Compose and resolve velocities, determine resultant vectors and use velocity vectors to solve problems in context such as motion and collisions

Vectors in Two Dimensions – Cambridge IGCSE 0606

Learning Objectives

• Write vectors using all notations accepted by the Cambridge syllabus (bold, arrow, component, ordered‑pair).
• Define and use position vectors and unit vectors.
• Resolve a vector into horizontal and vertical components and recombine components to obtain magnitude and direction.
• Perform vector operations: addition (graphical & algebraic), subtraction, scalar multiplication and find unit vectors.
• Apply vector methods to relative‑velocity problems, projectile motion and two‑dimensional collisions.
• Follow a systematic, exam‑ready problem‑solving routine.

1. Vector Notation (Syllabus 13.1)

The Cambridge syllabus recognises four equivalent ways of writing a vector:

• Boldface: \(\mathbf{v}\)
• Arrow over the symbol: \(\vec{v}\)
• Component (ordered‑pair) form: \(\langle vx,\;vy\rangle\)
• Ordered‑pair (exam‑style) form: \((vx,\;vy)\)

For a directed line segment \(\overline{AB}\) the vector is

\[

\vec{AB}= \langle xB-xA,\; yB-yA\rangle

= (xB-xA)\mathbf{i}+(yB-yA)\mathbf{j},

\]

where \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors along the positive \(x\)- and \(y\)-axes respectively.

1.1 Position Vectors

If the origin \(O\) is taken at \((0,0)\) and a point \(P\) has coordinates \((xP,\;yP)\), the position vector of \(P\) is

\[

\vec{OP}= \langle xP,\;yP\rangle = xP\mathbf{i}+yP\mathbf{j}.

\]

1.2 Unit Vectors

A unit vector has magnitude 1 and points in the same direction as the original vector:

\[

\hat{u}= \frac{\mathbf{v}}{|\mathbf{v}|}

= \frac{vx}{|\mathbf{v}|}\mathbf{i}+ \frac{vy}{|\mathbf{v}|}\mathbf{j}.

\]

2. Components, Magnitude and Direction (Syllabus 13.5)

3. Vector Operations (Syllabus 13.3, 13.6)

• Addition (Resultant) – algebraic: \(\mathbf{R}= \mathbf{A}+\mathbf{B}= (Ax+Bx)\mathbf{i}+(Ay+By)\mathbf{j}\).
  

  Graphical tip‑to‑tail method: draw \(\mathbf{A}\), then from its tip draw \(\mathbf{B}\); the vector from the start of \(\mathbf{A}\) to the tip of \(\mathbf{B}\) is \(\mathbf{R}\). (A small sketch is often required in the exam.)

• Subtraction (Relative velocity) – \(\mathbf{A}-\mathbf{B}= \mathbf{A}+(-\mathbf{B})\).
  

  The opposite vector \(-\mathbf{B}\) has components \((-Bx,\,-By)\). This reinforces that subtraction is simply addition of the opposite.

• Scalar multiplication – for a real number \(k\), \(k\mathbf{A}= (kAx)\mathbf{i}+(kAy)\mathbf{j}\).
• Unit vector – \(\hat{u}= \dfrac{\mathbf{A}}{|\mathbf{A}|}\).

4. Resolving and Composing Velocity Vectors (Syllabus 13.4)

Typical situations: a boat in a river, an aircraft in wind, a car on a sloping road.

1. Identify the speed \(V\) and the direction \(\theta\) (measure \(\theta\) from the positive \(x\)-axis, anticlockwise unless the question states otherwise).
2. Resolve each motion into components using \(vx = V\cos\theta\) and \(vy = V\sin\theta\). Pay attention to sign (east/right = + \(x\), north/up = + \(y\)).
3. Combine the horizontal components together and the vertical components together (algebraic addition).
4. Re‑calculate the resultant speed and direction from the combined components using the formulas in Section 2.

Checklist for Velocity‑Resolution Problems

• Draw a clear diagram; label axes (positive \(x\) = east, positive \(y\) = north).
• Write each velocity in component form, showing the vector symbol (e.g. \(\mathbf{v}_\text{boat}=0\mathbf{i}+10\mathbf{j}\)).
• Add/subtract components component‑wise.
• Find magnitude and direction; state the answer with the correct unit and bearing (e.g. “\(21.8^{\circ}\) east of north”).

5. Relative Velocity (Syllabus 13.8)

If object A moves with velocity \(\mathbf{v}A\) and object B with \(\mathbf{v}B\), the velocity of A relative to B is

\[

\mathbf{v}{A/B}= \mathbf{v}A-\mathbf{v}_B.

\]

Because subtraction is addition of the opposite, you may also write \(\mathbf{v}{A/B}= \mathbf{v}A+(-\mathbf{v}_B)\).

6. Momentum Conservation in Two Dimensions (Syllabus 13.9)

For a perfectly elastic (or perfectly inelastic) collision the total linear momentum is conserved separately in the \(x\)- and \(y\)-directions:

\[

m1\mathbf{v}{1i}+m2\mathbf{v}{2i}=m1\mathbf{v}{1f}+m2\mathbf{v}{2f}.

\]

• Mass is a scalar – it is carried unchanged through each component equation.
• Elastic collisions: kinetic energy is also conserved (not required for the basic IGCSE momentum question, but useful for extension).
• Inelastic collisions: the two bodies stick together; use the same component equations but with a single final velocity.

Procedure

1. Resolve all velocities into components.
2. Write the conservation equation for the \(x\)-components and for the \(y\)-components.
3. Solve the simultaneous equations for the unknown components.
4. Re‑combine the found components to give the final speed and direction of each object.

7. Worked Examples

Example 1 – Boat Crossing a River (Relative Velocity)

A boat can travel at \(10\;\text{km h}^{-1}\) in still water. It aims directly north across a river that flows east at \(4\;\text{km h}^{-1}\). Find the resultant speed and direction relative to the ground.

1. \(\mathbf{v}_\text{boat}=0\mathbf{i}+10\mathbf{j}\;\text{km h}^{-1}\)
2. \(\mathbf{v}_\text{river}=4\mathbf{i}+0\mathbf{j}\;\text{km h}^{-1}\)
3. \(\mathbf{v}_\text{ground}= (4+0)\mathbf{i}+(0+10)\mathbf{j}=4\mathbf{i}+10\mathbf{j}\)
4. Speed: \(|\mathbf{v}_\text{ground}|=\sqrt{4^{2}+10^{2}}=\sqrt{116}\approx10.8\;\text{km h}^{-1}\)
5. Direction (east of north): \(\theta=\tan^{-1}\!\left(\dfrac{4}{10}\right)\approx21.8^{\circ}\) east of north.

Example 2 – Projectile Motion Using Vectors

A ball is kicked with speed \(20\;\text{m s}^{-1}\) at an angle of \(30^{\circ}\) above the horizontal. Resolve the initial velocity into components, then find the time to reach the highest point.

1. \(vx = 20\cos30^{\circ}=17.3\;\text{m s}^{-1},\quad vy = 20\sin30^{\circ}=10.0\;\text{m s}^{-1}\)
2. At the highest point \(vy=0\). Using \(vy = u_y - gt\) with \(g=9.8\;\text{m s}^{-2}\):
   

   \(0 = 10.0 - 9.8t \;\Rightarrow\; t = 1.02\;\text{s}\).

3. Once the components are known, the usual kinematic formulas give range, maximum height, etc. (linking vectors to the separate “projectile motion” part of the syllabus).

Example 3 – Elastic Collision of Two Billiard Balls

Two identical balls (mass \(m\)) collide. Before impact ball A moves east at \(6\;\text{m s}^{-1}\); ball B is at rest. After the collision ball A moves north at \(4\;\text{m s}^{-1}\). Find the speed and direction of ball B.

1. Initial momentum components:
   

   \(x\): \(m(6)+m(0)=6m\)  \(y\): \(0+0=0\).

2. Let ball B’s final velocity be \(\mathbf{v}B=v{Bx}\mathbf{i}+v_{By}\mathbf{j}\). Conservation gives:
   

   \(x\): \(6m = m(0)+m v{Bx}\) ⟹ \(v{Bx}=6\;\text{m s}^{-1}\).
   

   \(y\): \(0 = m(4)+m v{By}\) ⟹ \(v{By}=-4\;\text{m s}^{-1}\) (south).

3. Speed of B: \(|\mathbf{v}_B|=\sqrt{6^{2}+(-4)^{2}}= \sqrt{52}\approx7.2\;\text{m s}^{-1}\).
4. Direction: \(\theta = \tan^{-1}\!\left(\dfrac{|v{By}|}{v{Bx}}\right)=\tan^{-1}\!\left(\dfrac{4}{6}\right)\approx33.7^{\circ}\) south of east.

Example 4 – Graphical Addition (Tip‑to‑Tail)

Two forces act on a point: \(\mathbf{F}1=5\mathbf{i}+2\mathbf{j}\) N and \(\mathbf{F}2=-3\mathbf{i}+4\mathbf{j}\) N. Sketch the tip‑to‑tail construction and state the resultant.

• Algebraically, \(\mathbf{R}= (5-3)\mathbf{i}+(2+4)\mathbf{j}=2\mathbf{i}+6\mathbf{j}\) N.
• Graphically, draw \(\mathbf{F}1\) from the origin, then from its tip draw \(\mathbf{F}2\); the vector from the origin to the final tip is \(\mathbf{R}\). (Examiners often award marks for a neat sketch.)

8. Summary of Key Formulas

9. Practice Questions

1. A car travels \(30\;\text{m s}^{-1}\) due east while a wind blows \(10\;\text{m s}^{-1}\) from the north. Find the car’s resultant speed and direction.
2. Particle 1 has velocity \(\mathbf{v}1=5\mathbf{i}+12\mathbf{j}\;\text{m s}^{-1}\). Particle 2 moves with \(\mathbf{v}2=-3\mathbf{i}+4\mathbf{j}\;\text{m s}^{-1}\). Determine the velocity of particle 1 relative to particle 2.
3. A projectile is fired with speed \(25\;\text{m s}^{-1}\) at \(40^{\circ}\) above the horizontal. Resolve the initial velocity into components and calculate the horizontal range (ignore air resistance, \(g=9.8\;\text{m s}^{-2}\)).
4. Two identical spheres collide elastically. Before impact sphere A moves at \(8\;\text{m s}^{-1}\) north‑east (\(45^{\circ}\) to the north). After the collision sphere A moves due south at \(3\;\text{m s}^{-1}\). Find the speed and direction of sphere B after the collision.
5. A plane flies north at \(200\;\text{km h}^{-1}\) while a wind blows from the west at \(50\;\text{km h}^{-1}\). Determine the ground speed and the bearing of the flight path.

10. Tips for Exam Success

• Diagram first: draw a clear picture, label axes (east = + \(x\), north = + \(y\)) and all vectors with their symbols.
• Check quadrants when using \(\tan^{-1}\); if the \(x\)-component is negative add \(180^{\circ}\), if only the \(y\)-component is negative subtract from \(360^{\circ}\) (or state “south of east”, etc.).
• Sign convention: keep east/right and north/up positive; west/left and south/down are negative.
• Work component‑wise for collisions and relative‑velocity problems – never add whole vectors directly.
• Units: keep them consistent (all m s\(^{-1}\) or all km h\(^{-1}\)). Write the final answer with the correct unit.
• Check your answer by recombining components; the magnitude you obtain should match the speed you calculated.
• Graphical addition may be required – a neat tip‑to‑tail sketch can earn marks even if the algebraic result is already shown.
• Remember the syllabus language: use terms such as “resultant”, “relative velocity”, “conservation of momentum in the x‑ and y‑directions”.