AP Physics C: Mechanics Practice Test 11 – 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 velocity v(t) = 8.00 + 2.00t + 2.00t2, where v is in m/s and t is in s. What is the velocity of the particle at t = 3 s?

  • (A) 25.6 m/s
  • (B) 38.4 m/s
  • (C) 28.0 m/s
  • (D) 32.0 m/s
Show Answer

Answer: D – Differentiate nothing: substitute t = 3 s: v = 8.00 + 2.00(3) + 2.00(3)2 = 32.0 m/s.

Q2. A particle starts with velocity 9.00 m/s and has acceleration a(t) = 5.00 + 3.00t, in m/s2. What is its velocity at t = 4 s?

  • (A) 45.0 m/s
  • (B) 60.9 m/s
  • (C) 29.0 m/s
  • (D) 53.0 m/s
Show Answer

Answer: D – Integrate: v(T) = v0 + ∪a dt = 9.00 + 5.00(4) + 3.00(4)2/2 = 53.0 m/s.

Q3. A particle has position x(t) = 0 + 4.00t + 0.500t3, in metres. What is its displacement between t = 0 and t = 2 s?

  • (A) 9.00 m
  • (B) 12.0 m
  • (C) 15.0 m
  • (D) 17.0 m
Show Answer

Answer: B – Displacement = x(2) – x(0) = 4.00(2) + 0.500(2)3 = 12.0 m.

Q4. A particle moves with velocity v(t) = 20.0 – 5.00t2, in m/s. At what time does it momentarily come to rest?

  • (A) 1.60 s
  • (B) 2.00 s
  • (C) 2.50 s
  • (D) 4.00 s
Show Answer

Answer: B – Set v = 0: 20.0 – 5.00t2 = 0, so t = √(20.0/5.00) = 2.00 s.

Q5. A particle has position x(t) = 5.00t3 + 2.00t, in metres. What is its acceleration at t = 2 s?

  • (A) 42.0 m/s2
  • (B) 78.0 m/s2
  • (C) 60.0 m/s2
  • (D) 30.0 m/s2
Show Answer

Answer: C – v = dx/dt = 15.0t2 + 2.00, a = dv/dt = 30.0t. At t = 2 s, a = 30.0(2) = 60.0 m/s2.

Q6. A particle starts at x = 1.00 m with velocity 2.00 m/s and acceleration a(t) = 2.00t, in m/s2. What is its position at t = 3 s?

  • (A) 16.0 m
  • (B) 13.6 m
  • (C) 18.4 m
  • (D) 20.0 m
Show Answer

Answer: A – Integrate twice: v = 2.00 + 2.00t2/2, x = 1.00 + 2.00t + 2.00t3/6. At t = 3 s, x = 16.0 m.

Q7. A projectile is launched with speed 30.0 m/s at 60 degrees above the horizontal. What is its maximum height? Take g = 9.8 m/s2.

  • (A) 34.4 m
  • (B) 27.6 m
  • (C) 41.3 m
  • (D) 45.9 m
Show Answer

Answer: A – Vertical component vy = 30.0 sin 60 degrees = 26.0 m/s. Max height = vy2/(2g) = 34.4 m.

Unit 2: Force and Translational Dynamics

Q8. A 12.0 kg block slides down a plane inclined at 30 degrees. The coefficient of kinetic friction is 0.100. What is the acceleration of the block down the plane?

  • (A) 3.24 m/s2
  • (B) 4.05 m/s2
  • (C) 4.86 m/s2
  • (D) 4.90 m/s2
Show Answer

Answer: B – a = g(sin θ – μk cos θ) = 9.8(sin 30 degrees – 0.100 cos 30 degrees) = 4.05 m/s2.

Q9. Two masses 3.00 kg and 6.00 kg are connected by a light string over a frictionless pulley. What is the magnitude of their acceleration?

  • (A) 2.45 m/s2
  • (B) 4.25 m/s2
  • (C) 3.27 m/s2
  • (D) 4.90 m/s2
Show Answer

Answer: C – a = (m2 – m1)g/(m1 + m2) = (6.00 – 3.00)(9.8)/(3.00 + 6.00) = 3.27 m/s2.

Q10. A 2.00 kg block is pushed against a spring of constant 100 N/m, compressing it by 0.200 m, and released from rest. What is the initial acceleration of the block?

  • (A) 8.00 m/s2
  • (B) 12.5 m/s2
  • (C) 20.0 m/s2
  • (D) 10.0 m/s2
Show Answer

Answer: D – Spring force F = kx = 100(0.200) = 20.0 N. a = F/m = 20.0/2.00 = 10.0 m/s2.

Q11. A 5.00 kg object on a frictionless surface experiences a horizontal force F(t) = 10.0 + 5.00t, in newtons. What is its acceleration at t = 2 s?

  • (A) 3.40 m/s2
  • (B) 4.80 m/s2
  • (C) 4.00 m/s2
  • (D) 2.00 m/s2
Show Answer

Answer: C – At t = 2 s, F = 10.0 + 5.00(2) = 20.0 N. a = F/m = 4.00 m/s2.

Q12. A 2.00 kg mass moves in a horizontal circle of radius 0.500 m at speed 4.00 m/s, held by a string. What is the tension in the string?

  • (A) 51.2 N
  • (B) 64.0 N
  • (C) 80.0 N
  • (D) 8.00 N
Show Answer

Answer: B – Centripetal force = mv2/r = 2.00(4.00)2/0.500 = 64.0 N.

Q13. An astronaut of mass 60.0 kg is at a height of half an Earth radius above the Earth surface. What is approximately the astronaut weight there? Take g = 9.8 m/s2 at the surface.

  • (A) 209 N
  • (B) 261 N
  • (C) 327 N
  • (D) 588 N
Show Answer

Answer: B – Weight scales as (R/(R+h))2 = 0.444. Weight = 60.0(9.8)(0.444) = 261 N.

Q14. A 10.0 kg block on a horizontal surface is pulled by a force of 100 N at 37 degrees above the horizontal. The coefficient of kinetic friction is 0.100. What is the acceleration of the block?

  • (A) 7.61 m/s2
  • (B) 6.09 m/s2
  • (C) 9.51 m/s2
  • (D) 7.99 m/s2
Show Answer

Answer: A – Normal force N = mg – F sin θ = 37.8 N. Friction = μN = 3.78 N. a = (F cos θ – friction)/m = 7.61 m/s2.

Q15. Two blocks of masses 4.00 kg and 10.0 kg are connected by a string on a horizontal table. The 4.00 kg block has coefficient of kinetic friction 0.100 with the table, the other block is frictionless. A horizontal force of 60.0 N pulls the system. What is the acceleration?

  • (A) 3.20 m/s2
  • (B) 4.81 m/s2
  • (C) 4.29 m/s2
  • (D) 4.01 m/s2
Show Answer

Answer: D – Net force = F – μm1g = 60.0 – 3.92 = 56.1 N. a = net/(m1+m2) = 4.01 m/s2.

Q16. A 3.00 kg object is driven by a constant force of 30.0 N while experiencing a drag force bv with b = 2.00 N·s/m. What is its acceleration when its speed is 5.00 m/s?

  • (A) 6.67 m/s2
  • (B) 5.33 m/s2
  • (C) 8.33 m/s2
  • (D) 10.0 m/s2
Show Answer

Answer: A – Net force = 30.0 – 2.00(5.00) = 20.0 N. a = net/m = 6.67 m/s2.

Q17. A mass on a string moves in a horizontal circle of radius 25.0 m at speed 10.0 m/s, forming a conical pendulum. What angle does the string make with the vertical?

  • (A) 17.8 degrees
  • (B) 26.6 degrees
  • (C) 22.2 degrees
  • (D) 67.8 degrees
Show Answer

Answer: C – tan θ = v2/(rg) = 10.02/(25.0(9.8)) = 0.408. θ = 22.2 degrees.

Q18. A 60.0 kg person stands on a scale in a lift accelerating upward at 2.00 m/s2. What is the scale reading in newtons?

  • (A) 588 N
  • (B) 468 N
  • (C) 850 N
  • (D) 708 N
Show Answer

Answer: D – Apparent weight = m(g + a) = 60.0(9.8 + 2.00) = 708 N.

Q19. A 2.00 kg block on a frictionless surface experiences a position-dependent force F(x) = 10.0 + 5.00x, in newtons. What is its acceleration when x = 4.00 m?

  • (A) 12.0 m/s2
  • (B) 18.8 m/s2
  • (C) 5.00 m/s2
  • (D) 15.0 m/s2
Show Answer

Answer: D – At x = 4.00 m, F = 10.0 + 5.00(4.00) = 30.0 N. a = F/m = 15.0 m/s2.

Unit 3: Work, Energy, and Power

Q20. A variable force F(x) = 10.0 + 4.00x, in newtons, acts on a particle as it moves from x = 1.00 m to x = 4.00 m. How much work is done by this force?

  • (A) 51.0 J
  • (B) 69.0 J
  • (C) 60.0 J
  • (D) 30.0 J
Show Answer

Answer: C – W = ∪F dx = 10.0(x2-x1) + 4.00/2 (x22-x12) = 60.0 J.

Q21. A spring of constant 200 N/m is stretched from 0.200 m to 0.400 m beyond its natural length. How much work is done by the spring force during this stretch? Give the magnitude.

  • (A) 9.60 J
  • (B) 15.0 J
  • (C) 4.00 J
  • (D) 12.0 J
Show Answer

Answer: D – Work by spring = (1/2)k(x12 – x22), magnitude (1/2)k(x22 – x12) = 12.0 J.

Q22. A constant force of 50.0 N acts on an object moving at 3.00 m/s, at 0 degrees to the velocity. What is the instantaneous power delivered by the force?

  • (A) 150 W
  • (B) 120 W
  • (C) 170 W
  • (D) 188 W
Show Answer

Answer: A – P = Fv cos θ = 50.0(3.00)cos 0 degrees = 150 W.

Q23. A 4.00 kg object moving at 3.00 m/s receives 40.0 J of net work. What is its final speed?

  • (A) 4.58 m/s
  • (B) 6.19 m/s
  • (C) 5.39 m/s
  • (D) 4.47 m/s
Show Answer

Answer: C – Work-energy: (1/2)m(v2 – v02) = W, so v = √(v02 + 2W/m) = 5.39 m/s.

Q24. A force F(x) = 6.00x2, in newtons, acts from x = 2.00 m to x = 3.00 m. How much work does it do?

  • (A) 30.4 J
  • (B) 45.6 J
  • (C) 38.0 J
  • (D) 15.0 J
Show Answer

Answer: C – W = ∪6.00x2 dx = 6.00/3 (x23 – x13) = 38.0 J.

Q25. A motor does 200 J of work in 10.0 s at a steady rate. What is its average power output?

  • (A) 16.0 W
  • (B) 20.0 W
  • (C) 25.0 W
  • (D) 2000 W
Show Answer

Answer: B – Average power = W/t = 200/10.0 = 20.0 W.

Q26. A 2.00 kg block is dropped from a height of 1.00 m above a vertical spring of constant 200 N/m. What is the maximum compression of the spring?

  • (A) 0.441 m
  • (B) 0.689 m
  • (C) 0.443 m
  • (D) 0.551 m
Show Answer

Answer: D – Energy: mg(h + x) = (1/2)kx2. Solving gives x = 0.551 m.

Q27. A 5.00 kg block slides at 10.0 m/s on a horizontal surface with coefficient of kinetic friction 0.200. What is its speed after sliding 5.00 m?

  • (A) 8.97 m/s
  • (B) 7.62 m/s
  • (C) 10.3 m/s
  • (D) 9.02 m/s
Show Answer

Answer: A – Work by friction = -μmgd. v2 = v02 – 2μgd = 80.4 , so v = 8.97 m/s.

Q28. A lift of total mass 50.0 kg rises at a constant speed of 2.00 m/s. What power must the motor supply against gravity?

  • (A) 784 W
  • (B) 980 W
  • (C) 1225 W
  • (D) 100 W
Show Answer

Answer: B – P = Fv = mgv = 50.0(9.8)(2.00) = 980 W.

Q29. An object moves at a constant speed of 2.00 m/s while a force F(t) = 10.0 + 4.00t, in newtons, acts in the direction of motion from t = 0 to t = 3 s. How much work is done?

  • (A) 81.6 J
  • (B) 96.0 J
  • (C) 110 J
  • (D) 60.0 J
Show Answer

Answer: B – W = ∪Fv dt = 2.00[10.0(3) + 4.00(3)2/2] = 96.0 J.

Q30. A ball is released from rest from a height of 5.00 m (initial speed 1.00 m/s). What is its speed just before hitting the ground? Neglect air resistance.

  • (A) 9.95 m/s
  • (B) 8.46 m/s
  • (C) 11.4 m/s
  • (D) 7.00 m/s
Show Answer

Answer: A – Energy: v2 = v02 + 2gh = 99.0, so v = 9.95 m/s.


Unit 4: Linear Momentum

Q31. A 4.00 kg object experiences a force F(t) = 20.0 + 3.00t, in newtons, from t = 0 to t = 2 s. If it starts from rest, what is its change in velocity?

  • (A) 11.5 m/s
  • (B) 9.20 m/s
  • (C) 14.4 m/s
  • (D) 46.0 m/s
Show Answer

Answer: A – Impulse J = ∪F dt = 20.0(2) + 3.00(2)2/2 = 46.0 N·s. Δv = J/m = 11.5 m/s.

Q32. A 4.00 kg object moving at 2.00 m/s is acted on by a constant net force of 40.0 N for 5.00 s in the direction of motion. What is its final velocity?

  • (A) 44.2 m/s
  • (B) 59.8 m/s
  • (C) 52.0 m/s
  • (D) 202 m/s
Show Answer

Answer: C – Impulse = Ft = 200 N·s. Δv = Ft/m = 50.0 m/s, so v = 2.00 + 50.0 = 52.0 m/s.

Q33. A 2.00 kg object moving at 8.00 m/s collides elastically head-on with a 3.00 kg object moving at -2.00 m/s. What is the final velocity of the 2.00 kg object?

  • (A) -2.00 m/s
  • (B) -4.00 m/s
  • (C) -6.00 m/s
  • (D) 4.00 m/s
Show Answer

Answer: B – For an elastic collision, v1f = [(m1-m2)v1 + 2m2v2]/(m1+m2) = -4.00 m/s.

Q34. A 3.00 kg object moving at 6.00 m/s collides with a 5.00 kg object moving at -3.00 m/s and they stick together. What is their common velocity?

  • (A) 1.88 m/s
  • (B) 0.375 m/s
  • (C) -1.12 m/s
  • (D) 2.25 m/s
Show Answer

Answer: B – Momentum conservation: v = (m1v1 + m2v2)/(m1+m2) = 0.375 m/s.

Q35. A stationary object of mass 10.0 kg explodes into two parts of masses 3.00 kg and 7.00 kg. The 3.00 kg part moves at 6.00 m/s. What is the velocity of the other part?

  • (A) -2.06 m/s
  • (B) -2.57 m/s
  • (C) 2.57 m/s
  • (D) 1.80 m/s
Show Answer

Answer: B – Total momentum is zero: 3.00(6.00) + 7.00v2 = 0, so v2 = -2.57 m/s.

Q36. A force pulse on a 5.00 kg object has impulse equal to the area of a triangle with peak force 80.0 N and base 4.00 s. What is the change in velocity of the object?

  • (A) 25.6 m/s
  • (B) 40.0 m/s
  • (C) 64.0 m/s
  • (D) 32.0 m/s
Show Answer

Answer: D – Impulse J = (1/2)(80.0)(4.00) = 160 N·s. Δv = J/m = 32.0 m/s.

Q37. A 2.00 kg object experiences a force F(t) = 4.00t2, in newtons, from t = 0 to t = 3 s. What is its change in velocity if it starts from rest?

  • (A) 18.0 m/s
  • (B) 15.3 m/s
  • (C) 21.6 m/s
  • (D) 21.0 m/s
Show Answer

Answer: A – J = ∪4.00t2 dt = 4.00(3)3/3 = 36.0 N·s. Δv = J/m = 18.0 m/s.

Q38. A 1.00 kg ball hits a wall at 12.0 m/s and rebounds at -10.0 m/s. The contact time is 0.0500 s. What is the magnitude of the average force on the ball?

  • (A) 352 N
  • (B) 550 N
  • (C) 440 N
  • (D) 22.0 N
Show Answer

Answer: C – Change in momentum = m(v2 – v1) = -22.0 kg·m/s. Average force = |Δp|/Δt = 440 N.

Unit 5: Torque and Rotational Dynamics

Q39. A force of 40.0 N is applied at a distance of 0.500 m from a pivot, at 60 degrees to the lever arm. What is the magnitude of the torque about the pivot?

  • (A) 13.9 N·m
  • (B) 17.3 N·m
  • (C) 21.7 N·m
  • (D) 20.0 N·m
Show Answer

Answer: B – Torque = rF sin θ = 0.500(40.0) sin 60 degrees = 17.3 N·m.

Q40. A uniform rod of mass 4.00 kg and length 2.00 m is pivoted at one end. A torque of 18.0 N·m is applied about the pivot. What is the angular acceleration?

  • (A) 3.38 rad/s2
  • (B) 2.70 rad/s2
  • (C) 4.22 rad/s2
  • (D) 1.12 rad/s2
Show Answer

Answer: A – I = (1/3)ML2 = 5.33 kg·m2. α = τ/I = 3.38 rad/s2.

Q41. A hanging mass of 4.00 kg is attached to a string wound around a solid disk pulley of mass 2.00 kg and radius 0.200 m. What is the acceleration of the hanging mass?

  • (A) 6.66 m/s2
  • (B) 9.02 m/s2
  • (C) 7.84 m/s2
  • (D) 9.80 m/s2
Show Answer

Answer: C – For the disk, I = (1/2)MR2, so I/R2 = 1.00 kg. a = mg/(m + I/R2) = 7.84 m/s2.

Q42. A wheel has angular acceleration α(t) = 2.00 + 2.00t, in rad/s2, and initial angular velocity 1.00 rad/s. What is its angular velocity at t = 3 s?

  • (A) 13.6 rad/s
  • (B) 18.4 rad/s
  • (C) 7.00 rad/s
  • (D) 16.0 rad/s
Show Answer

Answer: D – ω(T) = ω0 + ∪α dt = 1.00 + 2.00(3) + 2.00(3)2/2 = 16.0 rad/s.

Q43. A uniform rod of mass 8.00 kg and length 1.00 m has moment of inertia (1/12)ML2 about its centre. What is its moment of inertia about a parallel axis 0.500 m from the centre?

  • (A) 2.67 kg·m2
  • (B) 2.13 kg·m2
  • (C) 3.33 kg·m2
  • (D) 0.667 kg·m2
Show Answer

Answer: A – Parallel-axis theorem: I = Icm + Md2 = 0.667 + 8.00(0.500)2 = 2.67 kg·m2.

Q44. A uniform rod of mass 6.00 kg and length 2.00 m is pivoted at one end and held at 60 degrees above the horizontal, then released. What is its initial angular acceleration?

  • (A) 2.94 rad/s2
  • (B) 4.59 rad/s2
  • (C) 4.90 rad/s2
  • (D) 3.68 rad/s2
Show Answer

Answer: D – Torque from gravity = Mg(L/2)cos θ = 29.4 N·m. I = (1/3)ML2 = 8.00 kg·m2. α = τ/I = 3.68 rad/s2.

Q45. Two torques of 10.0 N·m and 6.00 N·m act in opposite directions on a wheel of moment of inertia 4.00 kg·m2. What is the angular acceleration of the wheel?

  • (A) 0.800 rad/s2
  • (B) 1.25 rad/s2
  • (C) 1.00 rad/s2
  • (D) 2.50 rad/s2
Show Answer

Answer: C – Net torque = 10.0 – 6.00 = 4.00 N·m. α = τnet/I = 1.00 rad/s2.

Q46. A thin rod of length 2.00 m has uniform linear mass density 4.00 kg/m. What is its moment of inertia about an axis through one end, perpendicular to the rod?

  • (A) 10.7 kg·m2
  • (B) 8.53 kg·m2
  • (C) 13.3 kg·m2
  • (D) 32.0 kg·m2
Show Answer

Answer: A – I = ∪x2 dm = ∪0L λx2 dx = λL3/3 = 10.7 kg·m2.

Unit 6: Energy and Momentum of Rotating Systems

Q47. A flywheel with moment of inertia 4.00 kg·m2 rotates at 5.00 rad/s. What is its rotational kinetic energy?

  • (A) 40.0 J
  • (B) 62.5 J
  • (C) 100 J
  • (D) 50.0 J
Show Answer

Answer: D – K = (1/2)Iω2 = 0.5(4.00)(5.00)2 = 50.0 J.

Q48. A rotating disk of moment of inertia 2.00 kg·m2 spinning at 8.00 rad/s drops a second disk of moment of inertia 2.00 kg·m2 onto it, and they rotate together. What is their final angular speed?

  • (A) 3.20 rad/s
  • (B) 5.00 rad/s
  • (C) 4.00 rad/s
  • (D) 8.00 rad/s
Show Answer

Answer: C – Angular momentum conservation: I1ω1 = (I1+I2)ωf, so ωf = 4.00 rad/s.

Q49. A satellite orbits Earth at a distance of 6.50 million metres from Earth centre. What is its orbital speed in km/s? Use GM for Earth = 3.986 × 1014 m3/s2.

  • (A) 6.66 km/s
  • (B) 9.01 km/s
  • (C) 7.83 km/s
  • (D) 11.1 km/s
Show Answer

Answer: C – v = √(GM/r) = √(3.986 × 1014 / 6500000) = 7831 m/s = 7.83 km/s.

Q50. A satellite orbits Earth at radius 7.50 million metres from Earth centre. What is its orbital period in minutes? Use GM = 3.986 × 1014 m3/s2.

  • (A) 91.6 minutes
  • (B) 124 minutes
  • (C) 108 minutes
  • (D) 215 minutes
Show Answer

Answer: C – T = 2π√(r3/GM) = 6464 s = 108 minutes.

Q51. What is the escape speed from Earth at a distance of 12.0 million metres from Earth centre, in km/s? Use GM = 3.986 × 1014 m3/s2.

  • (A) 8.15 km/s
  • (B) 6.93 km/s
  • (C) 5.76 km/s
  • (D) 9.78 km/s
Show Answer

Answer: A – Escape speed = √(2GM/r) = 8.15 km/s.

Q52. A 4.00 kg mass falls 2.00 m, unwinding a string from a solid disk pulley of mass 4.00 kg and radius 0.200 m. What is the angular speed of the pulley just as the mass has fallen 2.00 m, starting from rest?

  • (A) 21.7 rad/s
  • (B) 29.4 rad/s
  • (C) 31.3 rad/s
  • (D) 25.6 rad/s
Show Answer

Answer: D – Energy: mgh = (1/2)mv2 + (1/2)Iω2 with v = ωR and I = (1/2)MR2. This gives ω = 25.6 rad/s.

Q53. A 5.00 kg particle moves at 6.00 m/s in a circle of radius 3.00 m. What is the magnitude of its angular momentum about the centre?

  • (A) 72.0 kg·m2/s
  • (B) 112 kg·m2/s
  • (C) 90.0 kg·m2/s
  • (D) 30.0 kg·m2/s
Show Answer

Answer: C – L = mvr = 5.00(6.00)(3.00) = 90.0 kg·m2/s.

Unit 7: Oscillations

Q54. A 1.00 kg mass on a spring of constant 50.0 N/m oscillates in simple harmonic motion. What is its period?

  • (A) 0.889 s
  • (B) 0.755 s
  • (C) 1.02 s
  • (D) 44.4 s
Show Answer

Answer: A – T = 2π√(m/k) = 2π√(1.00/50.0) = 0.889 s.

Q55. A simple pendulum of length 1.00 m oscillates with small amplitude. What is its period? Take g = 9.8 m/s2.

  • (A) 2.01 s
  • (B) 1.71 s
  • (C) 2.31 s
  • (D) 19.7 s
Show Answer

Answer: A – T = 2π√(L/g) = 2π√(1.00/9.8) = 2.01 s.

Q56. A mass-spring system with mass 2.00 kg and spring constant 100 N/m oscillates with amplitude 0.200 m. What is its maximum speed?

  • (A) 1.13 m/s
  • (B) 1.41 m/s
  • (C) 1.77 m/s
  • (D) 0.200 m/s
Show Answer

Answer: B – ω = √(k/m) = 7.07 rad/s. vmax = Aω = 1.41 m/s.

Q57. A mass-spring oscillator has amplitude 0.300 m, mass 2.00 kg and spring constant 50.0 N/m. What is its speed when its displacement is 0.100 m?

  • (A) 1.20 m/s
  • (B) 1.63 m/s
  • (C) 1.50 m/s
  • (D) 1.41 m/s
Show Answer

Answer: D – ω = √(k/m) = 5.00 rad/s. v = ω√(A2 – x2) = 1.41 m/s.

Q58. A 2.00 kg mass on a spring of constant 40.0 N/m is displaced 0.100 m from equilibrium. What is the magnitude of its acceleration at that instant?

  • (A) 1.60 m/s2
  • (B) 2.50 m/s2
  • (C) 4.00 m/s2
  • (D) 2.00 m/s2
Show Answer

Answer: D – From d2x/dt2 = -ω2x, with ω2 = k/m = 20.0, |a| = ω2|x| = 2.00 m/s2.

Q59. A pendulum of length 1.00 m has period 2.01 s. If its length is increased to 2.00 m, what is its new period?

  • (A) 2.41 s
  • (B) 2.84 s
  • (C) 3.26 s
  • (D) 4.01 s
Show Answer

Answer: B – T is proportional to √L, so Tnew = 2.01 × √(2.00) = 2.84 s.

Q60. A 3.00 kg mass is attached to two springs in parallel with constants 40.0 N/m and 70.0 N/m. What is the frequency of oscillation?

  • (A) 0.819 Hz
  • (B) 0.964 Hz
  • (C) 1.11 Hz
  • (D) 0.581 Hz
Show Answer

Answer: B – Parallel springs add: k = 40.0 + 70.0 = 110 N/m. f = (1/2π)√(k/m) = 0.964 Hz.

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