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Newton's three laws and applications.
Friction and inclined planes.
Connected bodies and Atwood machine.
Pseudo forces and advanced problems.
A comprehensive mock covering all subtopics — Newton's laws, friction on inclined planes, connected body problems, Atwood machine, and pseudo force applications. Ideal after reading the notes and working through solved examples.
1. Newton's First Law of Motion is also called:
Explanation: Newton's First Law — a body at rest stays at rest and a body in motion stays in motion unless acted upon by an external force — describes inertia. So it is the Law of Inertia.
2. The SI unit of force is:
Explanation: The SI unit of force is the Newton (N). 1 N = 1 kg·m/s².
3. According to Newton's Third Law:
Explanation: Newton's Third Law: For every action, there is an equal and opposite reaction. These forces act on different bodies.
4. Static friction acts when:
Explanation: Static friction acts when a body is stationary but a force is applied that tries to set it in motion.
5. If the net force on a body is zero, it will:
Explanation: Newton's First Law: Zero net force means zero acceleration. The body continues at constant velocity or stays at rest.
6. The maximum static friction force is:
Explanation: Maximum (limiting) static friction = μₛN, where μₛ is the coefficient of static friction and N is the normal force.
7. Newton's Second Law states:
Explanation: Newton's Second Law: net force = mass × acceleration: F = ma.
8. A passenger in a bus tends to fall backward when the bus suddenly starts. This is due to:
Explanation: The passenger was at rest. When the bus starts, the body tends to stay at rest due to inertia of rest (Newton's First Law).
9. Kinetic friction is generally:
Explanation: Kinetic (sliding) friction is generally less than maximum static friction. It is easier to keep a body moving than to start it.
10. A force of 10 N acts on a body of mass 2 kg. The acceleration produced is:
Explanation: a = F/m = 10/2 = 5 m/s².
11. Momentum of a body is defined as:
Explanation: Momentum p = mass × velocity = mv. SI unit: kg·m/s or N·s.
12. Inertia of a body depends on:
Explanation: Mass is the measure of inertia. Greater mass means greater inertia.
13. The impulse of a force is defined as:
Explanation: Impulse J = Force × time = F·Δt. Impulse equals the change in momentum: J = Δp.
14. The coefficient of friction is the ratio of:
Explanation: μ = f/N = friction force / normal force. It is dimensionless.
15. The law of conservation of momentum states total momentum is conserved when:
Explanation: Total momentum is conserved when net external force is zero.
16. In an Atwood machine with masses 4 kg and 6 kg, the acceleration of the system is (g = 10 m/s²):
Explanation: a = (m₂−m₁)g/(m₁+m₂) = (6−4)(10)/(10) = 2 m/s².
17. A block of mass 5 kg on a frictionless surface is pulled by 20 N. The acceleration is:
Explanation: a = F/m = 20/5 = 4 m/s².
18. A block of mass 10 kg is on a rough surface (μₖ = 0.3). The friction force during motion (g = 10 m/s²) is:
Explanation: Friction = μₖmg = 0.3×10×10 = 30 N.
19. A 3 kg ball moving at 4 m/s collides with a stationary 1 kg ball and they stick together. Their combined velocity is:
Explanation: Conservation of momentum: 3×4 = (3+1)v → v = 12/4 = 3 m/s.
20. Blocks of 2 kg and 5 kg together on frictionless surface. F = 21 N on 5 kg block. Both move together. Acceleration is:
Explanation: a = F/(m₁+m₂) = 21/7 = 3 m/s².
21. The angle of friction λ and coefficient of friction μ are related by:
Explanation: At the angle of friction λ: tanλ = friction/normal = μ.
22. Two blocks (3 kg and 5 kg) connected by string on frictionless surface. Force 16 N pulls the 5 kg block. Tension in string is:
Explanation: a = 16/8 = 2 m/s². T = 3×2 = 6 N.
23. A gun (3 kg) fires a bullet (50 g) at 400 m/s. The recoil speed of the gun is:
Explanation: Conservation of momentum: 0 = 3V + 0.05×400. 3V = −20. V = 20/3 ≈ 20/3 m/s.
24. A body of mass 10 kg on surface with μₛ = 0.4. Minimum force to start moving it (g = 10 m/s²) is:
Explanation: Limiting friction = μₛmg = 0.4×100 = 40 N.
25. A lift accelerates upward at 2 m/s². Apparent weight of 60 kg person (g = 10 m/s²) is:
Explanation: Apparent weight = m(g+a) = 60(12) = 720 N.
26. In an Atwood machine, if masses are equal (m₁ = m₂ = m), the tension in the string is:
Explanation: Equal masses: system in equilibrium. Tension = weight of each mass = mg.
27. A rocket ejects gases at 500 m/s. If fuel consumption rate is 2 kg/s, the thrust force is:
Explanation: Thrust = v_exhaust × (dm/dt) = 500 × 2 = 1000 N.
28. A 2 kg block on frictionless incline (angle 30°). Acceleration along plane (g = 10 m/s²):
Explanation: a = g sinθ = 10 × sin30° = 10 × 0.5 = 5 m/s².
29. If the normal force doubles while friction coefficient stays same, the friction force:
Explanation: f = μN. If N doubles: f_new = μ(2N) = 2μN = twice the original.
30. The tension in the rope holding a 5 kg block at rest (g = 10 m/s²) is:
Explanation: Equilibrium: T = mg = 5×10 = 50 N.
31. A 10 kg block on rough incline (angle 30°, μₖ = 0.2) slides down. Acceleration (g = 10 m/s²) is:
Explanation: a = g(sinθ − μₖcosθ) = 10(0.5 − 0.2×0.866) = 10(0.5 − 0.173) = 3.27 m/s².
32. A box is in an accelerating truck (acceleration a). The angle of tilt of a pendulum inside the truck is:
Explanation: In the non-inertial frame, the pendulum experiences pseudo force ma horizontally and mg vertically. tanθ = a/g, so θ = tan⁻¹(a/g).
33. A ball (0.1 kg) moving at 10 m/s bounces back at 10 m/s. The impulse on the ball is:
Explanation: Impulse = |Δp| = m|v_f − v_i| = 0.1 × |−10−10| = 0.1×20 = 2 N·s.
34. Two masses M and m (M > m) on a pulley; system acceleration = g/3. The ratio M/m is:
Explanation: a = (M−m)g/(M+m) = g/3. 3(M−m) = M+m → 2M = 4m → M/m = 2.
35. A coin on a rotating disc stays due to:
Explanation: The coin needs centripetal force (toward center) for circular motion. This is provided by static friction.
36. Two bodies (4 kg at 3 m/s and 6 kg at 2 m/s in opposite directions) collide and stick. Velocity after collision:
Explanation: Taking 4 kg direction positive: p = 4×3 − 6×2 = 12 − 12 = 0. v = 0/(10) = 0 m/s.
37. A block of mass m pressed against wall by force F at angle θ to wall. Minimum F to prevent sliding (friction μ):
Explanation: Normal force N = Fcosθ. Friction = μFcosθ. Equilibrium: Fsinθ + μFcosθ = mg → F = mg/(sinθ + μcosθ).
38. The minimum force needed to move mass m on surface with μ, applied at optimal angle, is:
Explanation: Minimum force at optimal angle θ = tan⁻¹(μ): F_min = μmg/√(1+μ²).
39. An elevator descends with acceleration g/4. Apparent weight of 60 kg person (g = 10 m/s²):
Explanation: Apparent weight = m(g−a) = 60(10 − 2.5) = 60×7.5 = 450 N.
40. The angle of inclination at which a body starts to slide, with μₛ as coefficient of static friction, is called angle of repose. Its value is:
Explanation: At angle of repose θᵣ: tanθᵣ = μₛ. θᵣ = tan⁻¹(μₛ).
41. A 0.5 kg ball changes velocity from 20 m/s to −30 m/s. Contact time = 0.01 s. Average force is:
Explanation: Impulse = 0.5×50 = 25 N·s. F = 25/0.01 = 2500 N.
42. Rolling friction is generally ________ than sliding friction.
Explanation: Rolling friction is much less than sliding friction. This is why wheels reduce friction in transport.
43. A rocket of initial mass M burns fuel at rate r kg/s and ejects gas at speed v. The thrust is:
Explanation: Thrust = v_exhaust × (mass ejection rate) = v × r = rv.
44. A person stands on a scale in an elevator. The scale reads zero when:
Explanation: In free fall, apparent weight = m(g−g) = 0. The scale reads zero — this is weightlessness.
45. A body moves up an incline (angle θ) with deceleration a. The coefficient of kinetic friction μₖ is:
Explanation: Going up: net deceleration a = g sinθ + μₖg cosθ. So μₖ = (a−g sinθ)/(g cosθ).
46. The property of a body to resist change in its state is called:
Explanation: Inertia is the property of a body to resist any change in its state of rest or motion. It is measured by mass.
47. A bullet (10 g) moving at 400 m/s is stopped in 0.01 s. The retarding force is:
Explanation: F = m(Δv/Δt) = 0.01 × (400/0.01) = 0.01 × 40000 = 400 N.
48. Three blocks (2 kg, 3 kg, 5 kg) are connected in series on frictionless surface. Force F = 30 N on the 5 kg block. Tension between 3 kg and 2 kg blocks is:
Explanation: a = F/(2+3+5) = 30/10 = 3 m/s². Tension on 2 kg block: T = 2×3 = 6 N.
49. A car (1000 kg) moving at 20 m/s is brought to rest in 4 s. The braking force is:
Explanation: a = 20/4 = 5 m/s². F = ma = 1000×5 = 5000 N.
50. Newton's third law action-reaction pair acts on:
Explanation: Action-reaction forces always act on different bodies. They are equal in magnitude and opposite in direction.
51. A block (5 kg) pushed against rough wall (μ = 0.5) by horizontal force 100 N. Friction force (g = 10 m/s²):
Explanation: Weight = 5×10 = 50 N. Max friction = μN = 0.5×100 = 50 N. Block stays. Friction = weight = 50 N.
52. A body of mass m is pulled by force F at angle θ above horizontal on rough surface. Normal force is:
Explanation: Vertical equilibrium: N + Fsinθ = mg → N = mg − Fsinθ.
53. Mass 2 kg acted on by force 6î + 8ĵ N for 3 s (starting from rest). Velocity acquired:
Explanation: a = F/m = (6î+8ĵ)/2 = 3î+4ĵ m/s². v = at = (3î+4ĵ)×3 = 9î+12ĵ m/s.
54. A block hangs in elevator moving upward and decelerating at 3 m/s² (mass = 2 kg, g = 10 m/s²). Tension is:
Explanation: Elevator decelerating upward = acceleration downward at 3 m/s². Apparent weight = m(g−a) = 2(10−3) = 14 N.
55. A feather and a stone dropped simultaneously in vacuum reach the ground:
Explanation: In vacuum, all objects fall with same acceleration g. Both reach ground simultaneously.
56. A 60 kg person in a lift moving down at constant velocity reads on a weighing machine:
Explanation: At constant velocity, acceleration = 0. Apparent weight = mg = 60×10 = 600 N.
57. A 10 kg block on rough surface (μₛ = 0.5, μₖ = 0.4) is pushed by 40 N. Acceleration (g = 10 m/s²):
Explanation: Max static friction = μₛmg = 50 N > 40 N (applied force). Block does not move. Acceleration = 0 m/s².
58. Action and reaction forces are:
Explanation: By Newton's Third Law, action and reaction forces are always equal in magnitude (opposite in direction, different bodies).
59. Net force on a 5 kg body if velocity v = 3t² − 2t is:
Explanation: a = dv/dt = 6t − 2. F = ma = 5(6t−2) = 30t − 10 N.
60. A 50 g stone tied to a 1 m string whirled in horizontal circle at 2 m/s. Centripetal force is:
Explanation: F = mv²/r = (0.05)(4)/1 = 0.2 N.