✈️ Chapter 4: Motion in a Plane (Class 11 Physics) 🗺️
Hello Physics Explorers! Pichle chapter mein humne straight line motion (1-dimension) padha. Lekin duniya mein motion sirf seedhi line mein nahi hota! Cars turn karti hain, ball curve mein jaati hai. Ab hum **Motion in a Plane** (2-dimensional motion) ke concepts, especially vectors, projectile motion, aur circular motion ko samjhenge.
Hello Physics Explorers! Pichle chapter mein humne motion in a straight line (1-dimension) padha. Lekin duniya mein motion sirf seedhi line mein nahi hota! Cars mudti hain, ball curve mein jaati hai. Ab hum Samatl mein Gati (Motion in a Plane) (2-dimensional motion) ke concepts, khaaskar vectors, prakshepya gati (projectile motion), aur vrittiya gati (circular motion) ko samjhenge.📏🧭 Scalar and Vector Quantities (Adish Aur Sadish Rashiyan)
Physical quantities ko describe karne ke liye humein units ke saath kabhi direction batana padta hai aur kabhi nahi.
Scalar Quantity: A physical quantity that has only magnitude and no direction. It is specified completely by a single number, along with the proper unit.
Adish Rashi (Scalar): Ek physical quantity jisme sirf parimaan (magnitude) hota hai, disha nahi. Ise sirf ek number aur unit se poori tarah bataya jaata hai.Vector Quantity: A physical quantity that has both magnitude and direction and obeys the laws of vector addition (like Triangle Law or Parallelogram Law).
Sadish Rashi (Vector): Ek physical quantity jisme parimaan (magnitude) aur disha (direction) dono hote hain aur jo vector jod ke niyamon ka paalan karti hai.Key Difference Summary:
- Scalars just need a number and unit. Vectors need number, unit, AND direction.
- Scalars add using simple arithmetic. Vectors add using specific vector laws (considering direction).
📍 Position and Displacement Vectors (Sthiti Aur Visthapan Sadish)
To describe motion in a plane (e.g., x-y plane), we use vectors.
- Position Vector (r): A vector drawn from the origin (O) of a coordinate system to the position (P) of the object. It tells ‘where’ the object is. Sthiti Sadish (r): Origin (O) se vastu ki position (P) tak kheencha gaya vector. Batata hai ki vastu ‘kahan’ hai.
- In 2D (x-y plane), if point P has coordinates (x, y), its position vector is
r = x i + y j, where i and j are unit vectors along x and y axes. - Magnitude:
| r | = r = √(x² + y²). - Displacement Vector (Δr): Represents the change in position. It’s a vector drawn from the initial position (P₁) to the final position (P₂). ➡️ Visthapan Sadish (Δr): Sthiti mein parivartan ko darshata hai. Shuruaati sthiti (P₁) se antim sthiti (P₂) tak kheencha gaya vector.
- If initial position vector is r₁ and final is r₂, then
Δr = r₂ - r₁. - In 2D:
Δr = (x₂ - x₁) i + (y₂ - y₁) j = Δx i + Δy j. - Magnitude:
|Δr| = √(Δx² + Δy²).
↗️ General Vectors and Notations
- Vectors are usually represented by a bold letter (like A) or a letter with an arrow on top (A).
- The magnitude (or modulus) of a vector A is represented by |A| or simply A. It is always positive or zero.
- Graphically, a vector is shown by an arrow: the length represents magnitude (to scale), and the arrowhead indicates direction. 📏🧭
Equality of Vectors
Equal Vectors: Two vectors (A and B) are said to be equal if they have the same magnitude AND the same direction, regardless of their starting points.
Samaan Vectors: Do vectors (A aur B) barabar tabhi hote hain jab unka **parimaan (magnitude) aur disha (direction) dono samaan** hon, chahe unke starting points kahin bhi hon.Multiplication of Vector by a Real Number (Scalar)
- Multiplying a vector A by a positive real number k results in a new vector kA.
- Magnitude becomes k times: |kA| = k |A|.
- Direction remains the same as A.
- Multiplying by a negative number -k gives -kA.
- Magnitude becomes k times: |-kA| = k |A|.
- Direction becomes opposite to A.
➕➖ Addition and Subtraction of Vectors
Vectors add according to specific rules, considering direction.
Graphical Methods (Chitra Vidhi):
- Triangle Law of Vector Addition: If two vectors (A, B) are represented (in magnitude and direction) by two sides of a triangle taken in order, then their sum or resultant (R = A + B) is represented by the third side of the triangle taken in the opposite order. △ Vector Jod Ka Tribhuj Niyam: Agar do vectors ko ek triangle ki do sides se (kram mein) represent kiya jaye, toh unka jod (resultant) triangle ki teesri side dwara (vipreet kram mein) represent hota hai.
- Parallelogram Law of Vector Addition: If two vectors (A, B) starting from the same point are represented by two adjacent sides of a parallelogram, then their resultant (R = A + B) is represented by the diagonal of the parallelogram starting from the same point. ▱ Vector Jod Ka Samanantar Chaturbhuj Niyam: Agar ek hi point se shuru hone wale do vectors ko ek parallelogram ki do paas wali sides se represent kiya jaye, toh unka resultant usi point se shuru hone wale diagonal dwara represent hota hai.
- Vector Subtraction: Subtracting vector B from A (A – B) is the same as adding the negative of B to A. i.e.,
A - B = A + (-B). (Here, –B has same magnitude as B but opposite direction).
Vector Ghatav: A mein se B ghatana, A mein B ka negative (-B) jodne ke barabar hai.
☝️ Unit Vector (Ikayi Sadish)
Unit Vector: A vector having magnitude equal to one (unity) and pointing in a particular direction. It is used to specify direction.
Notation: Represented by a letter with a cap or hat (^). Ex: n.
Ikayi Sadish (Unit Vector): Ek vector jiska **magnitude one** hota hai aur jo ek vishesh disha mein point karta hai. Disha batane ke liye istemal hota hai. Notation: Letter ke upar cap (^).Unit vector in the direction of vector A:
a = A| A | or a = AA
(Vector divided by its magnitude)
- Special Unit Vectors:
- i: Unit vector along the positive x-axis.
- j: Unit vector along the positive y-axis.
- k: Unit vector along the positive z-axis.
🧩 Resolution of a Vector in a Plane
Any vector in a plane can be broken down (resolved) into two perpendicular components along the x and y axes. This is useful for calculations.
Plane mein kisi bhi vector ko x aur y axes ke anudish do perpendicular components mein toda (resolve) ja sakta hai. Yeh calculations ke liye upyogi hai.Rectangular Components ( आयताकार घटक):
- If a vector A makes an angle θ with the positive x-axis:
- Component along x-axis (Horizontal):
Ax = A cos θ - Component along y-axis (Vertical):
Ay = A sin θ - The vector can be written as:
A = Ax i + Ay j - Magnitude from components:
A = √( Ax² + Ay² ) - Direction (Angle θ):
tan θ = AyAx
Vector Addition using Components:
- If
A = Ax i + Ay jandB = Bx i + By j - Then Resultant
R = A + Bis: R = (Ax + Bx) i + (Ay + By) j
(Bas x-components ko jodo aur y-components ko jodo).
- Magnitude
R = √( Rx² + Ry² ), whereRx = Ax + BxandRy = Ay + By. - Subtraction is similar:
A - B = (Ax - Bx) i + (Ay - By) j.
✖️• Scalar and Vector Products of Vectors
There are two main ways to multiply vectors.
1. Scalar (Dot) Product (A ⋅ B):
The scalar product of two vectors A and B is a SCALAR quantity equal to the product of their magnitudes and the cosine of the angle (θ) between them.
Do vectors ka scalar product ek **SCALAR** quantity hai, jo unke magnitude aur unke beech ke angle (θ) ke cosine ke गुणनफल (product) ke barabar hota hai.A ⋅ B = |A| |B| cos θ = AB cos θ
Properties of Dot Product:
- Result is a scalar.
- It is commutative:
A ⋅ B = B ⋅ A. - It is distributive:
A ⋅ (B + C) = A ⋅ B + A ⋅ C. - If A ⊥ B (θ=90°), then A ⋅ B = AB cos 90° = 0. Dot product of perpendicular vectors is zero.
- If A || B (θ=0°), then A ⋅ B = AB cos 0° = AB (Max value).
- If A is anti-parallel to B (θ=180°), then A ⋅ B = AB cos 180° = -AB (Min value).
- Dot product with itself:
A ⋅ A = AA cos 0° = A². - For unit vectors:
i⋅i = j⋅j = k⋅k = (1)(1)cos 0° = 1. i⋅j = j⋅k = k⋅i = (1)(1)cos 90° = 0.- In component form: If
A = Axi+Ayj+AzkandB = Bxi+Byj+Bzk, then
A ⋅ B = AxBx + AyBy + AzBz.
Use Example: Work Done (W)
Work done by a constant force F causing a displacement d is given by their dot product: W = F ⋅ d = Fd cos θ.
2. Vector (Cross) Product (A × B):
The vector product of two vectors A and B is a VECTOR quantity (C = A × B).
- Magnitude:
| C | = |A| |B| sin θ = AB sin θ, where θ is the angle between A and B. - Direction: Direction of C is perpendicular to the plane containing A and B, given by the Right-Hand Rule (or Right-Hand Screw Rule). ✋
Right-Hand Rule: Curl fingers of your right hand from vector A towards vector B through the smaller angle. Your extended thumb points in the direction of A × B.
Properties of Cross Product:
- Result is a vector.
- NOT commutative:
A × B = - (B × A). - It is distributive:
A × (B + C) = (A × B) + (A × C). - If A || B (θ=0°) or anti-parallel (θ=180°), then sin θ = 0, so A × B = 0 (Null vector). Cross product of parallel vectors is zero.
- If A ⊥ B (θ=90°), then |A × B| = AB sin 90° = AB (Max magnitude).
- Cross product with itself:
A × A = AA sin 0° = 0. - For unit vectors:
i×i = j×j = k×k = 0. - Cyclic Rule:
i × j = k,j × k = i,k × i = j. - Anti-Cyclic:
j × i = -k,k × j = -i,i × k = -j. - In component form: Calculated using determinant method (usually taught in detail with vectors).
Use Example: Torque (τ)
Torque due to a force F applied at a position r from the axis is given by their cross product: τ = r × F.
✍️ Solved Practice Problems: Vectors
Problem 1 (Addition): If A = 2i + 3j and B = -i + 5j, find R = A + B and its magnitude.
Add respective components:
Rx = Ax + Bx = 2 + (-1) = 1
Ry = Ay + By = 3 + 5 = 8
So, R = 1i + 8j = i + 8j.
Magnitude |R| = √(Rx² + Ry²) = √(1² + 8²) = √(1 + 64) = √65.
Answer: R = i + 8j; |R| = √65
Problem 2 (Resolution): Find the x and y components of a force of 10 N acting at an angle of 60° with the positive x-axis.
Given: Force magnitude F = 10 N, Angle θ = 60°.
x-component Fx = F cos θ = 10 × cos 60°
cos 60° = 1/2
Fx = 10 × (1/2) = 5 N.
y-component Fy = F sin θ = 10 × sin 60°
sin 60° = √3 / 2
Fy = 10 × (√3 / 2) = 5√3 N.
Answer: Fx = 5 N, Fy = 5√3 N
Problem 3 (Unit Vector): Find the unit vector in the direction of A = 3i - 4j.
Unit vector a = A| A |.
First, find magnitude |A|:
|A| = √(Ax² + Ay²) = √(3² + (-4)²) = √(9 + 16) = √25 = 5.
Now, divide the vector by its magnitude:
a = (3i - 4j)5 = 35i - 45j.
Answer: a = (3/5)i – (4/5)j
Problem 4 (Dot Product): Find the angle between vectors A = i + j and B = i - j.
Use dot product definition: A ⋅ B = AB cos θ. So, cos θ = (A ⋅ B) / (AB).
1. Find A ⋅ B using components:
Ax=1, Ay=1; Bx=1, By=-1.
A ⋅ B = (1)(1) + (1)(-1) = 1 - 1 = 0.
2. Since the dot product is 0, and magnitudes A=√2, B=√2 are non-zero, we must have cos θ = 0.
3. Therefore, the angle θ = 90°.
Answer: The vectors are perpendicular (θ = 90°).
Problem 5 (Cross Product): Find a vector perpendicular to both A = 2i + j + k and B = i - j + 2k.
The cross product A × B gives a vector perpendicular to both A and B.
Using the determinant method for cross product:
A × B =
i
j
k
2
1
1
1
-1
2
= i [ (1)(2) - (1)(-1) ] - j [ (2)(2) - (1)(1) ] + k [ (2)(-1) - (1)(1) ]
= i [ 2 - (-1) ] - j [ 4 - 1 ] + k [ -2 - 1 ]
= i [ 3 ] - j [ 3 ] + k [ -3 ]
= 3i - 3j - 3k
Answer: 3i - 3j - 3k (or any scalar multiple of this vector)
✈️ Motion in a Plane (Samatl mein Gati)
Motion in 2D (or 3D) requires vectors to describe position, velocity, and acceleration.
- Position vector:
r(t) = x(t)i + y(t)j - Velocity vector (Instantaneous):
v(t) = drdt = dxdti + dydtj = vxi + vyj - Acceleration vector (Instantaneous):
a(t) = dvdt = dvxdti + dvydtj = axi + ayj
Motion in a plane can be treated as two independent motions along the x and y axes simultaneously.
Case 1: Uniform Velocity in a Plane
- Velocity vector v is constant (both magnitude and direction).
- Acceleration a = 0.
- Position vector at time t:
r(t) = r₀ + vt(where r₀ is initial position vector at t=0).
Case 2: Uniform Acceleration in a Plane
- Acceleration vector a is constant.
- Kinematic Equations in Vector Form:
v = u + atΔr = ut + 12at²(where u = initial velocity, Δr = displacement)- These can be broken into independent x and y components:
vx = ux + axtΔx = uxt + ½axt²vy = uy + aytΔy = uyt + ½ayt²
↗️↘️ Projectile Motion (Prakshepya Gati)
Projectile Motion: Motion of an object thrown or projected into the air, subject only to acceleration due to gravity (g), assuming air resistance is negligible.
Prakshepya Gati: Hawa mein phenki gayi vastu ki gati, jo sirf gravity ke karan acceleration mehsoos karti hai (hawa ke pratirodh ko nazarandaz karte hue).- It is an example of motion in a plane with **constant acceleration** (ax = 0, ay = -g, taking upward as positive).
- Path followed is a Parabola ⌒.
Key Formulas (Object projected with initial velocity u at angle θ with horizontal):
Initial Velocity Components: ux = u cos θ, uy = u sin θ
Acceleration Components: ax = 0, ay = -g
Time of Flight (T): T = 2u sin θg
Maximum Height (H): H = (u sin θ)²2g = u² sin²θ2g
Horizontal Range (R): R = ux T = (u cos θ) × (2u sin θg) = u² (2sinθcosθ)g = u² sin(2θ)g
Equation of Trajectory: y = (tan θ) x - (g2u² cos²θ) x² (Parabolic)
Note: Range R is maximum when sin(2θ) is maximum (i.e., 1), which occurs when 2θ = 90° or θ = 45°.
🔄 Uniform Circular Motion (Ek Saman Vrittiya Gati)
Uniform Circular Motion (UCM): Motion of an object travelling at a constant speed along a circular path.
Ek Saman Vrittiya Gati: Kisi vastu ka ek vrittakar path par **sthir chaal (constant speed)** se chalna.- Although speed is constant, the direction of motion continuously changes.
- Therefore, the velocity is continuously changing.
- Since velocity changes, the motion is accelerated. ⚠️ Haalaanki speed constant hai, gati ki disha lagatar badalti hai. Isliye, velocity lagatar badalti hai. Aur isliye, motion accelerated hota hai.
- This acceleration is called Centripetal Acceleration (ac). It is always directed towards the centre of the circle. 🎯 Is acceleration ko **Abhikendriy Tvaran (Centripetal Acceleration)** kehte hain. Iski disha hamesha circle ke kendra ki taraf hoti hai.
Magnitude of Centripetal Acceleration:
ac = v²r = ω²r
(where v = speed, r = radius, ω = angular velocity)
The force required to produce this acceleration (and keep the object in circular path) is called Centripetal Force, also directed towards the center.
🏏 Solved Practice Problems: Projectile & Circular Motion
Problem 1 (Projectile): A cricket ball is thrown at a speed of 28 m/s in a direction 30° above the horizontal. Calculate (a) max height, (b) time of flight, (c) horizontal range. (g=9.8 m/s²)
Given: u = 28 m/s, θ = 30°, g = 9.8 m/s².
uy = u sin θ = 28 sin 30° = 28 × (1/2) = 14 m/s.
ux = u cos θ = 28 cos 30° = 28 × (√3/2) = 14√3 m/s.
(a) Max Height (H):
H = uy²2g = (14)²2 × 9.8 = 19619.6 = 10 m.
(b) Time of Flight (T):
T = 2 uyg = 2 × 149.8 = 289.8 ≈ 2.86 s.
(c) Horizontal Range (R):
R = ux T = (14√3) × (289.8)
R ≈ (14 × 1.732) × 2.86 ≈ 24.25 × 2.86 ≈ 69.35 m.
(Alt: R = u²sin(2θ)/g = 28²sin(60°)/9.8 = 784 × (√3/2) / 9.8 ≈ 784 × 0.866 / 9.8 ≈ 678.7 / 9.8 ≈ 69.25 m)
Answer: H=10m, T≈2.86s, R≈69.3m
Problem 2 (Projectile): A projectile has a range of 50 m and reaches a max height of 10 m. Calculate the angle of projection.
Given: R = 50 m, H = 10 m.
Formulas: R = u² sin(2θ)g, H = u² sin²θ2g
Divide H by R: HR = (u² sin²θ / 2g)(u² sin(2θ) / g)
HR = sin²θ2g × gsin(2θ) (u² cancels)
HR = sin²θ2 sin(2θ) = sin²θ2 (2 sinθ cosθ) = sinθ4 cosθ = tan θ4
So, tan θ = 4HR.
Substitute values: tan θ = (4 × 10) / 50 = 40 / 50 = 4/5 = 0.8.
θ = tan⁻¹(0.8).
θ ≈ 38.7°.
Answer: θ ≈ 38.7°
Problem 3 (Projectile – Max Range): Find the maximum range of a projectile fired with an initial velocity of 49 m/s. (g=9.8 m/s²).
Range R = u² sin(2θ) / g.
Range is maximum when sin(2θ) is maximum, i.e., sin(2θ)=1.
This occurs when 2θ=90°, or θ=45°.
Max Range Rmax = u² (1) / g = u²/g.
Rmax = (49)² / 9.8 = (49 × 49) / 9.8 = 49 × (49/9.8) = 49 × 5 = 245 m.
Answer: 245 m
Problem 4 (UCM): An insect trapped in a circular groove of radius 12 cm moves steadily and completes 7 revolutions in 100 s. What is its angular speed (ω) and linear speed (v)?
Given: r = 12 cm = 0.12 m. Number of revolutions = 7. Time t = 100 s.
Frequency (ν) = Revolutions / Time = 7 / 100 Hz.
Angular Speed (ω):
ω = 2πν = 2π (7/100) = 14π / 100 rad/s.
Using π ≈ 3.14: ω ≈ (14 × 3.14) / 100 = 43.96 / 100 ≈ 0.44 rad/s.
Linear Speed (v):
v = ωr = (14π / 100) × 0.12 m/s.
v ≈ 0.44 × 0.12 ≈ 0.053 m/s (or 5.3 cm/s).
Answer: ω ≈ 0.44 rad/s, v ≈ 0.053 m/s
Problem 5 (UCM): Find the centripetal acceleration of a stone whirled in a horizontal circle of radius 50 cm with a constant speed of 5 m/s.
Given: radius r = 50 cm = 0.5 m. Speed v = 5 m/s.
Centripetal acceleration ac formula: ac = v²r.
ac = (5)²0.5 = 250.5 = 50 m/s².
The direction is towards the center of the circle.
Answer: 50 m/s² (towards the center)
— End of Part 1: Notes & Concepts —
Ready for Part 2 (Q&A and Quiz)? Let me know when!
❓ Sawal Jawab (Questions & Answers)
🤏 Very Short Answer Questions
1. Differentiate between scalar and vector quantities.
Scalars have only magnitude (e.g., speed, mass). Vectors have both magnitude and direction (e.g., velocity, force).
2. What is a position vector?
A vector drawn from the origin to the position of an object.
3. What is a displacement vector?
A vector representing the change in position (Final position vector – Initial position vector).
4. When are two vectors considered equal?
When they have the same magnitude AND the same direction.
5. What happens to a vector when multiplied by -2?
Its magnitude becomes double, and its direction becomes opposite.
6. What is a unit vector?
A vector with magnitude equal to one, used to specify direction.
7. Name the unit vectors along x, y, and z axes.
î, ĵ, k̂ (or i, j, k).
8. If A = Ax i + Ay j, what is its magnitude?
|A| = √( Ax² + Ay² ).
9. What is the dot product of two perpendicular vectors?
Zero (0).
10. What is the cross product of two parallel vectors?
Null vector (0).
11. Is dot product commutative?
Yes (A ⋅ B = B ⋅ A).
12. Is cross product commutative?
No (A × B = – B × A).
13. What quantity does the slope of a velocity-time graph give?
Instantaneous Acceleration.
14. What is the path of a projectile (neglecting air resistance)?
A parabola.
15. At what angle of projection is the horizontal range maximum?
45°.
16. In uniform circular motion, what remains constant?
Speed.
17. What is the direction of centripetal acceleration?
Towards the center of the circle.
18. Write the formula for centripetal acceleration.
ac = v²/r or ac = ω²r.
📝 Short Answer Questions
1. Define position vector and displacement vector.
- Position Vector (r): Vector from origin to object’s position.
- Displacement Vector (Δr): Vector from initial to final position (change in position vector).
2. Can displacement be greater than distance travelled? Explain.
- No.
- Displacement is the shortest distance between initial and final points.
- Distance (path length) is the actual path length covered.
- Magnitude of displacement is always less than or equal to the distance. Equality holds only for straight line motion in one direction.
3. State the triangle law of vector addition.
- If two vectors are represented by two sides of a triangle taken in order (magnitude & direction),
- Then their resultant (sum) is represented by the third side taken in the opposite order.
4. State the parallelogram law of vector addition.
- If two vectors starting from same point are represented by two adjacent sides of a parallelogram,
- Then their resultant is represented by the diagonal starting from the same point.
5. What is resolution of a vector? How do you find rectangular components?
- Resolution is splitting a vector into two or more components.
- Rectangular components are perpendicular components along x and y axes.
- If vector A makes angle θ with x-axis:
- Ax = A cos θ
- Ay = A sin θ
6. Define scalar (dot) product of two vectors. Write its formula.
- Dot product yields a scalar quantity.
- Defined as product of magnitudes of the two vectors and the cosine of the angle (θ) between them.
- Formula: A ⋅ B = AB cos θ.
7. Define vector (cross) product of two vectors. Write its formula for magnitude.
- Cross product yields a vector quantity perpendicular to the plane of the two vectors.
- Magnitude is product of magnitudes and the sine of the angle (θ) between them.
- Magnitude: |A × B| = AB sin θ.
- Direction is given by the Right-Hand Rule.
8. Calculate i ⋅ i, i ⋅ j, i × i, and i × j.
- i ⋅ i = |i||i|cos 0° = 1×1×1 = 1.
- i ⋅ j = |i||j|cos 90° = 1×1×0 = 0.
- i × i = |i||i|sin 0° n = 1×1×0 n = 0 (Null vector).
- i × j = |i||j|sin 90° k = 1×1×1 k = k.
9. Find the scalar and vector product of A=2i and B=3j.
- Vectors are perpendicular (along x and y axes). θ=90°.
- Scalar Product: A ⋅ B = AB cos 90° = A B (0) = 0. (Also AxBx + AyBy + AzBz = 2*0 + 0*3 + 0*0 = 0).
- Vector Product Magnitude: |A × B| = AB sin 90° = (2)(3)(1) = 6.
- Vector Product Direction: By right-hand rule (from i towards j), direction is along +z axis (k̂).
- Vector Product: A × B = 6k.
10. What is projectile motion? Why is the path parabolic?
- Motion of an object thrown in air, under gravity only (air resistance ignored).
- It’s a 2D motion with constant horizontal velocity (ax=0) and constant downward vertical acceleration (ay=-g).
- Horizontal motion: x = (u cos θ)t. Vertical motion: y = (u sin θ)t – ½gt².
- Eliminating ‘t’ between these two equations gives an equation of y in terms of x², which is the equation of a parabola.
11. Define Time of Flight, Maximum Height, and Horizontal Range for a projectile.
- Time of Flight (T): Total time the projectile remains in the air. (T = 2u sinθ / g).
- Maximum Height (H): Highest vertical position reached by the projectile. (H = u²sin²θ / 2g).
- Horizontal Range (R): Total horizontal distance covered during the time of flight. (R = u²sin(2θ) / g).
12. Is uniform circular motion an accelerated motion? Why?
- Yes.
- In UCM, the speed (magnitude of velocity) is constant, but the direction of velocity continuously changes.
- Since velocity is a vector and it’s changing (due to direction change), there must be an acceleration.
- This acceleration is centripetal acceleration, directed towards the center.
13. A particle moves from (2,3)m to (5,7)m. Find its displacement vector and magnitude.
- Initial position r₁ = 2i + 3j. Final position r₂ = 5i + 7j.
- Displacement Δr = r₂ – r₁ = (5i + 7j) – (2i + 3j).
- Δr = (5-2)i + (7-3)j = 3i + 4j m.
- Magnitude |Δr| = √(3² + 4²) = √(9 + 16) = √25 = 5 m.
14. Give an example where average speed is not zero, but average velocity is zero.
- An object completing a round trip and returning to its starting point.
- Example: Running one lap around a circular track and stopping at the start.
- Total displacement is zero (final position = initial position).
- Therefore, average velocity (Displacement/Time) is zero.
- But total path length (distance covered) is non-zero (circumference of track).
- Therefore, average speed (Distance/Time) is non-zero.
15. Find a unit vector parallel to the resultant of vectors A = 2i+4j-5k and B = i+2j+3k.
- Resultant R = A + B = (2+1)i + (4+2)j + (-5+3)k = 3i + 6j – 2k.
- Magnitude |R| = √(3² + 6² + (-2)²) = √(9 + 36 + 4) = √49 = 7.
- Unit vector r = R / |R| = (3i + 6j – 2k) / 7.
- Answer:
37i + 67j - 27k
📜 Long Answer Questions
1. Explain Scalar and Vector quantities giving sufficient examples. How are vectors represented?
- Scalars: Quantities having only magnitude. Ex: Distance(10m), Speed(20m/s), Mass(5kg), Time(2s), Temp(300K).
- Vectors: Quantities having both magnitude and direction, obey vector laws. Ex: Displacement(10m East), Velocity(20m/s North), Force(5N Downwards), Acceleration.
- Representation:
- Symbolically: Bold letter (A) or arrow (A). Magnitude by |A| or A.
- Graphically: Arrow. Length proportional to magnitude, arrowhead shows direction.
2. State and explain the Parallelogram Law of Vector Addition. Derive the formula for magnitude and direction of the resultant.
Statement: If two vectors acting simultaneously at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from that point, their resultant vector is represented in magnitude and direction by the diagonal of the parallelogram passing through that same point.
Derivation (Magnitude):
- Consider vectors P and Q at angle θ, forming parallelogram OACB. Resultant R=OA=P+Q. Diagonal OB=R.
- Draw BN ⊥ ON (extended OA).
- In ΔANB, AN = Q cos θ, BN = Q sin θ.
- In right ΔONB: OB² = ON² + BN² = (OA + AN)² + BN²
- R² = (P + Q cos θ)² + (Q sin θ)²
- R² = P² + Q²cos²θ + 2PQcosθ + Q²sin²θ
- R² = P² + Q²(cos²θ + sin²θ) + 2PQcosθ
- R² = P² + Q²(1) + 2PQcosθ
- Magnitude R = √(P² + Q² + 2PQ cos θ)
Derivation (Direction):
- Let β be the angle resultant R makes with P.
- In right ΔONB: tan β = BN / ON = BN / (OA + AN)
- Direction tan β = (Q sin θ) / (P + Q cos θ)
3. What is resolution of vectors? Explain how a vector is resolved into rectangular components in a plane. Find magnitude and direction from components.
Resolution: Splitting a vector into two or more components whose combined effect is same as the original vector.
Rectangular Components in a Plane (x-y):
- Consider vector A making angle θ with +x-axis. Draw perpendiculars from head of A to x and y axes, meeting at P and Q respectively.
- The x-component (projection on x-axis) is Ax = OP. In right ΔOP(Head), cos θ = OP / OA = Ax / A. So, Ax = A cos θ.
- The y-component (projection on y-axis) is Ay = OQ. In right Δ, sin θ = P(Head) / OA = OQ / A = Ay / A. So, Ay = A sin θ.
- Vector in component form: A = Ax i + Ay j.
Finding Magnitude and Direction from Components:
- If A = Ax i + Ay j.
- Magnitude: Using Pythagoras in the right triangle formed by Ax and Ay, A = √( Ax² + Ay² ).
- Direction (angle θ with x-axis): tan θ = Ay / Ax. (Careful with quadrant for θ).
4. Explain dot product and cross product with their properties.
Dot Product (A ⋅ B):
- Result is a Scalar. Formula: AB cos θ.
- Commutative: Yes.
- Distributive: Yes.
- Perpendicular vectors (θ=90°): Dot product = 0. (Important check)
- Parallel vectors (θ=0°): Dot product = AB (Max).
- Self-product: A ⋅ A = A².
- Component form: AxBx + AyBy + AzBz.
Cross Product (A × B):
- Result is a Vector perpendicular to plane of A and B.
- Magnitude Formula: |A × B| = AB sin θ.
- Direction: Right Hand Rule.
- NOT Commutative: (A × B = – B × A).
- Distributive: Yes.
- Parallel vectors (θ=0°/180°): Cross product = 0 (Null vector).
- Perpendicular vectors (θ=90°): Max magnitude |A × B| = AB.
- Self-product: A × A = 0.
- Unit vector products: i×j=k (cyclic), j×i=-k (anti-cyclic).
5. What is Projectile Motion? Derive expressions for Time of Flight, Maximum Height, and Horizontal Range for a projectile fired at angle θ with horizontal.
Projectile Motion: 2D motion under constant downward acceleration ‘g’ (neglecting air resistance).
Initial velocity u, angle θ.
- ux = u cos θ (constant); ax = 0.
- uy = u sin θ; ay = -g.
Derivations:
- Time of Flight (T): Time taken to return to initial height. Vertical displacement Δy=0. Use s = ut + ½at².
0 = uyT + ½ayT² = (u sin θ)T – ½gT².
Since T≠0, divide by T: (u sin θ) – ½gT = 0 ⇒ T = 2u sin θ / g. - Maximum Height (H): At max height, vertical velocity vy = 0. Use v²=u²+2as vertically:
0² = uy² + 2ayH = (u sin θ)² + 2(-g)H.
2gH = (u sin θ)² ⇒ H = u²sin²θ / 2g. - Horizontal Range (R): Horizontal distance covered in Time T. Use x = uxt (since ax=0):
R = ux T = (u cos θ) × (2u sin θ / g).
R = u² (2 sin θ cos θ) / g = u² sin(2θ) / g.
6. Explain Uniform Circular Motion. Show that it is an accelerated motion and derive the expression for centripetal acceleration.
UCM: Motion in a circle with constant **speed**.
- Accelerated Motion? Yes. Although speed is constant, the **direction** of velocity vector (v) continuously changes (it’s always tangent to the circle). Since velocity (vector) changes, there must be an acceleration. 🔄
- Centripetal Acceleration (ac): This acceleration is always directed **towards the center** of the circle. It is responsible for changing the direction of velocity, not its magnitude (speed). 🎯
Derivation of Magnitude (using calculus or geometry):
- Geometric Method (simplified idea): Consider small time Δt, particle moves from A to B. Change in velocity Δv = vB – vA. Form velocity triangle, similar to position triangle (OAB).
- For small angle Δθ, Arc length ≈ Chord length ≈ vΔt. Also, |Δv| ≈ vΔθ. And Δθ ≈ (vΔt)/r.
- So, |Δv| ≈ v * (vΔt / r) = v²Δt / r.
- Acceleration ac = |Δv| / Δt ≈ (v²Δt / r) / Δt = v²/r.
- Using v = ωr (ω=angular velocity): ac = (ωr)² / r = ω²r²/r = ω²r.
- Formulas: ac = v²/r = ω²r.
7. Discuss Relative Velocity in one and two dimensions (briefly).
Relative velocity is the velocity of an object as observed from another moving object (or frame of reference).
- One Dimension:
- Velocity of A relative to B: vAB = vA – vB.
- Example: If Car A moves at +50m/s and Car B moves at +30m/s (same direction), velocity of A wrt B is 50 – 30 = +20m/s.
- If Car B moves at -30m/s (opposite), velocity of A wrt B is 50 – (-30) = +80m/s.
- Two Dimensions (Plane):
- The same vector subtraction principle applies.
- vAB = vA – vB.
- This subtraction is done using vector rules (graphically or using components).
- If vA = vAxi + vAyj and vB = vBxi + vByj,
- Then vAB = (vAx – vBx)i + (vAy – vBy)j.
- Used in problems like rain falling vertically appears slanted to a moving observer (rain-man problems), or crossing a river.
8. Prove that the trajectory of a projectile fired horizontally from a height is a parabola.
Consider object fired horizontally with velocity ‘u’ from height ‘h’. Origin at point of projection, x-axis horizontal, y-axis vertical (downwards positive).
- Initial velocity: ux = u, uy = 0.
- Acceleration: ax = 0, ay = +g.
Motion Equations:
- Horizontal motion: x = uxt + ½axt² => x = ut + 0 => t = x/u.
- Vertical motion: y = uyt + ½ayt² => y = 0*t + ½gt² => y = ½gt².
Equation of Trajectory (y in terms of x):
- Substitute t = x/u from horizontal eqn into vertical eqn:
y = 12 g ( xu )² = 12 g x²u² = (g2u²) x².- Since g and u are constants for a given projection, let K = g / (2u²).
- Then y = K x².
- This equation represents a parabola opening downwards (if y was taken positive upwards initially) or opening upwards (if y taken positive downwards as here).
Thus, the path is parabolic.
9. What is the relation between linear velocity (v), angular velocity (ω), and radius (r) for an object in uniform circular motion?
Consider an object moving in a circle of radius ‘r’ with constant angular velocity ‘ω’ and linear speed ‘v’.
- Linear Speed (v): The magnitude of the instantaneous linear velocity (tangential velocity). It’s the distance covered along the circumference per unit time.
- Angular Velocity (ω): The rate at which the angular position changes. It’s the angle covered per unit time (measured in radians per second, rad/s).
- Relationship: In one full revolution (Time period T):
- Angle covered = 2π radians.
- Distance covered = Circumference = 2πr.
- Angular Velocity ω = Angle / Time = 2π / T.
- Linear Speed v = Distance / Time = 2πr / T.
- Substitute T = 2π / ω into the equation for v:
v = 2πr / (2π / ω) = 2πr × (ω / 2π) = ωr.
Relation: v = ωr. The linear speed is the product of the angular velocity and the radius of the circular path.
10. Explain vector addition using rectangular components. Find the magnitude and direction of the resultant vector.
Adding vectors A and B using components:
- Resolve vectors: Find x and y components of each vector.
- A = Ax i + Ay j
- B = Bx i + By j
- Add respective components: Find components of Resultant R = A + B.
- Rx = Ax + Bx
- Ry = Ay + By
- Write Resultant Vector: R = Rx i + Ry j.
- Find Magnitude of Resultant: Use Pythagoras theorem on components.
R = |R| = √( Rx² + Ry² ). - Find Direction of Resultant: Find the angle (α) the resultant vector makes with the positive x-axis.
tan α = Ry / Rx. The quadrant depends on the signs of Rx and Ry.
