y-axis.

R_x = +12\text{ km}

and

R_y = -9\text{ km} + 4\text{ km} = -5\text{ km}

.

North and South act in opposite directions along the y-axis, so their magnitudes subtract.

3

Calculate the magnitude of the resultant displacement vector using Pythagoras' theorem.

|\vec{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ km}

.

Perpendicular components combine geometrically to give the magnitude of the resultant vector.

Key Concept

Vector resolution and addition of non-collinear displacement vectors.

Estimated Time:1m 30s

Question 14Question →

Match each physical scenario or vector operation on the left with its corresponding physical property or resultant classification on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Particle in uniform circular motion pairs with a vector quantity of constant magnitude but changing direction; Dot product of force and displacement pairs with a scalar quantity representing work done; Two equal forces at 120 degrees pair with a resultant magnitude equal to P; Slope of displacement-time graph pairs with a vector quantity representing velocity.

Each physical scenario correctly corresponds to its fundamental vector/scalar classification or calculated resultant based on vector algebra.

Step-by-Step Solution

1

Analyze circular motion at constant speed

The velocity vector changes direction at every point, giving rise to centripetal acceleration directed toward the center.

Any physical quantity with direction is a vector quantity.

2

Evaluate the product of force and displacement vectors

W = \mathbf{F} \cdot \mathbf{d} = F d \cos\theta

, producing a scalar quantity (work done).

The dot product of two vectors yields a scalar.

3

Calculate the resultant of two equal vectors

P

at

120^\circ

R = \sqrt{P^2 + P^2 + 2P^2 \cos 120^\circ} = \sqrt{2P^2 - P^2} = P

.

Vector addition follows the law of cosines taking spatial angle into account.

4

Determine the physical quantity represented by the gradient of a displacement-time graph

\text{Slope} = \frac{\Delta s}{\Delta t} = v \text{ (velocity)}.

Displacement per unit time is velocity, a vector quantity.

Key Concept

Scalars, Vectors, and Vector Addition Laws

Scalars and Vectors

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