Subtopics - Motion In Two Dimension (NEET)
Four major blocks: oblique projectile (trajectory, T, R, H, complementary angles, max range), horizontal projectile (time of flight, range, velocity components), uniform circular motion (angular quantities, centripetal acceleration and force, centrifugal force), and applications of circular motion (skidding, banking, bending of cyclist, overturning, vertical circle).
1) Oblique Projectile Motion
A body projected at angle θ with horizontal speed u follows a parabolic trajectory governed by y = x tanθ − gx²/(2u²cos²θ). The horizontal (u cosθ) and vertical (u sinθ) components act independently. Key results: T = 2u sinθ/g; H = u²sin²θ/2g; R = u²sin2θ/g; R_max = u²/g at θ = 45° with H = R_max/4. Complementary angles θ and (90°−θ) give equal R. R = 4H cotθ; if R = nH then tanθ = 4/n. Relative motion between two projectiles tracks a straight line. KE at highest point = (1/2)mu²cos²θ.
2) Horizontal Projectile Motion
A body projected horizontally with speed u from height h. The horizontal velocity remains constant at u throughout. Vertical velocity builds from zero under gravity. Trajectory: y = gx²/2u² (downward parabola). Time of flight T = √(2h/g). Horizontal range R = u√(2h/g). Instantaneous speed v = √(u² + 2gy). Angle of velocity with horizontal: tanφ = √(2gy)/u = gt/u.
3) Uniform Circular Motion
A particle moving in a circle at constant speed. Angular displacement θ is an axial vector. Angular velocity ω = dθ/dt = 2π/T = 2πn (axial vector). Centripetal acceleration a = v²/r = ω²r, always directed towards centre. Centripetal force F = mv²/r = mω²r. Work done by centripetal force is always zero (force ⊥ displacement). Centrifugal force is a fictitious reaction force in the rotating frame. Angular acceleration α = dω/dt = 0 for uniform circular motion.
4) Applications of Circular Motion
Practical consequences of centripetal force requirements: (1) Skidding on level road — safe speed v ≤ √(μrg). (2) Banking of road — tan θ = v²/rg; with friction v_max = √(rg(μ+tanθ)/(1−μtanθ)). (3) Bending of cyclist — same relation tan θ = v²/rg because horizontal reaction provides centripetal force. (4) Overturning of vehicle — happens when v > √(gra/h) where a = half wheel-base and h = centre-of-gravity height. (5) Vertical circle — minimum speed at lowest point √(5gl); critical speed at top √(gl).
Motion In Two Dimension Download Notes & Weightage Plan
For each topic in the Motion In Two Dimension chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.
Derive trajectory equation, T, R, H from component-wise kinematics. Understand complementary-angle range equality and derivation of R_max = u²/g at 45°.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Know T, R, H cold. For complementary angles remember: same R, T ratio = tanθ, H ratio = tan²θ. For max range: 45°, R_max = u²/g, H = R/4.
- High-risk Area: Confusing H = u²sin²θ/2g with R = u²sin2θ/g — NEET frequently places both as options. Also, applying the horizontal range formula when the projectile lands at a different height from projection.
- Best Practice Style: Substitution numericals (given u, θ find R/H/T). Also short proof: show R = 4H cotθ. Complementary angle MCQs at higher difficulty.
Projectile launched horizontally from height h; trajectory y = gx²/2u²; T = √(2h/g); R = u√(2h/g); v = √(u²+2gy).
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: T = √(2h/g) and R = u√(2h/g) are the two direct-scoring formulas. For velocity, know v = √(u²+2gy).
- High-risk Area: Forgetting that the horizontal component of velocity stays constant (= u) throughout. Students sometimes erroneously apply Newton's equations to the horizontal direction.
- Best Practice Style: Short one-step or two-step numericals. NEET may combine with free-fall comparisons (e.g., which body reaches ground first).
Angular kinematics (ω, α, θ), centripetal acceleration a = v²/r = ω²r, centripetal force F = mv²/r, centrifugal (fictitious) force, work done = 0.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Know a = v²/r = ω²r and F = mv²/r. Understanding that centripetal force is always zero net work is a frequent True/False type MCQ.
- High-risk Area: Treating centrifugal force as a real force in inertial frame analysis. Also confusing centripetal acceleration direction (inward) with velocity direction (tangential).
- Best Practice Style: Identification MCQs (which force is centripetal in a given setting) plus numericals for a and F.
Applications of Circular Motion
Skidding (v ≤ √(μrg)), banking (tanθ = v²/rg), bending of cyclist (tanθ = v²/rg), overturning (v = √(gra/h)), vertical circle (u_min = √(5gl)).
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Banking formula tanθ = v²/rg appears most often. Safe speed √(μrg) and overturning speed √(gra/h) appear regularly.
- High-risk Area: Vertical circle — students often forget the minimum speed condition √(5gl) applies at the bottom, not the top. Also mixing up banking with flat-road conditions.
- Best Practice Style: Free-body-diagram derivation questions followed by numerical substitution. NEET often tests: 'what happens to safe speed if mass doubles?' (answer: unchanged — mass cancels).
Motion In Two Dimension Chapter NEET Traps & Common Mistakes (Topic-Wise)
Each subtopic below is of the Motion In Two Dimension chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.
Mistake Snapshot (What Students Do Wrong)
- Swapping sin2θ and sin²θ: R = u²sin2θ/g uses sin2θ = 2sinθcosθ (involves both components). H = u²sin²θ/2g uses sin²θ (only vertical component). Students swap the sin-squared vs sin-double-angle and get the wrong formula.
- Using oblique range formula when launch and landing heights differ: R = u²sin2θ/g only holds when the projectile returns to the same vertical height as launch. When landing on a cliff or slope, this formula is inapplicable.
u = 20 m/s, θ = 30°. R = 400 × sin60° / 10 = 400 × (√3/2) / 10 = 20√3 m ≈ 34.6 m. H = 400 × sin²30° / 20 = 400 × 0.25 / 20 = 5 m. Note R >> H at 30°.
How NEET Frames The Trap
NEET lists H and R as answer options A and B with values swapped to catch formula confusion.
Q. A ball is projected with initial velocity 20 m/s at 30° above horizontal. What is its maximum height? (g=10 m/s²)
A. 5 m B. 20√3 m C. 10 m D. √3 m
Trick: Correct answer is 5 m (H = u²sin²θ/2g). Option B (20√3 m) is the range and is deliberately placed to trap students who swap formulas.
Mistake Snapshot (What Students Do Wrong)
- Assuming time of flight is also equal for complementary angles: Range is equal for θ and (90°−θ), but time of flight T₁/T₂ = tanθ ≠ 1. NEET exploits this by asking about T after planting the range-equality concept.
- Forgetting that maximum height ratio is tan²θ: H₁/H₂ = tan²θ, not tanθ. Students who recall the T ratio often write the same for H.
θ = 30°, θ' = 60°. T₁/T₂ = tan30° = 1/√3. H₁/H₂ = tan²30° = 1/3. R₁/R₂ = 1.
How NEET Frames The Trap
Question states two projectiles have equal range. 'Which pair has equal time of flight?' — the trap is that equal range does NOT imply equal T.
Q. Two projectiles are thrown with the same speed at 30° and 60° to the horizontal. The ratio of their times of flight T₁ : T₂ is:
A. 1:√3 B. 1:3 C. 1:1 D. √3:1
Trick: Correct answer is 1:√3 (T = 2usinθ/g so ratio = sin30°/sin60° = (1/2)/(√3/2) = 1/√3). Students picking 1:1 confuse this with the equal-range result.
Mistake Snapshot (What Students Do Wrong)
- Thinking centripetal force does work because speed is maintained: Centripetal force keeps the particle on the circular path but is always perpendicular to velocity. Since W = F·d·cosθ and θ = 90°, work is always zero regardless of the speed or duration.
- Confusing centripetal force with centrifugal force in inertial frame: In the ground (inertial) frame, centrifugal force does not exist. Only centripetal force acts, directed inward, doing zero work.
An electron orbits a nucleus at constant speed. KE stays constant because the electrostatic force (centripetal) does zero work at every instant.
How NEET Frames The Trap
MCQ: 'How much work does the centripetal force do on a satellite in one complete orbit?' — distractor answers include non-zero values.
Q. A particle moves in a uniform circular path. The work done by the centripetal force in one complete revolution is:
A. Zero B. mv²/r × 2πr C. −mv²/r × 2πr D. mv²
Trick: Correct answer is Zero. Centripetal force is always perpendicular to displacement (tangent), so dot product F·ds = 0 at every point. No work is accumulated over any arc or full revolution.
Mistake Snapshot (What Students Do Wrong)
- Applying the banking formula to a flat (level) road: On a flat road, there is no banking angle; safe speed comes from friction alone: v ≤ √(μrg). Students incorrectly include a tanθ term when θ = 0 for a flat road.
- Forgetting that safe speed on level road is mass-independent: v_safe = √(μrg) has no m in it because centripetal force required and friction force available both scale identically with mass. NEET tests 'effect of doubling mass on safe speed' — answer is no change.
μ = 0.5, r = 50 m, g = 10 m/s². v_safe = √(0.5 × 50 × 10) = √250 = 5√10 m/s ≈ 15.8 m/s. Doubling mass: v_safe still = 5√10 m/s.
How NEET Frames The Trap
NEET asks max safe speed on a circular level road and lists mass-dependent options.
Q. A vehicle of mass 1000 kg takes a circular turn of radius 50 m on a flat road (μ = 0.5, g = 10 m/s²). The maximum safe speed is:
A. 5√10 m/s B. 50 m/s C. 5√10/1000 m/s D. 10√5 m/s and depends on mass
Trick: Correct answer is 5√10 m/s. Mass cancels completely. Option C (dividing by mass) is the main trap for students who write μmg = mv²/r without cancelling m.