Kinematics: JEE Advanced PYQ, 2007–2026

Answer: (C)

The angular speeds are ω₁ = 2π/3 and ω₂ = 2π/5. For motion in the same direction, the faster particle must gain one full turn (2π) on the slower one. Time = 2π/(ω₁ − ω₂) = 2π/[2π(1/3 − 1/5)] = 1/(2/15) = 7.5 s. By then the faster particle has made 2.5 turns and the slower one 1.5 turns, a difference of exactly one turn.

Answer: (A)

At latitude λ the person moves in a circle of radius r = R cos λ = 6400 km × 0.5 = 3.2 × 10⁶ m, about the Earth's axis. The centripetal acceleration is a = ω²r = (7.27 × 10⁻⁵)² × 3.2 × 10⁶ ≈ 1.7 × 10⁻² m/s², directed perpendicular to the axis. The line from the person to the Earth's centre makes angle λ = 60° with this horizontal direction. The component along it is a cos 60° ≈ 8.5 × 10⁻³ m/s².

Answer: (D)

The velocity–time graph is a triangle with the peak at speed v. The rising slope is 2, so t₁ = v/2. The falling slope has magnitude 3, so t₂ = v/3. The total time is t₁ + t₂ = v(1/2 + 1/3) = 5v/6 = 10 s, so v = 12 m s⁻¹. Check: t₁ = 6 s and t₂ = 4 s, which are not equal. The distance travelled is ½ × 12 × 10 = 60 m, and the average speed of 6 m s⁻¹ is half the peak speed.

Answer: (A)

Take v₁ = 10î and v₂ = 10(cos120° î + sin120° ĵ) = −5î + 5√3 ĵ. Then Δv = v₂ − v₁ = −15î + 5√3 ĵ, so |Δv| = √(225 + 75) = √300 = 10√3 m/s. The same result follows from |Δv|² = v² + v² − 2v² cos120° = 3v². The speed is |v| = 10 m/s both before and after, so the change in speed is zero. The magnitude of the change in a vector is not the change in its magnitude.

Answer: (D)

The area of a parallelogram with adjacent sides A and B equals the magnitude of the vector product |A × B|. For vectors in the x–y plane, A × B = (A_xB_y − A_yB_x) k̂ = (2·1 − 3·4) k̂ = −10 k̂. The magnitude is 10 m². The scalar product (11), half the area (5) and a sign slip (14) are the tempting errors.

Answer: (A)

The first fall takes t₀ = √(2h/g) = 2 s and the impact speed is 20 m/s. Each rebound at speed e·v lasts 2ev/g, so the rebound times are 2 s, 1 s, 0.5 s, …, a geometric series. The total time is 2 + 2(1 + ½ + ¼ + …) = 2 + 4 = 6 s. Rebound heights are 5 m, 1.25 m, … (ratio ¼). The total distance is 20 + 2(5 + 1.25 + …) = 20 + 2(5)/(1 − ¼) = 20 + 13.3 = 33.3 m. The average speed is 33.3/6 ≈ 5.6 m/s.

Answer: (D)

The speed is v = 2πr/T = 2π × 2/4 = π m/s. In a quarter of a revolution, which takes 1 s, the velocity turns through 90°, so the change in velocity has magnitude √(v² + v²) = √2 v = π√2 m/s. The average acceleration is π√2/1 = π√2 m/s², about 4.44 m/s².

Answer: (A)

Radius of curvature R = v²/a⊥. At launch the velocity makes 60° with the horizontal, so the angle between velocity and the vertical gravity vector is 30°. The component of g perpendicular to the velocity is g cos60°, and R₁ = u²/(g cos60°). At the highest point the velocity is horizontal with magnitude u cos60°, and g is entirely perpendicular to it, so R₂ = u²cos²60°/g. Dividing gives R₁/R₂ = 1/cos³60° = 8.

Answer: (C)

Since R = A + B, we have B = R − A. Because R and A are perpendicular, |R − A|² = |R|² + |A|² − 2R·A = 16 + 9 − 0 = 25. So |B| = 5 units. In the triangle formed by A, B and R, B is the hypotenuse because the right angle lies between A and R. The wrong options come from adding or subtracting magnitudes, or from putting the right angle in the wrong place.

Answer: (A)

The graph is a straight line, so v = 10 − 2x (SI units), with slope dv/dx = −2 s⁻¹. Acceleration is a = dv/dt = (dv/dx)(dx/dt) = v(dv/dx). At x = 2 m, v = 10 − 4 = 6 m/s. So a = 6 × (−2) = −12 m/s². The slope of a v–x graph is not the acceleration; it must be multiplied by v.