AS Level Physics chapter 3: Dynamics MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
69 exam-style questions on Dynamics, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. A block slides across a rough horizontal floor. Which statement about the frictional force on the block is correct?
A
it acts parallel to the floor, opposing the motion of the block relative to the floor
B
it acts in the direction of motion
C
it acts vertically upwards, balancing the weight
D
it is zero once the block is moving
Show answer
Answer: A. Friction acts along the surfaces in contact and opposes their relative motion. The vertical force balancing the weight is the normal contact force.
2. A ball is thrown vertically upwards. Air resistance is not negligible. What is its acceleration at the instant it reaches its highest point?
A
zero
B
less than g
C
equal to g
D
greater than g
Show answer
Answer: C. At the top the velocity is momentarily zero, so there is no air resistance and the only force is the weight: a = g. On the way up it is greater than g and on the way down less than g.
3. An object hung from a newton-meter on the Earth gives a reading of 12 N. What is its mass?
A
0.82 kg
B
1.2 kg
C
12 kg
D
118 kg
Show answer
Answer: B. W = mg, so m = 12 ÷ 9.81 = 1.2 kg. Multiplying instead of dividing gives 118 kg.
4. At 15 m s−1 the total drag force on a car is 400 N. Assume the drag is proportional to the square of the speed. What driving force is needed to keep the car moving at a constant 30 m s−1 on a level road?
A
400 N
B
800 N
C
1600 N
D
3200 N
Show answer
Answer: C. At constant velocity the driving force equals the drag. Doubling the speed multiplies the drag by 22 = 4, so the driving force = 4 × 400 = 1600 N. 800 N assumes drag is proportional to speed.
5. An object is dropped from rest and falls through air. Which describes its acceleration?
A
constant and equal to g
B
starts at zero and increases to g
C
starts at g and decreases to zero
D
starts at g and increases
Show answer
Answer: C. At release the speed is zero, so there is no drag and a = g. As speed rises, drag increases, the resultant force falls and the acceleration falls to zero at terminal velocity.
6. A ball of mass 0.20 kg hits a wall at 15 m s−1 at right angles and rebounds along the same line at 10 m s−1. What is the magnitude of the change in its momentum?
A
1.0 N s
B
2.0 N s
C
3.0 N s
D
5.0 N s
Show answer
Answer: D. Taking towards the wall as positive, Δp = 0.20 × (−10) − 0.20 × 15 = −5.0 N s, so the magnitude is 5.0 N s. Subtracting the speeds as if both were in the same direction gives 1.0 N s.
7. An object has a weight of 60 N on the Earth. The acceleration of free fall on the Moon is 1.6 m s−2. What is the weight of the object on the Moon?
A
6.1 N
B
9.8 N
C
38 N
D
96 N
Show answer
Answer: B. Mass = 60 ÷ 9.81 = 6.12 kg, which is the same on the Moon. Weight on the Moon = 6.12 × 1.6 = 9.8 N. 6.1 is the mass, not a weight.
8. A 2.0 kg trolley moving at 4.0 m/s collides with a stationary 6.0 kg trolley and they stick together. What is their common velocity after the collision?
A
0.5 m/s
B
1.0 m/s
C
1.3 m/s
D
2.0 m/s
Show answer
Answer: B. Momentum conserved: 2.0 × 4.0 = 8.0 × v, so v = 1.0 m/s.
9. Which statement of the principle of conservation of momentum is correct?
A
the momentum of each object in a collision stays the same
B
the total momentum of a system is always zero
C
the total kinetic energy of a system is constant when no external force acts
D
the total momentum of a system remains constant provided no resultant external force acts on it
Show answer
Answer: D. The principle applies to the total momentum of a closed system: it is constant if there is no resultant external force. Individual momenta can change, and kinetic energy need not be conserved.
10. A ball of mass 0.20 kg hits a wall at 6.0 m/s and rebounds at 4.0 m/s. The contact time is 0.050 s. What is the average force on the ball?
A
8.0 N
B
16 N
C
24 N
D
40 N
Show answer
Answer: D. Δp = 0.20 × (6.0 − (−4.0)) = 2.0 kg m/s. F = Δp/Δt = 2.0 ÷ 0.050 = 40 N.
69, each with the answer and a short explanation, across 3 topics: Momentum and Newton’s laws of motion, Non-uniform motion, Linear momentum and its conservation. Every chapter test draws ten of them, so you can take it many times.
Is the AS Level Physics chapter 3 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
