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Mechanics Practice (Interactive)

Intuition

Physics practice is like learning a musical instrument, you must play (solve problems) to develop fluency, not just read sheet music (study formulas): The key to mechanics problems is translating physical scenarios into mathematical equations, then solving and interpreting the results

Why it matters: Regular practice builds the intuition needed to tackle unfamiliar problems in exams and real-world applications

The key insight: The key to mechanics problems is translating physical scenarios into mathematical equations, then solving and interpreting the results

IB Physics, Mechanics Practice

8 auto-graded practice problems at medium to hard difficulty, aligned to the IB Physics syllabus (SL/HL). Select an answer, submit, and review the explanation.

Worked Examples

Example 1: Inclined Plane with Friction

Problem: A 2.0 kg block slides down a 35° incline with μk=0.25\mu_k = 0.25. Find the acceleration. (g=9.81g = 9.81 m/s²)

Solution: Step 1: Component of gravity along incline: mgsin⁡35°=2.0×9.81×0.574=11.26mg\sin 35° = 2.0 \times 9.81 \times 0.574 = 11.26 N

Step 2: Normal force: N=mgcos⁡35°=2.0×9.81×0.819=16.07N = mg\cos 35° = 2.0 \times 9.81 \times 0.819 = 16.07 N

Step 3: Friction force: f=μkN=0.25×16.07=4.02f = \mu_k N = 0.25 \times 16.07 = 4.02 N

Step 4: Net force along incline: Fnet=11.26−4.02=7.24F_{net} = 11.26 - 4.02 = 7.24 N

Step 5: a=Fnetm=7.242.0=3.62a = \frac{F_{net}}{m} = \frac{7.24}{2.0} = 3.62 m/s²

Key insight: Always resolve forces along and perpendicular to the incline, not horizontally and vertically.

Example 2: Work-Energy Theorem

Problem: A 5.0 kg object slides across a rough surface with μk=0.30\mu_k = 0.30. If it starts at 10 m/s, how far does it slide before stopping?

Solution: Step 1: Work-energy theorem: Wnet=ΔKEW_{net} = \Delta KE

Step 2: Work done by friction: W=−f×d=−μkmg×dW = -f \times d = -\mu_k mg \times d

Step 3: ΔKE=0−12mv2=−12(5.0)(10)2=−250\Delta KE = 0 - \frac{1}{2}mv^2 = -\frac{1}{2}(5.0)(10)^2 = -250 J

Step 4: −0.30×5.0×9.81×d=−250-0.30 \times 5.0 \times 9.81 \times d = -250

Step 5: d=25014.715=17.0d = \frac{250}{14.715} = 17.0 m

Key insight: Work-energy problems are often easier than using F=maF = ma because you don’t need to find time or acceleration first.

Example 3: Projectile from Height

Problem: A ball is thrown at 25 m/s at 30° above horizontal from a 20 m building. Find the time of flight. (g=9.81g = 9.81 m/s²)

Solution: Step 1: Vertical component: uy=25sin⁡30°=12.5u_y = 25\sin 30° = 12.5 m/s

Step 2: When the ball hits the ground, s=−20s = -20 m (below starting point)

Step 3: Use s=uyt−12gt2s = u_y t - \frac{1}{2}gt^2: −20=12.5t−4.905t2-20 = 12.5t - 4.905t^2

Step 4: 4.905t2−12.5t−20=04.905t^2 - 12.5t - 20 = 0

Step 5: t=12.5±156.25+392.49.81=12.5±23.429.81t = \frac{12.5 \pm \sqrt{156.25 + 392.4}}{9.81} = \frac{12.5 \pm 23.42}{9.81}

Step 6: t=35.929.81=3.66t = \frac{35.92}{9.81} = 3.66 s (reject negative root)

Key insight: For projectiles from heights, the displacement is negative when the landing point is below the starting point.

Kinematics

Dynamics

Energy

Momentum

Common Mistakes

Confusing elastic with perfectly inelastic collisions: Elastic collisions conserve both momentum AND kinetic energy. Perfectly inelastic collisions conserve momentum but stick together (maximum KE loss). Don’t assume all collisions are elastic.

Forgetting to define a sign convention: Always specify which direction is positive before solving momentum problems. Mixing up signs is the most common error.

Mixing up impulse with force: Impulse is force × time (Ns). Force is impulse/time (N). They’re related but different quantities.

Cross-References