JEE Physics Relative Motion MCQs Set 04

Answer: (d) 15 m/s, 5 m/s

Explanation:
1. Overtaking: relative speed = v1 - v2 = 200/20 = 10 m/s.
2. Crossing: v1 + v2 = 200/10 = 20 m/s.
3. So v1 = 15 m/s and v2 = 5 m/s.

Teacher's Note:
1. The relative distance in each case is the sum of the two lengths, 200 m.
2. Same direction: subtract speeds. Opposite directions: add speeds.

Answer: (a) 4 s

Explanation:
1. Relative speed = 10 + 15 = 25 m/s.
2. Relative distance = 50 + 50 = 100 m, so t = 100/25 = 4 s.

Teacher's Note:
1. For opposite directions the speeds add.
2. Add the two lengths.

Answer: (c) u

Explanation:
1. vA = u - gt (upward) and vB = -gt (downward, released from rest).
2. Relative velocity = vA - vB = u - gt + gt = u.

Teacher's Note:
1. Both balls have the same acceleration g.
2. So their relative velocity stays constant.

Answer: (a) \( \frac1{\sqrt2} \) s

Explanation:
1. Relative to the elevator the coin starts from rest with acceleration g.
2. 2.5 = (1/2)(10)t², so t = 1/√2 s.

Teacher's Note:
1. The constant speed of the elevator does not matter.
2. Use the frame of the elevator.

Answer: (c) 1250 m

Explanation:
1. The initial relative speed is zero, so the relative displacement in 50 s is (1/2)(1)(50)² = 1250 m.
2. This equals the original gap, 1250 m.

Teacher's Note:
1. Relative initial velocity is zero.
2. Use s = (1/2)at² for the relative motion.

Answer: (d) 36 s

Explanation:
1. Walking speed = L/90 and escalator speed = L/60.
2. Combined speed = L(1/90 + 1/60) = L/36.
3. So the time is 36 s.

Teacher's Note:
1. Add the speeds when they act in the same direction.
2. Use L as the escalator length.

Answer: (b) 1 m/s²

Explanation:
1. Speeds: A = 10 m/s, B = C = 15 m/s. Take A at x = 0, B at x = -1000 m moving forward, C at x = +1000 m moving backward.
2. C reaches A's position when 1000 - 15t = 10t, i.e. t = 40 s, at x = 400 m.
3. B must cover 1400 m in 40 s: 1400 = 15(40) + (1/2)a(40)², so a = 1 m/s².

Teacher's Note:
1. B must pass A before C arrives at A's position.
2. Find that time first, then the needed acceleration.

Answer: (c) \( \cos^{-1}\left(\frac vu\right) \)

Explanation:
1. To move along AB relative to the platform, the particle's component along AD must equal the platform's velocity.
2. u cos θ = v, so θ = cos⁻¹(v/u).

Teacher's Note:
1. Relative to the platform, the particle must have no velocity along AD.
2. Match the horizontal components.

Answer: (d) 3 km/h

Explanation:
1. Downstream drift = 300 m in 10 min gives a current of 30 m/min.
2. Crossing speed in still water = 200/5 = 40 m/min (5 minutes for one crossing).
3. Resultant speed = √(30² + 40²) = 50 m/min = 3 km/h.

Teacher's Note:
1. The time to cross depends only on the perpendicular speed.
2. Add the perpendicular and downstream velocities vectorially.

Answer: (a) 3

Explanation:
1. Speed across the river along the shortest path = 1 km/0.25 h = 4 km/h.
2. u² = v² + 4²: 25 = v² + 16, so v = 3 km/h.

Teacher's Note:
1. Shortest path means the boat's velocity is perpendicular to the banks relative to the ground.
2. Use Pythagoras.

Answer: (a) 60

Explanation:
1. Speed across = 320/4 = 80 m/min (perpendicular to the current).
2. (5c/3)² = c² + 80², so (16/9)c² = 6400 and c = 60 m/min.

Teacher's Note:
1. His still-water velocity has a component against the current.
2. Use Pythagoras with the swimming speed as the hypotenuse.

Answer: (c) due north to reach in the shortest time

Explanation:
1. Shortest time: head perpendicular to the banks, i.e. due north.
2. Shortest path: sin θ = 5/10, so θ = 30° west of north, so (b), (c) and (d) are wrong.

Teacher's Note:
1. Time depends only on the velocity component across the river.
2. Shortest path needs the drift to be cancelled.

Answer: (c) \( \frac{40}{3} \) m/s

Explanation:
1. The projectile must have no horizontal velocity relative to the cart, so it is fired vertically (relative to the cart).
2. Time = 80/30 = 8/3 s = 2u/g, so u = (10)(8/3)/2 = 40/3 m/s.

Teacher's Note:
1. Match the horizontal velocities to land back on the cart.
2. The time of flight of a vertical throw is 2u/g.

Answer: (d) \( \frac{av_1}{v^2} \)

Explanation:
1. Separation squared: d² = (a - v1t)² + (v2t)².
2. Minimising: -2v1(a - v1t) + 2v2²t = 0, so t = av1/(v1² + v2²) = av1/v².

Teacher's Note:
1. Differentiate the squared distance and set it to zero.
2. v² = v1² + v2².

Answer: (c) \( \frac va\cos\alpha \)

Explanation:
1. The relative velocity squared = v² + a²t² - 2vat cos α.
2. Its minimum occurs when 2a²t - 2va cos α = 0, so t = (v/a) cos α.

Teacher's Note:
1. Write the relative velocity as the vector difference.
2. Minimise its magnitude with respect to t.

Answer: (c) \( 4\hat i+6\hat j \)

Explanation:
1. Case 1: the boat's velocity is 4î; the wind relative to the boat is northward, so the wind's x-component is 4.
2. Case 2: the boat's velocity is î + 3ĵ; the relative wind is along north-east, so (4 - 1) = (wy - 3), giving wy = 6.
3. Wind velocity = 4î + 6ĵ.

Teacher's Note:
1. The flag points along the wind velocity relative to the boat.
2. Use the two cases to solve for the two components.

Answer: (d) i, ii, iii

Explanation:
1. Time = 100/4 = 25 s.
2. Absolute velocity = (4 - 2)î + 4ĵ = 2î + 4ĵ, with magnitude √20 = 2√5 m/s.
3. Drift = 2 × 25 = 50 m.

Teacher's Note:
1. The crossing time depends only on the y-component of the velocity.
2. Add the river velocity to get the absolute velocity.

Answer: (c) ii, iii

Explanation:
1. To fly due north the westward component of the plane's air velocity must cancel the wind: 200 sin θ = 90, so sin θ = 0.45.
2. Ground speed = 200 cos θ = √(200² - 90²) ≈ 179 km/h.

Teacher's Note:
1. The sideways components must cancel.
2. The northward component is the ground speed.

Answer: (d) 1/2 rad/s

Explanation:
1. x = d tan θ, so v = d sec²θ (dθ/dt).
2. ω = v cos²θ/d = 40(1/2)/40 = 1/2 rad/s.

Teacher's Note:
1. Relate the car's position to the angle.
2. Differentiate with respect to time.

Answer: (a) 960 km/h

Explanation:
1. The horizontal distance x = h cot θ, so v = h cosec²θ (dθ/dt).
2. v = 8 km × (4/3) × 0.025 = 0.2667 km/s = 960 km/h.

Teacher's Note:
1. Express the horizontal distance through the elevation angle.
2. Differentiate with respect to time and convert units.

Answer: (a) \( v\tan\alpha \)

Explanation:
1. The length is constant: x² + y² = L², so x vB + y vA = 0 (taking magnitudes, vB = (y/x)vA).
2. tan α = y/x, so vB = v tan α.

Teacher's Note:
1. The rod is rigid, so the constraint x² + y² = L² holds.
2. Differentiate with respect to time.