IB Diploma Physics chapter 1: Space, time and motion MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
47 exam-style questions on Space, time and motion, 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 small steel sphere falls through oil and reaches terminal speed v. A steel sphere of twice the radius is released in the same oil. Assuming Stokes' law applies, what is its terminal speed?
A
8v
B
4v
C
2v
D
v
Show answer
Answer: B. Weight minus buoyancy is proportional to r³, while drag = 6πηrv is proportional to rv. Equating them gives v ∝ r², so doubling r gives 4v.
2. A skydiver is falling vertically at her terminal speed. Which statement is correct?
A
her acceleration is equal to g
B
the drag force is greater than her weight, so she slows down
C
the drag force is zero because her speed is constant
D
the drag force equals her weight, so her acceleration is zero
Show answer
Answer: D. At terminal speed the resultant force is zero: upward drag balances weight, so there is no acceleration. Drag is not zero; it is exactly as large as the weight.
3. An electric motor lifts a 50 kg load through 6.0 m in 12 s while taking 400 W of electrical power. g = 9.8 m s−2. What is its efficiency?
A
25%
B
163%
C
82%
D
61%
Show answer
Answer: D. Useful power = mgh ÷ t = 50 × 9.8 × 6.0 ÷ 12 = 245 W. Efficiency = 245 ÷ 400 = 0.61 = 61%. 163% is the inverted ratio; efficiency can never exceed 100%.
4. A ball is thrown vertically upwards at 14 m s−1. Air resistance is negligible and g = 9.8 m s−2. What is the maximum height reached?
A
20 m
B
0.71 m
C
1.4 m
D
10 m
Show answer
Answer: D. At the top v = 0, so 0 = u² − 2gh and h = 14² ÷ (2 × 9.8) = 196 ÷ 19.6 = 10 m. 20 m comes from leaving out the factor 2.
5. A 5.0 kg box rests on a level floor. The coefficient of static friction is 0.40 and of dynamic friction 0.30. g = 9.8 m s−2. A horizontal push of 18 N is applied. What is the friction force on the box?
A
0 N
B
19.6 N
C
18 N
D
14.7 N
Show answer
Answer: C. Maximum static friction = 0.40 × 49 N = 19.6 N. The 18 N push is smaller, so the box stays still and static friction exactly matches the push: 18 N.
6. A stone is thrown horizontally at 8.0 m s−1 from the top of a cliff 20 m high. Air resistance is negligible and g = 9.8 m s−2. How far from the foot of the cliff does it land?
A
8.0 m
B
20 m
C
33 m
D
16 m
Show answer
Answer: D. Vertically: 20 = ½ × 9.8 × t², so t = 2.0 s. The horizontal velocity is unchanged, so distance = 8.0 × 2.0 = 16 m. 33 m uses t² instead of t.
7. A trolley of mass 2.0 kg moving at 3.0 m s−1 collides with a stationary trolley of mass 1.0 kg and they stick together. What is their common velocity?
A
6.0 m s−1
B
3.0 m s−1
C
2.0 m s−1
D
1.5 m s−1
Show answer
Answer: C. Momentum is conserved: 2.0 × 3.0 = (2.0 + 1.0)v, so v = 6.0 ÷ 3.0 = 2.0 m s−1. 1.5 m s−1 wrongly averages the two speeds.
8. A book rests on a table. According to Newton's third law, which force pairs with the gravitational pull of the Earth on the book?
A
the gravitational pull of the book on the Earth
B
the normal force of the table on the book
C
the weight of the table
D
the normal force of the book on the table
Show answer
Answer: A. Third-law pairs are the same type of force acting on two different bodies. The normal force on the book balances its weight (first law) but is not its third-law partner.
9. A satellite moves in a circular orbit at constant speed. How much work does the gravitational force do on it during one orbit?
A
a positive amount equal to its kinetic energy
B
zero, because the force is always perpendicular to the velocity
C
a negative amount, because gravity pulls inwards
D
a positive amount equal to force × circumference
Show answer
Answer: B. W = Fs cos θ with θ = 90° at every instant, so no work is done and the kinetic energy stays constant.
10. A ball of mass 0.40 kg is dropped from 5.0 m and reaches the ground at 9.0 m s−1. g = 9.8 m s−2. How much energy was transferred to the air by air resistance?
A
19.6 J
B
16.2 J
C
35.8 J
D
3.4 J
Show answer
Answer: D. Gravitational potential energy lost = 0.40 × 9.8 × 5.0 = 19.6 J. Kinetic energy gained = ½ × 0.40 × 9.0² = 16.2 J. The difference, 3.4 J, went to the air.
47, each with the answer and a short explanation, across 3 topics: A.1 Kinematics, A.2 Forces and momentum, A.3 Work, energy and power. Every chapter test draws ten of them, so you can take it many times.
Is the IB Diploma Physics chapter 1 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.
