Mechanics Practice Questions
Thanaweya Amma (Egypt) · Thanaweya Physics · 135 free MCQs with instant results and detailed explanations.
135
Total
43
Easy
66
Medium
26
Hard
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Sample Questions from Mechanics
Here are 10 sample questions. Start a quiz to get randomized questions with scoring.
Q1
Easy
Which of the following statements about an object in free fall is correct?
A.
It experiences zero net force.
B.
It accelerates downward due to gravity.
C.
It moves at a constant speed.
D.
It gains mass as it falls.
Show Answer & Explanation
Correct Answer: B
An object in free fall accelerates downward primarily due to the force of gravity, which is approximately 9.81 m/sยฒ on Earth.
Q2
Easy
What is the acceleration of an object in free fall near the Earth's surface?
A.
9.81 m/sยฒ
B.
0 m/sยฒ
C.
1.62 m/sยฒ
D.
15.9 m/sยฒ
Show Answer & Explanation
Correct Answer: A
The acceleration due to gravity near the Earth's surface is approximately 9.81 m/sยฒ, which is the standard value used in physics for free-fall calculations.
Q3
Easy
Which of the following best describes Newton's First Law of Motion?
A.
An object at rest stays at rest unless acted upon by an external force.
B.
Force equals mass times acceleration.
C.
For every action, there is an equal and opposite reaction.
D.
Momentum is conserved in a closed system.
Show Answer & Explanation
Correct Answer: A
Newton's First Law states that an object will remain at rest or in uniform motion unless acted upon by an external force, illustrating the concept of inertia.
Q4
Medium
A car accelerates from rest at a constant rate of 2 m/sยฒ. What is the distance traveled by the car after 5 seconds?
A.
25 meters
B.
20 meters
C.
30 meters
D.
15 meters
Show Answer & Explanation
Correct Answer: A
Using the formula for distance under constant acceleration: distance = 0.5 * acceleration * timeยฒ. Here, distance = 0.5 * 2 m/sยฒ * (5 s)ยฒ = 25 meters.
Q5
Medium
A 5 kg block is pulled across a horizontal surface with a constant velocity. If the frictional force opposing the motion is 10 N, what is the work done against friction if the block moves 8 meters?
A.
80 Joules
B.
50 Joules
C.
40 Joules
D.
70 Joules
Show Answer & Explanation
Correct Answer: A
Work done against friction is calculated by the formula: Work = Frictional Force ร Distance. Here, Work = 10 N ร 8 m = 80 Joules.
Q6
Medium
A ball is thrown vertically upward with an initial velocity of 20 m/s. What maximum height will it reach? (Take g = 10 m/sยฒ)
A.
20 meters
B.
30 meters
C.
40 meters
D.
10 meters
Show Answer & Explanation
Correct Answer: A
The maximum height can be calculated using the formula: h = (vยฒ)/(2g). Here, h = (20 m/s)ยฒ / (2 * 10 m/sยฒ) = 20 meters.
Q7
Medium
A projectile is launched from the ground with an angle of 45 degrees and initial speed of 30 m/s. What is the total time of flight until it hits the ground?
A.
4.5 seconds
B.
3 seconds
C.
6 seconds
D.
2 seconds
Show Answer & Explanation
Correct Answer: C
The time of flight for a projectile launched at angle ฮธ is given by: T = (2 * v * sin(ฮธ)) / g. Here, T = (2 * 30 * sin(45ยฐ)) / 10 = 6 seconds.
Q8
Hard
A block of mass 5 kg is placed on a frictionless surface and is subjected to a force of 10 N. What is the velocity of the block after 4 seconds?
A.
8 m/s
B.
4 m/s
C.
2 m/s
D.
10 m/s
Show Answer & Explanation
Correct Answer: A
According to Newton's second law, F = ma, where F is the force, m is mass, and a is acceleration. Given F = 10 N and mass m = 5 kg, we find the acceleration: a = F/m = 10 N / 5 kg = 2 m/sยฒ. The velocity after time t can be calculated using the formula: final velocity = initial velocity + acceleration * time. Since the initial velocity is 0, we get final velocity = 0 + (2 m/sยฒ * 4 s) = 8 m/s.
Q9
Hard
A 2000 kg car is moving at a speed of 15 m/s when the driver applies the brakes, bringing the car to a stop in 5 seconds. What is the average force exerted by the brakes on the car?
A.
6000 N
B.
3000 N
C.
4000 N
D.
5000 N
Show Answer & Explanation
Correct Answer: A
The average force can be calculated using the formula F = m * a, where 'a' is acceleration. First, calculate acceleration: a = (final velocity - initial velocity) / time = (0 - 15) / 5 = -3 m/sยฒ. Then, F = 2000 kg * -3 m/sยฒ = -6000 N. The negative sign indicates direction; hence, the force of 6000 N is exerted by the brakes in the opposite direction.
Q10
Hard
A projectile is launched with an initial velocity of 40 m/s at an angle of 30ยฐ to the horizontal. What is the maximum height reached by the projectile? (Use g = 10 m/sยฒ)
A.
80 m
B.
60 m
C.
40 m
D.
20 m
Show Answer & Explanation
Correct Answer: B
The maximum height H can be found using the formula H = (uยฒ * sinยฒ ฮธ) / (2g). Here, u = 40 m/s, ฮธ = 30ยฐ (sin 30ยฐ = 0.5), and g = 10 m/sยฒ. Substituting these values: H = (40ยฒ * (0.5)ยฒ) / (2 * 10) = (1600 * 0.25) / 20 = 200 m / 20 = 60 m.
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Mechanics โ Thanaweya Amma (Egypt) Thanaweya Physics Practice Questions Online
This page contains 135 practice MCQs for the chapter Mechanics in Thanaweya Amma (Egypt) Thanaweya Physics. The questions are organized by difficulty โ 43 easy, 66 medium, 26 hard โ so you can choose the right level for your preparation.
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