Chapter 04: Motion in a Plane
Physics XI : Complete NCERT Exercise Solutions
Practice these examination-oriented questions to master vectors, projectile motion, circular motion, and relative velocity. Each question includes a detailed solution to enhance your understanding.
Question 4.1
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.
Answer & Explanation:
Scalars: Volume, mass, speed, density, number of moles, angular frequency.
Vectors: Acceleration, velocity, displacement, angular velocity.
Explanation: Scalars have only magnitude; vectors have both magnitude and direction.
Vectors: Acceleration, velocity, displacement, angular velocity.
Explanation: Scalars have only magnitude; vectors have both magnitude and direction.
Question 4.2
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.
Answer & Explanation:
Scalar quantities: Work, Current.
Explanation: Work is energy transfer (scalar). Current is charge flow per time (scalar). All others are vectors.
Explanation: Work is energy transfer (scalar). Current is charge flow per time (scalar). All others are vectors.
Question 4.3
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.
Answer & Explanation:
Vector quantity: Impulse.
Explanation: Impulse = Force × time, and force is a vector. All others are scalars.
Explanation: Impulse = Force × time, and force is a vector. All others are scalars.
Question 4.4
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.
(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.
Answer & Explanation:
(a) Meaningful only if scalars have same units (e.g., mass + mass).
(b) Not meaningful – cannot add scalar to vector.
(c) Meaningful – scalar multiplication changes magnitude of vector.
(d) Meaningful – e.g., mass × density.
(e) Meaningful – vector addition follows triangle/parallelogram law.
(f) Meaningful – component is a vector; can be added to original vector.
(b) Not meaningful – cannot add scalar to vector.
(c) Meaningful – scalar multiplication changes magnitude of vector.
(d) Meaningful – e.g., mass × density.
(e) Meaningful – vector addition follows triangle/parallelogram law.
(f) Meaningful – component is a vector; can be added to original vector.
Question 4.5
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 is either greater or equal to the magnitude of average velocity over the same interval.
(e) Three vectors not lying in a plane can never add up to give a null vector.
(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 is either greater or equal to the magnitude of average velocity over the same interval.
(e) Three vectors not lying in a plane can never add up to give a null vector.
Answer & Explanation:
(a) True – magnitude is a scalar.
(b) False – components are scalars only with respect to axes; but \(A_x \hat{i}\) is a vector.
(c) False – path length ≥ displacement magnitude.
(d) True – speed ≥ magnitude of velocity.
(e) False – three non-coplanar vectors can sum to zero if they form a closed triangle in 3D.
(b) False – components are scalars only with respect to axes; but \(A_x \hat{i}\) is a vector.
(c) False – path length ≥ displacement magnitude.
(d) True – speed ≥ magnitude of velocity.
(e) False – three non-coplanar vectors can sum to zero if they form a closed triangle in 3D.
Question 4.6
Establish the following vector inequalities geometrically or otherwise:
(a) \(|a+b| \leq |a| + |b|\)
(b) \(|a+b| \geq |a| - |b|\)
(c) \(|a-b| \leq |a| + |b|\)
(d) \(|a-b| \geq ||a| - |b||\)
When does the equality sign apply?
(a) \(|a+b| \leq |a| + |b|\)
(b) \(|a+b| \geq |a| - |b|\)
(c) \(|a-b| \leq |a| + |b|\)
(d) \(|a-b| \geq ||a| - |b||\)
When does the equality sign apply?
Answer & Explanation:
Equality holds when:
(a) Vectors are in same direction.
(b) Vectors are in opposite direction.
(c) Vectors are in opposite direction.
(d) Vectors are in same direction.
These are triangle inequalities derived from geometry of vector addition.
(a) Vectors are in same direction.
(b) Vectors are in opposite direction.
(c) Vectors are in opposite direction.
(d) Vectors are in same direction.
These are triangle inequalities derived from geometry of vector addition.
Question 4.7
Given \(a + b + c + d = 0\), which of the following statements are correct?
(a) \(a, b, c, d\) 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 magnitudes of \(b, c, d\).
(d) \(b+c\) must lie in the plane of \(a\) and \(d\) if \(a\) and \(d\) are not collinear.
(a) \(a, b, c, d\) 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 magnitudes of \(b, c, d\).
(d) \(b+c\) must lie in the plane of \(a\) and \(d\) if \(a\) and \(d\) are not collinear.
Answer & Explanation:
(a) False – they can be non-zero but form a closed polygon.
(b) True – from \(a+c = -(b+d)\), magnitudes equal.
(c) True – follows from triangle inequality.
(d) True – \(b+c = -(a+d)\), so lies in same plane.
(b) True – from \(a+c = -(b+d)\), magnitudes equal.
(c) True – follows from triangle inequality.
(d) True – \(b+c = -(a+d)\), so lies in same plane.
Question 4.8
Three girls skating on a circular ice ground of radius 200 m start from point P and reach point Q diametrically opposite following different paths. What is the magnitude of displacement vector for each? For which girl is this equal to the actual path length?
Answer & Explanation:
Displacement magnitude: Diameter = 2 × 200 = 400 m for all.
Path length = displacement only for the girl moving along the diameter (straight line).
Path length = displacement only for the girl moving along the diameter (straight line).
Question 4.9
A cyclist starts from centre O of a circular park of radius 1 km, reaches edge P, cycles along circumference, returns to centre via QO. Round trip takes 10 min. Find:
(a) net displacement,
(b) average velocity,
(c) average speed.
(a) net displacement,
(b) average velocity,
(c) average speed.
Answer & Explanation:
(a) Net displacement = 0 (returns to start).
(b) Average velocity = 0 (displacement/time).
(c) Average speed = total distance/time = (OP + arc PQ + QO) / (10/60) hr.
(b) Average velocity = 0 (displacement/time).
(c) Average speed = total distance/time = (OP + arc PQ + QO) / (10/60) hr.
Question 4.10
On an open ground, a motorist turns left by 60° after every 500 m. Starting from a given turn, specify displacement at third, sixth, and eighth turn. Compare displacement magnitude with total path length.
Answer & Explanation:
Path is a hexagon (each side 500 m, internal angle 60°).
• After 3 turns: displacement = 500√3 m at 30°.
• After 6 turns: back to start, displacement = 0.
• After 8 turns: same as after 2 turns (pattern repeats every 6).
Path length > displacement unless net displacement is zero.
• After 3 turns: displacement = 500√3 m at 30°.
• After 6 turns: back to start, displacement = 0.
• After 8 turns: same as after 2 turns (pattern repeats every 6).
Path length > displacement unless net displacement is zero.
Question 4.11
A passenger is taken 23 km instead of 10 km straight road in 28 min. Find (a) average speed, (b) magnitude of average velocity. Are they equal?
Answer & Explanation:
(a) Average speed = total distance / time = 23 km / (28/60) hr ≈ 49.3 km/h.
(b) Average velocity magnitude = displacement / time = 10 km / (28/60) hr ≈ 21.4 km/h.
They are not equal because path is circuitous.
(b) Average velocity magnitude = displacement / time = 10 km / (28/60) hr ≈ 21.4 km/h.
They are not equal because path is circuitous.
Question 4.12
Rain falls vertically at 30 m/s. Woman rides bicycle 10 m/s north to south. What direction should she hold umbrella?
Answer & Explanation:
Relative velocity of rain w.r.t. cyclist: \( \vec{v}_{rc} = \vec{v}_r - \vec{v}_c \).
Direction θ from vertical: \( \tan θ = v_c / v_r = 10/30 ≈ 0.333 \) → \( θ ≈ 18.4° \) towards south.
Direction θ from vertical: \( \tan θ = v_c / v_r = 10/30 ≈ 0.333 \) → \( θ ≈ 18.4° \) towards south.
Question 4.13
Swimmer speed in still water = 4 km/h. River flows at 3 km/h, width = 1 km. He swims perpendicular to current. Find time to cross and downstream drift.
Answer & Explanation:
Time to cross = width / swimmer’s speed across = 1 km / 4 km/h = 0.25 hr = 15 min.
Downstream drift = river speed × time = 3 km/h × 0.25 hr = 0.75 km.
Downstream drift = river speed × time = 3 km/h × 0.25 hr = 0.75 km.
Question 4.14
Wind blows at 72 km/h N-E. Boat moves north at 51 km/h. What is direction of flag on mast?
Answer & Explanation:
Flag direction shows relative wind w.r.t. boat.
\( \vec{v}_{wind/boat} = \vec{v}_w - \vec{v}_b \).
Solve vector subtraction; direction will be east of north.
\( \vec{v}_{wind/boat} = \vec{v}_w - \vec{v}_b \).
Solve vector subtraction; direction will be east of north.
Question 4.15
Ceiling height = 25 m. Ball thrown at 40 m/s. Find max horizontal range without hitting ceiling.
Answer & Explanation:
For max height \( h_m = 25 \) m: \( 25 = \frac{(v_0 \sin θ)^2}{2g} \).
Also range \( R = \frac{v_0^2 \sin 2θ}{g} \).
Solve: \( \sin θ = \sqrt{2gh}/v_0 \), then find R. Approx R ≈ 150 m.
Also range \( R = \frac{v_0^2 \sin 2θ}{g} \).
Solve: \( \sin θ = \sqrt{2gh}/v_0 \), then find R. Approx R ≈ 150 m.
Question 4.16
Cricketer throws ball max horizontal distance 100 m. How high can he throw same ball vertically?
Answer & Explanation:
Max range \( R_{max} = v_0^2 / g = 100 \) m → \( v_0^2 = 100g \).
Max height if thrown vertically: \( h_{max} = v_0^2 / (2g) = (100g)/(2g) = 50 \) m.
Max height if thrown vertically: \( h_{max} = v_0^2 / (2g) = (100g)/(2g) = 50 \) m.
Question 4.17
Stone in horizontal circle, radius 80 cm, 14 rev in 25 s. Find magnitude and direction of acceleration.
Answer & Explanation:
Speed \( v = 2πr × (14/25) \).
Centripetal acceleration \( a_c = v^2 / r \) directed towards centre.
Calculation gives ~ 9.9 m/s².
Centripetal acceleration \( a_c = v^2 / r \) directed towards centre.
Calculation gives ~ 9.9 m/s².
Question 4.18
Aircraft loop radius 1 km, speed 900 km/h. Compare centripetal acceleration with g.
Answer & Explanation:
\( v = 900 \, \text{km/h} = 250 \, \text{m/s} \), \( r = 1000 \) m.
\( a_c = v^2 / r = 62.5 \, \text{m/s}^2 \), \( g = 9.8 \, \text{m/s}^2 \).
Ratio ≈ 6.38 times g.
\( a_c = v^2 / r = 62.5 \, \text{m/s}^2 \), \( g = 9.8 \, \text{m/s}^2 \).
Ratio ≈ 6.38 times g.
Question 4.19
State true/false with reasons:
(a) Net acceleration in circular motion always towards centre.
(b) Velocity vector always tangent to path.
(c) Average acceleration over one cycle in uniform circular motion is null.
(a) Net acceleration in circular motion always towards centre.
(b) Velocity vector always tangent to path.
(c) Average acceleration over one cycle in uniform circular motion is null.
Answer & Explanation:
(a) False – only if speed is constant; otherwise tangential component exists.
(b) True – velocity is always tangential.
(c) True – initial and final velocities same in UCM, so Δv = 0.
(b) True – velocity is always tangential.
(c) True – initial and final velocities same in UCM, so Δv = 0.
Question 4.20
Position \( \vec{r} = 3.0t \hat{i} - 2.0t^2 \hat{j} + 4.0 \hat{k} \) m. Find (a) \( \vec{v} \) and \( \vec{a} \), (b) magnitude and direction of velocity at t = 2 s.
Answer & Explanation:
(a) \( \vec{v} = d\vec{r}/dt = 3.0 \hat{i} - 4.0t \hat{j} \), \( \vec{a} = -4.0 \hat{j} \) m/s².
(b) At t=2: \( \vec{v} = 3\hat{i} - 8\hat{j} \), magnitude = √(9+64) ≈ 8.54 m/s, direction θ = tan⁻¹(-8/3) below x-axis.
(b) At t=2: \( \vec{v} = 3\hat{i} - 8\hat{j} \), magnitude = √(9+64) ≈ 8.54 m/s, direction θ = tan⁻¹(-8/3) below x-axis.
Question 4.21
Particle starts at origin with \( \vec{v} = 10.0 \hat{j} \) m/s, acceleration \( \vec{a} = 8.0\hat{i} + 2.0\hat{j} \) m/s². (a) When x-coordinate = 16 m, find time and y-coordinate. (b) Speed at that time.
Answer & Explanation:
(a) \( x = 0 + 0×t + ½ a_x t^2 \) → \( 16 = ½ × 8 × t^2 \) → t = 2 s.
\( y = v_{0y}t + ½ a_y t^2 = 10×2 + ½×2×4 = 24 \) m.
(b) \( v_x = a_x t = 16 \) m/s, \( v_y = v_{0y} + a_y t = 14 \) m/s.
Speed = √(16² + 14²) ≈ 21.26 m/s.
\( y = v_{0y}t + ½ a_y t^2 = 10×2 + ½×2×4 = 24 \) m.
(b) \( v_x = a_x t = 16 \) m/s, \( v_y = v_{0y} + a_y t = 14 \) m/s.
Speed = √(16² + 14²) ≈ 21.26 m/s.
Question 4.22
Magnitude and direction of \( \hat{i}+\hat{j} \) and \( \hat{i}-\hat{j} \)? Components of \( \vec{A} = 2\hat{i}+3\hat{j} \) along these directions.
Answer & Explanation:
\( \hat{i}+\hat{j} \): magnitude √2, direction 45° to x-axis.
\( \hat{i}-\hat{j} \): magnitude √2, direction -45°.
Component along \( \hat{i}+\hat{j} \): \( \frac{\vec{A}·(\hat{i}+\hat{j})}{|\hat{i}+\hat{j}|} = (2+3)/√2 = 5/√2 \).
Similarly for \( \hat{i}-\hat{j} \): (2-3)/√2 = -1/√2.
\( \hat{i}-\hat{j} \): magnitude √2, direction -45°.
Component along \( \hat{i}+\hat{j} \): \( \frac{\vec{A}·(\hat{i}+\hat{j})}{|\hat{i}+\hat{j}|} = (2+3)/√2 = 5/√2 \).
Similarly for \( \hat{i}-\hat{j} \): (2-3)/√2 = -1/√2.
Question 4.23
For arbitrary motion in space, which relations are true?
(a) \( \vec{v}_{avg} = ½ [\vec{v}(t_1)+\vec{v}(t_2)] \)
(b) \( \vec{v}_{avg} = [\vec{r}(t_2)-\vec{r}(t_1)]/(t_2-t_1) \)
(c) \( \vec{v}(t) = \vec{v}(0) + \vec{a}t \)
(d) \( \vec{r}(t) = \vec{r}(0) + \vec{v}(0)t + ½ \vec{a}t^2 \)
(e) \( \vec{a}_{avg} = [\vec{v}(t_2)-\vec{v}(t_1)]/(t_2-t_1) \)
(a) \( \vec{v}_{avg} = ½ [\vec{v}(t_1)+\vec{v}(t_2)] \)
(b) \( \vec{v}_{avg} = [\vec{r}(t_2)-\vec{r}(t_1)]/(t_2-t_1) \)
(c) \( \vec{v}(t) = \vec{v}(0) + \vec{a}t \)
(d) \( \vec{r}(t) = \vec{r}(0) + \vec{v}(0)t + ½ \vec{a}t^2 \)
(e) \( \vec{a}_{avg} = [\vec{v}(t_2)-\vec{v}(t_1)]/(t_2-t_1) \)
Answer & Explanation:
(a) False – only for constant acceleration.
(b) True – definition of average velocity.
(c) False – only for constant acceleration.
(d) False – only for constant acceleration.
(e) True – definition of average acceleration.
(b) True – definition of average velocity.
(c) False – only for constant acceleration.
(d) False – only for constant acceleration.
(e) True – definition of average acceleration.
Question 4.24
True/false: A scalar quantity is one that (a) is conserved, (b) never negative, (c) dimensionless, (d) same everywhere, (e) same for all observers.
Answer & Explanation:
(a) False – scalars need not be conserved.
(b) False – e.g., temperature can be negative.
(c) False – e.g., mass has dimension.
(d) False – e.g., temperature varies.
(e) True – scalar invariant under coordinate rotation.
(b) False – e.g., temperature can be negative.
(c) False – e.g., mass has dimension.
(d) False – e.g., temperature varies.
(e) True – scalar invariant under coordinate rotation.
Question 4.25
Aircraft height 3400 m, angle subtended in 10 s is 30°. Find speed.
Answer & Explanation:
Horizontal distance covered in 10 s: \( d = h × \tan 30° = 3400/√3 ≈ 1963 \) m.
Speed = d / time = 196.3 m/s ≈ 706.7 km/h.
Speed = d / time = 196.3 m/s ≈ 706.7 km/h.
Question 4.26
Does a vector have a location? Can it vary with time? Do equal vectors at different locations have same physical effects? Give examples.
Answer & Explanation:
• Vectors are free unless localised (e.g., force has point of application).
• Yes, vectors can vary with time (e.g., velocity).
• Not necessarily – e.g., same force at different points on a rod produces different torques.
• Yes, vectors can vary with time (e.g., velocity).
• Not necessarily – e.g., same force at different points on a rod produces different torques.
Question 4.27
Does anything with magnitude and direction become a vector? Is rotation a vector?
Answer & Explanation:
No – must obey vector addition (commutative).
Rotation: finite rotations are not vectors (non-commutative), but infinitesimal rotations/angular velocity are vectors.
Rotation: finite rotations are not vectors (non-commutative), but infinitesimal rotations/angular velocity are vectors.
Question 4.28
Can you associate vectors with (a) length of bent wire, (b) plane area, (c) sphere?
Answer & Explanation:
(a) No – length is scalar.
(b) Yes – area vector (magnitude = area, direction = normal).
(c) No – sphere is scalar; but surface area vector sum over closed surface = 0.
(b) Yes – area vector (magnitude = area, direction = normal).
(c) No – sphere is scalar; but surface area vector sum over closed surface = 0.
Question 4.29
Bullet fired at 30° hits 3 km away. Can same speed hit 5 km target?
Answer & Explanation:
Max range \( R_{max} = v_0^2/g \). Given \( R(30°) = v_0^2 \sin 60°/g = 3 \) km → \( v_0^2/g = 3/ \sin 60° ≈ 3.464 \) km.
\( R_{max} = v_0^2/g ≈ 3.464 \) km < 5 km → Not possible.
\( R_{max} = v_0^2/g ≈ 3.464 \) km < 5 km → Not possible.
Question 4.30
Plane at 1.5 km height, speed 720 km/h. Gun muzzle speed 600 m/s. Find firing angle from vertical and safe altitude.
Answer & Explanation:
Relative motion problem. Solve for angle where shell and plane meet.
Angle from vertical: \( \sin^{-1}(v_p / v_s) \) ≈ 20.4°.
Min safe altitude: derived from time to collide; ≈ 15.6 km.
Angle from vertical: \( \sin^{-1}(v_p / v_s) \) ≈ 20.4°.
Min safe altitude: derived from time to collide; ≈ 15.6 km.
Question 4.31
Cyclist speed 27 km/h, circular turn radius 80 m, brakes with deceleration 0.5 m/s². Find net acceleration magnitude and direction on turn.
Answer & Explanation:
Speed = 7.5 m/s.
Centripetal \( a_c = v^2/r ≈ 0.703 \) m/s².
Tangential \( a_t = -0.5 \) m/s².
Net \( a = √(a_c² + a_t²) ≈ 0.86 \) m/s², direction towards centre and backward.
Centripetal \( a_c = v^2/r ≈ 0.703 \) m/s².
Tangential \( a_t = -0.5 \) m/s².
Net \( a = √(a_c² + a_t²) ≈ 0.86 \) m/s², direction towards centre and backward.
Question 4.32
(a) Show for projectile: \( θ(t) = \tan^{-1}[(v_{0y} - gt)/v_{0x}] \).
(b) Show projection angle: \( θ_0 = \tan^{-1}(4h_m/R) \).
(b) Show projection angle: \( θ_0 = \tan^{-1}(4h_m/R) \).
Answer & Explanation:
(a) From \( v_x = v_{0x} \), \( v_y = v_{0y} - gt \), tan θ = \( v_y / v_x \).
(b) Using \( h_m = (v_0^2 \sin² θ_0)/(2g) \) and \( R = (v_0^2 \sin 2θ_0)/g \), eliminate \( v_0 \) and g to get relation.
(b) Using \( h_m = (v_0^2 \sin² θ_0)/(2g) \) and \( R = (v_0^2 \sin 2θ_0)/g \), eliminate \( v_0 \) and g to get relation.
📘 Exam Preparation Tip:
These exercise questions will help you understand two-dimensional motion and vector applications. You'll learn to resolve vectors, analyze projectile motion (range, height, time of flight), solve problems in uniform circular motion, calculate relative velocity in different scenarios, and apply vector addition/subtraction to real-world motion problems. Crucial for understanding complex motion patterns in physics.