Class 11 · Physics · Part 1

Chapter 3: Motion in a Plane

Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.

3.1–3.2Scalars and vectorsScalars, vectors, position, displacement and equality of vectors

Why vectors?

In a straight line the direction of position, velocity and acceleration can be shown by a + or − sign, because only two directions are possible.
In two or three dimensions we need vectors to describe these quantities. The ideas of this chapter extend easily to three dimensions.

Scalars

A scalar has magnitude only. It is specified by a number and a unit. Examples: distance, mass, temperature, time, volume, density, speed.
Scalars combine by the rules of ordinary algebra. Adding or subtracting scalars makes sense only if they have the same units; scalars of different units may be multiplied or divided (for example, mass/volume = density).

Vectors

A vector has magnitude and direction and obeys the triangle law (equivalently the parallelogram law) of addition. Examples: displacement, velocity, acceleration, force.
In print a vector is written in bold (A); by hand it is written with an arrow over the letter. Its magnitude, written |A| or A, is a scalar and is never negative.

Position and displacement vectors

The position vector r = OP joins the chosen origin O to the object at P.
If the object moves from P to P′, the displacement vector is Δr = r′ − r, the straight line from P to P′.
The displacement depends only on the end points, not on the actual path. The magnitude of displacement is therefore less than or equal to the path length.

Equality of vectors

Two vectors are equal only if they have the same magnitude and the same direction. Two vectors with the same magnitude but different directions are not equal.
In this chapter vectors are free vectors: shifting a vector parallel to itself leaves it unchanged.

3.3–3.4Multiplication by numbers and graphical additionλA, null vector, triangle and parallelogram methods, subtraction

Multiplying a vector by a real number

For λ > 0, λA has magnitude λ|A| and the same direction as A. For a negative number −λ, the direction is reversed.
If λ has a dimension, λA has the product of the dimensions. For example, velocity × time = displacement.

Adding vectors graphically

Head-to-tail (triangle) method: place the tail of B at the head of A. The resultant R = A + B is drawn from the tail of A to the head of B.
Parallelogram method: bring the tails of A and B together and complete the parallelogram. The diagonal from the common tail is R. The two methods give the same result.
Vector addition is commutative, A + B = B + A, and associative, (A + B) + C = A + (B + C).

Adding two vectors by the head-to-tail (triangle) method, and subtracting by adding the reversed vector.

Null vector and subtraction

A vector plus its negative is the null (zero) vector: A − A = 0, with |0| = 0. Its direction cannot be specified. Its properties: A + 0 = A, λ0 = 0, 0A = 0.
A body that goes from P to P′ and back to P has a null displacement.
Subtraction is defined by A − B = A + (−B): add the reversed vector. Note that A − B = −(B − A).

3.5Resolution of vectors and unit vectorsComponents along x and y, 3-D extension

Unit vectors

A unit vector has magnitude 1, no unit and no dimension. It only specifies a direction. A vector along A is n̂ = A/|A|, so A = |A| n̂.
Along the x-, y- and z-axes the unit vectors are î, ĵ, k̂. They are mutually perpendicular, and |î| = |ĵ| = |k̂| = 1.

Components in a plane

Any vector in the x–y plane can be written A = Aₓ î + A_y ĵ. The numbers Aₓ and A_y are the components; they are scalars (they can be positive, negative or zero), while Aₓ î and A_y ĵ are the component vectors.
Aₓ = A cos θ, A_y = A sin θ, where θ is the angle with the x-axis. Conversely A = √(Aₓ² + A_y²) and tan θ = A_y / Aₓ.
So a vector can be given either by (magnitude, direction) or by its components.

Resolving a vector A into components along the x- and y-axes.

Three dimensions

A = Aₓ î + A_y ĵ + A_z k̂, with A = √(Aₓ² + A_y² + A_z²). If α, β, γ are the angles with the three axes, Aₓ = A cos α, A_y = A cos β, A_z = A cos γ.
A position vector is r = x î + y ĵ + z k̂.

3.6Vector addition by componentsAnalytical method, law of cosines and law of sines

Adding by components

If A = Aₓ î + A_y ĵ and B = Bₓ î + B_y ĵ, then R = A + B = (Aₓ + Bₓ) î + (A_y + B_y) ĵ. Each component of the resultant is the sum of the corresponding components.
The same holds for three dimensions and for any number of vectors. For example, T = a + b − c has Tₓ = aₓ + bₓ − cₓ, and so on.

Magnitude and direction of the resultant

For two vectors of magnitudes A and B with angle θ between them (law of cosines): R = √(A² + B² + 2AB cos θ).
The direction of R relative to A is given by tan α = B sin θ / (A + B cos θ). Also R/sin(180° − θ) = A/sin β = B/sin α (law of sines).
Limits: maximum R = A + B at θ = 0°; minimum R = |A − B| at θ = 180°; for θ = 90°, R = √(A² + B²).
Example 3.1: rain falls vertically at 35 m s⁻¹ and the wind blows at 12 m s⁻¹ horizontally. The resultant is √(35² + 12²) ≈ 37 m s⁻¹ at tan⁻¹(12/35) ≈ 19° to the vertical, so the umbrella must be held tilted by about 19° from the vertical, towards the side from which the wind comes.

3.7–3.8Motion in a planePosition, velocity, acceleration, constant acceleration

Position, displacement and velocity

Position vector r = x î + y ĵ. Displacement Δr = r′ − r = Δx î + Δy ĵ.
Average velocity v̄ = Δr/Δt = (Δx/Δt) î + (Δy/Δt) ĵ; its direction is that of Δr.
Velocity v = lim(Δt→0) Δr/Δt = dr/dt = vₓ î + v_y ĵ, where vₓ = dx/dt and v_y = dy/dt. Its magnitude is v = √(vₓ² + v_y²) and its direction is tan θ = v_y/vₓ.
As Δt → 0 the displacement becomes tangent to the path: the velocity at any point is tangential to the path, in the direction of motion.

Acceleration

Average acceleration ā = Δv/Δt. Acceleration a = dv/dt = aₓ î + a_y ĵ, with aₓ = dvₓ/dt and a_y = dv_y/dt.
In one dimension velocity and acceleration lie on the same line. In two or three dimensions they can make any angle between 0° and 180°.
Example 3.4: for r = 3.0t î + 2.0t² ĵ + 5.0 k̂, v = 3.0 î + 4.0t ĵ and a = 4.0 ĵ m s⁻². At t = 1 s, v = 5.0 m s⁻¹ at tan⁻¹(4/3) ≈ 53° to the x-axis.

Constant acceleration in a plane

v = v₀ + at and r = r₀ + v₀t + ½at². These vector equations do not depend on the choice of axes.
In components: x = x₀ + v₀ₓt + ½aₓt², y = y₀ + v₀_yt + ½a_yt², vₓ = v₀ₓ + aₓt, v_y = v₀_y + a_yt.
Key idea: motion in a plane is the superposition of two independent one-dimensional motions along perpendicular directions. The choice of axes is a matter of convenience (for a projectile, horizontal and vertical).
Example 3.5: a particle starts at the origin with velocity 5.0 î m s⁻¹ and constant acceleration (3.0 î + 2.0 ĵ) m s⁻². Then x = 5t + 1.5t² and y = t². For x = 84 m, t = 6 s, y = 36 m and v = 23 î + 12 ĵ, speed ≈ 26 m s⁻¹.

3.9Projectile motionPath, time of flight, maximum height, range

Projectile and its two motions

A projectile is an object in flight after being thrown or projected. Neglecting air resistance, its acceleration is a = −g ĵ: aₓ = 0, a_y = −g.
Galileo's idea: the horizontal and vertical motions are independent. Horizontally the motion is uniform; vertically it is uniformly accelerated (like free fall).
With launch speed v₀ at angle θ₀, the starting components are v₀ₓ = v₀ cos θ₀ and v₀_y = v₀ sin θ₀.

Equations of motion

Taking the origin at the launch point: x = (v₀ cos θ₀) t and y = (v₀ sin θ₀) t − ½ g t².
Velocity: vₓ = v₀ cos θ₀ (constant) and v_y = v₀ sin θ₀ − g t.
Path: eliminating t gives y = (tan θ₀) x − g x² / [2 (v₀ cos θ₀)²], which is of the form y = ax + bx²: a parabola.
At the highest point v_y = 0, so the velocity there is horizontal, v = v₀ cos θ₀ (the minimum speed), and the acceleration is still g downward.

Path of a projectile. The horizontal component of velocity stays constant; the vertical component changes at the rate g.

Time of flight, maximum height and range

Time to maximum height: tₘ = v₀ sin θ₀ / g. Time of flight T_f = 2 v₀ sin θ₀ / g = 2tₘ (symmetry of the parabola).
Maximum height: hₘ = (v₀ sin θ₀)² / 2g.
Horizontal range: R = (v₀ cos θ₀) T_f = v₀² sin 2θ₀ / g.
For a given speed the range is greatest when sin 2θ₀ = 1, i.e. θ₀ = 45°, and then Rₘ = v₀²/g. At 45°, R = 4hₘ. In general R/hₘ = 4/tan θ₀.

Equal ranges (Galileo's statement)

Angles 45° + α and 45° − α give the same range, because sin(90° + 2α) = sin(90° − 2α) = cos 2α. Complementary angles (adding up to 90°) give equal ranges.
The times of flight and the maximum heights are different: for θ and 90° − θ, T₁/T₂ = tan θ and h₁/h₂ = tan²θ.

For the same speed, angles of projection that exceed or fall short of 45° by equal amounts give equal ranges.

Projectile from a height

A stone thrown horizontally at 15 m s⁻¹ from a cliff 490 m high (Example 3.7): from −490 = −½ (9.8) t², t = 10 s; then vₓ = 15, v_y = −98 and the speed is √(15² + 98²) ≈ 99 m s⁻¹.
Example 3.8: a ball thrown at 28 m s⁻¹ at 30° rises to 10 m, returns to its level in 2.9 s and lands 69 m away.

3.10Uniform circular motionCentripetal acceleration, angular speed, period, frequency

Acceleration in uniform circular motion

When an object moves along a circle at constant speed the motion is uniform circular motion. The direction of the velocity keeps changing, so the object is accelerated.
The velocity is along the tangent; the change in velocity Δv points towards the centre. The acceleration is the centripetal acceleration a_c = v²/R, always directed towards the centre.
Its magnitude is constant but its direction changes, so it is not a constant vector.

In uniform circular motion the acceleration is directed towards the centre; the change in velocity Δv points inward.

Angular speed, period and frequency

Angular speed ω = Δθ/Δt (rad s⁻¹); linear speed v = ωR; centripetal acceleration a_c = ω²R = v²/R.
Period T = time for one revolution; frequency ν = 1/T. Then ω = 2π/T = 2πν, v = 2πR/T = 2πRν and a_c = 4π²ν²R.
Example 3.9: an insect in a groove of radius 12 cm completes 7 revolutions in 100 s: ω = 2π × 7/100 = 0.44 rad s⁻¹, v = 5.3 cm s⁻¹, a_c = 2.3 cm s⁻².

Over a full revolution

The velocity returns to its initial value after one revolution, so the average acceleration over a complete revolution is zero, while the instantaneous acceleration is never zero.
The kinematic equations for constant acceleration do not apply, because the direction of the acceleration is changing.

PonderPoints to ponderCommon confusions

Remember

Path length ≥ |displacement|; the average speed ≥ |average velocity|. They are equal only when the object does not change direction.
A vector equation does not depend on the choice of axes; you can resolve it along any two independent axes.
Resultant velocity v = v₁ + v₂ is different from the velocity of 1 relative to 2, v₁₂ = v₁ − v₂.
In circular motion the net acceleration points to the centre only if the speed is constant.
The shape of a trajectory depends on the initial position and velocity as well as on the acceleration. Under the same g the path can be a straight line (thrown vertically) or a parabola.

Quick referenceFormulas at a glanceRevise before the exam

Key results

R = √(A² + B² + 2AB cos θ); tan α = B sin θ/(A + B cos θ); Aₓ = A cos θ, A_y = A sin θ; n̂ = A/|A|
v = dr/dt; a = dv/dt; v = v₀ + at; r = r₀ + v₀t + ½at²
Projectile: T_f = 2v₀ sin θ₀/g; hₘ = v₀² sin²θ₀/2g; R = v₀² sin 2θ₀/g; Rₘ = v₀²/g (45°)
Circular: a_c = v²/R = ω²R = 4π²ν²R; v = ωR; ω = 2πν

Common mistakes

Adding magnitudes of vectors instead of adding the vectors.
Taking the velocity of a projectile to be zero at the highest point (only v_y = 0).
Using the equations for constant acceleration in uniform circular motion.
Forgetting that equal ranges occur for complementary angles, but with different times of flight and heights.
Working with the calculator in the wrong angle mode (degrees or radians) when finding ω or sin 2θ₀.

Try each problem first, then open the solution to check your working.

3.1Vectors: addition, components, relative velocityResultants, components and unit vectors · 7 problems

Q1Perpendicular vectors

Two perpendicular vectors of magnitudes 8 units and 6 units are added. Find the magnitude of the resultant and its direction with respect to the 8-unit vector.

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Given: A = 8, B = 6, θ = 90°

Use: R = √(A² + B²), tan α = B/A

R = √(64 + 36) = 10 units.

tan α = 6/8 = 0.75, so α ≈ 36.9°.

Answer: 10 units, at about 36.9° to the 8-unit vector

Q2Resultant at 60°

Two forces of 5 N and 7 N act at an angle of 60° to each other. Find the magnitude of the resultant and its angle with the 5 N force.

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Given: A = 5 N, B = 7 N, θ = 60°

Use: R = √(A² + B² + 2AB cos θ), tan α = B sin θ/(A + B cos θ)

R² = 25 + 49 + 2 × 5 × 7 × 0.5 = 109, so R = 10.4 N.

tan α = 7 × 0.866/(5 + 3.5) = 6.06/8.5 = 0.713, so α ≈ 35.5°.

Answer: R ≈ 10.4 N, at 35.5° to the 5 N force

Q3Resolving a velocity

A velocity of 20 m s⁻¹ is directed at 37° above the x-axis. Find its x- and y-components.

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Given: v = 20 m s⁻¹, θ = 37°

Use: vₓ = v cos θ, v_y = v sin θ

vₓ = 20 cos 37° = 20 × 0.80 = 16.0 m s⁻¹.

v_y = 20 sin 37° = 20 × 0.60 = 12.0 m s⁻¹.

Check: √(16² + 12²) = 20 m s⁻¹.

Answer: vₓ ≈ 16 m s⁻¹, v_y ≈ 12 m s⁻¹

Q4Magnitude and direction from components

A vector is A = −3 î + 4 ĵ. Find its magnitude and the angle it makes with the positive x-axis.

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Given: Aₓ = −3, A_y = +4

Use: A = √(Aₓ² + A_y²), tan θ = A_y/Aₓ

A = √(9 + 16) = 5.

tan⁻¹(4/3) = 53.1°. Since Aₓ is negative and A_y positive, the vector lies in the second quadrant: θ = 180° − 53.1° = 126.9° from +x.

Answer: 5; 126.9° from the positive x-axis

Q5Adding and subtracting by components

If a = 2 î + 3 ĵ and b = î − 2 ĵ, find a + b, a − b, the magnitude of a − b, and 2a − 3b.

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Given: a = (2, 3), b = (1, −2)

Use: Add or subtract corresponding components

a + b = (2 + 1) î + (3 − 2) ĵ = 3 î + ĵ.

a − b = (2 − 1) î + (3 + 2) ĵ = î + 5 ĵ; its magnitude is √(1 + 25) = √26 ≈ 5.10.

2a − 3b = (4 − 3) î + (6 + 6) ĵ = î + 12 ĵ.

Answer: 3 î + ĵ; î + 5 ĵ (magnitude 5.10); î + 12 ĵ

Q6Umbrella in the rain

Rain falls vertically at 10 m s⁻¹. A man walks at 5 m s⁻¹ on a level road. At what angle to the vertical should he tilt his umbrella, and what is the speed of the rain relative to him?

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Given: v_rain = 10 m s⁻¹ (down), v_man = 5 m s⁻¹

Use: Velocity of rain relative to the man = v_rain − v_man

Relative to the man the rain has a horizontal component 5 m s⁻¹ (backwards) and a vertical component 10 m s⁻¹ (down).

tan θ = 5/10 = 0.5, so θ ≈ 26.6° from the vertical, tilted forward (towards the direction he is walking).

Speed relative to him = √(5² + 10²) = √125 ≈ 11.2 m s⁻¹.

Answer: about 26.6° from the vertical, forward; 11.2 m s⁻¹

Q7Boat crossing a river

A boat can move at 4 m s⁻¹ in still water. It heads straight across a river which flows at 3 m s⁻¹. Find the speed of the boat relative to the ground and the angle its path makes with the direction of the bank.

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Given: v_boat = 4 m s⁻¹ (across), v_river = 3 m s⁻¹ (along the bank)

Use: Perpendicular vectors add as √(a² + b²)

Resultant speed = √(4² + 3²) = 5 m s⁻¹.

Angle with the bank: tan α = 4/3, α ≈ 53.1° (or 36.9° from the direction straight across).

Answer: 5 m s⁻¹; 53° to the bank

3.2Motion in a planeUsing r(t), v(t) and constant acceleration · 5 problems

Q8Velocity and acceleration from r(t)

A particle has position r = (3t² + 2) î + 4t ĵ (metres, seconds). Find its velocity and acceleration at t = 2 s, and the speed.

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Given: x = 3t² + 2, y = 4t

Use: v = dr/dt, a = dv/dt

vₓ = dx/dt = 6t and v_y = dy/dt = 4, so v = 6t î + 4 ĵ.

At t = 2 s: v = 12 î + 4 ĵ m s⁻¹; speed = √(144 + 16) = 12.65 m s⁻¹ ≈ 12.7 m s⁻¹.

a = dv/dt = 6 î m s⁻² (constant, along x).

Answer: v = 12 î + 4 ĵ m s⁻¹ (12.7 m s⁻¹); a = 6 î m s⁻²

Q9A path with a turning point

A particle moves with r = 5t î + (10t − 5t²) ĵ. Find v and a, and the instant when the velocity is horizontal.

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Given: x = 5t, y = 10t − 5t²

Use: v = dr/dt, a = dv/dt

v = 5 î + (10 − 10t) ĵ; a = −10 ĵ m s⁻².

The velocity is horizontal when v_y = 0: 10 − 10t = 0, t = 1 s.

At that instant v = 5 î m s⁻¹ and the particle is at (5 m, 5 m): the highest point.

Answer: v = 5 î + (10 − 10t) ĵ; a = −10 ĵ m s⁻²; t = 1 s

Q10Constant acceleration in a plane

A particle starts from the origin with velocity 4 î m s⁻¹ and has a constant acceleration (2 î + 3 ĵ) m s⁻². Find its velocity, speed and position after 2 s.

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Given: v₀ = 4 î, a = 2 î + 3 ĵ, t = 2 s

Use: v = v₀ + at, r = v₀t + ½at²

v = 4 î + (2 î + 3 ĵ) × 2 = 8 î + 6 ĵ m s⁻¹; speed = √(64 + 36) = 10 m s⁻¹.

r = 4 î × 2 + ½ (2 î + 3 ĵ) × 4 = 8 î + (4 î + 6 ĵ) = 12 î + 6 ĵ m.

Answer: v = 8 î + 6 ĵ m s⁻¹ (10 m s⁻¹); r = 12 î + 6 ĵ m

Q11Where is the particle when x = 8 m?

A particle starts at the origin with velocity 3 î m s⁻¹ and acceleration (1 î + 2 ĵ) m s⁻². Find its y-coordinate when its x-coordinate is 8 m.

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Given: v₀ = 3 î, a = (1, 2)

Use: Use the two components separately

x = 3t + ½ (1) t² = 3t + 0.5t². Setting x = 8: t² + 6t − 16 = 0, so (t + 8)(t − 2) = 0 and t = 2 s.

y = 0 + ½ (2) t² = t² = 4 m.

Answer: y = 4 m (at t = 2 s)

Q12Average velocity and average acceleration

A particle moves from (1 m, 2 m) to (7 m, −4 m) in 3 s. Its velocity changes from 3 î + 4 ĵ to 6 î − 4 ĵ (m s⁻¹) in 2 s. Find the average velocity over the first interval and the average acceleration over the second.

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Given: Δr = (6, −6) m in 3 s; Δv = (3, −8) m s⁻¹ in 2 s

Use: v̄ = Δr/Δt, ā = Δv/Δt

v̄ = (6 î − 6 ĵ)/3 = 2 î − 2 ĵ m s⁻¹; magnitude 2.83 m s⁻¹, directed 45° below the x-axis.

ā = (3 î − 8 ĵ)/2 = 1.5 î − 4 ĵ m s⁻²; magnitude = √(2.25 + 16) = 4.27 m s⁻².

Answer: v̄ = 2 î − 2 ĵ m s⁻¹ (2.83 m s⁻¹, 45° below x); ā = 1.5 î − 4 ĵ m s⁻² (4.27 m s⁻²)

3.3Projectile motionTime of flight, height, range and equal ranges · 9 problems

Q13Ball thrown at 30°

A ball is thrown at 40 m s⁻¹ at 30° above the horizontal (g = 10 m s⁻²). Find the time of flight, the maximum height and the range.

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Given: v₀ = 40 m s⁻¹, θ₀ = 30°

Use: T_f = 2v₀ sin θ₀/g, hₘ = (v₀ sin θ₀)²/2g, R = v₀² sin 2θ₀/g

T_f = 2 × 40 × 0.5/10 = 4 s.

hₘ = (40 × 0.5)²/(2 × 10) = 400/20 = 20 m.

R = 1600 × sin 60°/10 = 160 × 0.866 = 138.6 m.

Answer: 4 s; 20 m; 138.6 m

Q14Horizontal throw from a cliff

A stone is thrown horizontally at 20 m s⁻¹ from the top of a 45 m high cliff (g = 10 m s⁻²). Find the time of fall, the horizontal distance, and the speed and direction of the stone when it hits the ground.

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Given: vₓ = 20 m s⁻¹, v₀_y = 0, h = 45 m

Use: h = ½ g t², x = vₓ t

45 = ½ × 10 × t², so t = 3 s.

x = 20 × 3 = 60 m.

v_y = −g t = −30 m s⁻¹; speed = √(20² + 30²) = 36.1 m s⁻¹.

tan θ = 30/20 = 1.5, so the stone hits the ground at about 56.3° below the horizontal.

Answer: 3 s; 60 m; 36.1 m s⁻¹ at 56.3° below the horizontal

Q15Maximum throwing distance

A player can throw a ball a maximum horizontal distance of 90 m. Find the speed of throw and the maximum height to which he can throw the ball vertically (g = 10 m s⁻²).

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Given: Rₘ = 90 m

Use: Rₘ = v₀²/g, h = v₀²/2g

v₀² = g Rₘ = 10 × 90 = 900, so v₀ = 30 m s⁻¹.

Vertically upward: h = v₀²/2g = 900/20 = 45 m (half the maximum range).

Answer: 30 m s⁻¹; 45 m

Q16Equal ranges

A projectile has a range of 80 m when projected at 20° to the horizontal. At what other angle will the range be the same, and what is the maximum range for the same speed?

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Given: R(20°) = 80 m

Use: Complementary angles give equal ranges; R = Rₘ sin 2θ₀

The range is equal for 20° and 90° − 20° = 70°.

R = Rₘ sin 2θ₀ gives 80 = Rₘ sin 40° = 0.6428 Rₘ, so Rₘ = 124.5 m.

Answer: 70°; about 124 m

Q17Ball at 60°

A ball is thrown at 20 m s⁻¹ at 60° to the horizontal (g = 10 m s⁻²). Find its speed at the highest point, the maximum height and the time to reach it.

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Given: v₀ = 20 m s⁻¹, θ₀ = 60°

Use: At the top v_y = 0

Speed at the top = v₀ cos 60° = 10 m s⁻¹ (horizontal).

hₘ = (20 sin 60°)²/(2 × 10) = (17.32)²/20 = 15 m.

tₘ = v₀ sin θ₀/g = 17.32/10 = 1.73 s.

Answer: 10 m s⁻¹; 15 m; 1.73 s

Q18Clearing the crossbar

A football is kicked at 20 m s⁻¹ at 45° (g = 10 m s⁻²). Will it clear a crossbar 2.4 m high at a horizontal distance of 30 m?

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Given: v₀ = 20 m s⁻¹, θ₀ = 45°, x = 30 m

Use: y = x tan θ₀ − g x²/[2 v₀² cos² θ₀]

y = 30 × 1 − (10 × 900)/(2 × 400 × 0.5) = 30 − 22.5 = 7.5 m.

7.5 m is greater than 2.4 m, so the ball passes over the crossbar. (The range is v₀²/g = 40 m, so at 30 m the ball is on its way down.)

Answer: Yes: height 7.5 m at x = 30 m

Q19When is the height equal to the range?

At what angle of projection is the maximum height of a projectile equal to its horizontal range?

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Given: hₘ = R

Use: hₘ = v₀² sin²θ₀/2g, R = 2 v₀² sin θ₀ cos θ₀/g

Equate: sin²θ₀/2 = 2 sin θ₀ cos θ₀.

Divide by sin θ₀: sin θ₀/2 = 2 cos θ₀, so tan θ₀ = 4.

θ₀ = tan⁻¹ 4 ≈ 76°.

Answer: θ₀ = tan⁻¹ 4 ≈ 76°

Q20Height from the landing point

A ball rolls off a horizontal platform at 12 m s⁻¹ and lands 24 m from the foot of the platform. How high is the platform (g = 10 m s⁻²)?

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Given: vₓ = 12 m s⁻¹, x = 24 m

Use: t = x/vₓ; h = ½ g t²

t = 24/12 = 2 s.

h = ½ × 10 × 2² = 20 m.

Answer: 20 m

Q21Velocity after 1 second

A ball is projected at 40 m s⁻¹ at 30° (g = 10 m s⁻²). Find its velocity (magnitude and direction) 1 s after projection.

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Given: vₓ = 40 cos 30° = 34.64 m s⁻¹; v₀_y = 20 m s⁻¹

Use: v_y = v₀_y − g t

After 1 s: vₓ = 34.64 m s⁻¹ (unchanged); v_y = 20 − 10 × 1 = 10 m s⁻¹.

Speed = √(34.64² + 10²) = 36.1 m s⁻¹.

Direction: tan θ = 10/34.64 = 0.289, θ = 16.1° above the horizontal.

Answer: 36.1 m s⁻¹, 16.1° above the horizontal

3.4Uniform circular motionAngular speed, period and centripetal acceleration · 7 problems

Q22Stone on a string

A stone tied to a 0.50 m string is whirled in a horizontal circle, making 4 revolutions per second. Find its speed and centripetal acceleration.

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Given: R = 0.50 m, ν = 4 Hz

Use: v = 2πRν, a_c = 4π²ν²R

v = 2π × 0.50 × 4 = 12.6 m s⁻¹.

a_c = 4π² × 4² × 0.50 = 316 m s⁻².

Answer: 12.6 m s⁻¹; 316 m s⁻² towards the centre

Q23Second hand of a clock

The tip of the second hand of a wall clock is 12 cm from the axis. Find the angular speed of the hand and the speed of its tip.

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Given: T = 60 s, R = 0.12 m

Use: ω = 2π/T, v = ωR

ω = 2π/60 = 0.105 rad s⁻¹.

v = ωR = 0.105 × 0.12 = 1.26 × 10⁻² m s⁻¹ = 1.26 cm s⁻¹.

Answer: 0.105 rad s⁻¹; 1.26 cm s⁻¹

Q24A car on a curve

A car travels at 36 km h⁻¹ on a circular curve of radius 50 m. Find its centripetal acceleration. What would it be at 72 km h⁻¹?

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Given: R = 50 m; v₁ = 10 m s⁻¹; v₂ = 20 m s⁻¹

Use: a_c = v²/R

At 36 km h⁻¹ = 10 m s⁻¹: a_c = 100/50 = 2 m s⁻².

At 72 km h⁻¹ = 20 m s⁻¹: a_c = 400/50 = 8 m s⁻², four times as large (a_c ∝ v²).

Answer: 2 m s⁻²; 8 m s⁻²

Q25Moon around the earth

The moon completes one revolution around the earth in 27.3 days at a distance of 3.84 × 10⁸ m. Find its orbital speed and its acceleration towards the earth (treat the orbit as circular).

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Given: T = 27.3 × 86400 s = 2.36 × 10⁶ s; R = 3.84 × 10⁸ m

Use: v = 2πR/T, a_c = 4π²R/T²

ω = 2π/T = 2.66 × 10⁻⁶ rad s⁻¹.

v = ωR = 2.66 × 10⁻⁶ × 3.84 × 10⁸ ≈ 1.02 × 10³ m s⁻¹ (about 1 km s⁻¹).

a_c = ω²R = (2.66 × 10⁻⁶)² × 3.84 × 10⁸ ≈ 2.7 × 10⁻³ m s⁻².

Answer: ≈ 1.02 km s⁻¹; ≈ 2.7 × 10⁻³ m s⁻²

Q26A spinning wheel

A wheel of radius 0.40 m rotates at 300 revolutions per minute. Find ω, the speed of a point on the rim and its centripetal acceleration.

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Given: ν = 300/60 = 5 Hz, R = 0.40 m

Use: ω = 2πν, v = ωR, a_c = ω²R

ω = 2π × 5 = 31.4 rad s⁻¹.

v = 31.4 × 0.40 = 12.6 m s⁻¹.

a_c = (31.4)² × 0.40 = 395 m s⁻².

Answer: 31.4 rad s⁻¹; 12.6 m s⁻¹; 395 m s⁻²

Q27From the acceleration to the speed

A particle moving in a circle of radius 4 m has a centripetal acceleration of 9 m s⁻². Find its speed, angular speed and period.

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Given: a_c = 9 m s⁻², R = 4 m

Use: a_c = v²/R, v = ωR, T = 2π/ω

v = √(a_c R) = √36 = 6 m s⁻¹.

ω = v/R = 6/4 = 1.5 rad s⁻¹.

T = 2π/ω = 4.19 s.

Answer: 6 m s⁻¹; 1.5 rad s⁻¹; 4.19 s

Q28Average and instantaneous acceleration

A particle moves at a constant speed of 5 m s⁻¹ in a circle of radius 2 m. Find the magnitude of its change in velocity in half a revolution, the time taken, and compare the average acceleration over this interval with the instantaneous acceleration.

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Given: v = 5 m s⁻¹, R = 2 m

Use: After half a revolution the velocity is reversed

Δv = v − (−v) = 2v = 10 m s⁻¹.

Time for half a revolution = πR/v = π × 2/5 = 1.26 s.

Average acceleration = Δv/Δt = 10/1.26 = 7.96 m s⁻².

Instantaneous acceleration = v²/R = 25/2 = 12.5 m s⁻². They are different; over a full revolution the average would be zero.

Answer: 10 m s⁻¹; 1.26 s; 7.96 m s⁻² compared with 12.5 m s⁻²

Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.

3.1Exercises 3.1 to 3.22Scalars and vectors, projectile motion, relative motion and circular motion · 22 questions

Q13.1 State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

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Scalars: volume, mass, speed, density, number of moles, angular frequency.
Vectors: acceleration, velocity, displacement, angular velocity.

Q23.2 Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

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Work and current are scalars. (Work is the dot product of force and displacement. Current, although it has a direction along the wire, adds as a scalar and does not obey the vector law of addition.) The others are vectors.

Q33.3 Pick out the only vector quantity in the following list: temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

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Impulse is the only vector (impulse = force × time interval, in the direction of the force). All the others are scalars.

Q43.4 State with reasons, whether the following algebraic operations with scalar and vector physical 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.

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(a) Meaningful only if the two scalars have the same dimensions (for example, adding a mass to a time makes no sense).
(b) Not meaningful. A scalar and a vector are different kinds of quantities; they cannot be added even if their dimensions are the same.
(c) Meaningful. The result is a vector with a new magnitude (and possibly a new dimension). Example: velocity × time = displacement.
(d) Meaningful. Any two scalars can be multiplied (for example, density × volume = mass).
(e) Meaningful only if the two vectors have the same dimensions (for example, a force cannot be added to a velocity).
(f) Meaningful if by 'component' we mean the component vector (such as Aₓ î): it is a vector with the same dimensions as A and can be added to it. The scalar component Aₓ alone cannot be added to a vector.

Q53.5 Read each statement below carefully and state with reasons, if it is true or false: (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 (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector.

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(a) True. The magnitude is a single non-negative number with no direction.
(b) True. Aₓ and A_y are numbers (with sign). The component vectors Aₓ î and A_y ĵ are vectors, but the components themselves are scalars.
(c) False. The path length is equal to the magnitude of the displacement only if the particle moves along a straight line without turning back; otherwise path length is greater.
(d) True. Average speed = path length/time ≥ |displacement|/time = magnitude of average velocity.
(e) True. If three vectors not in one plane could add to zero, the third would be the negative of the sum of the first two, which lies in their plane. This is a contradiction.

Q63.6 Establish the following vector inequalities geometrically or otherwise: (a) |a + b| ≤ |a| + |b|, (b) |a + b| ≥ ||a| − |b||, (c) |a − b| ≤ |a| + |b|, (d) |a − b| ≥ ||a| − |b||. When does the equality sign above apply?

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Draw a and b head to tail; the resultant a + b is the third side of the triangle whose other two sides have lengths |a| and |b|. In any triangle, one side is at most the sum and at least the difference of the other two sides.

(a) |a + b| ≤ |a| + |b|. Equality when a and b are in the same direction (angle 0°).
(b) |a + b| ≥ ||a| − |b||. Equality when a and b are in opposite directions (angle 180°).
(c) |a − b| ≤ |a| + |b|: apply (a) to a and (−b); |−b| = |b|. Equality when a and (−b) are in the same direction, i.e. a and b are antiparallel.
(d) |a − b| ≥ ||a| − |b||: apply (b) to a and (−b). Equality when a and b are in the same direction.

Algebraic check: |a + b|² = a² + b² + 2ab cos θ, which lies between (a − b)² (cos θ = −1) and (a + b)² (cos θ = +1).

Q73.7 Given a + b + c + d = 0, which of the following statements are correct: (a) a, b, c, and d must each be a null vector, (b) The magnitude of (a + c) equals the magnitude of (b + d), (c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d, (d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear?

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(a) Incorrect. Four non-zero vectors can add up to zero (they form a closed quadrilateral).
(b) Correct. a + c = −(b + d), so the two vectors have the same magnitude.
(c) Correct. a = −(b + c + d), so |a| = |b + c + d| ≤ |b| + |c| + |d| by the inequalities in Ex 3.6.
(d) Correct. b + c = −(a + d). The vector a + d lies in the plane of a and d (or on their line if collinear), and so does its negative.

Q83.8 Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 3.19. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?

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All three start at P and end at Q, so the displacement is the same for each: the straight line PQ, which is a diameter. Magnitude = 2 × 200 m = 400 m.

The displacement equals the length of the path only for girl B, who skates along the straight path PQ. Girls A and C follow curved paths whose lengths are greater than 400 m.

Q93.9 A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?

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From Fig. 3.20 the cyclist goes O → P (1 km along a radius), then along the circumference from P to Q (a quarter of the circle, since OP ⊥ OQ), then Q → O (1 km). The total time is 10 min = 1/6 h.

(a) The cyclist returns to the starting point O, so the net displacement is zero.

(b) Average velocity = 0 (displacement/time).

Q103.10 On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

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Turning left by 60° after every 500 m, the motorist goes round a regular hexagon of side 500 m. Take the starting turn as vertex 0.

Third turn (vertex 3, opposite to the start): displacement = the long diagonal = 2 × 500 = 1000 m, directed along the line through the centre. Path length = 1500 m. Displacement is 2/3 of the path length.
Sixth turn: the motorist is back at the starting point, so the displacement is zero; the path length is 3000 m.
Eighth turn: the position is the same as at the 2nd vertex. Displacement = 2 × 500 × cos 30° = 500√3 ≈ 866 m, at 30° to the direction of the first side. Path length = 4000 m; the displacement is 0.22 of the path length.

Ex 3.10: the track is a regular hexagon. Dashed: displacement at the 3rd turn; solid: at the 8th turn.

Q113.11 A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?

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Time = 28 min = 28/60 h = 0.467 h.

(a) Average speed = path length/time = 23/0.467 = 49.3 km h⁻¹.

(b) Average velocity = displacement/time = 10/0.467 = 21.4 km h⁻¹.

The two are not equal, because the path is longer than the displacement.

Q123.12 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s⁻¹ can go without hitting the ceiling of the hall?

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The ball must not rise higher than 25 m, so the maximum height hₘ = (v₀ sin θ₀)²/2g ≤ 25 m.

sin²θ₀ = 2 × 9.8 × 25/40² = 490/1600 = 0.306, so sin θ₀ = 0.553 and θ₀ = 33.6°.

Range R = v₀² sin 2θ₀/g = 1600 × sin 67.2°/9.8 = 1600 × 0.922/9.8 ≈ 150 m.

(The 45° throw would give a bigger range of 163 m but would rise to 40²/(4 × 9.8) = 40.8 m, above the ceiling.)

Q133.13 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?

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The maximum horizontal range is at 45°: Rₘ = v₀²/g = 100 m, so v₀² = 100 g.

Thrown vertically upward, the height is h = v₀²/2g = 100 g/(2g) = 50 m.

Q143.14 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

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R = 0.80 m; ν = 14/25 = 0.56 s⁻¹. ω = 2πν = 3.52 rad s⁻¹ and v = ωR = 2.81 m s⁻¹.

a = ω²R = (3.52)² × 0.80 = 9.9 m s⁻². Its direction is along the string towards the centre of the circle (centripetal).

Q153.15 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.

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v = 900 km h⁻¹ = 900 × 1000/3600 = 250 m s⁻¹; R = 1000 m.

a_c = v²/R = 250²/1000 = 62.5 m s⁻².

a_c/g = 62.5/9.8 = 6.38. The centripetal acceleration is about 6.4 times the acceleration due to gravity.

Q163.16 Read each statement below carefully and state, with reasons, if it is true or false: (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre, (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point, (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.

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(a) False. The net acceleration is towards the centre only for uniform circular motion. If the speed changes, there is also a tangential component, so the net acceleration is not along the radius.
(b) True. The velocity vector is the limit of Δr/Δt as Δt → 0, which is along the tangent to the path.
(c) True. After one cycle the velocity returns to its initial value, so Δv = 0 and the average acceleration over a cycle is zero. (The acceleration vector at each instant points to the centre, and these directions cancel when averaged over a full cycle.)

Q173.17 The position of a particle is given by r = 3.0t î − 2.0t² ĵ + 4.0 k̂ m, where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find the v and a of the particle. (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?

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(a) v = dr/dt = 3.0 î − 4.0t ĵ m s⁻¹ and a = dv/dt = −4.0 ĵ m s⁻² (constant, along −y).

(b) At t = 2.0 s: v = 3.0 î − 8.0 ĵ. Magnitude = √(9 + 64) = √73 = 8.54 m s⁻¹. Direction: tan θ = −8/3 = −2.67, so θ = −69.4°: about 69° below the +x axis (the k̂ term is constant, so it does not affect v).

Q183.18 A particle starts from the origin at t = 0 s with a velocity of 10.0 ĵ m/s and moves in the x–y plane with a constant acceleration of (8.0 î + 2.0 ĵ) m s⁻². (a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time?

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(a) x = ½ aₓ t² = 4t² (since v₀ₓ = 0). x = 16 m gives t² = 4, t = 2 s.

y = v₀_y t + ½ a_y t² = 10 × 2 + ½ × 2 × 4 = 24 m.

(b) vₓ = 8 × 2 = 16 m s⁻¹; v_y = 10 + 2 × 2 = 14 m s⁻¹. Speed = √(16² + 14²) = √452 = 21.3 m s⁻¹.

Q193.19 î and ĵ are unit vectors along x- and y-axis respectively. What is the magnitude and direction of the vectors î + ĵ, and î − ĵ? What are the components of a vector A = 2î + 3ĵ along the directions of î + ĵ and î − ĵ? [You may use graphical method]

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|î + ĵ| = √2 ≈ 1.41; it makes an angle tan⁻¹(1) = 45° with the x-axis. |î − ĵ| = √2; it makes an angle of −45° (45° below the x-axis).

The unit vectors along these directions are n̂₁ = (î + ĵ)/√2 and n̂₂ = (î − ĵ)/√2, which are perpendicular. Writing A = λ(î + ĵ) + μ(î − ĵ): λ + μ = 2 and λ − μ = 3, so λ = 2.5 and μ = −0.5.

Component along (î + ĵ): 2.5(î + ĵ), of magnitude 2.5√2 = 5/√2 = 3.54, in the direction of î + ĵ.
Component along (î − ĵ): −0.5(î − ĵ), of magnitude 0.5√2 = 1/√2 = 0.71, in the direction opposite to î − ĵ.

Q203.20 For any arbitrary motion in space, which of the following relations are true: (a) v_average = (1/2)(v(t₁) + v(t₂)), (b) v_average = [r(t₂) − r(t₁)]/(t₂ − t₁), (c) v(t) = v(0) + at, (d) r(t) = r(0) + v(0)t + (1/2)at², (e) a_average = [v(t₂) − v(t₁)]/(t₂ − t₁). (The 'average' stands for average of the quantity over the time interval t₁ to t₂.)

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(a) False in general. It holds only for constant acceleration.
(b) True. This is the definition of average velocity (displacement/time) and holds for any motion.
(c) False in general. It holds only for constant acceleration.
(d) False in general. It holds only for constant acceleration.
(e) True. This is the definition of average acceleration for any motion.

So, for arbitrary motion, only (b) and (e) are true.

Q213.21 Read each statement below carefully and state, with reasons and examples, if it is true or false: A scalar quantity is one that (a) is conserved in a process, (b) can never take negative values, (c) must be dimensionless, (d) does not vary from one point to another in space, (e) has the same value for observers with different orientations of axes.

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(a) False. Kinetic energy is a scalar but is not conserved in an inelastic collision.
(b) False. Temperature (in °C), electric charge and potential energy are scalars that can be negative.
(c) False. Mass, length, time and charge are scalars with dimensions.
(d) False. Temperature, density and gravitational potential are scalars that can vary from point to point.
(e) True. A scalar has no direction, so rotating the coordinate axes does not change it (the components of a vector change, a scalar does not).

Q223.22 An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?

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The printed question has no figure, so the usual reading is taken: the aircraft flies horizontally, and its two positions A and B are symmetric about the vertical through the observer O. Then triangle OAB is isosceles with apex angle 30° at O and the perpendicular height 3400 m.

AB = 2 × 3400 × tan 15° = 6800 × 0.268 = 1822 m.

Speed = AB/t = 1822/10.0 ≈ 182 m s⁻¹ (about 0.18 km s⁻¹, or 656 km h⁻¹).