Ap Physics C Mechanics Practice Questions

AP Physics C: Mechanics – A Comprehensive Practice Question Guide

Introduction

The AP Physics C: Mechanics exam is a rigorous test of both conceptual understanding and mathematical skill. Consider this: this guide presents a curated set of practice questions, organized by topic, with detailed solutions and explanatory commentary. Mastery comes from solving a wide range of problems that mirror the exam’s format: multiple‑choice, free‑response, and graph‑based questions. By working through these problems, students will sharpen problem‑solving strategies, reinforce core concepts, and build confidence for the actual exam.

---

1. Motion in One Dimension

1.1 Displacement, Velocity, and Acceleration

Question 1.
A car moves along a straight road with a velocity given by
(v(t)=4t^2-12t+8) (m/s), where (t) is time in seconds.
Find the car’s acceleration at (t=3) s and the displacement between (t=2) s and (t=4) s.

Solution
Acceleration is the derivative of velocity:
(a(t)=\frac{dv}{dt}=8t-12).
At (t=3) s:
(a(3)=8(3)-12=12;\text{m/s}^2).

Displacement is the integral of velocity:
(\Delta x=\int_{2}^{4} (4t^2-12t+8),dt).
\overline{3} - 2.In practice, \overline{6}-24+16\right])
(= 21. Evaluate from 2 to 4:
(\left[\frac{4}{3}(64)-6(16)+32\right] - \left[\frac{4}{3}(8)-6(4)+16\right])
(= \left[85.\overline{6} = 18.\overline{3}-96+32\right] - \left[10.Even so, compute the antiderivative:
(\frac{4}{3}t^3-6t^2+8t). 7;\text{m}) (rounded to one decimal).

---

1.2 Projectile Motion

Question 2.
A baseball is struck upward with an initial speed of 20 m/s at an angle of 45° above the horizontal from a height of 1 m above the ground.
(a) What is the maximum vertical height reached?
(b) How far from the point of launch does the ball land?

Solution

(a) Vertical component: (v_{0y}=20\sin45^\circ=14.Practically speaking, 14;\text{m/s}). Using (v^2=v_0^2-2g(h_{\text{max}}-h_0)) with (v=0):
(0=14.Think about it: 14^2-2(9. 8)(h_{\text{max}}-1)).
Solve: (h_{\text{max}}=1+\frac{14.14^2}{2(9.8)}=1+10.2=11.2;\text{m}).

(b) Time to reach ground: solve (y(t)=h_0+v_{0y}t-\frac{1}{2}gt^2=0).
Quadratic: (-4.That said, 9t^2+14. Think about it: 14t+1=0). Which means positive root: (t=\frac{14. 14+ \sqrt{14.That said, 14^2-4(-4. 9)(1)}}{9.Because of that, 8}\approx2. 34;\text{s}).
Which means horizontal component: (v_{0x}=20\cos45^\circ=14. Now, 14;\text{m/s}). Range: (x=v_{0x}t=14.14(2.This leads to 34)=33. 1;\text{m}).

---

2. Kinematics in Two Dimensions

2.1 Circular Motion

Question 3.
A car negotiates a flat curve of radius 50 m at 15 m/s.
(a) What is the required centripetal acceleration?
(b) If the coefficient of static friction is 0.30, is the car able to avoid slipping?

Solution

(a) (a_c=\frac{v^2}{r}=\frac{15^2}{50}=4.5;\text{m/s}^2).

(b) Maximum frictional force: (f_{\max}= \mu_s N = \mu_s mg).
Practically speaking, centripetal force needed: (F_c = m a_c). Requirement: (F_c \le f_{\max}) → (a_c \le \mu_s g).
On the flip side, (\mu_s g = 0. 30(9.Practically speaking, 8)=2. 94;\text{m/s}^2).
Since (4.In practice, 5 > 2. 94), the car will slip.

---

2.2 Relative Motion

Question 4.
Two trains depart from the same station simultaneously. Train A travels east at 80 km/h; Train B travels north at 60 km/h.
(a) What is the rate at which the distance between them increases after 2 h?
(b) After how many hours will they be 200 km apart?

Solution

(a) Separation vector components: (x=80t), (y=60t).
That said, separation speed: (\frac{d}{dt}\sqrt{x^2+y^2} = \frac{(x,dx/dt + y,dy/dt)}{\sqrt{x^2+y^2}}). At (t=2) h: (x=160), (y=120).
(dx/dt=80), (dy/dt=60).
(\frac{d}{dt}= \frac{160(80)+120(60)}{\sqrt{160^2+120^2}})
(= \frac{12800+7200}{200} = 100;\text{km/h}).

(b) Set (\sqrt{(80t)^2+(60t)^2}=200).
(\sqrt{6400t^2+3600t^2}=200) → (\sqrt{10000t^2}=200) → (100t=200) → (t=2) h.

---

3. Dynamics

3.1 Newton’s Laws

Question 5.
A 5 kg block rests on a rough horizontal surface (μₙ = 0.4). A horizontal force of 20 N is applied.
(a) Determine the block’s acceleration.
(b) If the force is increased to 30 N, will the block start sliding?

Solution

(a) Friction force: (f_{\max} = μₙ mg = 0.Here's the thing — 4(5)(9. 8)=19.And 6;\text{N}). Net force: (F_{\text{net}} = 20 - 19.6 = 0.4;\text{N}).
Acceleration: (a = F_{\text{net}}/m = 0.4/5 = 0.08;\text{m/s}^2).

(b) For 30 N: (F_{\text{net}} = 30-19.Now, 4/5=2. 4;\text{N}).
6 = 10.Since (F_{\text{net}} > 0), the block will slide with (a = 10.08;\text{m/s}^2).

---

3.2 Rotational Dynamics

Question 6.
A solid disk of mass 2 kg and radius 0.5 m rolls without slipping on a horizontal surface. A force of 3 N is applied tangentially at the rim.
(a) What is the angular acceleration?
(b) How far does the disk travel in 4 s?

Solution

Moment of inertia for solid disk: (I = \frac{1}{2} m r^2 = 0.25;\text{kg·m}^2).
Torque: (\tau = Fr = 3(0.5)=1.Even so, 5;\text{N·m}). Angular acceleration: (\alpha = \tau/I = 1.5/0.25 = 6;\text{rad/s}^2).

Linear acceleration (rolling without slipping): (a = \alpha r = 6(0.Now, 5)=3;\text{m/s}^2). Displacement: (s = \frac{1}{2} a t^2 = 0.5(3)(4^2)=24;\text{m}).

---

4. Work, Energy, and Power

4.1 Work–Energy Theorem

Question 7.
A 10 kg sled is pulled up a 30° incline by a constant force of 80 N over a distance of 20 m.
(a) Calculate the work done by the pulling force.
(b) Find the sled’s speed at the top, assuming it started from rest.

Solution

(a) Work: (W = F d \cos\theta = 80(20)\cos30^\circ = 1600(0.But 866)=1385. 6;\text{J}).

Want to learn more? We recommend who claimed that behavior is affected by reinforcement and which statement is an example of symmetric property of congruence for further reading.

(b) Work done equals change in kinetic energy: (\Delta K = \frac{1}{2}mv^2).
Practically speaking, 6) J → (v = \sqrt{2W/m} = \sqrt{2(1385. Day to day, 12} = 16. 6)/10} = \sqrt{277.Day to day, set (\Delta K = 1385. 65;\text{m/s}).

---

4.2 Power

Question 8.
A cyclist applies a constant force of 200 N to maintain a speed of 5 m/s on level ground.
What is the power output?

Solution
Power: (P = Fv = 200(5)=1000;\text{W}).

---

5. Conservation Laws

5.1 Linear Momentum

Question 9.
Two ice skaters, masses 70 kg and 50 kg, push off from each other on a frictionless surface. The lighter skater moves at 2 m/s. What is the speed of the heavier skater?

Solution
Conservation: (m_1 v_1 + m_2 v_2 = 0).
(70 v_1 + 50(2) = 0) → (70 v_1 = -100) → (v_1 = -1.43;\text{m/s}).
Negative sign indicates opposite direction.

---

5.2 Angular Momentum

Question 10.
A figure skater with a moment of inertia (I_1 = 0.5;\text{kg·m}^2) spins at 4 rad/s. She pulls her arms in, reducing (I) to 0.3 kg·m². What is her new angular speed?

Solution
(I_1 \omega_1 = I_2 \omega_2).
(\omega_2 = \frac{I_1}{I_2}\omega_1 = \frac{0.5}{0.3}(4) = 6.67;\text{rad/s}).

---

6. Harmonic Motion

6.1 Simple Harmonic Oscillator

Question 11.
A mass of 0.3 kg is attached to a spring with a constant of 120 N/m. If the mass is displaced 0.05 m from equilibrium and released from rest, compute:
(a) The angular frequency.
(b) The period.
(c) The maximum speed.

Solution

(a) (\omega = \sqrt{k/m} = \sqrt{120/0.3} = \sqrt{400} = 20;\text{rad/s}).

(b) (T = 2\pi/\omega = 2\pi/20 = 0.314;\text{s}).

(c) Maximum speed (v_{\max} = A\omega = 0.Here's the thing — 05(20) = 1. 0;\text{m/s}).

---

6.2 Damped Oscillations

Question 12.
A damped oscillator has a damping coefficient (b = 0.8;\text{kg/s}), spring constant (k = 4;\text{N/m}), and mass (m = 0.5;\text{kg}).
(a) Is the motion overdamped, critically damped, or underdamped?
(b) What is the damped angular frequency if underdamped?

Solution

(a) Compute (\beta = b/(2m) = 0.8/(1) = 0.Natural frequency (\omega_0 = \sqrt{k/m} = \sqrt{4/0.8).
5}=2).
Since (\beta < \omega_0), the motion is underdamped.

(b) Damped frequency: (\omega_d = \sqrt{\omega_0^2 - \beta^2} = \sqrt{4-0.36}=1.64}= \sqrt{3.83;\text{rad/s}).

---

7. Fluid Statics and Dynamics

7.1 Pressure and Buoyancy

Question 13.
A steel cylinder of radius 0.2 m and height 1 m is submerged in water. Its density is 7,800 kg/m³.
(a) What is the buoyant force?
(b) Is the cylinder floating or sinking?

Solution

Volume: (V = \pi r^2 h = \pi(0.Which means 2)^2(1)=0. 1257;\text{m}^3).
In real terms, weight: (W = \rho_{\text{steel}} V g = 7800(0. 1257)(9.8)=9,600;\text{N}).
Buoyant force: (B = \rho_{\text{water}} V g = 1000(0.Still, 1257)(9. 8)=1,232;\text{N}).
Since (B < W), the cylinder sinks.

---

7.2 Continuity and Bernoulli

Question 14.
Water flows through a horizontal pipe that narrows from a diameter of 0.5 m to 0.25 m. If the velocity in the larger section is 2 m/s, what is the velocity in the narrower section?

Solution
Cross‑sectional area:
(A_1 = \pi (0.25)^2 = 0.196;\text{m}^2).
(A_2 = \pi (0.125)^2 = 0.049;\text{m}^2).
Continuity: (A_1 v_1 = A_2 v_2).
(v_2 = \frac{A_1}{A_2} v_1 = \frac{0.196}{0.049} (2) = 8;\text{m/s}).

---

8. Common Exam Strategies

1. Identify the law or principle before manipulating equations.
2. Check units at every step; mismatched units often reveal mistakes.
3. Sketch diagrams for multi‑step problems—visual cues aid memory.
4. Estimate before calculating; a rough guess can flag unreasonable answers.
5. Use free‑response practice to improve timing and writing clarity.

---

FAQ

Q: How many practice problems should I solve daily?
A: Aim for 5–10 high‑quality problems per day, alternating topics to build breadth.

Q: Should I focus on multiple‑choice or free‑response?
A: Both. Multiple‑choice hones quick reasoning; free‑response develops detailed explanation skills.

Q: What is the best way to review mistakes?
A: Re‑solve the problem without looking at the solution, then compare and note the error source.

---

Conclusion

Mastering AP Physics C: Mechanics hinges on a solid grasp of core concepts and the ability to apply them flexibly to diverse problems. Which means by systematically working through these practice questions—ranging from kinematics to rotational dynamics, energy conservation to fluid mechanics—students can internalize the subject’s mathematical framework and develop the analytical mindset needed for exam success. Consistent practice, coupled with thoughtful review, transforms theoretical knowledge into reliable problem‑solving proficiency.