Mechanics Practice Questions

Matric (South Africa) · Matric Physical Sciences · 138 free MCQs with instant results and detailed explanations.

138
Total
43
Easy
67
Medium
28
Hard

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Sample Questions from Mechanics

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Q1
Easy
What is the result of a net force of zero acting on an object?
A. The object will accelerate.
B. The object will remain at rest or continue moving at constant velocity.
C. The object's speed will increase.
D. The object will change direction.
Show Answer & Explanation
Correct Answer: B
When a net force of zero acts on an object, it means there are no unbalanced forces acting on it. According to Newton's first law of motion, this means the object will either remain at rest or continue to move at a constant velocity.
Q2
Easy
If an object is thrown vertically upwards with an initial velocity of 20 m/s, what will be its velocity at the highest point of its trajectory?
A. 20 m/s
B. 0 m/s
C. -20 m/s
D. 10 m/s
Show Answer & Explanation
Correct Answer: B
At the highest point of its trajectory, the object's velocity is momentarily zero before it starts to fall back down due to gravity. This is a key concept in projectile motion.
Q3
Easy
A car accelerates from rest at a rate of 2 m/sยฒ. What is the speed of the car after 5 seconds?
A. 5 m/s
B. 10 m/s
C. 20 m/s
D. 25 m/s
Show Answer & Explanation
Correct Answer: B
Using the formula for acceleration, speed = initial speed + (acceleration ร— time). Here, initial speed is 0, acceleration is 2 m/sยฒ, and time is 5 s, resulting in a final speed of 10 m/s.
Q4
Medium
A car accelerates uniformly from rest to a speed of 20 m/s in 5 seconds. What is the distance covered by the car during this time?
A. 50 m
B. 100 m
C. 200 m
D. 25 m
Show Answer & Explanation
Correct Answer: A
The distance covered under uniform acceleration can be calculated using the formula: distance = initial velocity ร— time + (1/2) ร— acceleration ร— timeยฒ. Here, the initial velocity is 0, and acceleration can be found as (final velocity - initial velocity)/time = (20 m/s - 0 m/s)/5 s = 4 m/sยฒ. Thus, distance = 0 + (1/2) ร— 4 m/sยฒ ร— (5 s)ยฒ = 50 m.
Q5
Medium
A cyclist travels in a straight line at a constant speed of 15 m/s. If the cyclist maintains this speed for 10 seconds, how far does the cyclist travel?
A. 30 m
B. 150 m
C. 135 m
D. 120 m
Show Answer & Explanation
Correct Answer: B
Distance can be calculated using the formula: distance = speed ร— time. Here, speed = 15 m/s and time = 10 s, hence distance = 15 m/s ร— 10 s = 150 m.
Q6
Medium
A car accelerates uniformly from rest to a speed of 30 m/s in 10 seconds. What is the distance covered by the car during this time?
A. 150 m
B. 300 m
C. 75 m
D. 600 m
Show Answer & Explanation
Correct Answer: A
The distance covered is calculated using the formula: distance = initial velocity * time + 0.5 * acceleration * time^2. The initial velocity is 0, acceleration is 3 m/sยฒ, leading to 150 m.
Q7
Medium
A ball is thrown vertically upwards with an initial velocity of 20 m/s. How long will it take for the ball to reach its highest point? (Take g = 10 m/sยฒ)
A. 2 s
B. 1 s
C. 4 s
D. 3 s
Show Answer & Explanation
Correct Answer: A
At the highest point, the final velocity is 0. Using the formula: v = u - gt, we can calculate the time taken to reach that point.
Q8
Hard
A 2 kg block is sliding down a frictionless incline of 30 degrees. What is the acceleration of the block down the incline?
A. 9.8 m/sยฒ
B. 4.9 m/sยฒ
C. 5.5 m/sยฒ
D. 2.5 m/sยฒ
Show Answer & Explanation
Correct Answer: B
The acceleration down the incline is given by a = g * sin(ฮธ), where g = 9.8 m/sยฒ and ฮธ = 30 degrees. Thus, a = 9.8 * sin(30) = 9.8 * 0.5 = 4.9 m/sยฒ.
Q9
Hard
A block of mass 5 kg is sliding down a frictionless incline at an angle of 30 degrees to the horizontal. What is the acceleration of the block down the incline?
A. 4.9 m/sยฒ
B. 5.0 m/sยฒ
C. 3.0 m/sยฒ
D. 2.5 m/sยฒ
Show Answer & Explanation
Correct Answer: A
The acceleration of the block can be calculated using the formula a = g * sin(ฮธ), where g = 9.8 m/sยฒ and ฮธ = 30 degrees. Thus, a = 9.8 * sin(30ยฐ) = 9.8 * 0.5 = 4.9 m/sยฒ.
Q10
Hard
A 2 kg cart is moving with a velocity of 6 m/s when it collides elastically with a stationary cart of mass 3 kg. What is the velocity of the 2 kg cart after the collision?
A. 4 m/s
B. 2 m/s
C. 3 m/s
D. 5 m/s
Show Answer & Explanation
Correct Answer: A
In an elastic collision, momentum and kinetic energy are conserved. Solving the equations leads to the final velocity of the 2 kg cart being 4 m/s after the collision.

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Mechanics โ€” Matric (South Africa) Matric Physical Sciences Practice Questions Online

This page contains 138 practice MCQs for the chapter Mechanics in Matric (South Africa) Matric Physical Sciences. The questions are organized by difficulty โ€” 43 easy, 67 medium, 28 hard โ€” so you can choose the right level for your preparation.

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