Class 11Physics · MechanicsFull chapter

Motion in a Plane

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

In this chapter
Scalars, Vectors and Vector RepresentationVector Addition and Subtraction — Triangle and Parallelogram LawsResolution of Vectors and Unit VectorsMotion in a Plane with Constant AccelerationRelative Velocity in Two DimensionsProjectile MotionUniform Circular Motion and Centripetal Acceleration

Scalars, Vectors and Vector Representation

Quick answer A scalar needs only a magnitude to be fully described, while a vector needs both a magnitude and a direction, and is drawn as an arrow to scale.

Scalar quantities are completely described by a magnitude (a number with a unit) alone. Examples include mass, distance, speed, time, temperature and energy. Scalars are combined using the ordinary rules of algebra.

Vector quantities require both a magnitude and a direction to be completely specified. Examples include displacement, velocity, acceleration, force and momentum. A vector is represented geometrically by a straight line with an arrowhead, drawn to scale: the length of the line gives the magnitude and the arrowhead gives the direction.

A vector A is written in bold italics, and its magnitude is written as |A| or simply A. Two vectors are equal only if they have the same magnitude and the same direction, irrespective of where their initial points are drawn. The negative of a vector, −A, has the same magnitude as A but points in exactly the opposite direction. Multiplying a vector by a positive scalar changes only its magnitude; multiplying by a negative scalar also reverses its direction; in general, multiplying a vector A by a real number n gives a new vector of magnitude |n||A|.

The location of a point in a plane is described by a position vector drawn from a fixed origin to that point. If a particle moves from a point with position vector r1 to a point with position vector r2, its displacement vector is Δr = r2 − r1. Displacement is a vector while the path length (distance) travelled is a scalar; the magnitude of displacement can never exceed the distance travelled.

Worked example. A particle moves from point A(2 m, 3 m) to point B(6 m, 7 m) in the x-y plane. Find its displacement vector and the magnitude of the displacement.

Given: r1 = 2i + 3j (m), r2 = 6i + 7j (m)

Formula: Δr = r2 − r1; |Δr| = √(Δx2 + Δy2)

Substitution: Δr = (6−2)i + (7−3)j = 4i + 4j (m); |Δr| = √(42 + 42) = √32

Result: Displacement = 4i + 4j m, of magnitude 4√2 ≈ 5.66 m, directed at 45° to the x-axis.

Displacement vector Δr = r₂ − r₁
Magnitude of a vector |A| = √(Ax² + Ay²)
Scalar multiplication |nA| = |n| |A| direction unchanged if n > 0, reversed if n < 0
Remember
  • Scalars have only magnitude; vectors have both magnitude and direction.
  • Equal vectors have the same magnitude and direction, regardless of position.
  • Multiplying a vector by −1 reverses its direction while keeping the same magnitude.
  • Displacement is Δr = r₂ − r₁; its magnitude is never greater than the distance travelled.
  • A position vector locates a point relative to a fixed origin.

Vector Addition and Subtraction — Triangle and Parallelogram Laws

Quick answer Vectors are added geometrically using the triangle law (head-to-tail) or the parallelogram law; subtraction is simply addition of the negative vector.

Vectors cannot be combined by ordinary algebraic addition because direction must also be accounted for; they are combined using geometric laws.

Triangle law of vector addition: if two vectors are represented in magnitude and direction by two sides of a triangle taken in the same order, their resultant is represented in magnitude and direction by the third side of the triangle, taken in the reverse order. To add P and Q, the tail of Q is joined to the head of P; the vector from the free tail of P to the free head of Q gives the resultant R = P + Q.

Parallelogram law of vector addition: if two vectors acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from that point, their resultant is represented completely (in magnitude and direction) by the diagonal of the parallelogram passing through that same point.

If P and Q have magnitudes P and Q with an angle θ between them, the magnitude of the resultant is R = √(P2 + Q2 + 2PQ cosθ), and the angle α it makes with P satisfies tanα = Q sinθ/(P + Q cosθ). Vector addition is commutative (P+Q = Q+P) and associative. The resultant is largest, P+Q, when θ = 0°, and smallest, |P−Q|, when θ = 180°.

Vector subtraction is defined as addition of the negative vector: P − Q = P + (−Q). Geometrically, −Q is drawn equal in magnitude to Q but opposite in direction, and then added to P by the triangle law.

Worked example. Two forces of magnitude 3 N and 4 N act on a body at an angle of 60° to each other. Find the magnitude of the resultant force and the angle it makes with the 3 N force.

Given: P = 3 N, Q = 4 N, θ = 60°

Formula: R = √(P2 + Q2 + 2PQ cosθ); tanα = Q sinθ/(P + Q cosθ)

Substitution: R = √(32 + 42 + 2×3×4×cos60°) = √(9 + 16 + 12) = √37; tanα = (4×sin60°)/(3+4cos60°) = (4×0.866)/(3+2) = 3.464/5 = 0.693

Result: R = √37 ≈ 6.08 N, and α = tan−1(0.693) ≈ 34.7° from the 3 N force.

Resultant magnitude R = √(P² + Q² + 2PQ cosθ)
Direction of resultant tanα = Q sinθ / (P + Q cosθ)
Vector subtraction P − Q = P + (−Q)
Extreme cases Rmax = P + Q at θ=0°; Rmin = |P − Q| at θ=180°
Remember
  • Triangle law: place vectors head-to-tail; the resultant runs from the free tail to the free head.
  • Parallelogram law: the resultant is the diagonal of the parallelogram formed by the two vectors drawn from the same point.
  • R = √(P² + Q² + 2PQcosθ) gives the resultant magnitude for two vectors inclined at angle θ.
  • Vector addition is commutative and associative; subtraction is addition of the negative vector.
  • Resultant is maximum (P+Q) when θ = 0° and minimum (|P−Q|) when θ = 180°.

Resolution of Vectors and Unit Vectors

Quick answer Any vector can be split into perpendicular components (Ax = Acosθ, Ay = Asinθ) and written using unit vectors as A = Axi + Ayj.

Any vector lying in a plane can be split into two mutually perpendicular components, usually taken along the x-axis and y-axis; this process is called resolution of a vector. If a vector A makes an angle θ with the x-axis, its components are Ax = A cosθ (horizontal) and Ay = A sinθ (vertical).

A unit vector has a magnitude of exactly 1 and simply indicates direction. The unit vectors along the x, y and z axes are denoted i, j and k. Any vector can be written using its components and these unit vectors as A = Axi + Ayj. The unit vector along any vector A is obtained by dividing the vector by its own magnitude: Â = A/|A|.

Once vectors are written in component form, addition and subtraction become simple algebra — components along each axis are combined separately. If A = Axi + Ayj and B = Bxi + Byj, then A + B = (Ax+Bx)i + (Ay+By)j. The magnitude and direction of the original vector can always be recovered: A = √(Ax2 + Ay2) and θ = tan−1(Ay/Ax).

Worked example. A vector of magnitude 10 units makes an angle of 30° with the x-axis. Find its rectangular components and the unit vector along its direction.

Given: A = 10 units, θ = 30°

Formula: Ax = A cosθ, Ay = A sinθ, Â = A/A

Substitution: Ax = 10 cos30° = 10×0.866 = 8.66; Ay = 10 sin30° = 10×0.5 = 5

Result: A = 8.66i + 5j units; unit vector  = 0.866i + 0.5j (its magnitude is exactly 1, as required).

Rectangular components Ax = A cosθ, Ay = A sinθ
Vector in component form A = Ax i + Ay j
Unit vector  = A / |A|
Magnitude from components A = √(Ax² + Ay²)
Direction from components θ = tan⁻¹(Ay / Ax)
Remember
  • Any vector resolves into perpendicular components: Ax = Acosθ, Ay = Asinθ.
  • A = Axi + Ayj expresses a vector completely using unit vectors i and j.
  • A unit vector has magnitude 1 and only specifies direction: Â = A/|A|.
  • Vector addition/subtraction becomes simple algebra once vectors are in component form.
  • Magnitude and direction are recovered as A = √(Ax²+Ay²) and θ = tan⁻¹(Ay/Ax).

Motion in a Plane with Constant Acceleration

Quick answer For constant acceleration in a plane, motion along the x and y axes is independent, and v = v₀ + at, r = r₀ + v₀t + ½at² apply as vector equations.

When a particle moves in a plane (the x-y plane) with constant acceleration a, its motion can be analysed by treating the x and y directions completely independently — the acceleration, velocity and displacement along one axis have no effect on the motion along the other axis.

If the particle has initial position r0 and initial velocity v0 at t = 0, and experiences constant acceleration a, its velocity and position at time t are given by the vector equations v = v0 + at and r = r0 + v0t + ½at2. These are direct vector generalisations of the straight-line equations of motion, and each splits into separate x- and y-component equations, for example vx = v0x + axt and vy = v0y + ayt.

Worked example. A particle starts at the origin with initial velocity v0 = 5i m/s and moves with constant acceleration a = 2j m/s2. Find its velocity and position after t = 3 s.

Given: r0 = 0, v0 = 5i m/s, a = 2j m/s2, t = 3 s

Formula: v = v0 + at; r = v0t + ½at2

Substitution: v = 5i + (2×3)j = 5i + 6j m/s; r = (5×3)i + (½×2×32)j = 15i + 9j m

Result: After 3 s, the velocity is 5i + 6j m/s (magnitude √(25+36) = √61 ≈ 7.81 m/s) and the position is 15i + 9j m.

Velocity-time relation v = v₀ + at
Position-time relation r = r₀ + v₀t + ½at²
x-component equations vx = v0x + axt; x = x0 + v0xt + ½axt²
y-component equations vy = v0y + ayt; y = y0 + v0yt + ½ayt²
Remember
  • In 2D motion with constant acceleration, motion along x and y axes is independent of each other.
  • v = v₀ + at and r = r₀ + v₀t + ½at² are the vector kinematic equations for constant acceleration.
  • Each vector equation splits into two independent scalar equations, one per axis.
  • These equations reduce to the familiar one-dimensional equations of motion along a single axis.

Relative Velocity in Two Dimensions

Quick answer The velocity of one object as seen by another moving object is the vector difference of their velocities, vBA = vB − vA, found by subtracting components.

When two objects A and B move with velocities vA and vB with respect to a common (ground) frame of reference, the velocity of B relative to A is defined as the vector difference vBA = vB − vA. It is the velocity that B appears to have to an observer moving along with A. Similarly, the velocity of A relative to B is vAB = vA − vB = −vBA.

Since this is an ordinary vector subtraction, relative velocity problems are solved by resolving each velocity into rectangular components, subtracting the components along each axis separately, and then recombining the result to find the magnitude and direction of the relative velocity. If A and B move with exactly equal velocity, their relative velocity is zero and each appears stationary to the other.

Worked example. Rain falls vertically downward with a speed of 35 m/s. A woman cycles due east with a speed of 12 m/s. At what angle to the vertical should she hold her umbrella to protect herself from the rain?

Given: Taking j as vertically upward and i as the direction the woman cycles: velocity of rain vr = −35j m/s; velocity of woman vw = 12i m/s

Formula: Velocity of rain relative to the woman, vrw = vr − vw; tanθ = (horizontal component)/(vertical component), θ measured from the vertical

Substitution: vrw = −35j − 12i = −12i − 35j m/s; |vrw| = √(122 + 352) = √1369 = 37 m/s; tanθ = 12/35 = 0.343

Result: The rain appears to fall on the woman at 37 m/s, tilted at θ = tan−1(0.343) ≈ 19° from the vertical, toward the direction she is cycling. She should hold her umbrella tilted about 19° from the vertical, toward the front.

Relative velocity of B w.r.t. A vBA = vB − vA
Reciprocal relation vAB = −vBA
Rain-umbrella angle tanθ = v_person / v_rain θ measured from the vertical, tilted toward the direction of motion
Remember
  • Relative velocity of B with respect to A is vBA = vB − vA, an ordinary vector subtraction.
  • vAB = −vBA; two objects moving with equal velocity have zero relative velocity between them.
  • Relative velocity problems are solved by resolving each velocity into components, subtracting component-wise, then recombining.
  • Classic application: to stay dry while moving through vertically falling rain, tilt the umbrella toward the direction of motion by θ = tan⁻¹(v_person / v_rain) from the vertical.

Projectile Motion

Quick answer A projectile follows a parabolic path with constant horizontal velocity and uniformly accelerated vertical motion; its time of flight, height and range follow from u, θ and g.

A projectile is any object given an initial velocity and then left to move freely under gravity alone (air resistance is neglected). If a projectile is launched with speed u at an angle θ above the horizontal, its horizontal velocity component u cosθ stays constant throughout the flight, while its vertical component u sinθ decreases due to gravity, becomes zero at the highest point, and then increases in magnitude downward.

Taking the point of projection as the origin, with x horizontal and y vertically upward, the position at time t is x = (u cosθ)t and y = (u sinθ)t − ½gt2. Eliminating t between these equations gives the trajectory equation, y = x tanθ − gx2/(2u2cos2θ), which has the form y = ax + bx2 — the path of a projectile is therefore a parabola.

The time of flight T is the total time the projectile remains in the air; since the vertical displacement is zero at landing, T = 2u sinθ/g. The maximum height H is reached when the vertical velocity component becomes zero: H = u2sin2θ/(2g). The horizontal range R is the horizontal distance covered during the time of flight: R = u2sin2θ/g. The range is maximum when θ = 45°, giving Rmax = u2/g; complementary angles θ and (90°−θ) give the same range for a given speed.

Worked example. A ball is thrown with a speed of 20 m/s at an angle of 30° above the horizontal. Take g = 9.8 m/s2. Find the time of flight, maximum height and horizontal range.

Given: u = 20 m/s, θ = 30°, g = 9.8 m/s2

Formula: T = 2u sinθ/g; H = u2sin2θ/(2g); R = u2sin2θ/g

Substitution: T = (2×20×0.5)/9.8 = 20/9.8; H = (400×0.25)/(2×9.8) = 100/19.6; R = (400×sin60°)/9.8 = (400×0.866)/9.8

Result: T ≈ 2.04 s, H ≈ 5.10 m, and R ≈ 35.35 m.

Trajectory equation y = x tanθ − gx² / (2u² cos²θ)
Time of flight T = 2u sinθ / g
Maximum height H = u² sin²θ / (2g)
Horizontal range R = u² sin2θ / g
Maximum range Rmax = u² / g occurs at θ = 45°
Remember
  • A projectile has uniform horizontal velocity and uniformly accelerated (due to g) vertical motion, independent of each other.
  • The trajectory is a parabola: y = x tanθ − gx²/(2u²cos²θ).
  • Time of flight T = 2u sinθ/g, maximum height H = u²sin²θ/2g, horizontal range R = u²sin2θ/g.
  • Range is maximum at θ = 45°; complementary angles (θ, 90°−θ) give equal ranges for the same speed.

Uniform Circular Motion and Centripetal Acceleration

Quick answer In uniform circular motion, speed stays constant but velocity direction keeps changing, producing a centripetal acceleration ac = v²/r directed toward the centre.

When a particle moves along a circular path of radius r with constant speed, its motion is called uniform circular motion. Although the speed (magnitude of velocity) remains constant, the direction of the velocity vector keeps changing continuously, since velocity is always tangent to the circle. Because the velocity is changing, the motion is accelerated even though the speed is constant.

The rate of change of angular position is the angular speed, ω, measured in rad/s; it relates to the linear speed v by v = ωr. The time to complete one full revolution is the period T, and the number of revolutions per second is the frequency ν, with T = 2π/ω = 1/ν.

The acceleration in uniform circular motion is directed radially inward, toward the centre of the circle, and is called centripetal acceleration. Its magnitude is ac = v2/r, which can also be written ac = ω2r using v = ωr. This acceleration changes only the direction of the velocity, not its magnitude, which is why the speed stays constant. A net inward (centripetal) force, F = mv2/r, must act on the particle to produce this acceleration; without it, the particle would move off in a straight line, as required by Newton's first law.

Worked example. A stone tied to a string is whirled in a horizontal circle of radius 1.5 m with a constant speed of 6 m/s. Find its centripetal acceleration and angular speed.

Given: r = 1.5 m, v = 6 m/s

Formula: ac = v2/r; ω = v/r

Substitution: ac = 62/1.5 = 36/1.5; ω = 6/1.5

Result: ac = 24 m/s2, directed toward the centre of the circle; ω = 4 rad/s.

Linear-angular speed relation v = ωr
Centripetal acceleration ac = v² / r = ω² r
Centripetal force F = mv² / r
Period and frequency T = 2π / ω = 1 / ν
Remember
  • In uniform circular motion, speed is constant but velocity direction continuously changes, so the motion is accelerated.
  • Angular speed relates to linear speed by v = ωr.
  • Centripetal acceleration ac = v²/r = ω²r is directed radially inward, toward the centre.
  • A centripetal force F = mv²/r must act toward the centre to sustain the circular path.
  • Period T = 2π/ω and frequency ν = 1/T describe the time per revolution.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

Δr = r₂ − r₁
Displacement vector
|A| = √(Ax² + Ay²)
Magnitude of a vector
|nA| = |n| |A|
Scalar multiplication
R = √(P² + Q² + 2PQ cosθ)
Resultant magnitude
tanα = Q sinθ / (P + Q cosθ)
Direction of resultant
P − Q = P + (−Q)
Vector subtraction
Rmax = P + Q at θ=0°; Rmin = |P − Q| at θ=180°
Extreme cases
Ax = A cosθ, Ay = A sinθ
Rectangular components
A = Ax i + Ay j
Vector in component form
 = A / |A|
Unit vector
A = √(Ax² + Ay²)
Magnitude from components
θ = tan⁻¹(Ay / Ax)
Direction from components
v = v₀ + at
Velocity-time relation
r = r₀ + v₀t + ½at²
Position-time relation
vx = v0x + axt; x = x0 + v0xt + ½axt²
x-component equations
vy = v0y + ayt; y = y0 + v0yt + ½ayt²
y-component equations
vBA = vB − vA
Relative velocity of B w.r.t. A
vAB = −vBA
Reciprocal relation
tanθ = v_person / v_rain
Rain-umbrella angle
y = x tanθ − gx² / (2u² cos²θ)
Trajectory equation
T = 2u sinθ / g
Time of flight
H = u² sin²θ / (2g)
Maximum height
R = u² sin2θ / g
Horizontal range
Rmax = u² / g
Maximum range
v = ωr
Linear-angular speed relation
ac = v² / r = ω² r
Centripetal acceleration
F = mv² / r
Centripetal force
T = 2π / ω = 1 / ν
Period and frequency

Test yourself

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0 correct · 0/12 answered
Q1 Scalars and Vectors easy

Which of the following is a vector quantity?

Q2 Vector Addition medium

The minimum number of unequal vectors required to produce a zero resultant is

Q3 Vector Addition medium

Two forces of magnitude 3 N and 4 N act on a body perpendicular to each other. What is the magnitude of the resultant force?

Q4 Scalars and Vectors easy

What is the angle between a vector A and its negative, −A?

Q5 Resolution of Vectors medium

A vector of magnitude 10 units makes an angle of 60° with the x-axis. What is its x-component?

Q6 Unit Vectors easy

What is the magnitude of a unit vector?

Q7 Motion in a Plane medium

A particle has velocity v = (3i + 4j) m/s. What is its speed?

Q8 Projectile Motion easy

At the highest point of a projectile's trajectory, the vertical component of its velocity is

Q9 Projectile Motion medium

A ball is thrown with a speed of 10 m/s at 30° above the horizontal. Taking g = 10 m/s², what is its time of flight?

Q10 Projectile Motion medium

For a projectile launched with a fixed speed, the horizontal range is maximum when the angle of projection is

Q11 Circular Motion medium

A stone tied to a string of length 1 m moves in a circle with a constant speed of 2 m/s. What is its centripetal acceleration?

Q12 Circular Motion easy

In uniform circular motion, which of the following quantities remains constant?

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 State, with reasons, whether the following algebraic operations with scalar and vector quantities are meaningful: (a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions, (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector.Scalars and Vectors

(a) Meaningful, provided the two scalars represent the same physical quantity (same dimensions), for example adding two masses or two time intervals.

(b) Not meaningful. A scalar and a vector are fundamentally different types of quantities and can never be added, even if their dimensions match, because a vector carries direction information that a scalar does not.

(c) Meaningful. Multiplying a vector by a scalar gives another vector; the direction stays the same if the scalar is positive and reverses if the scalar is negative, while the magnitude scales by the magnitude of the scalar.

(d) Meaningful. The product of any two scalars is always another scalar, regardless of their individual dimensions, for example speed multiplied by time gives distance.

(e) Meaningful only if the two vectors represent the same physical quantity (same dimensions); for example, two velocities can be added, but a velocity cannot be added to a force.

(f) Meaningful. A component of a vector is itself a vector of the same dimensions as the original vector, so it can always be added to the vector using the triangle law of vector addition.

2 State whether the following statements are true or false, giving reasons: (a) The magnitude of a vector is always a scalar. (b) Each component of a vector is always a scalar. (c) The total path length is always equal to the magnitude of the displacement vector of a particle. (d) The average speed of a particle is greater than or equal to the magnitude of its average velocity over a given time interval.Scalars and Vectors

(a) True. The magnitude of a vector is just a number (with an appropriate unit) representing its size; it has no direction attached to it, so it is a scalar.

(b) False. A component of a vector along an axis is itself a vector quantity (it carries a sign/direction along that axis), not a pure scalar.

(c) False. This holds only for motion along a straight line without any reversal of direction. In general, the path length (distance) is greater than or equal to the magnitude of displacement, since displacement is the straight-line distance between the initial and final points.

(d) True. Average speed equals total path length divided by total time, while the magnitude of average velocity equals |displacement| divided by time. Since path length ≥ |displacement|, average speed ≥ magnitude of average velocity; equality holds only when the particle moves in a straight line without reversing direction.

3 A cricketer can throw a ball to a maximum horizontal distance of 100 m. Assuming the ball is thrown with the same speed in every case and neglecting air resistance, how much high above the ground can the cricketer throw the same ball?Projectile Motion

Given: Maximum horizontal range, Rmax = 100 m

Formula: For a given launch speed u, the range is maximum when the ball is thrown at 45°, so Rmax = u2/g. The greatest height is reached when the same ball is thrown straight up with the same speed u, giving Hmax = u2/(2g).

Substitution: From Rmax = u2/g = 100 m, we get u2 = 100g. Therefore Hmax = u2/(2g) = 100g/(2g).

Result: Hmax = 50 m. The cricketer can throw the ball to a maximum height of 50 m.

4 A ball is thrown at an angle of 30° above the horizontal with a speed of 15 m/s. Taking g = 9.8 m/s², calculate (i) the time of flight, (ii) the maximum height, and (iii) the horizontal range.Projectile Motion

Given: u = 15 m/s, θ = 30°, g = 9.8 m/s2

Formula: T = 2u sinθ/g; H = u2sin2θ/(2g); R = u2sin2θ/g

Substitution:

  • T = (2 × 15 × sin30°)/9.8 = (2 × 15 × 0.5)/9.8 = 15/9.8
  • H = (152 × sin230°)/(2 × 9.8) = (225 × 0.25)/19.6 = 56.25/19.6
  • R = (152 × sin60°)/9.8 = (225 × 0.866)/9.8 = 194.85/9.8

Result: T ≈ 1.53 s, H ≈ 2.87 m, R ≈ 19.88 m.

5 An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.Circular Motion

Given: r = 1.00 km = 1000 m, v = 900 km/h

Formula: ac = v2/r

Substitution: Converting speed, v = 900 × (1000/3600) m/s = 250 m/s. Then ac = (250)2/1000 = 62500/1000

Result: ac = 62.5 m/s2. Comparing with g = 9.8 m/s2, ac/g = 62.5/9.8 ≈ 6.38, so the centripetal acceleration is about 6.4 times the acceleration due to gravity.

6 Two vectors A and B of magnitudes 5 units and 12 units respectively are perpendicular to each other. Find the magnitude of their resultant and the angle it makes with vector A.Vector Addition

Given: A = 5 units, B = 12 units, θ = 90°

Formula: R = √(A2 + B2 + 2AB cosθ); tanα = B sinθ/(A + B cosθ)

Substitution: Since θ = 90°, cosθ = 0 and sinθ = 1, so R = √(52 + 122) = √(25+144) = √169; tanα = 12/5 = 2.4

Result: R = 13 units, and α = tan−1(2.4) ≈ 67.4° from vector A.

Previous-year board questions 4

Q1 Can two vectors of unequal magnitude add up to give a zero resultant? Can three? CBSE 2020 1 mark

No, two vectors of unequal magnitude can never add up to give a zero resultant. For the resultant of two vectors to be zero, they must be exactly equal in magnitude and exactly opposite in direction; if their magnitudes differ, some part of the larger vector always remains uncancelled.

Yes, three vectors can add up to give a zero resultant, provided they can be arranged head to tail to form a closed triangle; the magnitudes and directions must be such that each vector can be represented as a side of a closed triangle.

Q2 Derive expressions for the time of flight and the maximum height attained by a projectile launched with speed u at an angle θ above the horizontal. CBSE 2022 3 marks

Consider a projectile launched with speed u at angle θ to the horizontal, with g the acceleration due to gravity acting vertically downward. Taking the point of projection as the origin, the vertical component of velocity at time t is vy = u sinθ − gt, and the vertical displacement is y = (u sinθ)t − ½gt2.

Time of flight: The projectile returns to the same horizontal level when y = 0, i.e. (u sinθ)t − ½gt2 = 0, so t[(u sinθ) − ½gt] = 0. This gives t = 0 (point of projection) or t = 2u sinθ/g (point of landing). Hence the time of flight is T = 2u sinθ/g.

Maximum height: The projectile is at its highest point when the vertical velocity component is zero: vy = u sinθ − gt = 0, giving t = u sinθ/g. Substituting this time into the expression for y: H = (u sinθ)(u sinθ/g) − ½g(u sinθ/g)2 = u2sin2θ/g − u2sin2θ/(2g) = u2sin2θ/(2g).

Thus T = 2u sinθ/g and H = u2sin2θ/(2g).

Q3 A projectile is fired with a velocity of 40 m/s at an angle of 60° above the horizontal. Taking g = 10 m/s², calculate (i) the time of flight, (ii) the maximum height attained, and (iii) the horizontal range of the projectile. CBSE 2023 5 marks

Given: u = 40 m/s, θ = 60°, g = 10 m/s2

Formula: T = 2u sinθ/g; H = u2sin2θ/(2g); R = u2sin2θ/g

Substitution:

  • sin60° = 0.866, sin260° = 0.75, sin120° = 0.866
  • T = (2 × 40 × 0.866)/10 = 69.28/10
  • H = (402 × 0.75)/(2 × 10) = 1200/20
  • R = (402 × 0.866)/10 = 1385.6/10

Result: T ≈ 6.93 s, H = 60 m, R ≈ 138.56 m.

Q4 A cyclist moves on a circular track of radius 25 m at a constant speed of 9 m/s. Find (i) the angular speed and (ii) the centripetal acceleration of the cyclist. CBSE 2019 2 marks

Given: r = 25 m, v = 9 m/s

Formula: ω = v/r; ac = v2/r

Substitution: ω = 9/25; ac = 92/25 = 81/25

Result: ω = 0.36 rad/s, ac = 3.24 m/s2, directed toward the centre of the circular track.

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