Problem Types & Solved Examples
6 categories. Examiner perspective on every problem. Think first, calculate second.
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0/0 (0%)Direct substitution into one formula. CBSE 2-mark and NEET single-step MCQs. Speed matters here — solve in under 60 seconds or you're wasting time.
Use sin30° = 1/2, cos30° = √3/2. v₀sinθ = 20×½ = 10, v₀cosθ = 20×(√3/2) = 10√3.
T = 2v₀sinθ/g = 2×20×sin30°/10 = 2×20×0.5/10 = 2 s
H = v₀²sin²θ/2g = (20)²×(0.5)²/(2×10) = 400×0.25/20 = 5 m
R = v₀²sin2θ/g = (20)²×sin60°/10 = 400×(√3/2)/10 = 20√3 ≈ 34.6 m
For θ=30°: T=v₀/g, H=v₀²/8g, R=v₀²√3/2g. Memorize the 30-45-60 results and you'll save 40 seconds per problem.
ω = v/r = 6/2 = 3 rad/s
T = 2π/ω = 2π/3 ≈ 2.09 s
aᶜ = v²/r = 36/2 = 18 m/s² (toward center)
Understanding without numbers. JEE uses this to separate mugged formulae from real understanding. Read carefully — the trap is in the wording.
Horizontal: Ball 1 has vₓ = 2u, Ball 2 has vₓ = u. Vertical (same height h): vy = √(2gh) for both.
Speed₁ = √(4u² + 2gh), Speed₂ = √(u² + 2gh). Unless h is specified, ratio is NOT simple.
Students pick 2:1 thinking only horizontal matters. WRONG — vertical speed adds too. If h → ∞, ratio → 1:1; if h → 0, ratio → 2:1. Context matters!
R = v₀²sin2θ/g, so v₀²/g = R/sin2θ
H' = v₀²sin²(90°−θ)/2g = v₀²cos²θ/2g
H' = (v₀²/g) × cos²θ/2 = R·cos²θ/(2sin2θ) = R·cos²θ/(4sinθcosθ) = R·cosθ/4sinθ = R·cotθ/4
For θ=45°: H' = R/4. This is a standard result. For complementary angles, H₁+H₂ = R/4 (when θ=45°). Memorize this for MCQ speed.
JEE Main and NEET multi-step problems require 3–4 steps. The order of steps matters — do it wrong and you'll get a wrong intermediate result that cascades.
Take launch point as origin. Down is negative. At ground: y = −40 m (not 0). This is where most students make an error.
v₀y = 20sin60° = 20×(√3/2) = 10√3 m/s upward
v₀x = 20cos60° = 20×0.5 = 10 m/s
Use y = v₀y·t − ½gt²: −40 = 10√3·t − 5t²
5t² − 10√3·t − 40 = 0 → t² − 2√3·t − 8 = 0
t = [2√3 ± √(12+32)]/2 = [2√3 ± √44]/2 = [2√3 ± 2√11]/2 = √3 + √11 ≈ 1.73 + 3.32 = 5.05 s
Range = v₀x × t = 10 × 5.05 ≈ 50.5 m
Take POSITIVE root only (t cannot be negative). The quadratic gives two roots — discard the negative one. This costs marks when students forget.
Required centripetal force = mv²/r = 1000×400/50 = 8000 N
Max friction available = μmg = 0.4×1000×10 = 4000 N
Required (8000N) > Available (4000N) → Car is NOT safe. It will skid outward.
Max speed: μmg = mv_max²/r → v_max = √(μgr) = √(0.4×10×50) = √200 = 10√2 ≈ 14.1 m/s
v_max = √(μgr) — This is the maximum speed on a flat circular road. For banked road: v_max = √(rg·tan θ) without friction. These two come in NEET repeatedly.
Graph-based problems appear in CBSE (3-mark), JEE Main (MCQ), and JEE Advanced (matching). The key: extract physics from the graph shape, slope, and intercepts.
Graph slopes = rate of change = velocity. For x-t: slope = vₓ = constant. For y-t: slope = vy = changes with time.
Slope of x-t graph = dx/dt = horizontal velocity (vₓ = v₀cosθ) = constant. That's why x-t is a straight line.
y-intercept of y-t graph = initial y position. If launched from ground: y-intercept = 0. If from height h: y-intercept = h.
Slope of y-t graph = vy = 0 when t = v₀sinθ/g (at maximum height).
• x-t: straight line (uniform x-velocity)
• y-t: upside-down parabola (opens downward)
• vₓ-t: horizontal line (constant)
• vy-t: straight line with slope = −g
• y-x (trajectory): parabola (shape of path)
Speed-time: Horizontal straight line (speed = constant in UCM). Graph is a flat line at v = constant.
KE-time: Horizontal straight line. KE = ½mv² = constant since speed is constant.
aᶜ-time: Horizontal straight line. aᶜ = v²/r = constant since v and r are constant.
Don't confuse "speed = constant" with "displacement = constant". Displacement changes every moment in UCM. The displacement-time graph for UCM is NOT a straight line.
Both A and R can be true/false, independently. Key: even if both are true, R must be the correct explanation of A. This format appears in 2–3 questions per CBSE paper and occasionally in NEET.
(A) Both A and R are true, R explains A
(B) Both A and R are true, R doesn't explain A
(C) A is true, R is false
(D) A is false, R is true
Assertion (A): The horizontal range of a projectile is maximum when the angle of projection is 45°.
Reason (R): The horizontal range is given by R = v₀²sin2θ/g, which is maximum when sin2θ = 1, i.e., 2θ = 90°, θ = 45°.
Both A and R are correct. A is a standard result. R provides the correct mathematical justification — sin2θ is maximum at 2θ=90°, so θ=45°. R correctly explains A. Answer: (A).
Assertion (A): In uniform circular motion, the particle undergoes acceleration.
Reason (R): The speed of the particle changes continuously in UCM.
A is TRUE — In UCM, direction changes → velocity changes → acceleration exists (centripetal).
R is FALSE — Speed is CONSTANT in UCM; only direction changes. R incorrectly claims speed changes. Answer: (C).
A paragraph or scenario is given, followed by 4–5 related questions. Read the passage carefully first — all data is hidden inside it. This appears as a 5-mark question in CBSE boards.
📋 Case Study Passage
A cricket player hits a ball at an angle of 60° with the horizontal with a velocity of 30 m/s. The ground is level and g = 10 m/s². The ball is caught by a fielder at the same height as it was hit. The fielder starts running at the moment the ball is hit.
T = 2v₀sinθ/g = 2×30×sin60°/10 = 2×30×(√3/2)/10 = 60√3/20 = 3√3 ≈ 5.2 s
H = v₀²sin²θ/2g = (30)²×(√3/2)²/(2×10) = 900×0.75/20 = 675/20 = 33.75 m