FREE Gaokao Physics: Mechanics and Kinematics Questions and Answers

Question 1

A car starts from rest and accelerates uniformly on a straight road. It travels a distance x in the first 2 seconds. What is the distance it travels in the subsequent 2 seconds (from t=2s to t=4s)?

Accepted Answer

3x

Answer

For an object in uniformly accelerated motion starting from rest, the distance traveled is proportional to the square of time (s = ½at²). The ratio of distances covered in equal consecutive time intervals is 1:3:5:7... Since the distance in the first interval (0-2s) is x, the distance in the second interval (2-4s) will be 3x.

Question 2

An object of mass 2 kg rests on a horizontal surface with a coefficient of kinetic friction μ=0.2. A horizontal force F=10 N is applied. What is the acceleration of the object? (Assume g = 10 m/s²)

Accepted Answer

3 m/s²

Answer

First, calculate the maximum static friction and the kinetic friction. The normal force N = mg = 2kg * 10m/s² = 20N. The kinetic friction force is f_k = μN = 0.2 * 20N = 4N. Since the applied force (10N) is greater than the friction force (4N), the object accelerates. The net force is F_net = F - f_k = 10N - 4N = 6N. Using Newton's second law, a = F_net / m = 6N / 2kg = 3 m/s².

Question 3

A small ball is released from rest at the top edge of a smooth hemispherical bowl of radius R. What is its speed when it reaches the bottom of the bowl?

Accepted Answer

√(2gR)

Answer

Use the principle of conservation of mechanical energy. At the top, the ball has gravitational potential energy PE = mgR (relative to the bottom) and zero kinetic energy. At the bottom, it has zero potential energy and kinetic energy KE = ½mv². Equating the initial and final energies: mgR = ½mv², which simplifies to v² = 2gR, so v = √(2gR).

Question 4

A block of mass 2 kg starts from rest and slides 4 m down a rough inclined plane with an angle of 30°. Its final speed is 5 m/s. How much work is done by the friction force? (Assume g = 10 m/s²)

Accepted Answer

-15 J

Answer

According to the work-energy theorem, the net work done on the object equals its change in kinetic energy (W_net = ΔKE). The net work is the sum of work done by gravity (W_g) and work done by friction (W_f). ΔKE = ½mv_f² - ½mv_i² = ½(2)(5)² - 0 = 25 J. The work done by gravity is W_g = mgh = mg(d sin30°) = (2)(10)(4 * 0.5) = 40 J. Therefore, W_net = W_g + W_f => 25 J = 40 J + W_f, which gives W_f = -15 J.

Question 5

An object is thrown horizontally with an initial velocity of 15 m/s from the top of a cliff. It lands at a horizontal distance of 45 m from the base of the cliff. How high is the cliff? (Ignore air resistance, g = 10 m/s²)

Accepted Answer

45 m

Answer

The motion can be separated into horizontal and vertical components. The time of flight is determined by the horizontal motion: t = distance / velocity = 45 m / 15 m/s = 3 s. The vertical motion is free fall from rest. The height (h) of the cliff is the vertical distance traveled in this time: h = ½gt² = ½(10 m/s²)(3 s)² = 5 * 9 = 45 m.

Gaokao Exam Practice Test

Gaokao Exam Practice Test

FREE Gaokao Physics: Mechanics and Kinematics Questions and Answers