Class 11 Physics Notes Chapter 4 (Chapter 4) – Examplar Problems (English) Book

Alright class, let's delve into Chapter 4: Motion in a Plane from your NCERT Exemplar. This chapter is crucial for understanding how objects move in two dimensions, laying the groundwork for many physics concepts you'll encounter, especially in competitive government exams. Pay close attention to the vector nature of quantities here.

Chapter 4: Motion in a Plane - Detailed Notes

1. Scalars and Vectors

• Scalars: Physical quantities having only magnitude and no direction. They are specified by a single number, along with the proper unit.
  • Examples: Distance, Speed, Mass, Temperature, Time, Work, Energy.
  • Scalars are added, subtracted, multiplied, or divided using ordinary rules of algebra.
• Vectors: Physical quantities having both magnitude and direction. They must also obey the laws of vector addition.
  • Examples: Displacement, Velocity, Acceleration, Force, Momentum, Torque.
  • Representation:
    • Graphically: An arrow whose length is proportional to the magnitude and whose direction indicates the vector's direction.
    • Symbolically: Usually denoted by a bold letter (e.g., V) or a letter with an arrow above it (e.g., V⃗). Magnitude is denoted by |V| or simply V.
• Key Vector Types:
  • Position Vector (r⃗): A vector drawn from the origin of a coordinate system to the position of a particle. r⃗ = x î + y ĵ + z k̂.
  • Displacement Vector (Δr⃗): The change in position vector. If a particle moves from position r⃗₁ to r⃗₂, then Δr⃗ = r⃗₂ - r⃗₁.
  • Unit Vector: A vector with a magnitude of 1, used to specify direction. A unit vector in the direction of A is denoted by  (read as 'A cap' or 'A hat') and is given by  = A / |A|. Unit vectors along x, y, z axes are î, ĵ, k̂ respectively.
  • Zero Vector (Null Vector): A vector with zero magnitude and an arbitrary direction. Represented as 0.
  • Equal Vectors: Two vectors are equal if they have the same magnitude and the same direction.

2. Vector Operations

• Addition:
  • Triangle Law: If two vectors are represented by two sides of a triangle taken in order (magnitude and direction), then their sum (resultant) is represented by the third side of the triangle taken in the opposite order.
  • Parallelogram Law: If two vectors are represented by two adjacent sides of a parallelogram drawn from a point, then their resultant is represented by the diagonal of the parallelogram passing through that point.
  • Analytical Method (Resultant of A⃗ and B⃗ with angle θ between them):
    • Magnitude: R = |A + B| = √(A² + B² + 2AB cos θ)
    • Direction: If R makes an angle α with A, then tan α = (B sin θ) / (A + B cos θ)
  • Properties: Vector addition is commutative (A + B = B + A) and associative (A + (B + C) = (A + B) + C).
• Subtraction: Subtraction of B from A is defined as the addition of -B to A. A - B = A + (-B). The magnitude is |A - B| = √(A² + B² - 2AB cos θ).
• Multiplication by a Scalar: Multiplying a vector A by a scalar 'λ' gives a new vector λA. Its magnitude is |λ| times the magnitude of A. Its direction is the same as A if λ > 0, and opposite to A if λ < 0.
• Resolution of Vectors (Components): A vector can be resolved into components along perpendicular axes.
  • In 2D: A = Aₓ î + Ay ĵ, where Aₓ = A cos θ and Ay = A sin θ (θ is the angle A makes with the x-axis). Magnitude |A| = √(Aₓ² + Ay²).
  • In 3D: A = Aₓ î + Ay ĵ + Az k̂. Magnitude |A| = √(Aₓ² + Ay² + Az²).

3. Motion in a Plane

• Position and Displacement: Described by position vector r and displacement vector Δr.
• Velocity:
  • Average Velocity (v⃗_avg): Δr / Δt = (r⃗₂ - r⃗₁) / (t₂ - t₁)
  • Instantaneous Velocity (v⃗): The limit of average velocity as Δt approaches zero. v⃗ = lim (Δt→0) [Δr / Δt] = dr/dt.
  • Direction of v is tangent to the path at that point.
  • v = vₓ î + vy ĵ + vz k̂, where vₓ = dx/dt, vy = dy/dt, vz = dz/dt.
• Acceleration:
  • Average Acceleration (a⃗_avg): Δv / Δt = (v⃗₂ - v⃗₁) / (t₂ - t₁)
  • Instantaneous Acceleration (a⃗): a⃗ = lim (Δt→0) [Δv / Δt] = dv/dt = d²r/dt².
  • a = aₓ î + ay ĵ + az k̂, where aₓ = dvₓ/dt, ay = dvy/dt, az = dvz/dt.
• Motion with Constant Acceleration: If acceleration a is constant:
  • v⃗ = u⃗ + a t (where u⃗ is initial velocity)
  • r⃗ = r⃗₀ + u⃗ t + ½ a t² (where r⃗₀ is initial position)
  • These can be resolved into x and y components for 2D motion.

4. Relative Velocity

• Velocity of object A relative to object B: V_AB = V_A - V_B
• Velocity of object B relative to object A: V_BA = V_B - V_A = -V_AB
• Used in problems like:
  • Rain falling vertically appears slanted to a moving observer.
  • Crossing a river (shortest time vs shortest path).

5. Projectile Motion

• Motion of an object thrown into space, subject only to the acceleration due to gravity (g, acting downwards). Air resistance is usually neglected.
• It's a combination of:
  • Horizontal motion: Constant velocity (acceleration aₓ = 0).
  • Vertical motion: Constant downward acceleration (ay = -g).
• Assumptions: Object projected with initial velocity u at an angle θ with the horizontal.
• Equations of Motion:
  • Horizontal: x = (u cos θ) t
  • Vertical: y = (u sin θ) t - ½ gt²
  • Horizontal velocity (vₓ) = u cos θ (constant)
  • Vertical velocity (vy) = u sin θ - gt
• Trajectory Equation: Path followed is a parabola: y = (tan θ) x - [g / (2 u² cos² θ)] x²
• Time of Flight (T): Total time the projectile is in the air. T = (2 u sin θ) / g
• Maximum Height (H): Highest vertical point reached. H = (u² sin² θ) / (2g)
• Horizontal Range (R): Total horizontal distance covered. R = (u² sin 2θ) / g
• Maximum Range: Occurs when θ = 45°. R_max = u² / g
• Range is the same for complementary projection angles (θ and 90° - θ).

6. Uniform Circular Motion (UCM)

• Motion of an object along a circular path with constant speed.
• Velocity: Magnitude (speed) is constant, but direction continuously changes (tangential to the circle). Hence, velocity is not constant.
• Acceleration: Since velocity changes, there must be acceleration. This acceleration is directed towards the center of the circle and is called Centripetal Acceleration (a_c).
  • Magnitude: a_c = v² / r = ω² r (where r is the radius of the circle).
• Angular Velocity (ω): Rate of change of angular displacement (θ). ω = dθ/dt. Unit: rad/s.
• Relation between linear speed (v) and angular speed (ω): v = ω r
• Time Period (T): Time taken for one complete revolution. T = 2π / ω.
• Frequency (f or ν): Number of revolutions per second. f = 1 / T = ω / 2π. Unit: Hz.
• Centripetal Force (F_c): The net force required to produce centripetal acceleration, directed towards the center. F_c = m a_c = m v² / r = m ω² r. (This is not a new type of force, but rather the role played by forces like tension, gravity, friction, etc., in causing circular motion).

Multiple Choice Questions (MCQs)

1. A particle moves from position r₁ = (3î + 2ĵ - 6k̂) m to r₂ = (14î + 13ĵ + 9k̂) m under the action of a constant force F = (4î + ĵ + 3k̂) N. The work done by the force is:
   a) 100 J
   b) 75 J
   c) 200 J
   d) 50 J
   (Hint: Work Done = F ⋅ Δr)

2. For a projectile, the ratio of maximum height reached to the square of time of flight is:
   a) 5 : 4
   b) 5 : 2
   c) g : 8
   d) g : 4
   (Hint: Use formulas for H and T)

3. A particle is moving in a circle of radius 'r' with constant speed 'v'. The magnitude of the change in velocity after it has described an angle θ is:
   a) 2v sin(θ/2)
   b) 2v cos(θ/2)
   c) 2v sin θ
   d) v tan(θ/2)
   (Hint: Use vector subtraction |Δv| = |v₂ - v₁| and geometry/trigonometry)

4. A boat is sent across a river with a velocity of 8 km/h. If the resultant velocity of the boat is 10 km/h, then the velocity of the river flow is:
   a) 12.8 km/h
   b) 6 km/h
   c) 8 km/h
   d) 10 km/h
   (Hint: Velocities form a right-angled triangle if the boat heads perpendicular to the flow)

5. The horizontal range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be:
   a) 60 m
   b) 71 m
   c) 100 m
   d) 141 m
   (Hint: R = (u²/g) sin(2θ))

6. Which of the following quantities remains constant during uniform circular motion?
   a) Velocity
   b) Acceleration
   c) Speed
   d) Displacement

7. Two vectors A and B are such that |A + B| = |A - B|. The angle between vectors A and B is:
   a) 0°
   b) 60°
   c) 90°
   d) 180°
   (Hint: Square both sides and use the formula for resultant magnitude)

8. A particle has an initial velocity (3î + 4ĵ) m/s and an acceleration (0.4î + 0.3ĵ) m/s². Its speed after 10 s is:
   a) 7 units
   b) 7√2 units
   c) 8.5 units
   d) 10 units
   (Hint: Find final velocity vector v = u + at, then find its magnitude)

9. The trajectory of a projectile near the surface of the Earth is:
   a) Straight line
   b) Parabolic
   c) Circular
   d) Hyperbolic

10. If A = 2î + 3ĵ and B = î + ĵ, the component of vector A along the vector B is:
    a) 5/√2
    b) 4/√2
    c) 3/√2
    d) 1/√2
    (Hint: Component of A along B is (A ⋅ B) / |B|)

Answers to MCQs:

1. a) 100 J (Δr = r₂ - r₁ = 11î + 11ĵ + 15k̂; W = F ⋅ Δr = (4)(11) + (1)(11) + (3)(15) = 44 + 11 + 45 = 100 J)
2. c) g : 8 (H = u²sin²θ / 2g; T = 2usinθ / g => T² = 4u²sin²θ / g²; H/T² = (u²sin²θ / 2g) / (4u²sin²θ / g²) = g / 8)
3. a) 2v sin(θ/2) (Use vector diagram and cosine rule on the triangle formed by v₁, -v₂, and Δv)
4. b) 6 km/h (Resultant² = BoatVel² + RiverVel² => 10² = 8² + RiverVel² => RiverVel² = 100 - 64 = 36 => RiverVel = 6 km/h)
5. c) 100 m (R₁ = (u²/g) sin(30°) = 50; R₂ = (u²/g) sin(90°). Since sin(30°) = 1/2 and sin(90°) = 1, R₂ = 2 * R₁ = 2 * 50 = 100 m)
6. c) Speed
7. c) 90° (|A + B|² = |A - B|² => A² + B² + 2ABcosθ = A² + B² - 2ABcosθ => 4ABcosθ = 0. Since A≠0, B≠0, cosθ = 0 => θ = 90°)
8. b) 7√2 units (v = (3î + 4ĵ) + (0.4î + 0.3ĵ)(10) = (3î + 4ĵ) + (4î + 3ĵ) = 7î + 7ĵ. Speed = |v| = √(7² + 7²) = √(49 + 49) = √98 = 7√2 units)
9. b) Parabolic
10. a) 5/√2 (A ⋅ B = (2)(1) + (3)(1) = 5. |B| = √(1² + 1²) = √2. Component = 5/√2)

Make sure you understand the concepts behind these formulas and applications. Practice solving numerical problems based on these topics, as they are frequently asked in exams. Good luck!