Scalars vectors and projectile motion

Scalars and Vectors

A scalar quantity has only magnitude, while a vector quantity has both magnitude and direction. This distinction determines how the quantity is added, represented on diagrams, and used in calculations.

A vector is drawn as an arrow. The length of the arrow represents the magnitude of the vector, and the direction of the arrow represents the direction of the vector. In printed text, vectors are shown in bold (e.g. F) or with an arrow above the symbol (e.g. F⃗). The same symbol in italic without bold (e.g. F) refers to the magnitude only.

Two vectors are equal only when both their magnitude and their direction match. Two scalars are equal when their magnitudes match.

Misconception“Speed” and “velocity” are not interchangeable. Speed is a scalar and only has magnitude. Velocity is a vector and must always include direction.

Exam TipIf the question names a vector, your answer must include a direction or sign.

Resolving Vectors into Components

A vector at an angle can always be resolved into two perpendicular components that, when added, reproduce the original vector. Splitting a vector this way makes it possible to apply equations of motion or force balance independently in each direction.

Key Equations

• Horizontal component:

$${F}_{x}=Fcos\theta $$

• Vertical component:

$${F}_{y}=Fsin\theta $$

Variables:

• F = magnitude of the vector
• θ = angle measured from the horizontal
• $F_{\mathrm{x}}$ = horizontal component
• $F_{\mathrm{y}}$ = vertical component

SI unit: same as the original vector (e.g. N for force, m s⁻¹ for velocity)

Rearrangements: $F=\frac{{F}_{x}}{cos\theta }$ and $\theta ={tan}^{-1}\left(\frac{{F}_{y}}{{F}_{x}}\right)$

ProportionalityEach component is directly proportional to the magnitude F when the angle is fixed.

The horizontal component uses cosine when the angle is measured from the horizontal axis, and the vertical component uses sine in this case. If the angle is instead measured from the vertical axis, the roles of sine and cosine swap. Always check which axis the angle is measured from before substituting.

By scale drawing, the same result is obtained graphically. The vector is drawn to scale at the correct angle, and perpendicular dashed lines are dropped from its tip to each axis. The lengths along the axes are then measured and converted back using the chosen scale.

Worked Example: Resolving a Towing Force

Scenario

A rope pulls a sledge with a force of 80 N at an angle of 25° above the horizontal ground. Find the horizontal and vertical components of this force.

Finding the horizontal component

Equation used

$${F}_{x}=Fcos\theta $$

Given

$$F=80 \text{N}$$

$$\theta =25^{\circ}$$

Working

$${F}_{x}=80cos25^{\circ}$$

$${F}_{x}=80\times 0.9063$$

Answer

$${F}_{x}=72.5 \text{N}$$

Finding the vertical component

Equation used

$${F}_{y}=Fsin\theta $$

Working

$${F}_{y}=80sin25^{\circ}$$

$${F}_{y}=80\times 0.4226$$

Answer

$${F}_{y}=33.8 \text{N}$$

Interpretation

The horizontal component pulls the sledge forwards, while the vertical component lifts it slightly off the ground, reducing the normal contact force.

Finding the Resultant of Two Vectors

The resultant of two vectors is the single vector that produces the same effect as the two combined. The method used depends on whether the vectors are perpendicular or at a general angle.

For two perpendicular vectors, the magnitude of the resultant is found using Pythagoras’ theorem and the angle is found using the tangent ratio. For two coplanar vectors at any other angle, an accurate scale drawing is used: each vector is drawn to scale, tip to tail, with correct angles measured by protractor. The resultant is drawn from the tail of the first to the tip of the second, and its length and direction are measured and converted back using the scale.

Key Equations

• Magnitude of resultant (perpendicular vectors):

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

• Direction of resultant:

$$\theta ={tan}^{-1}\left(\frac{B}{A}\right)$$

Variables:

• A, B = magnitudes of the two perpendicular vectors
• R = magnitude of the resultant
• θ = angle of the resultant from vector A

SI unit: same as the input vectors

ProportionalityR is not directly proportional to either A or B alone, because the relationship is non-linear.

Worked Example: Resultant of Two Forces

Scenario

A boat experiences a forward thrust of 450 N from its engine and a sideways current force of 600 N at right angles to the thrust. Find the magnitude and direction of the resultant force on the boat.

Finding the magnitude of the resultant

Equation used

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

Given

$$A=450 \text{N}$$

$$B=600 \text{N}$$

Working

$$R=\sqrt{{450}^{2}+{600}^{2}}$$

$$R=\sqrt{202500+360000}$$

$$R=\sqrt{562500}$$

Answer

$$R=750 \text{N}$$

Finding the direction of the resultant

Equation used

$$\theta ={tan}^{-1}\left(\frac{B}{A}\right)$$

Working

$$\theta ={tan}^{-1}\left(\frac{600}{450}\right)$$

$$\theta ={tan}^{-1}(1.333)$$

Answer

$$\theta =53.1^{\circ}$$

Interpretation

The boat experiences a single equivalent force of 750 N at 53° from its intended forward direction, deflecting it sideways from its course.

Examiner InsightA vector answer must always include both magnitude AND direction. Stating only “750 N” loses the direction mark even when the calculation is correct.

Exam TipAlways finish with “…at [angle]° to [reference direction].”

Projectile Motion Under Gravity

Projectile motion describes any object moving freely under gravity after being launched, where horizontal and vertical motions are independent of each other. This independence means each direction can be analysed separately using the suvat equations.

The horizontal velocity remains constant throughout the flight because no horizontal force acts when air resistance is neglected. The vertical motion accelerates downwards at g = 9.81 m s⁻² due to the weight of the projectile. At the highest point, the vertical component of velocity is momentarily zero, while the horizontal component is unchanged.

To analyse projectile problems, resolve the initial velocity into horizontal and vertical components first. Apply the suvat equations to the vertical motion to find time of flight or maximum height, then use the time in the horizontal motion to find the range. Taking upwards as positive, the vertical acceleration is –g, and the horizontal acceleration is zero.

Key Equations

• Initial velocity components: ${u}_{x}=ucos\theta $ and ${u}_{y}=usin\theta $

suvat equations applied to each direction:

$$v=u+at$$

$$s=ut+1/2a{t}^{2}$$

$${v}^{2}={u}^{2}+2as$$

$$s=1/2(u+v)t$$

Variables:

• u = initial velocity component
• v = final velocity component
• a = acceleration in that direction
• s = displacement in that direction
• t = time of flight

SI unit: m s⁻¹ for velocity, m s⁻² for acceleration, m for displacement, s for time

ProportionalityIn vertical motion, displacement at constant acceleration is directly proportional to t² when starting from rest. Doubling the time falling from rest quadruples the distance fallen.

MisconceptionAt the highest point of the trajectory, the projectile is not stationary. Only the vertical component of velocity is zero at that point. The horizontal component continues unchanged, so the projectile is still moving horizontally.

Exam TipAt maximum height, $v_{\mathrm{y}}$ = 0 but $v_{\mathrm{x}}$ = u cos θ.

Worked Example: Projectile Launched at an Angle

Scenario

A ball is kicked from level ground with an initial speed of 25 m s⁻¹ at an angle of 35° above the horizontal. Calculate the maximum height reached and the horizontal range. Take g = 9.81 m s⁻² and ignore air resistance.

Step 1: Resolving the initial velocity

Horizontal component:

$${u}_{x}=25cos35^{\circ}=20.48 {\text{m s}}^{-1}$$

Vertical component:

$${u}_{y}=25sin35^{\circ}=14.34 {\text{m s}}^{-1}$$

Step 2: Finding the maximum height

Taking upwards as positive, at maximum height $v_{\mathrm{y}}$ = 0.

Equation used

$${v}^{2}={u}^{2}+2as$$

Rearranging for s:

$$s=\frac{{v}^{2}-{u}^{2}}{2a}$$

Substituting with u = $u_{\mathrm{y}}$, v = 0, a = –9.81 m s⁻²:

$$s=\frac{0-{14.34}^{2}}{2\times (-9.81)}$$

$$s=\frac{-205.6}{-19.62}$$

Answer

$$s=10.5 \text{m}$$

Step 3: Finding the time of flight

Time to reach the highest point uses v = u + at with v = 0:

$$0=14.34+(-9.81)t$$

$$t=\frac{14.34}{9.81}=1.462 \text{s}$$

Total time of flight is twice this value because the trajectory is symmetric:

$$T=2\times 1.462=2.924 \text{s}$$

Step 4: Finding the horizontal range

Horizontal motion has zero acceleration, so:

$$R={u}_{x}T$$

$$R=20.48\times 2.924$$

Answer

$$R=59.9 \text{m}$$

Interpretation

The ball reaches a peak height of about 10 m and lands roughly 60 m from the launch point. The horizontal velocity of 20.5 m s⁻¹ is unchanged throughout the flight.