Free Gaokao Physics: Mechanics and Kinematics Questions and Answers — 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)?
- x
- 2x
- 3x (Correct answer)
- 4x
Correct answer: 3x
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)
- 5 m/s²
- 3 m/s² (Correct answer)
- 4 m/s²
- 0 m/s²
Correct answer: 3 m/s²
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?
- √(gR)
- √(2gR) (Correct answer)
- gR
- 2gR
Correct answer: √(2gR)
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)
- -25 J
- 15 J
- -40 J
- -15 J (Correct answer)
Correct answer: -15 J
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)
- 30 m
- 45 m (Correct answer)
- 75 m
- 90 m
Correct answer: 45 m
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.
Question 6: A car of mass 1000 kg travels at a constant speed of 20 m/s over a convex bridge with a radius of curvature of 100 m. What is the normal force exerted by the bridge on the car at the highest point? (Assume g = 10 m/s²)
- 10000 N
- 4000 N
- 6000 N (Correct answer)
- 14000 N
Correct answer: 6000 N
At the highest point, the net force provides the centripetal force, which is directed downwards towards the center of the circle. The forces acting on the car are gravity (mg, downwards) and the normal force (N, upwards). The net force is mg - N. Therefore, mg - N = mv²/R. Solving for N: N = mg - mv²/R = (1000)(10) - (1000)(20²)/100 = 10000 - 4000 = 6000 N.
Free Gaokao Physics: Mechanics and Kinematics Questions and Answers
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)?