How to Use This Page
Attempt every question before checking the solution. For vector problems, draw a diagram. For projectile and circular motion numericals, write the formula, substitute, compute, and give the answer with correct SI units.
Key Concepts (Quick Recap)
- Vector addition: Parallelogram law — R = √(A² + B² + 2AB cos θ), direction tan α = B sin θ/(A + B cos θ)
- Unit vector: Â = A/|A|
- Resolution: Aₓ = A cos θ, Aᵧ = A sin θ
- Dot product: A·B = AB cos θ = AₓBₓ + AᵧBᵧ (scalar)
- Cross product: |A×B| = AB sin θ; direction by right-hand rule
- Projectile motion:
- Horizontal: x = (u cos θ)t, vₓ = u cos θ
- Vertical: y = (u sin θ)t − ½gt², vᵧ = u sin θ − gt
- Max height: H = u² sin²θ / 2g
- Time of flight: T = 2u sin θ / g
- Range: R = u² sin 2θ / g
- Circular motion: a_c = v²/r = ω²r, v = ωr
Intext Questions
Page 63
Q1. State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, angular frequency, displacement, angular velocity.
Answer
| Quantity | Type |
|---|
| Volume | Scalar |
| Mass | Scalar |
| Speed | Scalar |
| Acceleration | Vector |
| Density | Scalar |
| Number of moles | Scalar |
| Angular frequency | Scalar |
| Displacement | Vector |
| Angular velocity | Vector |
Q2. Pick out the two scalar quantities in the list: force, angular momentum, work, current, linear momentum, electric field, average velocity.
Answer
Work and current are the two scalar quantities. Work has only magnitude (W = F·s = Fs cos θ), and current has only magnitude and direction but not vector direction (it does not obey vector laws of addition).
Page 67
Q3. Pick out the only vector quantity in the following list: temperature, pressure, impulse, time, power, total path length, energy, gravitational potential.
Answer
Impulse is the only vector quantity. Impulse = F × Δt, and since force is a vector, impulse is also a vector. All others are scalars.
Q4. A vector A has magnitude 4 and is directed north-east. Resolve A into two rectangular components.
Answer
A = 4, θ = 45° (north-east from east)
Aₓ = A cos 45° = 4 × (1/√2) = 4/√2 = 2√2 ≈ 2.83 (eastward) Aᵧ = A sin 45° = 4 × (1/√2) = 4/√2 = 2√2 ≈ 2.83 (northward)
Page 70
Q5. A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? Justify.
Answer
No. A quantity having both magnitude and direction is not necessarily a vector. To qualify as a vector, the quantity must also obey the laws of vector addition (commutative and associative). For example, electric current has magnitude and a sense of direction, but it does not obey vector addition laws — it is a scalar. Rotational angles have magnitude and direction but do not add like vectors (finite rotations don't commute). Only quantities that obey vector algebra are vectors.
Q6. A mosquito is sitting on the origin. It flies along the path r = t²î + 2tĵ. Find the velocity and acceleration of the mosquito at t = 3 s.
Answer
r = t²î + 2tĵ
Velocity: v = dr/dt = 2tî + 2ĵ
| At t = 3: v = 6î + 2ĵ, | v | = √(36 + 4) = √40 = 2√10 m s⁻¹ |
|---|
Acceleration: a = dv/dt = 2î
| a | = 2 m s⁻² (in the x-direction) |
|---|
Page 74
Q7. State whether the following are true or false: (a) A·(B×C) = (A×B)·C (b) A×(B×C) = (A·C)B − (A·B)C
(a)
AnswerTrue — This is the scalar triple product identity. A·(B×C) = (A×B)·C. Both give the volume of the parallelepiped.
(b)
AnswerTrue — This is the vector triple product identity (BAC-CAB rule): A×(B×C) = B(A·C) − C(A·B).
Chapter-End Exercise
Q4.1. State, for each of the following, whether it is a scalar or a vector: (a) 5 s (b) 30 m s⁻¹ towards north (c) 3 kg (d) 250 J (e) 0.005 m²
Answer
(a) 5 s — Scalar (time) (b) 30 m s⁻¹ towards north — Vector (velocity with direction) (c) 3 kg — Scalar (mass) (d) 250 J — Scalar (energy) (e) 0.005 m² — Scalar (area, no direction)
Q4.2. Pick out the only vector quantity in the following: (a) speed (b) weight (c) kinetic energy (d) density (e) charge
Answer
(b) Weight — Weight is a force (W = mg) and is directed downward. All others are scalars.
Q4.3. A cyclist starts from the centre O of a circular park of radius 1 km and moves along the perimeter with a speed of 10 m s⁻¹. After 2 minutes, he is at point P on the circumference. What is his (a) net displacement (b) average velocity?
Answer
Radius r = 1 km = 1000 m, v = 10 m s⁻¹, t = 2 min = 120 s
Distance covered = vt = 10 × 120 = 1200 m Circumference = 2πr = 2π × 1000 = 6283.2 m Angle subtended = 1200/1000 = 1.2 rad ≈ 68.75°
(a) Net displacement = chord length OP Chord = 2r sin(θ/2) = 2(1000) sin(0.6) = 2000 × 0.5646 = 1129.3 m (from O to P)
(b) Average velocity = displacement/time = 1129.3/120 = 9.41 m s⁻¹ (directed from O to P along the chord)
Q4.4. A particle moves in a circle of radius 10 cm with a constant speed of 20 m s⁻¹. Find the (a) angular speed (b) centripetal acceleration.
Answer
r = 10 cm = 0.1 m, v = 20 m s⁻¹
(a) Angular speed ω = v/r = 20/0.1 = 200 rad s⁻¹
(b) Centripetal acceleration a_c = v²/r = (20)²/0.1 = 4000 = 4 × 10³ m s⁻² (or v²/r = 40 m s⁻² × 100 = let me recalculate)
a_c = v²/r = (20)²/0.1 = 400/0.1 = 4000 m s⁻²
Q4.5. A particle starts from the origin at t = 0 s with a velocity of 10.0 ĵ m s⁻¹ 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 at that time? (b) What is the speed of the particle at that time?
Answer
u = 10 ĵ, a = 8î + 2ĵ
x-direction: uₓ = 0, aₓ = 8 x = uₓt + ½aₓt² = 0 + 4t² When x = 16: 4t² = 16 → t² = 4 → t = 2 s
y-direction: uᵧ = 10, aᵧ = 2 y = uᵧt + ½aᵧt² = 10(2) + ½(2)(4) = 20 + 4 = 24 m
(b) At t = 2 s: vₓ = uₓ + aₓt = 0 + 8(2) = 16 m s⁻¹ vᵧ = uᵧ + aᵧt = 10 + 2(2) = 14 m s⁻¹ Speed = √(16² + 14²) = √(256 + 196) = √452 = 21.26 m s⁻¹
Q4.6. Rain is falling vertically with a speed of 12 m s⁻¹. A woman rides a bicycle with a speed of 12 m s⁻¹ in east to west direction. What is the direction in which she should hold her umbrella?
Answer
v_rain = 12 m s⁻¹ (downward), v_woman = 12 m s⁻¹ (westward)
Relative velocity of rain w.r.t. woman: v_rm = v_rain − v_woman
In component form: Vertical component = 12 m s⁻¹ (downward) Horizontal component = 0 − (−12) = 12 m s⁻¹ (rain appears to come from east toward woman moving west)
Actually: v_rm_x = 0 − (−12) = 12 (rain appears coming from the front, i.e., east) v_rm_y = −12 (downward)
tan θ = horizontal/vertical = 12/12 = 1 θ = 45° from the vertical, toward the east
She should hold her umbrella at 45° from the vertical, inclined toward the direction of motion (east).
Q4.7. A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 m s⁻¹. Neglecting air resistance, find the time taken by the stone to reach the ground and the speed with which it hits the ground. (g = 9.8 m s⁻²)
Answer
H = 490 m, uₓ = 15 m s⁻¹, uᵧ = 0
Time of flight: H = ½gt² → 490 = ½(9.8)t² → t² = 100 → t = 10 s
At t = 10 s: vₓ = uₓ = 15 m s⁻¹ (no horizontal acceleration) vᵧ = gt = 9.8 × 10 = 98 m s⁻¹
Speed = √(15² + 98²) = √(225 + 9604) = √9829 = 99.1 m s⁻¹
Q4.8. A cricket ball is thrown at a speed of 28 m s⁻¹ in a direction 30° above the horizontal. Calculate (a) the maximum height (b) the time taken to return to the same level (c) the distance from the thrower to the point where the ball returns to the same level.
Answer
u = 28 m s⁻¹, θ = 30°
(a) H = u² sin²θ / 2g = (28)² × (0.5)² / (2 × 9.8) = 784 × 0.25 / 19.6 = 196/19.6 = 10 m
(b) T = 2u sin θ / g = 2(28)(0.5)/9.8 = 28/9.8 = 2.86 s
(c) R = u² sin 2θ / g = (28)² × sin 60° / 9.8 = 784 × 0.866/9.8 = 678.9/9.8 = 69.3 m
Q4.9. An aircraft executes a horizontal loop of radius 1 km with a steady speed of 900 km h⁻¹. Compare its centripetal acceleration with the acceleration due to gravity.
Answer
r = 1 km = 1000 m, v = 900 km h⁻¹ = 250 m s⁻¹
a_c = v²/r = (250)²/1000 = 62500/1000 = 62.5 m s⁻²
g = 9.8 m s⁻²
Ratio = a_c/g = 62.5/9.8 = 6.38 (centripetal acceleration is about 6.4 times g)
Q4.10. A bullet fired at an angle of 30° with the horizontal hits the ground 3 km away. By adjusting its angle of projection, can one hope to hit a target 5 km away? Assume the initial speed to be fixed and neglect air resistance.
Answer
R = u² sin 2θ / g. At θ = 30°, R = 3000 m: 3000 = u² sin 60° / g = u²(0.866)/g u²/g = 3000/0.866 = 3464
Maximum range (at θ = 45°): R_max = u²/g = 3464 m = 3.464 km
Since R_max = 3.464 km < 5 km, the bullet cannot hit a target 5 km away regardless of the angle of projection.
Q4.11. A fighter plane flying with a constant speed of 150 m s⁻¹ in a vertical circle of radius 1000 m. Find the (a) centripetal acceleration and (b) the force acting on the pilot of mass 80 kg at the highest point.
Answer
v = 150 m s⁻¹, r = 1000 m
(a) a_c = v²/r = (150)²/1000 = 22500/1000 = 22.5 m s⁻²
(b) At the highest point, both gravity and centripetal force act downward: N + mg = mv²/r (where N is the normal reaction from seat) N = mv²/r − mg = 80(22.5) − 80(9.8) = 1800 − 784 = 1016 N (downward from seat on pilot)
Force on pilot = N + mg = 80 × 22.5 = 1800 N (toward the centre)
Q4.12. Two vectors A and B have equal magnitude of 10 units each. If A + B = 5√2, find the angle between A and B.
Answer
(5√2)² = 10² + 10² + 2(10)(10) cos θ 50 = 100 + 100 + 200 cos θ 50 = 200 + 200 cos θ 200 cos θ = 50 − 200 = −150 cos θ = −150/200 = −3/4 θ = cos⁻¹(−3/4) ≈ 138.6°
| A + B | ² = A² + B² + 2AB cos θ |
|---|
Q4.13. For any vector A, show that A·A = A². What does A·A represent geometrically?
Answer
A·A = AₓAₓ + AᵧAᵧ + A_zA_z = Aₓ² + Aᵧ² + A_z² = A²
| Geometrically, A·A = | A | ² cos 0° = A². It represents the square of the magnitude of the vector. |
|---|
| **Q4.14. If | A + B | = | A − B | , what is the angle between A and B?** |
|---|
| A + B | ² = A² + B² + 2AB cos θ |
|---|
| A − B | ² = A² + B² − 2AB cos θ |
A² + B² + 2AB cos θ = A² + B² − 2AB cos θ 4AB cos θ = 0 cos θ = 0 θ = 90°
The vectors are perpendicular to each other.
Internal Links
- [Chapter 4 Guide — Motion in a Plane](/school/cbse/class/11/physics/cbse-motion-in-a-plane)
- [All Class 11 Physics Solutions](/school/cbse/class/11/physics/solutions)
- [Class 11 Physics Subject Guide](/school/cbse/class/11/physics)
