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Projectile motion basics.
Circular motion.
Relative velocity in 2D.
Mixed applications.
A comprehensive mock covering all subtopics — projectile motion, uniform circular motion, relative velocity in 2D, and mixed 2D applications. Ideal after reading the notes and working through solved examples.
1. The path of a projectile (in absence of air resistance) is:
Explanation: In absence of air resistance, a projectile follows a parabolic path. Horizontal velocity is constant while vertical velocity changes under gravity.
2. At the highest point of a projectile's trajectory, the vertical component of velocity is:
Explanation: At the highest point, the vertical velocity becomes zero momentarily. The horizontal component remains constant throughout.
3. In uniform circular motion, which quantity remains constant?
Explanation: In uniform circular motion, speed remains constant. Direction changes continuously, so velocity and acceleration are not constant.
4. The horizontal range of a projectile is maximum when the angle of projection is:
Explanation: R = u²sin2θ/g is maximum when sin2θ = 1, i.e., 2θ = 90°, so θ = 45°.
5. Centripetal acceleration in circular motion is directed:
Explanation: Centripetal acceleration = v²/r is always directed towards the centre. This is why it is called centripetal (center-seeking).
6. For a projectile launched with speed u at angle θ, the maximum height is:
Explanation: At maximum height, vertical velocity = 0. v² = (usinθ)² − 2gH → H = u²sin²θ/2g.
7. The angular velocity ω and frequency f are related by:
Explanation: In one revolution, angle traversed = 2π radians. So ω = 2π/T = 2πf.
8. The time of flight of a projectile launched at angle θ with speed u is:
Explanation: Time of flight = 2 × (usinθ/g) = 2usinθ/g.
9. For a particle in circular motion with radius r and speed v, the centripetal acceleration is:
Explanation: Centripetal acceleration aₑ = v²/r. It can also be written as rω².
10. A projectile is fired horizontally from height h. The time to reach ground is:
Explanation: h = ½gt² → t = √(2h/g).
11. The period T and angular velocity ω are related by:
Explanation: ω = 2π/T, therefore T = 2π/ω.
12. The horizontal component of velocity of a projectile during flight:
Explanation: In absence of air resistance, no horizontal force acts. By Newton's first law, horizontal velocity remains constant.
13. A river flows east at 3 m/s. A boat travels at 4 m/s in still water heading north. The resultant speed is:
Explanation: Resultant = √(3² + 4²) = √25 = 5 m/s.
14. A car moves along a circular path of radius 50 m at 10 m/s. Its centripetal acceleration is:
Explanation: aₑ = v²/r = 100/50 = 2 m/s².
15. At what pairs of angles does a projectile have the same range?
Explanation: R = u²sin2θ/g. Since sin(180°−2θ) = sin2θ, both θ and (90°−θ) give the same range. E.g., 30° and 60°.
16. A ball is projected at 30° to horizontal with velocity 20 m/s. The horizontal range (g = 10 m/s²) is:
Explanation: R = u²sin2θ/g = 400×sin60°/10 = 400×(√3/2)/10 = 20√3 m.
17. A satellite orbits Earth at radius r with speed v. The time period T is:
Explanation: Circumference = 2πr, Speed = v. Time = distance/speed = 2πr/v.
18. A cricket ball is hit at 45° with velocity 20 m/s. Time of flight (g = 10 m/s²) is:
Explanation: T = 2usinθ/g = 2×20×(1/√2)/10 = 40/(10√2) = 4/√2 = 2√2 s.
19. A stone (radius 0.5 m, 2 rev/s) has centripetal acceleration:
Explanation: ω = 2πf = 4π rad/s. a = rω² = 0.5 × 16π² = 8π² m/s².
20. Rain falls vertically at 10 m/s. A person runs at 10 m/s horizontally. The rain appears to fall at angle θ where tanθ is:
Explanation: Relative velocity has vertical component 10 m/s and horizontal component 10 m/s. tanθ = 10/10 = 1, so θ = 45°.
21. A ball projected horizontally from height 80 m with speed 20 m/s (g = 10 m/s²) hits ground at horizontal distance:
Explanation: t = √(2h/g) = √16 = 4 s. Horizontal distance = 20×4 = 80 m.
22. For a particle in circular motion, centripetal force is:
Explanation: Centripetal force = mv²/r is always directed toward the center.
23. A projectile's maximum range is R when fired at 45°. The maximum height at this range is:
Explanation: At 45°: R = u²/g, H = u²/(4g) = R/4. So maximum height = R/4.
24. Two ships sail at 90° to each other at 30 km/h and 40 km/h. The relative speed is:
Explanation: Relative speed = √(30²+40²) = √2500 = 50 km/h.
25. The banking angle θ for a road of radius r at speed v satisfies:
Explanation: For a banked curve: tanθ = v²/rg.
26. Minimum speed at the top of a vertical loop of radius r for maintaining contact is:
Explanation: At top of loop: mg = mv²/r (minimum condition). v_min = √(rg).
27. A swimmer crosses a 100 m wide river (current 5 m/s) by swimming at 10 m/s perpendicular to the bank. Time to cross is:
Explanation: Time = width / swimmer's speed = 100/10 = 10 s. Current doesn't affect crossing time.
28. At what angle should a body be projected so the horizontal range equals the maximum height?
Explanation: R = H → 4cosθ = sinθ → tanθ = 4. θ = tan⁻¹(4).
29. A particle moves in a circle of radius 2 m completing 3 revolutions per second. The linear speed is:
Explanation: v = rω = r(2πf) = 2 × 2π × 3 = 12π m/s.
30. For two complementary angles (θ and 90°−θ), the product of times of flight T₁×T₂ is:
Explanation: T₁×T₂ = (2usinθ/g)×(2ucosθ/g) = 4u²sinθcosθ/g² = 2u²sin2θ/g² = 2R/g.
31. For a projectile fired at angle θ, the ratio of range to maximum height is:
Explanation: R = 2u²sinθcosθ/g; H = u²sin²θ/2g. R/H = 4cosθ/sinθ = 4/tanθ.
32. The angular momentum of a particle of mass m in a circle of radius r with angular velocity ω is:
Explanation: L = mvr = m(rω)r = mr²ω. Also L = Iω where I = mr².
33. Two projectiles thrown at same speed u at angles (45° + α) and (45° − α). Their ranges are:
Explanation: R₁ = u²sin(90°+2α)/g = u²cos2α/g. R₂ = u²sin(90°−2α)/g = u²cos2α/g. Both ranges are equal.
34. A particle moves in a circle of radius r. In half a revolution, the displacement is:
Explanation: In half revolution, the particle goes from one end to the other of a diameter. Displacement = 2r.
35. Condition for a vehicle not to skid on curved road of radius r with friction coefficient μ is:
Explanation: For circular motion: μmg ≥ mv²/r → v² ≤ μrg → v ≤ √(μrg).
36. In a vertical circle of radius R, the minimum speed at the bottom to complete the loop is:
Explanation: Minimum speed at top = √(gR). Energy conservation: v_bottom² = gR + 4gR = 5gR → v_bottom = √(5gR).
37. A particle is projected with speed u at angle θ. The speed at the highest point is:
Explanation: At highest point, vertical velocity = 0. Only horizontal component remains: u cosθ.
38. The ratio of angular velocities of the hour hand and minute hand of a clock is:
Explanation: ω_h/ω_m = (2π/12 h)/(2π/1 h) = 1:12.
39. A gun fires a bullet at 45° and hits a target 1 km away (same level). The muzzle speed is (g = 10 m/s²):
Explanation: R = u²sin90°/g = u²/g. 1000 = u²/10 → u² = 10000 → u = 100 m/s.
40. For circular motion, the work done by centripetal force in one revolution is:
Explanation: Centripetal force is always perpendicular to velocity. Work = F·d·cos90° = zero.
41. A projectile is fired at 60° with speed 40 m/s (g=10 m/s²). Time to reach maximum height is:
Explanation: t = usinθ/g = 40×sin60°/10 = 40×(√3/2)/10 = 2√3 s.
42. When a stone whirled in a circle breaks free from the string, it:
Explanation: When centripetal force disappears, no horizontal force acts. The stone flies in the direction of its instantaneous velocity: tangentially.
43. At what angle to horizontal should a cliff projectile (height H) be launched so it lands at greatest horizontal distance?
Explanation: For projectiles from height, the optimal angle for maximum range is less than 45°. The exact angle depends on the height H and initial speed u.
44. For a particle moving in horizontal circle of radius r with period T, the centripetal acceleration is:
Explanation: v = 2πr/T. a = v²/r = (2πr/T)²/r = 4π²r/T².
45. A ball is thrown with velocity 40 m/s at 30° to horizontal. The horizontal range (g = 10 m/s²) is:
Explanation: R = u²sin2θ/g = 1600×sin60°/10 = 1600×(√3/2)/10 = 80√3 m.
46. The acceleration in uniform circular motion is called:
Explanation: In uniform circular motion, speed is constant. The acceleration toward center is centripetal acceleration.
47. A ball is kicked at 60° to the ground with speed 20 m/s (g = 10 m/s²). The maximum height is:
Explanation: H = u²sin²θ/2g = 400×(3/4)/20 = 15 m.
48. A stone on a 1 m string whirled in vertical circle. Minimum speed at top to maintain circular motion (g = 10 m/s²) is:
Explanation: v_min = √(gr) = √(10×1) = √10 m/s.
49. The SI unit of angular velocity is:
Explanation: The SI unit of angular velocity is radian per second (rad/s).
50. A particle travels in a semicircle of diameter 10 m. Its displacement is:
Explanation: Displacement = diameter = 10 m. Distance = πr = 5π m.
51. For a particle in circular motion, which vector is always tangent to the path?
Explanation: Linear velocity is always directed tangentially to the circular path. Centripetal force and acceleration point toward the center.
52. A bullet fired horizontally at 200 m/s drops 0.1 m in time t. The value of t is (g = 10 m/s²):
Explanation: 0.1 = ½gt² → t² = 0.02 → t = 0.14 s.
53. The angular displacement in 5 seconds of a wheel starting from rest and accelerating at 2 rad/s² is:
Explanation: θ = ω₀t + ½αt² = 0 + ½(2)(25) = 25 rad.
54. A monkey drops from a tree and a hunter fires a bullet aimed at the monkey at the same instant. The bullet:
Explanation: Both bullet and monkey fall the same vertical distance under gravity. The bullet always hits the monkey regardless of angle or speed.
55. Which of the following is NOT true for uniform circular motion?
Explanation: In uniform circular motion, speed, |acceleration|, and kinetic energy remain constant. However, velocity changes continuously since direction changes.
56. If the radius of a circular path is doubled while angular velocity remains same, the centripetal force is:
Explanation: F = mω²r. If r doubles: F = mω²(2r) = 2mω²r. Centripetal force is doubled.
57. A projectile has the same range for angles α and β where α + β = 90°. If H₁ and H₂ are max heights, then H₁ + H₂ equals:
Explanation: H₁ = u²sin²α/2g, H₂ = u²cos²α/2g. H₁+H₂ = u²(sin²α+cos²α)/2g = u²/2g. At 45°, R = u²/g. So H₁+H₂ = R/2 (the range at 45°). Answer: R/2.
58. A particle is projected at 60° to horizontal with 30 m/s (g = 10 m/s²). Velocity at maximum height is:
Explanation: At max height, only horizontal component remains: v = ucos60° = 30×(1/2) = 15 m/s.
59. A body moves in a circle of radius 4 m with speed 8 m/s. The centripetal force if mass is 2 kg is:
Explanation: F = mv²/r = 2×64/4 = 32 N.
60. A projectile is launched at 45° with speed u. The time at which the velocity vector makes 30° with horizontal is:
Explanation: At angle 30°: vₓ = ucosθ = u/√2, vy = vₓtan30° = u/(√2√3) = u/√6. But initially vy = u/√2. vy = u/√2 − gt. At 30°: vy/vₓ = tan30° = 1/√3. vy = vₓ/√3 = (u/√2)/√3 = u/√6. So u/√6 = u/√2 − gt → t = (u/√2 − u/√6)/g = u(1/√2 − 1/√6)/g = u(1−1/√3)/g (factoring out 1/√2).