Laws of Motion

Correct Answer: B

Aristotle coded practical experience — objects need force to keep moving — but failed to recognize friction. In the absence of friction (frictionless surface), no force is needed to maintain uniform motion. Galileo identified this flaw.

Correct Answer: B — Resistance to change

Inertia means "resistance to change." A body does not change its state of rest or uniform motion unless compelled by an external force. Inertia is a property of matter directly related to mass — greater mass = greater inertia.

Correct Answer: B — kg m s⁻¹

p = mv → [kg][m s⁻¹] = kg m s⁻¹. This is also equivalent to N·s (since 1 N = 1 kg m s⁻²). Impulse (= change in momentum) also has the same unit.

Correct Answer: B

F = dp/dt (for constant mass: F = ma). The second law is a vector law — force and acceleration are in the same direction. It is also a local law: F at an instant determines a at that instant only.

Correct Answer: C — Zero

With no nearby stars and negligible gravitational attraction from the small spaceship, the net force on the astronaut = 0. By Newton's First Law, a = 0. He will continue at whatever velocity he had when separated.

Correct Answer: B — Force × time = Δp

J = F·Δt = Δp. Impulse is useful when large forces act for very short times (bat hitting ball, bullet entering block). The individual force and time may be hard to measure but their product (= change in momentum) is measurable.

Correct Answer: B

Key points: (1) Action and reaction act on different bodies — they cannot cancel. (2) They are simultaneous — no cause-effect. (3) They are always equal in magnitude and opposite in direction. F_AB = −F_BA always.

Correct Answer: B

Weight (by Earth on book) and Normal force (by table on book) both act on the same body (the book) — they are NOT an action-reaction pair. They are equal and opposite only because the book is in equilibrium (Newton's First Law). The actual reaction to weight is the gravitational pull of the book on the Earth.

Correct Answer: B

For an isolated system (zero net external force): p_total = constant. This follows from Newton's 2nd and 3rd laws. It applies to both elastic AND inelastic collisions. Individual momenta can change, but their sum remains constant.

Correct Answer: C — μs > μk

Experimentally confirmed: μs > μk. This is why it takes more force to start moving an object than to keep it moving. Once relative motion begins, friction decreases from the static maximum (fs)_max to the kinetic value fk.

Correct Answer: B

fs ≤ μsN. When F < (fs)_max, static friction = F (self-adjusting). When F = 0, fs = 0 (no static friction without applied force). Only when F exceeds (fs)_max = μsN does the body start sliding and kinetic friction takes over.

Correct Answer: B

F·Δt = Δp (impulse = change in momentum). The change in momentum (Δp) is fixed (ball must stop). By increasing Δt (longer time of contact by pulling hands back), the force F = Δp/Δt is reduced, preventing injury.

Correct Answer: C — Inertia of motion

When the bus stops, feet stop (due to friction with floor) but the upper body continues to move forward due to inertia (tendency to maintain its state of motion). This is a classic application of Newton's First Law.

Correct Answer: C — Static friction

On a level road, the only horizontal force is friction. Static friction provides the centripetal force = mv²/R. Maximum speed: v_max = √(μsRg). If speed exceeds this, friction is insufficient and the car skids outward.

Correct Answer: B

Initially at rest: p_total = 0. After firing: m_b·v_b + m_g·v_g = 0
v_gun = −(m_b/m_g)·v_bullet (negative = opposite direction). Since m_gun >> m_bullet, recoil speed is small but non-zero.

Correct Answer: C

For banked road: v_max = √[Rg(μs + tanθ)/(1 − μs·tanθ)]. This is greater than flat road (v_max = √(μsRg)) since banking provides additional centripetal force component. Optimum speed (no friction needed): v₀ = √(Rg·tanθ).

Correct Answer: C — 1050 N

Applying Newton's 2nd law (taking upward as +ve):
N − mg = ma → N = m(g + a) = 70(10 + 5) = 70 × 15 = 1050 N
When accelerating upward, apparent weight increases. For downward acceleration: N = m(g − a).

Correct Answer: C — θ = tan⁻¹(μs)

At limiting angle: mg sinθ = fs = μs·N = μs·mg cosθ
→ tanθ = μs → θ_max = tan⁻¹(μs). This is called the "angle of friction" and is independent of the mass of the block.

Correct Answer: C — Tangentially

When the string breaks, the centripetal force vanishes. By Newton's First Law, the stone continues in a straight line at the velocity it had at that instant — which is tangential to the circle. This is a direct application of inertia.

Correct Answer: B — Vector sum = zero

F₁ + F₂ + F₃ = 0 (vector sum). This means components in each direction sum to zero: ΣFx = 0, ΣFy = 0. The three forces can be represented by sides of a closed triangle with arrows in the same sense.

Correct Answer: B — mv²/R toward centre

F_c = mv²/R directed toward the centre. Note: "centripetal force" is not a new type of force — it is simply the name for whatever provides the inward acceleration (tension, friction, gravity, etc.).

Correct Answer: A — 2 m s⁻²

a = (m₂ − m₁)g / (m₁ + m₂) = (12 − 8) × 10 / (12 + 8) = 40/20 = 2 m s⁻²
Tension: T = 2m₁m₂g/(m₁+m₂) = 2×8×12×10/20 = 96 N

Correct Answer: C — Weightlessness

In free fall, a = g (downward). N − mg = m(−g) → N = m(g − g) = 0. The scale reads 0. This is the condition of weightlessness — the same experienced by astronauts in orbit (they're in free fall around Earth).

Correct Answer: C — Conservation of momentum

Rocket expels gases backward at high speed → by conservation of momentum (or Newton's 3rd law: gas pushes rocket forward as rocket pushes gas backward), the rocket moves forward. Works even in vacuum — no air needed to "push against."

Correct Answer: C — v₀ = √(Rg tanθ)

At optimum speed: friction = 0. Normal force's horizontal component provides all centripetal force: N sinθ = mv²/R and N cosθ = mg. Dividing: tanθ = v²/Rg → v₀ = √(Rg tanθ). Driving at this speed causes least tyre wear.

u = 90 m/s, v = 0, s = 0.6 m, m = 0.04 kg

Retardation: v² = u² + 2as → 0 = 90² + 2a(0.6) a = −8100/1.2 = −6750 m s⁻² Resistive force: F = ma = 0.04 × 6750 = 270 N Average resistive force = 270 N

Taking direction toward bowler as +ve:
Initial velocity = −12 m/s (coming from bowler), Final velocity = +12 m/s

Impulse = Δp = m(v_f − v_i) = 0.15 × (12 − (−12)) = 0.15 × 24 = 3.6 N·s Impulse = 3.6 N·s (toward the bowler)

F = 50 N (retarding), m = 20 kg, u = 15 m/s, v = 0

Retardation: a = F/m = 50/20 = 2.5 m s⁻² Time: v = u − at → 0 = 15 − 2.5t t = 15/2.5 = 6 s Time to stop = 6 s

Net force: F_net = √(8² + 6²) = √(64 + 36) = √100 = 10 N Acceleration: a = F_net/m = 10/5 = 2 m s⁻² Direction: θ = tan⁻¹(6/8) = tan⁻¹(0.75) = 36.87° ≈ 37° from the 8 N force direction
a = 2 m s⁻² at 37° to the 8 N force

v₀ = 36 km/h = 10 m/s, v = 0, t = 4 s
Total mass = 400 + 65 = 465 kg

Retardation: a = (v − v₀)/t = (0 − 10)/4 = −2.5 m s⁻² Retarding force: F = ma = 465 × 2.5 = 1162.5 N Retarding force = 1162.5 N ≈ 1163 N

m = 20,000 kg, a = 5 m/s² (upward), g = 10 m/s²

Applying Newton's 2nd Law (upward positive): Thrust − Weight = ma F_thrust − mg = ma F_thrust = m(g + a) = 20000 × (10 + 5) F_thrust = 20000 × 15 = 3,00,000 N Initial Thrust = 3 × 10⁵ N = 300 kN

m₁ = 8 kg, m₂ = 12 kg (heavier, goes down)

Equations (12 kg going down): For 12 kg: 12g − T = 12a ... (1) For 8 kg: T − 8g = 8a ... (2) Adding: 4g = 20a → a = 4×10/20 = 2 m s⁻² Tension: From (2): T = 8(g + a) = 8(10 + 2) = 8 × 12 = 96 N (a) a = 2 m s⁻²   (b) T = 96 N

Both initially at rest → p_initial = 0
By conservation of momentum: m_shell × v_shell + m_gun × v_gun = 0 0.02 × 80 + 100 × v_gun = 0 1.6 + 100 v_gun = 0 v_gun = −1.6/100 = −0.016 m s⁻¹ Recoil speed of gun = 0.016 m s⁻¹ (backward)

Apparent weight N = m(g ± a):

(a) At rest: a = 0 → N = 70 × 10 = 700 N (b) Const. velocity: a = 0 → N = 700 N (c) Upward accel: N = 70(10+5) = 1050 N (d) Free fall: a = g → N = 70(10−10) = 0 N (a) 700 N   (b) 700 N   (c) 1050 N   (d) 0 N (weightless)

At limiting angle θ_max = 15°: tan θ_max = μs μs = tan 15° = tan(45° − 30°) = (tan45° − tan30°)/(1 + tan45°·tan30°) = (1 − 1/√3)/(1 + 1/√3) ≈ 0.268 μs = tan 15° ≈ 0.27
Note: mass of block is irrelevant — angle of friction depends only on μs.

Kinetic friction: fk = μk × N = 0.04 × 20 × 10 = 8 N

Equations of motion: For hanging block (3 kg): 30 − T = 3a ...(1) For trolley (20 kg): T − 8 = 20a ...(2) Adding: 22 = 23a → a = 22/23 ≈ 0.96 m s⁻² Tension: From (1): T = 30 − 3(22/23) = 30 − 2.87 ≈ 27.1 N (a) a ≈ 0.96 m s⁻²   (b) T ≈ 27.1 N

ω = 40 rev/min = 40 × 2π/60 = 4π/3 rad/s
v = ωr = (4π/3) × 1.5 = 2π ≈ 6.28 m/s

(a) Tension (= centripetal force): T = mv²/r = 0.25 × (6.28)²/1.5 = 0.25 × 39.44/1.5 ≈ 6.57 N (a) T ≈ 6.57 N

(b) Max speed for T_max = 200 N: T_max = mv²_max/r → v²_max = T_max × r/m = 200 × 1.5/0.25 = 1200 v_max = √1200 ≈ 34.6 m s⁻¹ (b) v_max ≈ 34.6 m s⁻¹

v = 18 km/h = 5 m/s, R = 3 m, μs = 0.1, g = 9.8 m/s²

Maximum safe speed: v²_max = μs × R × g = 0.1 × 3 × 9.8 = 2.94 m²/s² v_max = √2.94 ≈ 1.71 m/s Check: v² = (5)² = 25 m²/s² > v²_max = 2.94 m²/s² Since v² > μsRg, friction is insufficient.
Yes, the cyclist will slip. Safe speed is only ≈ 1.71 m/s ≪ 5 m/s

R = 300 m, θ = 15°, μs = 0.2, g = 9.8 m/s²

(a) Optimum speed (μs = 0): v₀ = √(Rg tanθ) = √(300 × 9.8 × tan15°) tan15° = 0.2679 v₀ = √(300 × 9.8 × 0.2679) = √(787.6) v₀ ≈ 28.1 m s⁻¹ (a) v₀ ≈ 28.1 m s⁻¹

(b) Maximum speed: v_max = √[Rg(μs + tanθ)/(1 − μs tanθ)] = √[300×9.8×(0.2+0.2679)/(1−0.2×0.2679)] = √[2940 × 0.4679/0.9464] = √[1453.6] ≈ 38.1 m s⁻¹ (b) v_max ≈ 38.1 m s⁻¹

Total mass = 30 kg. Acceleration: a = F/(m₁+m₂) = 600/30 = 20 m s⁻²

Case (i) F applied to 10 kg body (A), 20 kg body (B) is pulled: For B (20 kg): T = m_B × a = 20 × 20 = 400 N T = 400 N

Case (ii) F applied to 20 kg body (B), 10 kg body (A) is pulled: For A (10 kg): T = m_A × a = 10 × 20 = 200 N T = 200 N

Note: The tension is smaller when force is applied to the heavier body — the lighter body (10 kg) needs less force to accelerate at the same rate.