Motion in 2D
🟢 Lite — Quick Review (1h–1d)
Rapid summary for last-minute revision before your exam.
Motion in 2D — Key Facts for JEE Main • Projectile motion: Position vector r = xî + yĵ; velocity components: vₓ = v₀cosθ (constant), vᵧ = v₀sinθ − gt; displacement: x = v₀cosθ·t, y = v₀sinθ·t − ½gt² • Time of flight: T = 2v₀sinθ/g; Range: R = v₀²sin2θ/g = v₀²sin2θ/g; Max height: H = v₀²sin²θ/2g • Relative motion: v_AB = v_A − v_B; used for boat-river problems (crossing shortest path vs fastest path) and rain-man problems • Circular motion: ω = dθ/dt (rad/s); v = ωr; Centripetal acceleration: a_c = v²/r = ω²r; Centrifugal (pseudo): F_cf = mv²/r = mω²r • Angular kinematics: θ = ω₀t + ½αt²; ω = ω₀ + αt; ω² = ω₀² + 2αθ ⚡ Exam tip: Range R = v₀²/g when θ = 45° gives maximum range in horizontal ground — but angle changes for elevated/depressed targets; in JEE, check the surface before applying R_max formula!
🟡 Standard — Regular Study (2d–2mo)
Standard content for students with a few days to months.
Motion in 2D — JEE Main / Advanced Study Guide Vector basics: Position vector r traces trajectory; instantaneous velocity v = dr/dt (tangent to path); acceleration a = dv/dt. Components: vₓ = dx/dt, vᵧ = dy/dt. In 2D, treat x and y independently (Principle of Independence).
Projectile motion: Split into horizontal (aₓ = 0, vₓ = v₀cosθ) and vertical (aᵧ = −g, vᵧ = v₀sinθ − gt). At any time t: x = v₀cosθ·t; y = v₀sinθ·t − ½gt². Trajectory equation: y = xtanθ − gx²/(2v₀²cos²θ) = xtanθ − (gx²)/(2u²)(1 + tan²θ). Time to reach maximum height: t_h = v₀sinθ/g. Total time: T = 2t_h = 2v₀sinθ/g. Range on level ground: R = v₀²sin2θ/g.
Important cases: (i) Same range for θ and (90°−θ); (ii) Maximum range at θ = 45° → R_max = v₀²/g; (iii) For given u and H, two angles possible when target not at same level; (iv) For a given speed v₀, maximum height attained = v₀²/2g (when thrown vertically).
Circular motion: Angular velocity ω = v/r, direction by right-hand rule. Tangential acceleration a_t = rα (change in speed); centripetal/Normal acceleration a_n = v²/r (change in direction). Net acceleration: a = √(a_t² + a_n²). Banking of roads: tanθ = v²/rg (no friction needed); conical pendulum: Tcosθ = mg, Tsinθ = mω²r.
Relative motion: v_AB = velocity of A relative to B = v_A − v_B. River-swimmer problem: downstream speed = v_swim + v_river; upstream = v_swim − v_river. Shortest path across river: head directly opposite point (perpendicular to banks) if v_river < v_swim.
Solved Example 1: A ball thrown at 40 m/s at 30° hits a platform 30 m high. Find range. y = 30 = 40·sin30°·t − ½·10·t² → 5t² − 20t + 30 = 0 → t = 3 s (or t = 2 s below platform). x = 40·cos30°·3 = 40·0.866·3 ≈ 104 m.
Solved Example 2: A car accelerates uniformly from rest at 2 m/s² on a circular track of radius 100 m. Find speed after 10 s and angle of resultant acceleration. v = at = 20 m/s. a_t = 2 m/s², a_n = v²/r = 400/100 = 4 m/s². tanφ = a_n/a_t = 4/2 = 2 → φ = tan⁻¹(2) ≈ 63.4°.
⚡ Exam tip: In projectile questions, always check whether the landing surface is horizontal or not — level ground simplifies to R = v₀²sin2θ/g; non-level surfaces require solving the trajectory equation for the specific y-value.
🔴 Extended — Deep Study (3mo+)
Comprehensive coverage for students on a longer timeline.
Motion in 2D — Comprehensive JEE Notes Advanced projectile analysis: For launch from height h above ground: y = xtanθ − gx²/(2u²cos²θ) + h. Range R = [u cosθ/g][u sinθ + √(u²sin²θ + 2gh)]. When h > 0 (launch from height), optimal angle < 45°. When h < 0 (depression), optimal angle > 45°. Projectile on inclined plane: Set x-axis along plane; equations transform with modified g component along plane.
Kinematics in polar coordinates: r = rr̂ + zẑ; velocity v = ṙr̂ + rθ̇θ̂; acceleration a = (r̈ − rθ̇²)r̂ + (rθ̈ + 2ṙθ̇)θ̂. Radial component = (r̈ − rθ̇²), transverse component = (rθ̈ + 2ṙθ̇). For uniform circular motion: θ̇ = ω, θ̈ = 0 → a_r = −ω²r = −v²/r (centripetal inward).
Non-uniform circular motion: a_t = rα (tangential) causes change in speed; a_n = v²/r causes change in direction. Total acceleration = √(a_t² + a_n²). Relationship: a = √[(dr̈/dt − rθ̇²)² + (rθ̈ + 2ṙθ̇)²]. Motion in vertical circle: For a mass on a string in vertical circle: minimum speed at top v_min = √(gR) to keep tension non-negative; at bottom v = √(5gR) to complete circle. Tension varies: T = m(v²/R ± g) at bottom/top respectively.
Relative motion in 2D: For rain falling vertically at v_r with man moving horizontally at v_m: apparent rain velocity = v_r − v_m; apparent direction = tan⁻¹(v_m/v_r) from vertical. For two moving bodies A and B: r_AB = r_A − r_B; differentiate to get velocities and accelerations. Closest approach: d(r_AB)/dt = 0 → components perpendicular to relative position.
Solved Example (Advanced): A projectile is fired at 100 m/s at 30° from a 50 m cliff. Find horizontal distance from cliff base when it hits the ground below. y = −50 = 100·sin30°·t − 5t² → 5t² − 50t − 50 = 0 → t = [50 ± √(2500 + 1000)]/10 = [50 ± 59.2]/10 → t = 10.9 s. x = 100·cos30°·10.9 = 86.6·10.9 ≈ 944 m.
⚡ JEE Advanced tip: In rotating reference frame problems, use Coriolis acceleration = 2Ω × v for particles moving on rotating bodies (Earth’s rotating frame, turntable problems). Direction: Coriolis deflects moving objects right in Northern Hemisphere, left in Southern.
📊 JEE Main Exam Essentials
| Detail | Value |
|---|---|
| Questions | 90 (30 per subject) |
| Sections | Physics, Chemistry, Mathematics |
| Type | MCQ + Numerical Value (NAT) |
| Time | 3 hours |
| Marking | +4 correct, −1 wrong (MCQ); +4 correct, 0 wrong (NAT) |
| Sessions | January + April per year; best score considered |
🎯 High-Yield Topics for JEE Main
- Projectile Motion — 3–4 marks
- Circular Motion & Banking — 3–4 marks
- Relative Velocity — 2–3 marks
- Vector Algebra (basics for all physics) — 5+ marks
📝 Previous Year Question Patterns
- Q: “A particle is projected at 60° to the horizontal with velocity 30 m/s…” [2025 Physics — 3 marks]
- Q: “A car moves on a circular road of radius 100 m. The angle of banking is…” [2024 Physics — 3 marks]
- Q: “Two boats A and B move with velocities 5 km/h and 12 km/h respectively…” [2023 Physics — 4 marks]
💡 Pro Tips
- Projectile motion: 60% of questions test the independence of horizontal and vertical motion
- Always resolve initial velocity along and perpendicular to the incline in inclined plane problems
- In relative velocity, the key is to correctly subtract the reference frame velocity
- Remember: centripetal force = m×v²/r points toward center; centripetal is NOT a new force (tension, gravity, normal can provide it)
🔗 Official Resources
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📐 Diagram Reference
Clean educational diagram showing projectile motion with trajectory, velocity components, range, maximum height annotated, angle of projection, white background, exam-style illustration
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