Exercises
1. State if each physical quantity is scalar or vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
Answer
Scalar: volume, mass, speed, density, number of moles, angular frequency
Vector: acceleration, velocity, displacement, angular velocity
2. Pick out the two scalar quantities from: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
Answer
The two scalar quantities are: work and current.
All others are vectors. (Current is scalar because it has no direction in space, only positive/negative sign along a conductor.)
3. Pick out the only vector quantity from: Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.
Answer
The only vector quantity is: impulse (impulse = change in momentum = force × time, which is a vector).
All others are scalars.
Q3.4. State whether the following operations are meaningful
(a) adding two scalars (b) adding a scalar to a vector of same dimensions (c) multiplying a vector by a scalar (d) multiplying two scalars (e) adding two vectors (f) adding a component of a vector to the same vector
Ans.
(a) Meaningful — two scalars of the same type can always be added (e.g., mass + mass).
(b) Not meaningful — a scalar and a vector are fundamentally different objects; their sum is not defined even if dimensions match.
(c) Meaningful — a vector multiplied by a scalar gives another vector (e.g., force = mass × acceleration).
(d) Meaningful — product of two scalars is a scalar (e.g., density × volume = mass).
(e) Meaningful — vectors of the same type and dimension can be added using the triangle/parallelogram law.
(f) Not meaningful — a component of a vector is a scalar magnitude; adding a scalar to a vector is not defined.
Q3.5. State true or false
(a) Magnitude of a vector is always a scalar (b) Each component of a vector is always a scalar (c) Total path length always equals magnitude of displacement (d) Average speed ≥ magnitude of average velocity (e) Three vectors not lying in a plane can never add to give null vector
Ans.
(a) True — magnitude is a number (with units), which is a scalar.
(b) False — components of a vector are themselves vectors (e.g., Aₓ = Aₓ î is a vector). The scalar value of the component is called the scalar component.
(c) False — total path length = magnitude of displacement only when the particle moves in a straight line without reversing. In general, path length ≥ |displacement|.
(d) True — average speed = (total path length)/t ≥ |displacement|/t = |average velocity|.
(e) True — for three vectors to add to zero, they must form a closed triangle. Three non-coplanar vectors cannot form a closed triangle (which is a planar figure), so they cannot sum to zero.
6. Establish vector inequalities: (a) |a+b| ≤ |a|+|b| (b) |a+b| ≥ ||a|−|b|| (c) |a−b| ≤ |a|+|b| (d) |a−b| ≥ ||a|−|b||. When does equality hold?
Answer
(a) By the triangle law, |a+b| ≤ |a| + |b| (the sum of two sides of a triangle ≥ the third side).
Equality holds when a and b are in the same direction (θ = 0°).
(b) From (a), |b| = |(a+b) − a| ≤ |a+b| + |a|
⇒ |a+b| ≥ |b| − |a|. Similarly |a+b| ≥ |a| − |b|. Hence |a+b| ≥ ||a| − |b||.
Equality holds when a and b are in opposite directions.
(c) |a−b| = |a + (−b)| ≤ |a| + |−b| = |a| + |b| (from (a), with b replaced by −b).
Equality holds when a and b are in opposite directions.
(d) |a−b| ≥ ||a| − |b|| (from (b), with b replaced by −b).
Equality holds when a and b are in the same direction.
Q3.7. Given a + b + c + d = 0, which statements are correct
(a) each must be null (b) |a+c| = |b+d| (c) |a| never greater than sum of magnitudes of b, c, d (d) b+c lies in plane of a and d if not collinear
Ans.
(a) Incorrect — individual vectors need not be zero; they can cancel each other. Example: a = î, b = −î + ĵ, c = −ĵ, d = 0.
(b) Correct — a + b + c + d = 0 → a + c = −(b + d) → |a + c| = |b + d|.
(c) Correct — from a + b + c + d = 0 → a = −(b + c + d) → |a| = |b + c + d| ≤ |b| + |c| + |d| (triangle inequality).
(d) Correct — a + d = −(b + c). If a and d are not collinear, they define a plane. The sum (a + d) lies in this plane. Hence −(b + c) lies in this plane, so (b + c) lies in the plane of a and d. If collinear, all vectors lie on the same line.
8. Three girls start from P and reach Q (diametrically opposite) on a circular ground of radius 200 m. What is the magnitude of displacement? For which girl is it equal to path length?
Answer
P and Q are diametrically opposite on a circle of radius 200 m.
Displacement = straight-line distance from P to Q = diameter = 2 × 200 = 400 m for all three girls (displacement depends only on initial and final positions).
Path length = displacement only when the girl travels in a straight line (the diameter). If Girl 3 travels directly along the diameter (straight path), her path length = 400 m = displacement.
Girls 1 and 2 travel along curved paths (arcs), so their path lengths > 400 m.
9. A cyclist starts from centre O of circular park (radius 1 km), goes to edge P, cycles along circumference to Q, returns via QO. Round trip = 10 min. Find (a) net displacement (b) average velocity (c) average speed.
Answer
The cyclist returns to centre O at the end. Therefore:
(a) Net displacement = 0 (starts and ends at O)
⇒ 0 km
(b) Average velocity = displacement/time = 0/t = 0 km h⁻¹
(c) For average speed, find total path length:
OP = 1 km (radius); arc PQ = (1/4) × circumference = (1/4) × 2π × 1 = π/2 km (assuming a quarter circle); QO = 1 km
Total path = 1 + π/2 + 1 = 2 + π/2 ≈ 2 + 1.571 = 3.571 km
Time = 10 min = 1/6 h
Average speed = 3.571/(1/6) ≈ 21.4 km h⁻¹
10. A motorist turns left by 60° after every 500 m. Find displacement at 3rd, 6th, 8th turns.
Answer
Each turn is left by 60°. The motorist follows a regular hexagonal path (since 60° × 6 = 360°).
At 3rd turn: Three sides of the hexagon. Net displacement = 500√3 m ≈ 866 m. Total path = 3 × 500 = 1500 m. Ratio = 1500/866 ≈ 1.73.
At 6th turn: Six sides complete a hexagon — back to start. Net displacement = 0 m. Total path = 6 × 500 = 3000 m.
At 8th turn: 8 sides = 1 complete hexagon + 2 more sides. Net displacement = displacement after 2 sides of hexagon = 500 m (magnitude, as the angle after 2 sides gives a net displacement = 500 m). Total path = 8 × 500 = 4000 m. Net displacement = 500 m. Ratio = 4000/500 = 8.
11. A hotel is 10 km from station. Cabman takes a path 23 km long in 28 min. Find (a) average speed (b) magnitude of average velocity. Are they equal?
Answer
Given: path length = 23 km; displacement = 10 km (straight-line distance); time = 28 min = 28/60 h
(a) Average speed = path length/time = 23/(28/60) = 23 × 60/28 = 49.3 km h⁻¹
(b) |Average velocity| = |displacement|/time = 10/(28/60) = 10 × 60/28 = 21.4 km h⁻¹
They are not equal, because path length ≠ displacement.
12. Ceiling of a hall is 25 m high. Find maximum horizontal distance a ball thrown at 40 m s⁻¹ can go without hitting the ceiling.
Answer
The maximum height of the projectile = ceiling height = 25 m
At maximum height, v_y = 0:
v_y² = u² sin²θ − 2g × H_max
0 = (40)² sin²θ − 2 × 10 × 25 (using g = 10 m/s²)
1600 sin²θ = 500
⇒ sin²θ = 500/1600 = 5/16
⇒ sinθ = √(5/16) = √5/4
cosθ = √(1 − 5/16) = √(11/16) = √11/4
Horizontal range = (u² sin2θ)/g = (u²/g) × 2 sinθ cosθ = (1600/10) × 2 × (√5/4) × (√11/4)
= 160 × 2 × √55/16 = 160 × √55/8 = 20√55 ≈ 20 × 7.416 ≈ 148.3 m
13. A cricketer can throw a ball to maximum horizontal distance of 100 m. How high can the same ball be thrown?
Answer
Maximum range R = u²/g = 100 m
⇒ u² = 100g = 100 × 10 = 1000 m² s⁻²
For maximum height, ball is thrown vertically (θ = 90°):
H_max = u²/(2g) = 1000/(2 × 10) = 50 m
14. A stone on a string of length 80 cm is whirled horizontally. It makes 14 revolutions in 25 s. Find magnitude and direction of acceleration.
Answer
Given: r = 80 cm = 0.80 m; frequency f = 14/25 = 0.56 Hz
Angular velocity ω = 2πf = 2π × 0.56 = 3.52 rad s⁻¹
Centripetal acceleration = ω²r = (3.52)² × 0.80 = 12.39 × 0.80 ≈ 9.91 m s⁻² ≈ 9.9 m s⁻²
Direction: directed toward the centre of the circle (centripetal — radially inward) at all times.
15. An aircraft executes a horizontal loop of radius 1.00 km at 900 km h⁻¹. Compare centripetal acceleration with g.
Answer
v = 900 km h⁻¹ = 900/3.6 = 250 m s⁻¹; r = 1.00 km = 1000 m
Centripetal acceleration a_c = v²/r = (250)²/1000 = 62500/1000 = 62.5 m s⁻²
a_c/g = 62.5/9.8 ≈ 6.38
The centripetal acceleration is about 6.38 times the acceleration due to gravity.
Q3.16. State true or false
(a) Net acceleration in circular motion is always toward centre
(b) Velocity vector is always along tangent at that point
(c) Acceleration vector in uniform circular motion averaged over one cycle is a null vector
Ans.
(a) False — in uniform circular motion, net acceleration is centripetal (toward centre). But in non-uniform circular motion, there is also a tangential component, so net acceleration is not purely radial.
(b) True — at any point, the instantaneous velocity is tangent to the path (by definition of velocity as lim Δr/Δt which gives the direction of the path).
(c) True — in uniform circular motion, the centripetal acceleration always points toward the center. Over one complete cycle, the acceleration vectors point in all directions with equal magnitude, and their vector average = 0 (null vector).
17. Position: r = (3.0t î − 2.0t² ĵ + 4.0 k̂) m. Find (a) v and a (b) magnitude and direction of v at t = 2.0 s.
Answer
r = 3.0t î − 2.0t² ĵ + 4.0 k̂
(a) v = dr/dt = 3.0 î − 4.0t ĵ m s⁻¹
a = dv/dt = −4.0 ĵ m s⁻²
(b) At t = 2.0 s: v = 3.0 î − 4.0(2.0) ĵ = 3.0 î − 8.0 ĵ m s⁻¹
Magnitude: |v| = √(3.0² + 8.0²) = √(9 + 64) = √73 ≈ 8.54 m s⁻¹
Direction: θ = arctan(8.0/3.0) = arctan(2.67) ≈ 69.4° below the x-axis (in the −ĵ direction)
18. Particle starts from origin with velocity 10.0 ĵ m s⁻¹ and acceleration (8.0 î + 2.0 ĵ) m s⁻². (a) When is x = 16 m? y at that time? (b) Speed at that time?
Answer
Initial velocity: u = 10.0 ĵ m s⁻¹; acceleration: a = 8.0 î + 2.0 ĵ m s⁻²
Position: x = uₓt + ½aₓt² = 0 + ½(8.0)t² = 4.0t²
y = u_y t + ½a_y t² = 10.0t + ½(2.0)t² = 10.0t + t²
(a) x = 16 m
⇒ 4.0t² = 16
⇒ t² = 4
⇒ t = 2 s
y at t = 2 s: y = 10(2) + (2)² = 20 + 4 = 24 m
(b) vₓ = aₓt = 8.0 × 2 = 16 m s⁻¹
v_y = u_y + a_y t = 10.0 + 2.0 × 2 = 14 m s⁻¹
Speed = √(vₓ² + v_y²) = √(256 + 196) = √452 ≈ 21.3 m s⁻¹
19. î and ĵ are unit vectors along x and y. Find magnitude and direction of (î + ĵ) and (î − ĵ). Find components of A = 2î + 3ĵ along these directions.
Answer
|î + ĵ| = √(1² + 1²) = √2; direction: 45° above x-axis (northeast)
|î − ĵ| = √(1² + (−1)²) = √2; direction: 45° below x-axis (southeast)
Unit vector along (î + ĵ): n̂₁ = (î + ĵ)/√2
Unit vector along (î − ĵ): n̂₂ = (î − ĵ)/√2
Component of A = (2î + 3ĵ) along n̂₁ = A · n̂₁ = (2 + 3)/√2 = 5/√2 = 5/√2 ≈ 3.54
Component along n̂₂ = A · n̂₂ = (2 − 3)/√2 = −1/√2 = −1/√2 ≈ −0.707
Q3.20. For arbitrary motion, which relations are true
(a) v_avg = ½[v(t₁)+v(t₂)] (b) v_avg = [r(t₂)−r(t₁)]/(t₂−t₁) (c) v(t) = v(0) + at (d) r(t) = r(0) + v(0)t + ½at² (e) a_avg = [v(t₂)−v(t₁)]/(t₂−t₁)
Ans.
(a) False — this holds only for constant acceleration (uniformly accelerated motion).
(b) True — this is the definition of average velocity: v_avg = Δr/Δt. Always true.
(c) False — this holds only for constant acceleration a.
(d) False — this holds only for constant acceleration a.
(e) True — this is the definition of average acceleration: a_avg = Δv/Δt. Always true.
Q3.21. A scalar quantity is one that — state true/false
(a) is conserved in a process (b) can never take negative values (c) must be dimensionless (d) does not vary from point to point (e) has same value for observers with different orientations
Ans.
(a) False — kinetic energy is scalar but not always conserved.
(b) False — temperature, charge, potential can be negative.
(c) False — mass, energy are scalars with dimensions.
(d) False — temperature is scalar but varies from point to point (scalar field).
(e) True — this is the defining property of a scalar: its value does not depend on the orientation of the coordinate axes.
22. An aircraft is at height 3400 m. Angle subtended at a ground point by two aircraft positions 10 s apart is 30°. Find the speed of the aircraft.
Answer
Given: height h = 3400 m; angle θ = 30°; time Δt = 10 s
The aircraft travels horizontally. Let the observation point on the ground be directly below the midpoint of the path.
Distance from observation point to each position: d = h/cos(θ/2) ≈ h (for small angles, but angle is 30°)
Actually, if angle subtended = 30°, and h = 3400 m:
Distance traveled by aircraft = d = 2h × tan(θ/2) = 2 × 3400 × tan(15°)
tan(15°) ≈ 0.2679
d = 2 × 3400 × 0.2679 ≈ 1822 m
Speed = d/Δt = 1822/10 ≈ 182 m s⁻¹ ≈ 655 km h⁻¹
Important Extra Questions
Q. State and prove the parallelogram law of vector addition.
Ans. If two vectors A and B are represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant R is represented by the diagonal of the parallelogram drawn from the same vertex.
Proof: Let A = OA and B = OB. Complete parallelogram OACB. OC = diagonal = A + B.
|R| = |A + B| = √(A² + B² + 2AB cosθ) where θ is the angle between A and B.
Direction: tanα = (B sinθ)/(A + B cosθ), where α is angle R makes with A.
Q. Derive the equation for range and maximum height of a projectile. At what angle is range maximum?
Answer
For a projectile with initial speed u at angle θ:
Horizontal: x = u cosθ × t
⇒ t = x/(u cosθ)
Vertical: y = u sinθ × t − ½gt²
Substituting t: y = x tanθ − gx²/(2u² cos²θ) — equation of parabola
Time of flight: T = 2u sinθ/g
Maximum height: H = u² sin²θ/(2g)
Range: R = u² sin2θ/g
Range is maximum when sin2θ = 1
⇒ 2θ = 90°
⇒ θ = 45°; R_max = u²/g
Q. Derive an expression for centripetal acceleration in uniform circular motion.
Ans.
Consider a particle moving in a circle of radius r with speed v. At time t, position vector makes angle θ with x-axis.
Position: r = r cosθ î + r sinθ ĵ
Velocity: v = dr/dt = r(−sinθ)(dθ/dt) î + r cosθ (dθ/dt) ĵ = ω × r(−sinθ î + cosθ ĵ)
|v| = ωr → ω = v/r
Acceleration: a = dv/dt = −ω²r cosθ î − ω²r sinθ ĵ = −ω²r
|a| = ω²r = v²/r
Direction is −r̂ (radially inward = toward centre)
⇒ centripetal acceleration a_c = v²/r, directed toward centre
Q. A ball is projected with velocity 20 m s⁻¹ at 30° with horizontal. Find time of flight, maximum height, and horizontal range. (g = 10 m s⁻²)
Ans.
u = 20 m s⁻¹, θ = 30°
u sinθ = 20 × 0.5 = 10 m s⁻¹; u cosθ = 20 × (√3/2) = 10√3 m s⁻¹
Time of flight: T = 2u sinθ/g = 2 × 10/10 = 2 s
Maximum height: H = u² sin²θ/(2g) = 400 × 0.25/(20) = 5 m
Range: R = u² sin2θ/g = 400 × sin60°/10 = 400 × (√3/2)/10 = 20√3 ≈ 34.6 m
Q. Two vectors A and B have magnitudes 3 and 4 units respectively and are perpendicular. Find |A + B| and the angle it makes with A.
Ans.
|A| = 3, |B| = 4, angle between them θ = 90°
|A + B| = √(A² + B² + 2AB cosθ) = √(9 + 16 + 0) = √25 = 5 units
Angle with A: tanα = (B sinθ)/(A + B cosθ) = (4 × 1)/(3 + 0) = 4/3
α = arctan(4/3) ≈ 53.1° with vector A