Laws of Motion: JEE Advanced PYQ, 2007–2026

Answer: (A)

Only vertical forces act on the cart-plus-water system, so its horizontal momentum is conserved. The rain falls vertically and has zero horizontal momentum initially. Before: 20 × 3 + 10 × 0 = 60 kg m/s. After: (20 + 10)v = 60, so v = 2.0 m/s.

Answer: (C)

Horizontally, the wall's normal reaction is the only radial force, so N = mω²R. Vertically, static friction must support the weight, so f = mg ≤ μN. Then mg ≤ μmω²R, so ω² ≥ g/(μR) = 10/(0.25 × 1.6) = 25 and ω_min = 5 rad/s. The mass cancels, so the 60 kg is irrelevant.

Answer: (D)

By symmetry, each half of the rope has length 5 m and horizontal projection 4 m, so the vertical sag is 3 m (3-4-5 triangle). The rope makes an angle θ with the horizontal where sinθ = 3/5. The horizontal components of the two tensions cancel. The vertical components support the weight: 2T sinθ = mg = 60 N. So T = 60/(2 × 0.6) = 50 N.

Answer: (A)

Without F the block would slide down, so limiting friction acts up the slope. Resolve along the slope: F cosθ + f = mg sinθ. Perpendicular to the slope: N = mg cosθ + F sinθ, and f = μN. Substituting gives F cosθ + μ(mg cosθ + F sinθ) = mg sinθ. Hence F = mg(sinθ − μcosθ)/(cosθ + μ sinθ). At θ = 45° this becomes F = mg(1 − μ)/(1 + μ) = 30 × 0.5/1.5 = 10 N.

Answer: (A)

For a rope with mass, the tension varies along its length. Treat the whole rope first: a = F/M = 20/4 = 5 m/s². At 0.5 m from the pulled end, the tension is the only force accelerating the rest of the rope. That part is 1.5 m long, so its mass is (1.5/2) × 4 = 3 kg. T = 3 × 5 = 15 N. Equivalently, T = F(1 − x/L) = 20 × (1 − 0.25) = 15 N.

Answer: (A)

No external force acts horizontally. The floor is smooth, and gravity and the normal force from the floor are vertical. So the horizontal component of the system's momentum stays zero, and the horizontal position of the centre of mass does not change. If the block moves a distance x_b one way and the wedge moves x_w the other way, then m·x_b = M·x_w. The block's displacement relative to the wedge is the base length, so x_b + x_w = 2.0 m. Substituting x_w = (m/M)x_b gives x_b(1 + 1/3) = 2.0, so x_b = 1.5 m. The wedge moves 0.5 m in the opposite direction. The height of 1.5 m is not needed, and the normal force between block and wedge is internal to the system, so it does not change the total horizontal momentum.

Answer: (A)

First check that the system moves. The hanging weight is 20 N, and the maximum friction is 0.3 × 30 = 9 N, which is smaller, so the blocks accelerate. For the hanging block, 20 − T = 2a. For the table block, T − 9 = 3a. Adding the two equations gives 11 = 5a, so a = 2.2 m s⁻².

Answer: (A)

In the ground frame the bob moves with the carriage, so its acceleration is a = 7.5 m s⁻² horizontally and zero vertically. The only forces on it are gravity mg downward and the string tension T along the string. Resolving the tension: vertically, T cosθ = mg = 4 N. Horizontally, T sinθ = ma = 3 N. These two components are perpendicular, so T = √(4² + 3²) = 5 N (the string makes θ = 37° with the vertical, since tanθ = 3/4). Adding the components directly (7 N) or using only one of them (4 N or 3 N) is wrong.

Answer: (A)

Two conditions apply: momentum conservation, 1(8) = 1v₁ + 3v₂, and kinetic energy conservation, which for a head-on elastic collision is equivalent to a relative speed of approach equal to the relative speed of separation, v₂ − v₁ = 8. Substituting v₂ = v₁ + 8 into the first equation gives 8 = v₁ + 3v₁ + 24, so v₁ = −4 m/s and v₂ = +4 m/s. The light ball rebounds.

Answer: (C)

The floor is smooth, so the horizontal momentum of cart, person and balls is conserved. The relative speed u is measured from the cart, so a ball's ground velocity is (cart velocity − u). Take forward as the cart's direction of motion. First throw: the total mass is M + 2m = 20 kg, initially at rest. The cart with the remaining ball (M + m = 18 kg) moves at v₁. Then 0 = 18v₁ + 2(v₁ − 9), so 20v₁ = 18 and v₁ = 0.9 m/s. Second throw: before it the momentum is 18 × 0.9 = 16.2. After it the cart (16 kg) moves at v₂ and the ball has velocity v₂ − 9. So 16.2 = 16v₂ + 2(v₂ − 9), giving 18v₂ = 34.2 and v₂ = 1.9 m/s.