Gaokao Exam Physics: Mechanics and Kinematics 1

Question 1

A ball is thrown horizontally from a cliff with an initial speed of 10 m/s. What is the magnitude of its vertical velocity component after 2 seconds? (g = 10 m/s², ignore air resistance)

Accepted Answer

20 m/s

Answer

In projectile motion, the horizontal and vertical components are independent. The vertical component starts at zero and increases due to gravity: v_y = g × t = 10 × 2 = 20 m/s. The horizontal component (10 m/s) does not affect the vertical calculation.

Question 2

A 5 kg box is pushed across a horizontal floor by a 30 N horizontal force. The coefficient of kinetic friction between the box and floor is 0.4. What is the acceleration of the box? (g = 10 m/s²)

Accepted Answer

2 m/s²

Answer

Normal force N = mg = 5 × 10 = 50 N. Friction force f = μN = 0.4 × 50 = 20 N. Net force = 30 − 20 = 10 N. By Newton's Second Law, a = F_net / m = 10 / 5 = 2 m/s².

Question 3

A train moving at 72 km/h applies its brakes and decelerates uniformly to rest in 10 seconds. What distance does the train cover while braking?

Accepted Answer

100 m

Answer

Convert speed: 72 km/h = 20 m/s. Using s = (u + v)/2 × t = (20 + 0)/2 × 10 = 100 m. Alternatively, find deceleration a = (0 − 20)/10 = −2 m/s², then s = ut + ½at² = 20(10) + ½(−2)(100) = 200 − 100 = 100 m.

Question 4

An object rests on a horizontal table. Which statement correctly describes the weight of the object and the normal force exerted by the table on the object?

Accepted Answer

They are equal in magnitude but act on the same object and are NOT an action-reaction pair

Answer

Weight and normal force are equal in magnitude and opposite in direction, but both act on the same object — they are a balanced force pair (Newton's First Law), not an action-reaction pair. An action-reaction pair (Newton's Third Law) always acts on two different objects. The true reaction to the object's weight is the gravitational pull the object exerts on the Earth.

Question 5

A 500 g ball traveling at 4 m/s is acted on by a constant net force for 2 seconds, accelerating it to 8 m/s in the same direction. What is the magnitude of the net force?

Accepted Answer

1 N

Answer

Mass m = 0.5 kg. Change in velocity Δv = 8 − 4 = 4 m/s over t = 2 s, so a = 4/2 = 2 m/s². By Newton's Second Law, F = ma = 0.5 × 2 = 1 N.

Gaokao Exam Practice Test

Gaokao Exam Practice Test

Gaokao Exam Physics: Mechanics and Kinematics 1