Laws of Motion: JEE Main PYQ, 2020–2025

Answer: (A)

Friction acts on the shoe at the point of contact with the ground, opposing the shoe's tendency to slide relative to the ground. When walking, the foot pushes the ground backward, so the tendency is backward and the static friction from the ground acts forward on the shoe. This forward friction is what moves the person forward. Static friction here is essential, not absent.

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: (C)

Newton's second law gives F = ma, where a is the rate of change of velocity. Here the velocity changes from 0 to 20 m/s in 10 s, so a = 20/10 = 2 m/s^2. Then F = 1500 kg x 2 m/s^2 = 3000 N. Always compute acceleration first as change in velocity divided by time, then multiply by mass.

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: (B)

A rocket moves by momentum conservation, which follows from Newton's third law. The engine pushes exhaust gas backward, so the gas pushes the rocket forward with an equal and opposite force. This needs no air, so the rocket works in vacuum. Action and reaction act on different bodies, so the rocket is accelerated by the force from the gas.

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)

Static friction acts on the block to accelerate it along with the truck. Apply Newton's second law in the horizontal direction: f = m·a = 2 × 3 = 6 N. Check that the required friction does not exceed the maximum μs·N = 0.5 × 2 × 10 = 10 N. Since 6 N < 10 N, the block stays at rest relative to the truck and the friction value is 6 N.