Newton's Laws of Motion – Force and inertia: NEET MCQs with Answers

Explanation: The spring force stays 4mg; for A: 4mg − 3mg = 3m·a_A ⇒ a_A = g/3 (up); for B: mg = m·a_B ⇒ a_B = g (down).

Explanation: Magnitude of the force = √(6² + 8² + 10²) = √200 = 10√2 N. From F = ma, m = 10√2/1 = 10√2 kg.

Explanation: Here the thrust is required only to balance the weight — the rocket is just supported, not accelerated: F = mg = 600 × 10 = 6000 N Using F = v_r(−dm/dt): −dm/dt = mg/v_r = 6000/1000 = 6 kg s⁻¹. (Taking g = 9.8 m s⁻² instead would give 5.88 kg s⁻¹, still closest to 6 kg s⁻¹.)

Explanation: Newton's first law states that a body continues in its state of rest or of uniform motion in a straight line unless it is acted upon by an external force. It therefore establishes force as the independent physical agency that changes a body's state of motion — this is what is meant by the 'physical independence' of force. The second law provides the quantitative measure of force (F = ma) and the third law describes forces as mutual interactions, but it is the first law that establishes the very concept of force independently.

Explanation: In force-free space no external force acts, so the momentum of the satellite changes only because it is accreting mass. The retarding force on the satellite is F = −v (dM/dt) = −v(αv) = −αv² Hence the acceleration is a = F/M = −αv²/M. The negative sign shows the satellite decelerates as it sweeps up the stationary dust.

Explanation: Using the thrust relation F = v_r(−dm/dt), where the negative sign only indicates that the thrust acts opposite to the direction of the escaping gases: −dm/dt = F/v_r = 345/300 = 1.15 kg/s.

Explanation: From Newton's second law, F = ma: m = F/a = 10/1 = 10 kg.

Explanation: The thrust must both support the rocket's weight and provide the upward acceleration: F_thrust − mg = ma F_thrust = m(g + a) = 5000(10 + 20) = 150000 N Since F_thrust = v_r(−dm/dt): −dm/dt = m(g + a)/v_r = 150000/800 = 187.5 kg s⁻¹. Forgetting the weight term would give 5000 × 20/800 = 125 kg s⁻¹, close to the distractor 127.5.

Explanation: The thrust on a rocket is F = v_r(−dm/dt) where v_r is the exhaust velocity relative to the rocket and (−dm/dt) the rate of combustion of fuel. Rearranging: −dm/dt = F/v_r = 210/300 = 0.7 kg/s.

Explanation: Force is the rate of change of momentum: F = Δp/Δt = m(v − u)/Δt Here m = 150 g = 0.15 kg, u = 20 m/s, v = 0 (the ball is brought to rest) and Δt = 0.1 s: F = 0.15 × (20 − 0)/0.1 = 3/0.1 = 30 N. This is why a player draws his hands back while catching — increasing Δt reduces the force for the same change in momentum.

Explanation: Since the force varies with time, we integrate rather than use v = u + at. From F = m dv/dt: dv = (1/3)(6t²î + 4tĵ) dt Integrating from t = 0 (where v = 0) to t = 3 s: v = (1/3)∫₀³ (6t²î + 4tĵ) dt = (1/3)[2t³î + 2t²ĵ]₀³ = (1/3)[2(27)î + 2(9)ĵ] = (1/3)(54î + 18ĵ) = 18î + 6ĵ m s⁻¹.

Explanation: Work in the frame of the accelerating wedge, where a pseudo force ma acts on the block horizontally, directed opposite to the acceleration. For the block not to slip, the components along the incline must balance: mg sin θ = ma cos θ ⇒ a = g tan θ The wedge pushes on the block through the normal reaction R, perpendicular to the incline. Resolving perpendicular to the incline: R = mg cos θ + ma sin θ = mg cos θ + m(g tan θ) sin θ = mg cos θ + mg sin²θ/cos θ = (mg/cos θ)(cos²θ + sin²θ) = mg/cos θ = mg sec θ.

Explanation: First find the magnitude of the force: |F| = √(6² + (−8)² + 10²) = √(36 + 64 + 100) = √200 = 10√2 N From Newton's second law, F = ma: m = |F|/a = 10√2/1 = 10√2 kg ≈ 14.1 kg. Since the acceleration is a scalar 1 m s⁻² here, only the magnitude of the force is needed.

Explanation: While descending, with B the (constant) buoyant force: mg − B = ma ⇒ B = m(g − a) …(i) Let m′ be the new mass after removing some ballast. The volume — and hence the buoyant force B — is unchanged. Moving up with acceleration a: B − m′g = m′a ⇒ B = m′(g + a) …(ii) Equating (i) and (ii): m(g − a) = m′(g + a) ⇒ m′ = m(g − a)/(g + a) Mass removed: Δm = m − m′ = m[1 − (g − a)/(g + a)] = m[(g + a − g + a)/(g + a)] = 2ma/(g + a).

Explanation: When three vectors are represented by the three sides of a triangle taken in order, they form a closed figure and their vector sum is zero: F_net = F_PQ + F_QR + F_RP = 0 By Newton's second law, F_net = ma = m(dv/dt) = 0 dv/dt = 0, so v = constant. With zero net force the particle is in equilibrium, and its velocity — both magnitude and direction — remains unchanged.