AP Physics C: Mechanics Practice Test 16 – 60 Questions with Answers

Test yourself with 60 original AP Physics C: Mechanics practice questions covering all 7 units of the course: kinematics, force and translational dynamics, work, energy, and power, linear momentum, torque and rotational dynamics, energy and momentum of rotating systems, and oscillations. Every answer includes the full working. All questions are calculus-based. Take g = 9.8 m/s2 unless stated otherwise.

Unit 1: Kinematics

Q1. A particle moves along the x-axis with position x(t) = 0.8 t2 + 3.0 t, where x is in metres and t is in seconds. What is its velocity at t = 2.2 s?

  • (A) 3.52 m/s
  • (B) 4.76 m/s
  • (C) 6.52 m/s
  • (D) 0.520 m/s
Show Answer

Answer: C – v = dx/dt = 2×0.8 t + 3.0. At t = 2.2 s, v = 2×0.8×2.2 + 3.0 = 6.52 m/s.

Q2. A particle has velocity v(t) = 2.9 + 2.5 t + 0.6 t2, with v in m/s and t in seconds. What is its displacement from t = 0 to t = 3.1 s?

  • (A) 50.89 m
  • (B) 21.00 m
  • (C) 14.95 m
  • (D) 26.96 m
Show Answer

Answer: D – Displacement = ∪ v dt = 2.9T + 2.5T2/2 + 0.6T3/3. At T = 3.1 s this gives 26.96 m.

Q3. A cart starts with velocity 3.5 m/s and has acceleration a(t) = 3.6 + 0.6 t, with a in m/s2 and t in seconds. What is its velocity at t = 3.2 s?

  • (A) 18.09 m/s
  • (B) 21.16 m/s
  • (C) 15.02 m/s
  • (D) 14.59 m/s
Show Answer

Answer: A – v = v0 + ∪ a dt = 3.5 + 3.6T + 0.6T2/2 = 18.09 m/s at T = 3.2 s.

Q4. An object starts from rest at the origin with acceleration a(t) = 1.5 t2, with a in m/s2 and t in seconds. What is its position at t = 3.0 s?

  • (A) 13.50 m
  • (B) 30.38 m
  • (C) 10.12 m
  • (D) 5.06 m
Show Answer

Answer: C – Integrating twice, v = 1.5t3/3 and x = 1.5t4/12. At t = 3.0 s, x = 10.12 m.

Q5. A projectile is launched with horizontal velocity 7.4 m/s and vertical velocity 8.0 m/s. What is its speed at t = 1.8 s? Take g = 9.8 m/s2.

  • (A) 12.15 m/s
  • (B) 17.04 m/s
  • (C) 10.90 m/s
  • (D) 9.64 m/s
Show Answer

Answer: A – vy = 8.0 – 9.8×1.8 = -9.64 m/s, while vx stays 7.4 m/s. Speed = √(7.42 + -9.642) = 12.15 m/s.

Q6. A particle has position x(t) = 1.1 t3 – 2.8 t2, with x in metres and t in seconds. At what positive time is its velocity zero?

  • (A) 0.848 s
  • (B) 1.70 s
  • (C) 2.55 s
  • (D) 5.09 s
Show Answer

Answer: B – v = dx/dt = 3×1.1t2 – 2×2.8t = t(3×1.1t – 2×2.8). Apart from t = 0, v = 0 when t = 2×2.8/(3×1.1) = 1.70 s.

Q7. A ball is thrown vertically upward with speed 12.2 m/s from ground level. What is its maximum height? Take g = 9.8 m/s2.

  • (A) 15.19 m
  • (B) 7.59 m
  • (C) 1.24 m
  • (D) 3.80 m
Show Answer

Answer: B – At the top v = 0, so 0 = 12.22 – 2×9.8×h, giving h = 12.22/(2×9.8) = 7.59 m.

Unit 2: Force and Translational Dynamics

Q8. A block of mass 2.2 kg on a frictionless surface is acted on by a horizontal force F(t) = 22 + 4 t, with F in newtons and t in seconds. What is its acceleration at t = 3.4 s?

  • (A) 16.18 m/s2
  • (B) 8.09 m/s2
  • (C) 10.00 m/s2
  • (D) 35.60 m/s2
Show Answer

Answer: A – F = 22 + 4×3.4 = 35.60 N, so a = F/m = 35.60/2.2 = 16.18 m/s2.

Q9. A block slides down a rough plane inclined at 35 degrees to the horizontal. The coefficient of kinetic friction is 0.25. What is its acceleration down the plane? Take g = 9.8 m/s2.

  • (A) 5.62 m/s2
  • (B) 7.63 m/s2
  • (C) 8.03 m/s2
  • (D) 3.61 m/s2
Show Answer

Answer: D – a = g(sinθ – μcosθ) = 9.8×(sin35 degrees – 0.25×cos35 degrees) = 3.61 m/s2.

Q10. A horizontal force of 83 N pushes a 11.0 kg crate across a floor with coefficient of kinetic friction 0.2. What is the acceleration of the crate? Take g = 9.8 m/s2.

  • (A) 7.55 m/s2
  • (B) 9.51 m/s2
  • (C) 1.96 m/s2
  • (D) 5.59 m/s2
Show Answer

Answer: D – Friction = 0.2×11.0×9.8 = 21.56 N. Net force = 83 – 21.56 = 61.44 N, so a = 61.44/11.0 = 5.59 m/s2.

Q11. A 1.3 kg mass is whirled in a horizontal circle of radius 1.0 m at speed 5.5 m/s by a string. What is the tension in the string?

  • (A) 39.33 N
  • (B) 44.81 N
  • (C) 7.15 N
  • (D) 12.74 N
Show Answer

Answer: A – Tension provides the centripetal force: T = mv2/r = 1.3×5.52/1.0 = 39.33 N.

Q12. A satellite of mass 852 kg orbits Earth (mass 6.0×1024 kg) at a distance of 9.6 million metres from Earth’s centre. What is the gravitational force on the satellite? Use G = 6.67×10-11 N m2/kg2.

  • (A) 1850 N
  • (B) 7400 N
  • (C) 36998 N
  • (D) 3700 N
Show Answer

Answer: D – F = GMm/r2 = 6.67×10-11×6.0×1024×852/(9.6×106)2 = 3700 N.

Q13. A 1.7 kg block is held against a spring of spring constant 165 N/m compressed by 0.16 m on a frictionless surface. What is the magnitude of its acceleration the instant it is released?

  • (A) 26.40 m/s2
  • (B) 15.53 m/s2
  • (C) 44.88 m/s2
  • (D) 0.094 m/s2
Show Answer

Answer: B – Spring force = kx = 165×0.16 = 26.40 N, so a = F/m = 26.40/1.7 = 15.53 m/s2.

Q14. A person of mass 67.5 kg stands on a scale in a lift accelerating upward at 1.7 m/s2. What is the scale reading in newtons? Take g = 9.8 m/s2.

  • (A) 776.2 N
  • (B) 661.5 N
  • (C) 546.8 N
  • (D) 114.8 N
Show Answer

Answer: A – Normal force N = m(g + a) = 67.5×(9.8 + 1.7) = 776.2 N.

Q15. Two masses 3 kg and 11 kg are connected by a light string over a frictionless pulley. What is the magnitude of their acceleration? Take g = 9.8 m/s2.

  • (A) 7.13 m/s2
  • (B) 17.15 m/s2
  • (C) 5.60 m/s2
  • (D) 9.80 m/s2
Show Answer

Answer: C – a = (m2 – m1)g/(m1 + m2) = (11 – 3)×9.8/(3 + 11) = 5.60 m/s2.

Q16. A car travels around a banked curve of radius 5.8 m banked at 30 degrees with no reliance on friction. What speed is required?

  • (A) 5.73 m/s
  • (B) 7.54 m/s
  • (C) 32.82 m/s
  • (D) 9.92 m/s
Show Answer

Answer: A – For a frictionless banked curve, tanθ = v2/(rg), so v = √(rg tanθ) = √(5.8×9.8×tan30 degrees) = 5.73 m/s.

Q17. A 7 kg block on a horizontal table (coefficient of kinetic friction 0.15) is connected by a string over a pulley to a hanging 4 kg block. What is the acceleration of the system? Take g = 9.8 m/s2.

  • (A) 3.56 m/s2
  • (B) 2.63 m/s2
  • (C) 4.50 m/s2
  • (D) 1.47 m/s2
Show Answer

Answer: B – Net force = 4×9.8 – 0.15×7×9.8 = 28.91 N. Total mass = 11 kg, so a = 28.91/11 = 2.63 m/s2.

Q18. A skydiver of mass 62.8 kg experiences air resistance F = bv with b = 10.0 kg/s. What is the terminal velocity? Take g = 9.8 m/s2.

  • (A) 6154 m/s
  • (B) 0.016 m/s
  • (C) 7.84 m/s
  • (D) 61.54 m/s
Show Answer

Answer: D – At terminal velocity mg = bv, so v = mg/b = 62.8×9.8/10.0 = 61.54 m/s.

Q19. A planet has mass 3.8×1023 kg and radius 4.2 million metres. What is the gravitational field strength at its surface? Use G = 6.67×10-11 N m2/kg2.

  • (A) 2.87 m/s2
  • (B) 1.44 m/s2
  • (C) 0.718 m/s2
  • (D) 6034762 m/s2
Show Answer

Answer: B – g = GM/R2 = 6.67×10-11×3.8×1023/(4.2×106)2 = 1.44 m/s2.

Unit 3: Work, Energy, and Power

Q20. A variable force F(x) = 14 + 11 x, with F in newtons and x in metres, acts on an object as it moves from x = 0 to x = 3.2 m. How much work is done?

  • (A) 44.80 J
  • (B) 157.4 J
  • (C) 101.1 J
  • (D) 80.00 J
Show Answer

Answer: C – W = ∪ F dx = 14L + 11L2/2 = 14×3.2 + 11×3.22/2 = 101.1 J.

Q21. A force F(x) = 1.5 x2, with F in newtons and x in metres, pushes an object from x = 0 to x = 2.6 m. How much work is done?

  • (A) 8.79 J
  • (B) 26.36 J
  • (C) 5.07 J
  • (D) 13.18 J
Show Answer

Answer: A – W = ∪ 1.5x2 dx = 1.5L3/3 = 1.5×2.63/3 = 8.79 J.

Q22. A spring of constant 249 N/m is stretched from 0.12 m to 0.3 m. How much work is done by the spring force during this stretch? Give the signed work.

  • (A) 9.41 J
  • (B) 11.21 J
  • (C) -9.41 J
  • (D) -11.21 J
Show Answer

Answer: C – W = -Δ(0.5kx2) = -0.5×249×(0.32 – 0.122) = -9.41 J, since the spring force opposes the stretch.

Q23. A 4.6 kg block moving at 9.8 m/s slides 3.0 m while a constant friction force of 20 N opposes it. What is its final speed?

  • (A) 11.05 m/s
  • (B) 8.36 m/s
  • (C) -3.24 m/s
  • (D) 9.80 m/s
Show Answer

Answer: B – Work-energy: 0.5m(v2 – 9.82) = -20×3.0, so v = √(9.82 – 2×20×3.0/4.6) = 8.36 m/s.

Q24. A constant force of 45 N acts in the direction of motion on an object moving at 5.5 m/s. What is the instantaneous power delivered?

  • (A) 247.5 W
  • (B) 8.18 W
  • (C) 50.50 W
  • (D) 123.8 W
Show Answer

Answer: A – P = Fv = 45×5.5 = 247.5 W.

Q25. A motor delivers power P(t) = 115 + 25 t, with P in watts and t in seconds. How much work does it do from t = 0 to t = 2.4 s?

  • (A) 348.0 J
  • (B) 276.0 J
  • (C) 420.0 J
  • (D) 336.0 J
Show Answer

Answer: A – W = ∪ P dt = 115T + 25T2/2 = 115×2.4 + 25×2.42/2 = 348.0 J.

Q26. A particle has potential energy U(x) = 2.8 x2 + 1.0 x3, with U in joules and x in metres. What is the force on the particle at x = 1.0 m?

  • (A) 8.60 N
  • (B) -8.60 N
  • (C) -5.60 N
  • (D) -3.00 N
Show Answer

Answer: B – F = -dU/dx = -(2×2.8x + 3×1.0x2). At x = 1.0 m, F = -(5.60 + 3.00) = -8.60 N.

Q27. A 2.4 kg block attached to a spring of constant 102 N/m is released from rest with the spring compressed by 0.33 m on a frictionless surface. What is its speed as it passes equilibrium?

  • (A) 0.051 m/s
  • (B) 14.03 m/s
  • (C) 6.52 m/s
  • (D) 2.15 m/s
Show Answer

Answer: D – Energy: 0.5k0.332 = 0.5mv2, so v = 0.33×√(102/2.4) = 2.15 m/s.

Q28. A 13.5 kg crate is lifted vertically at constant speed through a height of 5.6 m. How much work is done by the lifting force? Take g = 9.8 m/s2.

  • (A) 132.3 J
  • (B) 740.9 J
  • (C) 75.60 J
  • (D) 370.4 J
Show Answer

Answer: B – At constant speed the lifting force equals mg, so W = mgh = 13.5×9.8×5.6 = 740.9 J.

Q29. A motor lifts a 50.6 kg load through 3.5 m in 8.6 s at constant speed. What is its average power output? Take g = 9.8 m/s2.

  • (A) 1736 W
  • (B) 14926 W
  • (C) 201.8 W
  • (D) 57.66 W
Show Answer

Answer: C – Work = mgh = 1736 J. Average power = W/t = 1736/8.6 = 201.8 W.

Q30. A force F(x) = 11 + 6 x + 1.5 x2, with F in newtons and x in metres, acts from x = 0 to x = 2.1 m. How much work is done?

  • (A) 36.33 J
  • (B) 27.73 J
  • (C) 40.96 J
  • (D) 23.10 J
Show Answer

Answer: C – W = ∪ F dx = 11L + 6L2/2 + 1.5L3/3 = 40.96 J.


Unit 4: Linear Momentum

Q31. A time-varying force F(t) = 26 + 7 t, with F in newtons and t in seconds, acts from t = 0 to t = 3.5 s. What is the impulse delivered?

  • (A) 91.00 N s
  • (B) 176.8 N s
  • (C) 133.9 N s
  • (D) 115.5 N s
Show Answer

Answer: C – J = ∪ F dt = 26T + 7T2/2 = 26×3.5 + 7×3.52/2 = 133.9 N s.

Q32. A force-time graph shows a triangular pulse rising to 87 N and returning to zero over 0.2 s. What is the impulse delivered?

  • (A) 17.40 N s
  • (B) 4.35 N s
  • (C) 8.70 N s
  • (D) 5.80 N s
Show Answer

Answer: C – Impulse is the area under the F-t graph: J = 0.5×87×0.2 = 8.70 N s.

Q33. A 2.7 kg object moving at 2.8 m/s receives an impulse of 23 N s in the direction of motion. What is its final velocity?

  • (A) 25.80 m/s
  • (B) 64.90 m/s
  • (C) 11.32 m/s
  • (D) 8.52 m/s
Show Answer

Answer: C – J = mΔv, so Δv = J/m = 23/2.7 = 8.52 m/s. Final velocity = 2.8 + 8.52 = 11.32 m/s.

Q34. A 2 kg cart moving at 7.2 m/s collides with and sticks to a stationary 6 kg cart. What is their common velocity after the collision?

  • (A) 7.20 m/s
  • (B) 2.40 m/s
  • (C) 1.80 m/s
  • (D) 28.80 m/s
Show Answer

Answer: C – Momentum is conserved: 2×7.2 = (2 + 6)v, so v = 14.40/8 = 1.80 m/s.

Q35. An object of mass 8 kg at rest explodes into two fragments of masses 4 kg and 4 kg. The 4 kg fragment moves at 5.6 m/s. What is the velocity of the 4 kg fragment?

  • (A) 5.60 m/s
  • (B) -5.96 m/s
  • (C) 6.70 m/s
  • (D) -5.60 m/s
Show Answer

Answer: D – Total momentum stays zero: 4×5.6 + 4v2 = 0, so v2 = -22.40/4 = -5.60 m/s (opposite direction).

Q36. Two carts of masses 2 kg and 2 kg collide and stick together. Their x-velocities before the collision are 5.7 m/s and 0.0 m/s. What is the x-velocity of the combined carts after the collision?

  • (A) 5.70 m/s
  • (B) 6.81 m/s
  • (C) 3.59 m/s
  • (D) 2.85 m/s
Show Answer

Answer: D – px is conserved: vx = (2×5.7 + 2×0.0)/(2 + 2) = 2.85 m/s.

Q37. Masses 4 kg at x = 0.5 m and 5 kg at x = 3.4 m lie on the x-axis. What is the x-coordinate of their centre of mass?

  • (A) 9.50 m
  • (B) 2.11 m
  • (C) 1.95 m
  • (D) 4.25 m
Show Answer

Answer: B – xcm = (4×0.5 + 5×3.4)/(4 + 5) = 2.11 m.

Q38. A 0.4 kg ball hits a wall at 5.7 m/s and rebounds at 4.6 m/s in the opposite direction. What is the magnitude of the impulse delivered by the wall?

  • (A) -0.440 N s
  • (B) 1.84 N s
  • (C) 4.12 N s
  • (D) 2.28 N s
Show Answer

Answer: C – Taking the rebound direction as positive, Δp = m(vf – vi) = 0.4×(4.6 – (-5.7)) = 0.4×10.30 = 4.12 N s in magnitude.

Unit 5: Torque and Rotational Dynamics

Q39. A force of 20 N is applied at a distance 0.8 m from a pivot, at an angle of 45 degrees to the lever arm. What is the magnitude of the torque about the pivot?

  • (A) 11.31 N m
  • (B) 16.00 N m
  • (C) 13.15 N m
  • (D) 22.63 N m
Show Answer

Answer: A – τ = rF sinθ = 0.8×20×sin45 degrees = 11.31 N m.

Q40. What is the moment of inertia of a uniform rod of mass 5.9 kg and length 1.9 m about an axis through one end perpendicular to the rod?

  • (A) 1.77 kg m2
  • (B) 7.10 kg m2
  • (C) 21.30 kg m2
  • (D) 10.65 kg m2
Show Answer

Answer: B – For a rod about its end, I = (1/3)ML2 = (1/3)×5.9×1.92 = 7.10 kg m2.

Q41. What is the moment of inertia of a thin hoop of mass 4.6 kg and radius 0.7 m about its central axis?

  • (A) 2.92 kg m2
  • (B) 1.13 kg m2
  • (C) 2.25 kg m2
  • (D) 4.51 kg m2
Show Answer

Answer: C – For a thin hoop, I = 1.0MR2 = 1.0×4.6×0.72 = 2.25 kg m2.

Q42. A net torque of 22 N m acts on an object with moment of inertia 3.0 kg m2. What is its angular acceleration?

  • (A) 7.33 rad/s2
  • (B) 66.00 rad/s2
  • (C) 25.00 rad/s2
  • (D) 0.136 rad/s2
Show Answer

Answer: A – α = τ/I = 22/3.0 = 7.33 rad/s2.

Q43. A wheel has angular acceleration α(t) = 2.5 + 0.6 t, with α in rad/s2 and t in seconds, and initial angular velocity 3.6 rad/s. What is its angular velocity at t = 2.1 s?

  • (A) 11.50 rad/s
  • (B) 8.85 rad/s
  • (C) 10.17 rad/s
  • (D) 6.57 rad/s
Show Answer

Answer: C – ω = ω0 + ∪ α dt = 3.6 + 2.5T + 0.6T2/2 = 10.17 rad/s at T = 2.1 s.

Q44. An object has moment of inertia 3.0 kg m2 about its centre of mass. What is its moment of inertia about a parallel axis 0.8 m away if its mass is 4.0 kg?

  • (A) 5.56 kg m2
  • (B) 6.20 kg m2
  • (C) 3.00 kg m2
  • (D) 2.56 kg m2
Show Answer

Answer: A – Parallel axis theorem: I = Icm + Md2 = 3.0 + 4.0×0.82 = 5.56 kg m2.

Q45. A horizontal beam is pivoted at its left end. A weight of 143 N hangs 1.1 m from the pivot, and an upward force is applied at the right end 3.1 m from the pivot. What upward force balances the beam?

  • (A) 403.0 N
  • (B) 143.0 N
  • (C) 157.3 N
  • (D) 50.74 N
Show Answer

Answer: D – Torque balance about the pivot: F×3.1 = 143×1.1, so F = 143×1.1/3.1 = 50.74 N.

Q46. A disc with moment of inertia 5.1 kg m2 undergoes angular acceleration 3.0 rad/s2. What net torque acts on it?

  • (A) 1.70 N m
  • (B) 8.10 N m
  • (C) 30.60 N m
  • (D) 15.30 N m
Show Answer

Answer: D – τ = Iα = 5.1×3.0 = 15.30 N m.

Unit 6: Energy and Momentum of Rotating Systems

Q47. A rotating object has moment of inertia 2.6 kg m2 and angular velocity 5.1 rad/s. What is its rotational kinetic energy?

  • (A) 67.63 J
  • (B) 6.63 J
  • (C) 13.26 J
  • (D) 33.81 J
Show Answer

Answer: D – Krot = 0.5Iω2 = 0.5×2.6×5.12 = 33.81 J.

Q48. A spinning skater with moment of inertia 4.6 kg m2 rotates at 3.2 rad/s, then pulls in her arms so her moment of inertia becomes 1.4 kg m2. What is her new angular velocity? Neglect external torques.

  • (A) 10.51 rad/s
  • (B) 0.974 rad/s
  • (C) 3.20 rad/s
  • (D) 14.72 rad/s
Show Answer

Answer: A – Angular momentum is conserved: I1ω1 = I2ω2, so ω2 = 4.6×3.2/1.4 = 10.51 rad/s.

Q49. A solid cylinder of mass 3.7 kg rolls without slipping at speed 6.7 m/s. What is its total kinetic energy? For a solid cylinder I = 0.5MR2.

  • (A) 124.6 J
  • (B) 83.05 J
  • (C) 166.1 J
  • (D) 18.59 J
Show Answer

Answer: A – K = 0.5mv2 + 0.5Iω2 = 0.5mv2 + 0.25mv2 = 0.75mv2 = 0.75×3.7×6.72 = 124.6 J.

Q50. What is the orbital speed of a satellite in a circular orbit 11.4 million metres from Earth’s centre? Earth mass = 6.0×1024 kg, G = 6.67×10-11 N m2/kg2.

  • (A) 8887 m/s
  • (B) 35105263 m/s
  • (C) 2962 m/s
  • (D) 5925 m/s
Show Answer

Answer: D – v = √(GM/r) = √(6.67×10-11×6.0×1024/(11.4×106)) = 5925 m/s.

Q51. What is the orbital period of a satellite in a circular orbit 8.9 million metres from Earth’s centre? Earth mass = 6.0×1024 kg, G = 6.67×10-11 N m2/kg2.

  • (A) 4170 s
  • (B) 8339 s
  • (C) 16678 s
  • (D) 0.001 s
Show Answer

Answer: B – T = 2π√(r3/(GM)) = 2π×√((8.9×106)3/(6.67×10-11×6.0×1024)) = 8339 s.

Q52. What is the escape velocity from a distance of 11.1 million metres from Earth’s centre? Earth mass = 6.0×1024 kg, G = 6.67×10-11 N m2/kg2.

  • (A) 6005 m/s
  • (B) 8492 m/s
  • (C) 4246 m/s
  • (D) 12009 m/s
Show Answer

Answer: B – vesc = √(2GM/r) = √(2×6.67×10-11×6.0×1024/(11.1×106)) = 8492 m/s.

Q53. A 1.0 kg particle moves at 6.7 m/s on a line whose perpendicular distance from a fixed point is 0.7 m. What is its angular momentum about that point?

  • (A) 6.70 kg m2/s
  • (B) 4.69 kg m2/s
  • (C) 0.700 kg m2/s
  • (D) 5.67 kg m2/s
Show Answer

Answer: B – L = mvr = 1.0×6.7×0.7 = 4.69 kg m2/s.

Unit 7: Oscillations

Q54. A 0.8 kg mass on a spring of constant 78 N/m oscillates on a frictionless surface. What is its period?

  • (A) 62.04 s
  • (B) 0.101 s
  • (C) 0.636 s
  • (D) 0.064 s
Show Answer

Answer: C – T = 2π√(m/k) = 2π×√(0.8/78) = 0.636 s.

Q55. A simple pendulum of length 0.7 m undergoes small oscillations. What is its period? Take g = 9.8 m/s2.

  • (A) 23.51 s
  • (B) 1.68 s
  • (C) 0.267 s
  • (D) 0.449 s
Show Answer

Answer: B – T = 2π√(L/g) = 2π×√(0.7/9.8) = 1.68 s.

Q56. A mass of 0.6 kg on a spring of constant 30 N/m oscillates with amplitude 0.1 m. What is its maximum speed?

  • (A) 0.354 m/s
  • (B) 1.41 m/s
  • (C) 0.014 m/s
  • (D) 0.707 m/s
Show Answer

Answer: D – ω = √(k/m) = √(30/0.6) = 7.07 rad/s. vmax = Aω = 0.1×7.07 = 0.707 m/s.

Q57. An oscillator with amplitude 0.26 m and angular frequency 3.77 rad/s is at displacement 0.13 m from equilibrium. What is its speed there?

  • (A) 0.980 m/s
  • (B) 0.490 m/s
  • (C) 1.10 m/s
  • (D) 0.849 m/s
Show Answer

Answer: D – v = ω√(A2 – x2) = 3.77×√(0.262 – 0.132) = 0.849 m/s.

Q58. A mass of 1.1 kg on a spring of constant 73 N/m oscillates with amplitude 0.17 m. What is its maximum acceleration?

  • (A) 11.28 m/s2
  • (B) 1.38 m/s2
  • (C) 66.36 m/s2
  • (D) 0.692 m/s2
Show Answer

Answer: A – ω = √(k/m) = 8.15 rad/s. amax = ω2A = 8.152×0.17 = 11.28 m/s2.

Q59. A physical pendulum has moment of inertia 1.4 kg m2 about its pivot, mass 1.6 kg, and its centre of mass is 0.3 m from the pivot. What is its period for small oscillations? Take g = 9.8 m/s2.

  • (A) 11.52 s
  • (B) 0.546 s
  • (C) 1.87 s
  • (D) 3.43 s
Show Answer

Answer: D – T = 2π√(I/(Mgd)) = 2π×√(1.4/(1.6×9.8×0.3)) = 3.43 s.

Q60. A mass of 1.3 kg on a spring of constant 62 N/m oscillates as x = A sin(ωt), starting from equilibrium. How long does it first take to reach x = A/2?

  • (A) 0.227 s
  • (B) 0.076 s
  • (C) 0.152 s
  • (D) 0.455 s
Show Answer

Answer: B – x = A/2 when sin(ωt) = 0.5, so ωt = π/6 and t = T/12. Here T = 2π√(1.3/62) = 0.910 s, so t = 0.910/12 = 0.076 s.

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