Gaokao Exam Physics: Mechanics and Kinematics 3 — Questions and Answers

Question 1: A ball is thrown horizontally from a cliff of height 80 m with initial speed 20 m/s. How far from the base of the cliff does it land? (g = 10 m/s²)

80 m (Correct answer)
40 m
160 m
20 m

Correct answer: 80 m

Time to fall: h = ½gt² → 80 = ½(10)t² → t = 4 s. Horizontal distance: x = v₀t = 20 × 4 = 80 m.

This is a standard horizontal projectile (平抛运动) problem. Vertical: starting from rest vertically, h = ½gt² → 80 = ½(10)t² → t² = 16 → t = 4 s. Horizontal: no horizontal force, so vₓ = 20 m/s (constant). Horizontal distance: x = vₓt = 20 × 4 = 80 m. The ball lands 80 m from the base of the cliff. This type of calculation appears frequently on Gaokao physics papers and tests the student's ability to separate horizontal and vertical motions.

Question 2: Two objects collide in a perfectly elastic collision on a frictionless surface. What is conserved?

Both momentum and kinetic energy (Correct answer)
Only momentum
Only kinetic energy
Neither momentum nor kinetic energy

Correct answer: Both momentum and kinetic energy

In a perfectly elastic collision: (1) momentum is conserved (all collisions), and (2) kinetic energy is also conserved (elastic only). In inelastic collisions, momentum is conserved but kinetic energy is not.

Collision types in Chinese physics (必修): (1) Perfectly elastic (完全弹性碰撞): both momentum (动量) and kinetic energy (动能) are conserved. (2) Inelastic (非完全弹性碰撞): momentum is conserved, kinetic energy decreases. (3) Perfectly inelastic (完全非弹性碰撞): momentum is conserved, objects stick together, maximum kinetic energy loss. Momentum is always conserved (closed system), while kinetic energy is only conserved in elastic collisions. The elastic collision condition gives two equations (momentum + energy), allowing calculation of both final velocities.

Question 3: A satellite orbits Earth at radius R with period T. If it moves to orbit at radius 4R, what is its new period?

8T (Correct answer)
4T
2T
16T

Correct answer: 8T

Kepler's Third Law: T² ∝ R³. If R increases by factor 4, then T² increases by factor 4³ = 64, so T increases by factor 8. New period = 8T.

Kepler's Third Law (开普勒第三定律) states T² ∝ R³ (or T²/R³ = constant) for objects orbiting the same central body. If the radius changes from R to 4R: (T_new/T)² = (4R/R)³ = 4³ = 64. Therefore T_new/T = √64 = 8, so T_new = 8T. This type of problem is common in Gaokao physics sections on circular motion and gravitation (万有引力与航天). Understanding the relationship between orbital radius, period, velocity, and gravitational acceleration is essential.

Question 4: An object of mass 5 kg is pulled along a horizontal surface at constant velocity by a force of 20 N applied at an angle of 30° above horizontal. What is the coefficient of kinetic friction? (g = 10 m/s²)

≈ 0.40 (Correct answer)
0.50
0.20
0.25

Correct answer: ≈ 0.40

At constant velocity, net force = 0. Horizontal: Fcos30° = f = μN. Vertical: N = mg - Fsin30° = 50-10 = 40 N. μ = Fcos30°/N = 20×(√3/2)/40 = 10√3/40 ≈ 0.433 ≈ 0.43.

Constant velocity → ΣF = 0. Applied force F = 20 N at θ = 30°. Horizontal equilibrium: Fcos30° = friction f → 20 × (√3/2) = 10√3 ≈ 17.3 N. Vertical equilibrium: N + Fsin30° = mg → N = mg - Fsin30° = 5×10 - 20×0.5 = 50 - 10 = 40 N. Coefficient of kinetic friction: μ = f/N = 10√3/40 = √3/4 ≈ 0.433. This problem tests vector decomposition, static equilibrium conditions, and friction force calculations — all key Gaokao mechanics topics.

Question 5: In simple harmonic motion (简谐运动), when is the acceleration of the oscillating object at maximum?

At the maximum displacement (amplitude position, turning points) (Correct answer)
At the equilibrium position
When velocity is maximum
When kinetic energy is maximum

Correct answer: At the maximum displacement (amplitude position, turning points)

In SHM, acceleration a = -ω²x. Acceleration is maximum when displacement x is maximum (at the turning points/amplitude). At equilibrium (x=0), acceleration = 0.

In simple harmonic motion, the restoring force is F = -kx (Hooke's Law), so acceleration a = F/m = -(k/m)x = -ω²x. This means acceleration is directly proportional to displacement and opposite in direction. Acceleration is maximum when |x| is maximum — i.e., at the turning points (maximum displacement, ±A, the amplitude). At the equilibrium position (x = 0), acceleration = 0 but velocity is maximum. This inverse relationship between acceleration and velocity in SHM is a key concept in Chinese physics (选修 or 必修 depending on curriculum version).

Question 6: A 10 kg block falls freely from rest. After 3 seconds, what is its kinetic energy? (g = 10 m/s²)

4500 J (Correct answer)
450 J
300 J
900 J

Correct answer: 4500 J

Velocity after 3 s: v = gt = 30 m/s. KE = ½mv² = ½ × 10 × 900 = 4500 J.

Step 1 — Find velocity: in free fall, v = v₀ + gt = 0 + 10×3 = 30 m/s. Step 2 — Calculate kinetic energy: KE = ½mv² = ½ × 10 kg × (30 m/s)² = ½ × 10 × 900 = 4500 J. Alternatively, using the work-energy theorem: the work done by gravity equals the change in kinetic energy. Distance fallen: h = ½gt² = 45 m. Work by gravity: W = mgh = 10×10×45 = 4500 J. Both methods confirm KE = 4500 J. The work-energy theorem is a fundamental Gaokao physics topic.

Gaokao Exam Physics: Mechanics and Kinematics 3

A ball is thrown horizontally from a cliff of height 80 m with initial speed 20 m/s.
How far from the base of the cliff does it land? (g = 10 m/s²)