Motion in a Plane | ISC Class 11 Physics Notes

This note covers scalars and vectors, position and displacement, vector addition and subtraction, rectangular components, scalar and vector products, velocity and acceleration in a plane, projectile trajectories, time of flight, maximum height, horizontal range and uniform circular motion.

What distinguishes scalars, vectors and displacement?

How are quantities represented?

A scalar is specified by its magnitude, meaning its numerical size, together with its unit. Distance is path length, time measures duration, and speed is distance travelled per unit time; these are scalars. A vector has magnitude and direction and obeys the triangle law of addition, described below. Displacement (change in position), velocity (rate of change of position), acceleration (rate of change of velocity) and force (an interaction that can change motion) are vectors.

Bold letters such as A denote vectors; the corresponding plain letter A denotes magnitude, also written |A|. A drawn arrow represents a vector: its length represents magnitude on a chosen scale, and its arrowhead gives direction. Handwritten vectors may instead carry an arrow above the letter.

A unit is a reference standard used in measurement. SI means the International System of Units. The SI unit of distance is the metre, symbol m. The SI unit of time is the second, symbol s. Quantities must have compatible units before they are added.

How do position and displacement differ?

Choose an origin, a fixed reference point O. The position vector r points from O to the particle's position P. If the particle reaches P′, its new position vector is r′. A prime distinguishes the later value; the symbol Δ means change.

The displacement Δr is the directed straight line from the initial position to the final position: Δr = r′ − r. Its magnitude does not exceed the distance travelled, which is the actual path length. The SI unit of displacement is the metre.

Different paths between the same endpoints give the same displacement. On returning to the starting point, displacement is a null vector, a vector with zero magnitude whose direction cannot be specified. Distance travelled during that journey need not be zero.

Two vectors are equal when both their magnitudes and directions agree. For free vectors, translating an arrow parallel to itself leaves it unchanged. A localised vector also requires attention to its location or line of application. Vectors lying in the same plane are called coplanar vectors.

How are vectors added, subtracted and multiplied by numbers?

What do the triangle and parallelogram laws show?

The resultant is the vector sum. For vectors A and B, the triangle law places the tail of B at the head of A, without changing either direction. The arrow from the first tail to the final head represents their resultant R.

The parallelogram law places the two tails together. Complete the parallelogram using lines parallel to the given vectors. Its diagonal directed from the common tail to the opposite corner is R = A + B. Both constructions give the same vector.

Separate arrows A and B are followed by triangle constructions for the two orders of addition. Blue resultant arrows connect the first tail to the last head. A further construction illustrates grouping three vectors.

See Fig. 3.4 in your NCERT textbook

Vector addition is commutative, meaning that order does not change the sum: A + B = B + A. It is also associative: (A + B) + C = A + (B + C), where C is a third vector.

How do reversal and scaling work?

The vector −B has the magnitude of B and the opposite direction. Subtraction therefore means addition after reversal: A − B = A + (−B). In a parallelogram, the other diagonal represents a difference; its arrow direction determines which vector is subtracted.

If A + B = C, then C − A = B. The triangle construction can therefore express subtraction as well as addition. Equal and opposite vectors add to the null vector, written 0.

Multiplying A by a real number λ, pronounced lambda, gives magnitude |λ|A. A positive λ preserves direction; a negative λ reverses it. A zero multiplier gives 0. Multiplication by a physical scalar also changes units, as when constant velocity multiplied by time gives displacement.

How do rectangular components give a resultant's magnitude and direction?

What are unit vectors and components?

A unit vector has magnitude one and specifies direction; it has no physical unit or dimension. A physical dimension describes how a quantity depends on base quantities such as length and time. The symbols î, ĵ and k̂ denote mutually perpendicular unit vectors along the positive x, y and z coordinate axes. Mutually perpendicular directions are also called orthogonal directions.

Resolution expresses a vector as the sum of component vectors along chosen directions. Write A = Aₓî + Aᵧĵ in a plane, where Aₓ and Aᵧ are signed scalar components. The expressions Aₓî and Aᵧĵ are component vectors. In space, add A₃k̂, where A₃ is the z-component; the index 3 denotes the third coordinate axis.

Let θ, pronounced theta, be the angle from the positive x-axis to A. The functions sin, cos and tan mean sine, cosine and tangent. Rectangular components satisfy Aₓ = A cos θ and Aᵧ = A sin θ. They can be positive, negative or zero.

The magnitude is A = √(Aₓ² + Aᵧ²), where √ denotes the positive square root. Direction follows from tan θ = Aᵧ/Aₓ when Aₓ is non-zero. Use the signs of both components to identify the correct quadrant, one of the four regions separated by the axes.

Part (a) shows unit vectors on three axes. Parts (b) and (c) show a blue vector A in the x-y plane, perpendicular projections, and component arrows along the axes. Part (c) labels the angle θ at the origin.

See Fig. 3.9 in your NCERT textbook

Components along the same axis add algebraically: Rₓ = Aₓ + Bₓ and Rᵧ = Aᵧ + Bᵧ. Here Bₓ, Bᵧ and Rₓ, Rᵧ are the components of B and R. Subtraction uses corresponding differences. The same method applies to the z-components in three dimensions.

Derivation: Parallelogram magnitude and direction

Let A and B be the magnitudes of two vectors with angle θ between them. Choose the positive x-axis along A. Let α, pronounced alpha, be the angle the resultant makes with A.

1. Resolve the second vector: its components are B cos θ along x and B sin θ along y.
2. Add corresponding components to obtain Rₓ = A + B cos θ and Rᵧ = B sin θ.
3. Square and add: R² = (A + B cos θ)² + (B sin θ)². Use sin²θ + cos²θ = 1.
4. Obtain direction from the component ratio: tan α = B sin θ/(A + B cos θ), with the quadrant fixed by component signs.

R = √(A² + B² + 2AB cos θ)

The symbol ° denotes degrees, with 360° in one complete turn.

Worked example 1. Rain has a downward velocity component of 35 m/s and a westward component of 12 m/s. Find its resultant speed and the umbrella's tilt for a stationary boy.

Formula: R = √(vᵣ² + vₕ²); tan α = vₕ/vᵣ, where vᵣ and vₕ are the downward and westward speeds and α is the angle from the downward vertical. Substitute: R = √(35² + 12²); tan α = 12/35.

Answer: R = 37 m/s. The rain travels about 19° west of the downward vertical; hold the umbrella tilted about 19° towards the east from the upward vertical.

Worked example 2. A motorboat moves north at 25 km/h relative to water; km/h means kilometres per hour. The current is 10 km/h, directed 60° east of south. Find the boat's resultant velocity.

Formula: Rₓ = B sin 60°; Rᵧ = A − B cos 60°; R = √(Rₓ² + Rᵧ²). Here x points east, y points north, A is the boat's speed relative to water and B is the current speed. Substitute: Rₓ = 10 sin 60°; Rᵧ = 25 − 10 cos 60°.

Answer: The resultant speed is approximately 22 km/h, directed about 23.4° east of north. Use the unrounded component values when calculating the direction.

What is the scalar product of two vectors?

How does a dot product use a projection?

The scalar product, also called the dot product, combines two vectors to give a scalar. For vectors A and B with magnitudes A and B and angle θ between them, A · B = AB cos θ. The centred dot denotes this product.

The projection B cos θ is the signed component of B along A. Thus, the dot product multiplies the magnitude of one vector by the component of the other along it. The result has no direction, although its value can be positive, negative or zero.

The product is commutative: A · B = B · A. It is also distributive: multiplying a sum gives the sum of the separate products. Thus A · (B + C) = A · B + A · C.

For orthogonal unit vectors, î · î = ĵ · ĵ = k̂ · k̂ = 1, while î · ĵ = ĵ · k̂ = k̂ · î = 0. Consequently, A · B = AₓBₓ + AᵧBᵧ + A₃B₃, where B₃ is the z-component of B. Also, A · A = A².

How is work a scalar product?

A force is an interaction that can change motion. Denote a constant force by F and the displacement over which it acts by s, with magnitudes F and s. Work, W, is their scalar product: W = Fs cos θ, where θ is now the angle between force and displacement.

The SI unit of force is the newton, symbol N. The SI unit of work is the joule, symbol J, with 1 J = 1 N m. Perpendicular force and displacement give zero work; opposite directions give negative work. The angle and the constant-force condition are essential to this expression.

What is the vector product and how is its direction found?

What does the right-hand rule specify?

The vector product, or cross product, gives a vector perpendicular to the plane of two non-parallel vectors. It is written A × B = AB sin θ n̂. Here n̂ is a unit vector normal, meaning perpendicular, to their plane, with direction set by the right-hand rule.

For the right-hand rule, curl the fingers of the right hand from the first vector towards the second through the smaller angle. The extended thumb indicates the product's direction. The order of the vectors matters because reversing that order reverses the direction.

Accordingly, B × A = −(A × B). The cross product is not commutative. Its magnitude is AB sin θ: it vanishes for parallel vectors pointing in the same direction or antiparallel vectors pointing in opposite directions and equals AB for perpendicular vectors. A zero product has no specified direction.

These rules use a right-handed coordinate system, in which curling from positive x towards positive y points the thumb towards positive z. Cross products distribute over addition. Expand products of component expressions, apply the unit-vector rules, and collect terms along each axis.

How does torque illustrate the cross product?

Torque, or moment of force, describes the turning effect of a force about a chosen origin. Its vector symbol is τ, pronounced tau; the plain symbol τ denotes its magnitude. If r runs from the origin to the point of application of F, then τ = r × F.

Its magnitude is τ = rF sin θ, where r is the position-vector magnitude and θ is the angle between r and F. The SI unit of torque is the newton metre, N m. Torque is a vector; work is a scalar, despite their matching physical dimensions.

How are velocity and acceleration described in a plane?

How are average and instantaneous quantities distinguished?

For position coordinates x and y, r = xî + yĵ. Over the time interval Δt, changes Δx and Δy give Δr = Δxî + Δyĵ. Average velocity, vₐᵥ, is displacement divided by that time interval: vₐᵥ = Δr/Δt.

Average speed is total distance divided by elapsed time. It is greater than or equal to the magnitude of average velocity over the same interval. Equality requires the path length to equal displacement magnitude. The SI unit of speed is the metre per second, m/s.

Instantaneous velocity, v, is the limiting average velocity as Δt approaches zero. It points along the tangent to the path in the direction of motion. A tangent gives the local direction of the curve at that point. The SI unit of velocity is m/s.

Write v = dr/dt = vₓî + vᵧĵ, where d/dt denotes differentiation, the instantaneous rate of change with time t. The components are vₓ = dx/dt and vᵧ = dy/dt. v = √(vₓ² + vᵧ²) gives the instantaneous speed v.

Acceleration measures change in velocity per unit time. Average acceleration is Δv/Δt, where Δv is the final velocity minus the initial velocity. Instantaneous acceleration is a = dv/dt = aₓî + aᵧĵ, with aₓ = dvₓ/dt and aᵧ = dvᵧ/dt.

The SI unit of acceleration is the metre per second squared, m/s². In a plane, velocity and acceleration need not be parallel. Acceleration can change speed, direction, or both. A calculation must therefore use the change in the complete velocity vector.

How does differentiation give the motion?

Worked example 3. A particle has position r = (3.0 m/s)tî + (2.0 m/s²)t²ĵ + (5.0 m)k̂. Find its velocity and acceleration, and its speed and direction at t = 1.0 s.

Formula: v = dr/dt; a = dv/dt. Substitute: differentiating each component gives v = (3.0 m/s)î + (4.0 m/s²)tĵ and a = (4.0 m/s²)ĵ.

Answer: At 1.0 s the speed is √(3.0² + 4.0²) = 5.0 m/s, about 53° from positive x towards positive y. The constant z-coordinate contributes no velocity.

How is motion with constant acceleration solved using components?

Which equations can be used?

Constant acceleration means that the acceleration vector remains constant in both magnitude and direction. Let r₀ and v₀ be initial position and initial velocity at t = 0. The subscript zero denotes an initial value. Then v = v₀ + at.

Position is r = r₀ + v₀t + ½at². The symbol ½ means one half. These vector equations resolve into separate equations along perpendicular axes, but each component describes the same particle over the same elapsed time.

Here x₀ and y₀ are initial coordinates, while v₀ₓ and v₀ᵧ are initial velocity components. The acceleration components aₓ and aᵧ are constant. Choose positive directions before substituting, since the signs describe physical directions rather than the sizes of the quantities alone.

How does a shared time connect the components?

1. Resolve the given position, velocity and acceleration along fixed perpendicular axes.
2. Use the condition on one coordinate to determine the elapsed time.
3. Substitute that time into the other coordinate equation.
4. Find both velocity components at that time and combine them to obtain speed.

Worked example 4. A particle starts at the origin with velocity (5.0 m/s)î and constant acceleration (3.0î + 2.0ĵ) m/s². Find its y-coordinate and speed when x = 84 m.

Formula: x = v₀ₓt + ½aₓt²; y = v₀ᵧt + ½aᵧt²; vₓ = v₀ₓ + aₓt; vᵧ = v₀ᵧ + aᵧt. Substitute: 84 = 5.0t + 1.5t², with t measured in seconds, giving t = 6 s.

Answer: y = 1.0 × 6² = 36.0 m. The velocity components are 23.0 m/s and 12.0 m/s, so the speed is √(23² + 12²), approximately 26 m/s.

Note: Constant acceleration magnitude alone is insufficient for these equations. In uniform circular motion the acceleration direction changes continuously, so its acceleration vector is not constant.

How are a projectile's trajectory and instantaneous velocity obtained?

What assumptions define the motion?

A projectile is an object in flight after being thrown or projected. Its trajectory is the path it follows. Assume that air resistance has negligible effect and that gravitational acceleration is constant and vertically downward. A thrown cricket ball is an example of projectile motion under these assumptions.

Put the origin at launch, choose x horizontal and y vertically upward, and let initial speed be u at angle θ₀ above the horizontal. Here θ₀ is the projection angle. Let g be the positive magnitude of gravitational acceleration, measured in m/s²; then aₓ = 0 and aᵧ = −g.

The launch components are uₓ = u cos θ₀ and uᵧ = u sin θ₀, where uₓ and uᵧ mean horizontal and vertical initial velocity. At elapsed time t, x = u cos θ₀ t and y = u sin θ₀ t − ½gt².

Velocity components are vₓ = u cos θ₀ and vᵧ = u sin θ₀ − gt. The horizontal component remains constant; the vertical component decreases during ascent, reaches zero at the highest point, and becomes negative during descent.

Derivation: The equation of the trajectory

1. For a non-zero horizontal launch component, rearrange x = u cos θ₀ t to obtain t = x/(u cos θ₀).
2. Substitute this expression for time into y = u sin θ₀ t − ½gt².
3. Simplify the first term using sin θ₀/cos θ₀ = tan θ₀, retaining the square of the entire denominator in the second term.

y = x tan θ₀ − gx²/(2u² cos²θ₀)

The equation is quadratic in x and represents a parabola, the curved path described by this quadratic relation. A vertical projection instead follows a straight line. The trajectory therefore depends on initial conditions as well as acceleration.

A blue parabolic path carries velocity arrows on ascent and descent. Horizontal component arrows point right; vertical components point upward on ascent and downward on descent. At the highest point the vertical component is labelled zero.

See Fig. 3.17 in your NCERT textbook

Instantaneous speed is √[(u cos θ₀)² + (u sin θ₀ − gt)²]. Its direction angle θ relative to the horizontal satisfies tan θ = vᵧ/vₓ. At the top, the velocity is horizontal for an oblique launch, while the acceleration remains downward with magnitude g.

How are time of flight, maximum height and range calculated?

Derivation: The principal projectile results

Let tₘ be the time to the highest point, H the maximum height above launch, T the time of flight, and R the horizontal range. Here T and R refer to a projectile returning to its launch level. Assume negligible air resistance and constant g.

1. At the highest point, vᵧ = 0, so 0 = u sin θ₀ − gtₘ and tₘ = u sin θ₀/g.
2. Substitute tₘ into the vertical position equation to obtain H = u² sin²θ₀/(2g).
3. Set y = 0 for the return to launch level. Besides the launch instant t = 0, the solution is T = 2u sin θ₀/g.
4. Multiply the constant horizontal velocity by T: R = u cos θ₀ T = u² sin 2θ₀/g, using 2 sin θ₀ cos θ₀ = sin 2θ₀.

T = 2u sin θ₀/g

Thus T = 2tₘ for a return to launch level. At fixed u and g, R is greatest at θ₀ = 45°, giving maximum range u²/g. Complementary angles, whose sum is 90°, give equal ranges under these same conditions, but need not give equal heights or flight times.

How do the formulae apply to a thrown ball?

Worked example 5. A cricket ball is projected at 28 m/s, at 30° above horizontal, and returns to launch level. Neglect air resistance and take g = 9.8 m/s². Find maximum height, flight time and range.

Formula: H = u² sin²θ₀/(2g); T = 2u sin θ₀/g; R = u² sin 2θ₀/g. Substitute: u = 28 m/s and θ₀ = 30°, so the initial vertical component is 14 m/s.

Answer: H = 14²/(2 × 9.8) = 10.0 m; T = 28/9.8 ≈ 2.9 s; R = 28² sin 60°/9.8 ≈ 69 m.

What changes for a horizontal launch from a height?

A horizontally projected object has zero initial vertical velocity. If it lands below launch level, find flight time from its actual vertical displacement. The same-level time-of-flight formula would describe a different condition and cannot give the fall time from the elevated starting point.

Worked example 6. A hiker throws a stone horizontally at 15 m/s from a cliff 490 m above the ground. Neglect air resistance and take g = 9.8 m/s². Find the fall time and impact speed.

Formula: y = −½gt²; vₓ = u; vᵧ = −gt; v = √(vₓ² + vᵧ²). Here u is horizontal launch speed and y is displacement measured upward from the cliff edge. Substitute: −490 = −½ × 9.8 × t².

Answer: t = 10 s. The impact components are 15 m/s horizontally and −98 m/s vertically, so impact speed is √(15² + 98²) ≈ 99 m/s.

Why does uniform circular motion involve acceleration?

Which quantities remain constant?

Uniform circular motion is motion around a circle at constant speed. Velocity changes because its direction follows the tangent at successive points. The resulting centripetal acceleration points towards the circle's centre. Write its magnitude as aₙ and the circle's radius as r.

For speed v, aₙ = v²/r. This acceleration is perpendicular to velocity. Its magnitude remains constant when speed and radius are constant, but its direction changes. The complete acceleration vector is therefore not constant. For circular motion with changing speed, total acceleration need not point directly towards the centre.

Successive panels show circular arcs, radius vectors and tangential velocity arrows. Small velocity triangles show the velocity difference. The final panel has an acceleration arrow pointing from the particle towards the centre C.

See Fig. 3.18 in your NCERT textbook

How are angular speed, period and frequency connected?

Angular displacement is the angle swept by the radius. Measure it in radians, symbol rad, where the angle equals arc length divided by radius. Angular speed ω, pronounced omega, is angular distance per unit time. For uniform motion, ω = Δθ/Δt.

The SI unit of angular speed is rad/s. Since an arc length equals radius multiplied by its angle in radians, v = rω and aₙ = ω²r. The factor π, pronounced pi, is the ratio of a circle's circumference to its diameter.

The period, here denoted Tₚ, is the time for one complete revolution. Frequency f is the number of revolutions per second: f = 1/Tₚ. The SI unit of frequency is the hertz, symbol Hz, meaning one cycle per second.

One revolution covers circumference 2πr, so v = 2πr/Tₚ and ω = 2πf. These expressions connect the directly counted revolutions with linear speed. Distinguish circular period Tₚ from the projectile flight time T defined earlier.

Worked example 7. An insect moves steadily around a circular groove of radius 12 cm (centimetres), completing seven revolutions in 100 s. Find angular speed, linear speed and acceleration magnitude. Is acceleration a constant vector?

Formula: f = 7/100; ω = 2πf; v = ωr; aₙ = ω²r. Substitute: r = 12 cm = 0.12 m. The symbol cm means centimetre, one hundredth of a metre.

Answer: ω ≈ 0.44 rad/s; v ≈ 5.3 cm/s, or 0.053 m/s; aₙ ≈ 2.3 cm/s², or 0.023 m/s². Acceleration points inward and changes direction, so it is not a constant vector.

Glossary

• Scalar — A physical quantity specified by magnitude and unit without an associated direction.
• Vector — A quantity possessing magnitude and direction and obeying the triangle law of addition.
• Displacement — The directed straight line from an initial position to a final position, independent of path.
• Null vector — A vector of zero magnitude for which a direction cannot be specified.
• Unit vector — A dimensionless vector of magnitude one used to specify a particular direction.
• Resultant — The single vector obtained by adding the vectors under consideration.
• Resolution — Expressing a vector as the sum of component vectors along selected directions.
• Scalar product — The scalar obtained by multiplying two vector magnitudes and the cosine of their included angle.
• Vector product — A vector with magnitude equal to the two magnitudes times the sine of their included angle.
• Projectile — An object that is in flight after being thrown or projected.
• Trajectory — The path traced by a moving object as its position changes with time.
• Horizontal range — The horizontal distance from launch to the point where a projectile returns to launch level.
• Centripetal acceleration — The acceleration directed towards the centre during motion around a circle at constant speed.
• Period — The time taken to complete one revolution during circular motion.

Common errors and misconceptions

• Misconception: Equal vector magnitudes imply equal vectors. Correct: Directions must also agree; equal lengths alone do not establish vector equality.
• Misconception: Vector subtraction means subtracting magnitudes. Correct: Reverse the vector being subtracted and add it using a vector construction or components.
• Misconception: A scalar component is itself a vector. Correct: Aₓ is a signed scalar; Aₓî is the corresponding component vector.
• Misconception: Reversing a cross product leaves it unchanged. Correct: Reversal changes its sign; the dot product is commutative.
• Misconception: A projectile stops at its highest point. Correct: Its vertical velocity vanishes there, but an obliquely projected object's horizontal velocity remains non-zero.
• Misconception: The same flight-time expression applies to any landing height. Correct: T = 2u sin θ₀/g assumes return to launch level; otherwise use the actual vertical displacement.
• Misconception: Constant speed means zero acceleration. Correct: Uniform circular motion changes velocity direction continuously, producing inward acceleration.
• Misconception: Constant acceleration magnitude permits the constant-vector kinematic equations. Correct: Acceleration direction must also remain constant for those equations to apply.

Exam-style questions with model answers

Q1. State the two conditions for equality of two non-zero free vectors. [2 marks]

1. The two vectors must have equal magnitudes, represented by arrows of equal length on the same scale.
2. They must have the same direction; parallel translation can then make both their tails and their heads coincide.

Q2. For vectors A and B of magnitudes A and B, separated by angle θ, derive the magnitude of their resultant and an expression for its direction relative to A. [4 marks]

1. Choose the x-axis along vector A. Vector B has components B cos θ along x and B sin θ along y.
2. The resultant components are Rₓ = A + B cos θ and Rᵧ = B sin θ, where Rₓ and Rᵧ denote the x- and y-components.
3. Its magnitude R satisfies R² = Rₓ² + Rᵧ², giving R = √(A² + B² + 2AB cos θ).
4. If α is the resultant's angle from A, tan α = B sin θ/(A + B cos θ). Use the component signs to choose the quadrant; direction is unspecified for a null resultant.

Q3. Define the dot and cross products of two non-zero vectors A and B with magnitudes A and B and included angle θ. Explain what reversing their order does. [3 marks]

1. The dot product is A · B = AB cos θ, a scalar. It equals one magnitude multiplied by the projection of the other vector along it.
2. The cross product is A × B = AB sin θ n̂, where n̂ is the normal unit vector given by the right-hand rule for non-parallel vectors.
3. The dot product is unchanged on reversal: B · A = A · B. The cross product reverses sign: B × A = −(A × B). For parallel vectors the cross product is zero.

Q4. A projectile is launched from the origin with speed u at an angle θ₀ between 0° and 90° above the horizontal. With negligible air resistance and constant downward gravitational acceleration of magnitude g, derive its trajectory, maximum height, flight time and range on returning to launch level. [6 marks]

1. Take x horizontal and y upward, with t the elapsed time. The initial components are u cos θ₀ and u sin θ₀; acceleration components are zero and −g.
2. The position equations are x = u cos θ₀ t and y = u sin θ₀ t − ½gt², describing simultaneous horizontal and vertical motion.
3. Eliminate t using t = x/(u cos θ₀). This gives y = x tan θ₀ − gx²/(2u² cos²θ₀), the parabolic trajectory.
4. The vertical velocity is u sin θ₀ − gt. Setting it to zero gives highest-point time tₘ = u sin θ₀/g and maximum height H = u² sin²θ₀/(2g).
5. Set y = 0 again and take the non-zero time solution. The time of flight is T = 2u sin θ₀/g, twice the ascent time.
6. The horizontal range is R = u cos θ₀ T = u² sin 2θ₀/g. This expression uses return to launch level and the constant horizontal velocity.

Q5. A ball is thrown at 28 m/s, 30° above the horizontal, and returns to launch level. Neglect air resistance and use g = 9.8 m/s². Calculate its maximum height, time of flight and horizontal range. [3 marks]

1. The initial vertical speed is 28 sin 30° = 14 m/s. Maximum height above launch is H = 14²/(2 × 9.8) = 10.0 m.
2. For return to the same level, the time of flight is T = 2 × 14/9.8 ≈ 2.9 s. Both ascent and descent are included.
3. The horizontal range is R = 28² sin 60°/9.8 ≈ 69 m. This is the horizontal displacement from release to the return point, not the curved path length.

Q6. A stone is thrown horizontally at 15 m/s from a cliff 490 m above the ground. Neglect air resistance and take g = 9.8 m/s². Find its fall time, both impact velocity components and impact speed. [4 marks]

1. Choose the launch point as origin and upward as positive y. Initial vertical velocity is zero. From −490 = −½ × 9.8 × t², fall time t = 10 s.
2. The horizontal impact component is vₓ = 15 m/s because horizontal acceleration is zero throughout the flight.
3. The vertical impact component is vᵧ = −9.8 × 10 = −98 m/s. The negative sign specifies downward motion.
4. Impact speed is the magnitude of the combined velocity: v = √(15² + 98²) ≈ 99 m/s.

Q7. An insect completes seven revolutions in 100 s at constant speed around a circular groove of radius 12 cm. Calculate its angular speed, linear speed and acceleration magnitude, and explain the acceleration's direction and constancy. [5 marks]

1. The frequency f, meaning revolutions per second, is 7/100 s⁻¹. Angular speed ω = 2πf ≈ 0.44 rad/s, where π is the circle constant.
2. The radius is r = 12 cm = 0.12 m. Linear speed is v = ωr ≈ 0.053 m/s, equivalent to 5.3 cm/s.
3. The acceleration magnitude is aₙ = ω²r ≈ 0.023 m/s², equivalent to 2.3 cm/s². It remains constant because both radius and speed are constant.
4. Acceleration points towards the centre at each position. It is perpendicular to the tangential velocity and changes the direction of that velocity.
5. The acceleration vector is not constant because its direction changes as the insect moves around the circle. Constant magnitude does not imply a constant vector.

Key takeaways

• Vectors require magnitude, direction and the vector addition rule; equal magnitudes alone do not establish equality.
• Resolve vectors along fixed perpendicular axes, combine matching components, and use their signs to determine direction.
• A dot product gives a scalar using cosine; a cross product gives a vector using sine and the right-hand rule.
• Planar motion can be analysed through simultaneous perpendicular components that share the same elapsed time.
• With negligible air resistance, a projectile has constant horizontal velocity and constant downward gravitational acceleration.
• Projectile flight-time and range expressions for return to launch level require that landing-height condition explicitly.
• At a projectile's highest point, vertical velocity vanishes while its downward acceleration remains unchanged.
• Uniform circular motion has constant speed but changing velocity, with acceleration directed towards the circle's centre.

Test yourself

Can a particle travel a non-zero distance with zero displacement?

Yes. A return to its starting point gives zero displacement despite a non-zero distance travelled along the path.

What distinguishes Aₓ from Aₓî?

Aₓ is a signed scalar component; multiplying it by the unit vector î gives a component vector.

What does a negative real multiplier do to a non-zero vector?

It reverses the direction and multiplies the magnitude by the absolute value of that multiplier.

What are î · ĵ and î × ĵ in a right-handed coordinate system?

The dot product is zero because the unit vectors are perpendicular. Their cross product is k̂.

Why does an obliquely projected ball still move at its highest point?

Its vertical velocity is zero there, but its horizontal velocity remains non-zero when air resistance is negligible.

Under what conditions is 45° the angle for maximum projectile range?

The launch speed and gravitational acceleration must be fixed, air resistance negligible, and landing at the same level as launch.

Why are constant-acceleration vector equations unsuitable for uniform circular motion?

The inward acceleration changes direction continuously. Those equations require the complete acceleration vector to remain constant.

Is total acceleration towards the centre for every kind of circular motion?

No. It is directed towards the centre for constant speed; with changing speed, total acceleration need not be purely inward.