Mechanics Practice
Mechanics Practice
15 practice questions covering Newton’s laws, energy, momentum, rotational motion, and oscillations.
Newton’s Laws
Question 1. A 10 kg block rests on a horizontal surface with a coefficient of static friction of 0.4. What is the minimum horizontal force required to start the block moving? (Take g = 10 m/s^2)
(A) 25 N (B) 40 N (C) 50 N (D) 100 N
Correct Answer: (B)
Explanation: The maximum static friction force is f_s = mu_s N = mu_s mg = 0.4 10 10 = 40 N. The applied force must exceed this value to initiate motion. Once the block starts moving, kinetic friction (in standard practice lower) applies. The normal force equals the weight on a horizontal surface with no vertical acceleration.
Question 2. According to Newton’s third law, when a hammer strikes a nail:
(A) The hammer exerts a greater force on the nail than the nail exerts on the hammer (B) The nail exerts a greater force on the hammer (C) Both forces are equal in magnitude and opposite in direction (D) No force acts on the hammer
Correct Answer: (C)
Explanation: Newton’s third law states that for every action force, there is an equal and opposite reaction force. The force the hammer exerts on the nail is equal in magnitude and opposite in direction to the force the nail exerts on the hammer. These forces act on different objects, which is why they do not cancel. The nail accelerates because the force on it exceeds its resistance.
Question 3. A 5 kg object experiences a net force of 20 N to the right and 12 N to the left simultaneously. What is its acceleration?
(A) 0.64 m/s^2 to the right (B) 1.6 m/s^2 to the right (C) 6.4 m/s^2 to the right (D) 4.0 m/s^2 to the right
Correct Answer: (B)
Explanation: Net force = 20 - 12 = 8 N to the right. By Newton’s second law, a = F/m = 8/5 = 1.6 m/s^2 to the right. The direction of acceleration is the same as the direction of the net force.
Energy
Question 4. A ball is thrown vertically upward with an initial velocity of 25 m/s. What is its maximum height? (Take g = 10 m/s^2, ignore air resistance)
(A) 25 m (B) 31.25 m (C) 50 m (D) 62.5 m
Correct Answer: (B)
Explanation: At maximum height, the ball’s velocity is zero. Using conservation of energy: (1/2)mv^2 = mgh. Therefore h = v^2/(2g) = 625/20 = 31.25 m. Alternatively, using kinematics: v^2 = u^2 - 2gh gives 0 = 625 - 20h, so h = 31.25 m.
Question 5. A roller coaster car of mass 500 kg starts from rest at a height of 35 m. What is its speed at the bottom (ground level)? (Ignore friction, g = 10 m/s^2)
(A) 15 m/s (B) 20 m/s (C) 26.5 m/s (D) 35 m/s
Correct Answer: (C)
Explanation: By conservation of energy, potential energy at top equals kinetic energy at bottom: mgh = (1/2)mv^2. Therefore v = sqrt(2gh) = sqrt(2 10 35) = sqrt(700) = 26.46 m/s. The mass cancels out, so the speed is independent of mass.
Momentum
Question 6. A 2 kg ball moving at 4 m/s to the right collides elastically with a 2 kg ball at rest. What are the velocities after collision?
(A) Both balls move at 2 m/s to the right (B) First ball stops, second ball moves at 4 m/s to the right (C) First ball rebounds at 4 m/s to the left, second ball stays at rest (D) Both balls move at 4 m/s in opposite directions
Correct Answer: (B)
Explanation: For an elastic collision between two objects of equal mass, they exchange velocities. Before: v_1 = 4 m/s, v_2 = 0. After: v_1 = 0, v_2 = 4 m/s. This can be verified using conservation of momentum (2 4 = 0 + 2 4 = 8) and conservation of kinetic energy (both before and after equal 16 J).
Question 7. A 1500 kg car moving at 20 m/s collides with a wall and rebounds at 5 m/s in the opposite direction. What is the impulse delivered to the car?
(A) 15,000 Ns (B) 22,500 Ns (C) 37,500 Ns (D) 7,500 Ns
Correct Answer: (C)
Explanation: Impulse equals change in momentum: J = m (v_final - v_initial). Taking the initial direction as positive: v_initial = 20 m/s, v_final = -5 m/s. J = 1500 (-5 - 20) = 1500 * (-25) = -37,500 Ns. The magnitude is 37,500 Ns. The negative sign indicates the impulse is in the direction opposite to the car’s initial motion.
Question 8. A 0.5 kg hockey puck sliding on ice at 10 m/s is struck by a stick and moves in the opposite direction at 15 m/s. If the stick was in contact for 0.01 s, what was the average force exerted?
(A) 500 N (B) 1,250 N (C) 5,000 N (D) 12,500 N
Correct Answer: (B)
Explanation: Impulse J = F t = m (v_final - v_initial). Taking the original direction as positive: v_initial = 10 m/s, v_final = -15 m/s. F 0.01 = 0.5 (-15 - 10) = 0.5 * (-25) = -12.5. F = -12.5 / 0.01 = -1250 N. The magnitude of the average force is 1,250 N, directed opposite to the puck’s initial motion.
Rotational Motion
Question 9. A solid cylinder of mass 2 kg and radius 0.3 m rolls without slipping down an incline. What fraction of its total kinetic energy is rotational?
(A) 1/3 (B) 1/2 (C) 2/3 (D) 1/4
Correct Answer: (A)
Explanation: For a solid cylinder, I = (1/2)mR^2. The rotational KE = (1/2)I*omega^2 = (1/2)(1/2)mR^2(v/R)^2 = (1/4)mv^2. The translational KE = (1/2)mv^2. Total KE = (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2. The fraction that is rotational = (1/4)/(3/4) = 1/3.
Question 10. A figure skater spins with arms extended and then pulls them in close to her body. What happens to her angular velocity?
(A) It decreases because she loses energy (B) It increases because angular momentum is conserved and moment of inertia decreases (C) It stays the same because mass is constant (D) It decreases because the radius decreases
Correct Answer: (B)
Explanation: Angular momentum L = I*omega is conserved in the absence of external torques. When the skater pulls her arms in, her moment of inertia I decreases (mass is distributed closer to the axis). Since L is constant, omega must increase: omega_new = L / I_new > L / I_old = omega_old. This is why spinning skaters speed up as they pull their limbs inward.
Question 11. A uniform rod of length L and mass M is pivoted at one end. What is the moment of inertia about the pivot?
(A) (1/12)ML^2 (B) (1/3)ML^2 (C) ML^2 (D) (1/2)ML^2
Correct Answer: (B)
Explanation: For a uniform rod about its center of mass, I_cm = (1/12)ML^2. Using the parallel axis theorem: I = I_cm + Md^2 where d = L/2 (distance from center to end). I = (1/12)ML^2 + M(L/2)^2 = (1/12)ML^2 + (1/4)ML^2 = (1/3)ML^2. The moment of inertia is larger about the end because mass is distributed farther from the axis.
Oscillations
Question 12. A mass-spring system oscillates with a period of 2 seconds. If the mass is doubled, what is the new period?
(A) 1 s (B) sqrt(2) s (C) 2*sqrt(2) s approximately 2.83 s (D) 4 s
Correct Answer: (C)
Explanation: The period of a mass-spring system is T = 2pisqrt(m/k). If mass is doubled (m -> 2m), the new period is T_new = 2pisqrt(2m/k) = sqrt(2) 2pisqrt(m/k) = sqrt(2) T = sqrt(2) * 2 = 2.83 s. The period increases by a factor of sqrt(2) when mass is doubled.
Question 13. At which point in simple harmonic motion does an oscillating object have maximum kinetic energy?
(A) At the maximum displacement (amplitude) (B) At the equilibrium position (C) At the quarter-period position (D) Kinetic energy is constant throughout
Correct Answer: (B)
Explanation: In simple harmonic motion, energy oscillates between kinetic and potential. At maximum displacement, velocity is zero (all energy is potential). At the equilibrium position, velocity is maximum (all energy is kinetic). The total mechanical energy is constant: KE + PE = constant. Maximum KE = (1/2)mv_max^2 = (1/2)kA^2 where A is the amplitude.
Question 14. A simple pendulum has a period T on Earth. If the pendulum is taken to a planet where the gravitational acceleration is 4 times that of Earth, what is the new period?
(A) T/4 (B) T/2 (C) T (D) 2T
Correct Answer: (B)
Explanation: The period of a simple pendulum is T = 2pisqrt(L/g). If g becomes 4g, the new period is T_new = 2pisqrt(L/(4g)) = (1/2) * 2pisqrt(L/g) = T/2. The period is inversely proportional to the square root of gravitational acceleration. Stronger gravity means the pendulum oscillates faster.
Question 15. A damped harmonic oscillator has its amplitude decrease by 50 percent after 10 oscillations. What can be said about the damping?
(A) The system is overdamped (B) The system is critically damped (C) The system is underdamped (D) The system has no damping
Correct Answer: (C)
Explanation: An underdamped oscillator exhibits decaying oscillations where the amplitude decreases exponentially over time while the system oscillates. The fact that the system completes 10 oscillations with gradually decreasing amplitude confirms underdamping. Overdamped and critically damped systems do not oscillate; they return to equilibrium without overshooting. A decrease of 50 percent over 10 oscillations indicates relatively light damping.
Intuition
Building physical intuition through practice: Working through mechanics problems builds the intuition that lets you see solutions before you compute them, you develop a sense for how forces combine, how energy transforms, and how momentum is conserved.
Why it matters: Mechanics is the foundation of all physics and engineering. Practicing these skills makes you a better physicist, engineer, or scientist.
The key insight: The hardest part of mechanics is not the math, it is translating a physical situation into equations. Practice builds this translation skill, which is the real payoff of physics education.
Common Mistakes
Confusing damping regimes: Underdamped: oscillates with decay. Critically damped: fastest return without oscillation. Overdamped: slow return without oscillation. The discriminant b² − 4mk determines the regime.
Forgetting that simple harmonic motion requires small angles: The pendulum formula T = 2π√(L/g) only works for small angles. For large angles, the period increases and the formula breaks down.
Mixing up frequency and angular frequency: f is in Hz; ω = 2πf is in rad/s. Using f where ω is required (or vice versa) gives answers off by 2π.
Cross-References
- Site Home: Main landing page for physics notes.
- Practice: Practice problems for revision.