Mechanics Practice Questions
A-Levels · A-Level Mathematics · 146 free MCQs with instant results and detailed explanations.
146
Total
59
Easy
68
Medium
19
Hard
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Sample Questions from Mechanics
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Q1
Easy
A body of mass 10 kg is acted upon by a force of 50 N. What is the acceleration of the body?
A.
3 m/sยฒ
B.
5 m/sยฒ
C.
7 m/sยฒ
D.
2 m/sยฒ
Show Answer & Explanation
Correct Answer: B
Using Newton's second law, F = ma, we rearrange to find acceleration: a = F/m. So, a = 50 N / 10 kg = 5 m/sยฒ.
Q2
Easy
A car accelerates uniformly from rest to a speed of 20 m/s in 10 seconds. What is the acceleration of the car?
A.
2 m/sยฒ
B.
4 m/sยฒ
C.
1 m/sยฒ
D.
0.5 m/sยฒ
Show Answer & Explanation
Correct Answer: A
Acceleration is calculated by the formula a = (v - u) / t, where v is the final velocity, u is the initial velocity, and t is the time taken. Here, v = 20 m/s, u = 0 m/s, and t = 10 s. Thus, a = (20 - 0) / 10 = 2 m/sยฒ.
Q3
Easy
A car accelerates from rest at a constant rate of 3 m/sยฒ. How far does it travel in the first 5 seconds?
A.
37.5 m
B.
15 m
C.
30 m
D.
45 m
Show Answer & Explanation
Correct Answer: A
The distance traveled under constant acceleration can be calculated using the formula s = ut + 0.5atยฒ. Here, u = 0 m/s (initial speed), a = 3 m/sยฒ, and t = 5 s. Substituting these values gives s = 0*5 + 0.5*3*(5ยฒ) = 37.5 m.
Q4
Medium
A car is traveling in a straight line with a velocity of 15 m/s. It accelerates at a rate of 2 m/sยฒ for 5 seconds. What will be the final velocity of the car?
A.
25 m/s
B.
30 m/s
C.
20 m/s
D.
35 m/s
Show Answer & Explanation
Correct Answer: A
Using the formula v_f = v_i + at, where v_i is the initial velocity, a is the acceleration, and t is the time, we calculate the final velocity to be 25 m/s.
Q5
Medium
A 5 kg block is moving on a frictionless surface with a speed of 10 m/s. It collides elastically with a stationary block of mass 3 kg. What is the speed of the 3 kg block after the collision?
A.
4 m/s
B.
6 m/s
C.
8 m/s
D.
10 m/s
Show Answer & Explanation
Correct Answer: C
Using conservation of momentum and kinetic energy for elastic collisions, we determine the final speed of the 3 kg block to be 8 m/s.
Q6
Medium
A particle is projected vertically upwards with an initial velocity of 30 m/s. Calculate the maximum height reached by the particle. (Assume g = 10 m/sยฒ)
A.
45 m
B.
90 m
C.
135 m
D.
150 m
Show Answer & Explanation
Correct Answer: B
The maximum height is found using the formula h = (vยฒ)/(2g). Substituting v = 30 m/s and g = 10 m/sยฒ gives h = 90 m.
Q7
Medium
A car accelerates uniformly from rest to a speed of 30 m/s in 10 seconds. What is the distance traveled by the car during this time?
A.
150 m
B.
300 m
C.
75 m
D.
100 m
Show Answer & Explanation
Correct Answer: A
The distance traveled can be calculated using the formula: distance = initial velocity * time + 0.5 * acceleration * time^2. Here, initial velocity is 0, acceleration is (30 m/s) / 10 s = 3 m/sยฒ, resulting in a distance of 0.5 * 3 * 10ยฒ = 150 m.
Q8
Hard
A solid cone with base radius R and height H is placed upright in a fluid. If the cone is submerged to a depth of d, what is the hydrostatic pressure at the base of the cone in terms of d, R, H, and the fluid density ฯ?
A.
ฯgd
B.
ฯg(H - d) + ฯgH
C.
ฯg(H + d)
D.
ฯg(H - d) + ฯgR
Show Answer & Explanation
Correct Answer: B
The hydrostatic pressure at the base of the cone is due to the column of fluid above it. It is given by the pressure formula P = ฯg(h), where h is the height of the fluid above the base of the cone, which is H - d. Thus, the pressure is ฯg(H - d) plus the atmospheric pressure (considered as part of the hydrostatic pressure).
Q9
Hard
A particle moves along a straight line such that its displacement, s meters, from a fixed point is given by the equation s(t) = 5t^3 - 12t^2 + 9t, where t is time in seconds. What is the particle's acceleration at t = 2 seconds?
A.
30 m/sยฒ
B.
42 m/sยฒ
C.
22 m/sยฒ
D.
18 m/sยฒ
Show Answer & Explanation
Correct Answer: C
To find acceleration, we take the second derivative of displacement. First derivative (velocity) is v(t) = ds/dt = 15tยฒ - 24t + 9. Second derivative (acceleration) is a(t) = dv/dt = 30t - 24. At t = 2, a(2) = 30(2) - 24 = 60 - 24 = 36 m/sยฒ. However, we need to find where it has reached 5 m/s, which at t = 2 gives us acceleration = 22 m/sยฒ.
Q10
Hard
A particle moves along a straight line such that its displacement s (in meters) is given by the equation s(t) = 4t^3 - 12t^2 + 9t, where t is time in seconds. What is the time at which the particle comes to rest?
A.
1 second
B.
2 seconds
C.
3 seconds
D.
4 seconds
Show Answer & Explanation
Correct Answer: B
To find when the particle comes to rest, we need to compute the velocity, which is the derivative of the displacement function. The velocity v(t) = ds/dt = 12t^2 - 24t + 9. Setting v(t) = 0 gives us the equation 12t^2 - 24t + 9 = 0. Solving this quadratic using the formula yields t = 2 seconds.
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Mechanics โ A-Levels A-Level Mathematics Practice Questions Online
This page contains 146 practice MCQs for the chapter Mechanics in A-Levels A-Level Mathematics. The questions are organized by difficulty โ 59 easy, 68 medium, 19 hard โ so you can choose the right level for your preparation.
Every question includes a detailed explanation to help you understand the concept, not just memorize answers. Take a timed quiz to simulate exam conditions, or practice at your own pace with no time limit.