Motion in a Plane

1. Scalars and Vectors

Scalar: Magnitude only (Mass, Distance).
Vector: Magnitude + Direction + Vector Laws (Displacement, Velocity).

Resultant (R): √(A² + B² + 2ABcosθ).
Direction (α): tanα = (Bsinθ) / (A + Bcosθ).
Resolution: Splitting vector into components (Acosθ, Asinθ).

(Asked in NEET 2017, 2019)

2. Projectile Motion

Motion under gravity with constant horizontal velocity.

Trajectory: Parabolic path. y = (tanθ)x - (g/2u²cos²θ)x².
Max Height (H): u²sin²θ / 2g.
Range (R): u²sin2θ / g. (Max at 45°).
Time of Flight (T): 2usinθ / g.

(Asked in NEET 2018, 2021, 2022)

3. Circular Motion

Motion in a circle at constant speed (Uniform).

Angular Velocity (ω): dθ/dt = v/r.
Centripetal Acceleration (a_c): v²/r = rω². (Directed towards center).
Tangential Acceleration (a_t): Zero in UCM. Non-zero in Non-UCM (dv/dt).

4. Relative Velocity (2D)

Velocity of A wrt B: v_AB = v_A - v_B.
Rain-Man Problem: Vertical rain appears inclined to moving man.
River-Boat Problem:
Shortest Path: Swim at angle > 90° to flow.
Shortest Time: Swim perpendicular to flow.

(Asked in NEET 2016, 2020, 2023)

Numericals: Motion in a Plane

Q1. Two forces 3 N and 4 N act at 90°. Find resultant magnitude and direction.

Solution:
R = √(3² + 4²) = √(9+16) = √25 = 5 N.
Direction θ w.r.t 3 N force: tanθ = 4/3.
θ = tan^-1(1.33) = 53°.

Q2. A ball is projected at 30° with velocity 20 m/s. Find Range (g=10 m/s²).

Solution:
R = u²sin(2θ) / g.
θ = 30°, 2θ = 60°.
R = (20)² sin 60° / 10.
R = (400 × √3/2) / 10 = 20 √3 = 20 × 1.732 = 34.64 m.

Q3. Rain falls vertically at 30 m/s. Man walks at 10 m/s. What angle should he hold umbrella?

Solution:
Relative velocity v_RM = v_R - v_M.
tanθ = v_M / v_R (w.r.t vertical).
tanθ = 10 / 30 = 1/3.
θ = tan^-1(1/3) with vertical.

Q4. River width 500m flows at 3 km/h. Swimmer speed 5 km/h. Find shortest time to cross.

Solution:
Shortest time implies heading perpendicular to bank.
t = Width / v_swimmer.
t = 0.5 km / 5 km/h = 0.1 hour.
t = 0.1 × 60 = 6 minutes.

Q5. A particle moves in a circle of radius 2 m at constant speed 10 m/s. Find centripetal acceleration.

Solution:
a_c = v² / r.
a_c = (10)² / 2 = 100 / 2 = 50 m/s².

Q6. A body projected at 45° reaches max height H. What is range R?

Solution:
R = 4H cotθ.
At θ = 45°: R = 4H cot 45° = 4H (1).
R = 4H.

Q7. If A = 2i + 3j and B = i - j, find angle between them.

Solution:
A ⋅ B = |A||B| cosθ.
Dot Prod = (2)(1) + (3)(-1) = 2 - 3 = -1.
|A| = √(4+9) = √13. |B| = √(1+1) = √2.
cosθ = -1 / √26.
θ = cos^-1(-1 / √26).

Q8. A bomb is dropped from a plane flying horizontally at 360 km/h at height 490 m. How far horizontally will it hit?

Solution:
u = 360 km/h = 100 m/s.
Time to fall t = √(2h/g) = √(2×490 / 9.8) = √100 = 10 s.
Range = u × t = 100 × 10 = 1000 m (1 km).

Q9. A car turns 90° maintaining speed 10 m/s. Find change in velocity.

Solution:
Δv = v_f - v_i.
Magnitidue |Δv| = √(v² + v² - 2v²cos 90) = √(2v²) = v√2.
Change = 10 √2 14.14 m/s.

Q10. Calculate angular speed of seconds hand of a clock.

Solution:
ω = 2π / T.
Seconds hand period T = 60 s.
ω = 2π / 60 = π / 30 rad/s ≈ 0.105 rad/s.

Important Formulae

1. Vector Algebra

Dot Product (Scalar):

A ⋅ B = AB cosθ

Cross Product (Vector):

|A × B| = AB sinθ

Resultant R² = A² + B² + 2ABcosθ
Unit Vector n̂ = A / |A|

2. Projectile Motion

Time of Flight (T):

T = 2usinθ / g

Max Height (H):

H = u²sin²θ / 2g

Horizontal Range (R):

R = u²sin2θ / g

3. Circular Motion

Linear vs Angular:

v = rω | a_t = rα

Centripetal Acceleration:

a_c = v²/r = rω²

Net Acceleration a = √(a_c² + a_t²)

4. Relative Motion

Basic Equation:

v_AB = v_A - v_B

Rain-Man:

v_RM = v_R - v_M

River-Boat: v_actual = v_boat + v_river

20 NEET Golden Facts

1. Orthogonal Vectors: Dot product is zero (cos 90° = 0). A ⋅ B = 0.
2. Area of Triangle: Half magnitude of cross product of two side vectors. Area = ½ |A × B|.
3. Max Range: Projectile range is maximum at 45°. R_max = u²/g. Max Height at 45° is R/4.
4. Same Range: Angles θ and (90° - θ) give same range. (e.g., 30° and 60°).
5. Top of Trajectory: Velocity is ucosθ (Horizontal). Acceleration is g (Vertical). Angle is 90°.
6. Centripetal Force: Does zero work in Uniform Circular Motion (Force is perpendicular to velocity).
7. River Crossing: To cross in shortest time, head straight across perpendicular to flow (Drift will occur).
8. Horizontal Projectile: Height fallen depends only on time. h = ½gt².
9. Commutative: Vector Addition and Dot Product are commutative. Cross product is not (A×B = -B×A).
10. Null Vector: Magnitude zero, arbitrary direction. Result of A - A.
11. Tangential Accel: Changes magnitude of velocity (speed). Centripetal accel changes direction.
12. Rain Umbrella: Umbrella should be held opposite to relative velocity of rain with respect to man.
13. Equal Vectors: Same magnitude and same direction.
14. Change in Velocity: In UCM, speed is constant but velocity changes. Change in V in half rotation is 2v.
15. Trajectory Equation: Useful to find y at given x without calculating time.
16. Parallel Vectors: Cross product is zero (sin 0° = 0).
17. Speed at Height: At height h, speed v = √(u² - 2gh).
18. Angular Speed: Second hand of clock: ω = 2π/60 rad/s. Earth rotation: 2π/24 hr.
19. Range Relation: R = 4H cotθ. If R=H, then tanθ=4 (θ=76°).
20. Lami's Theorem: For 3 concurrent forces in equilibrium, F₁/sinα = F₂/sinβ = F₃/sinγ.

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