IB · Physics SL & HL · A Space, time & motion

A.2 Forces & momentum: IB Physics questions with answers

Free A.2 Forces & momentum questions for IB Physics (SL & HL), part of A Space, time & motion. Each question has the answer and the marking points an examiner would look for.

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A.2 Forces & momentum practice questions

Question 1Multiple choice · 1 markA* challenge

A ball of mass 0.5 kg moves in a vertical circle of radius 10 m on the end of a string. At the lowest point its speed is 10 m s−1.
What is the tension in the string at the lowest point?

  1. A 0.095 N
  2. B 4.9 N
  3. C 5.0 N
  4. D 9.9 N
Show answer

Answer: D

Question 2Multiple choice · 1 mark

A skydiver reaches terminal speed and then opens her parachute. What happens to the drag force and the speed immediately after the parachute opens?

drag forcespeed
Astays the samedecreases
Bincreasesincreases
Cincreasesdecreases
Ddecreasesdecreases
Show answer

Answer: C

Question 3Structured · 9 marksA* challenge

A steel ball of radius 1.5 mm and density 7800 kg m−3 is released from rest in a tall column of glycerol of density 1260 kg m−3 and viscosity 1.4 Pa s.

(a)

Show that the weight of the ball is about 1 × 10−3 N.

[2]
(b)

Calculate the initial acceleration of the ball.

[2]
(c)

Explain why the ball reaches a constant velocity.

[2]
(d)

Determine the terminal speed of the ball.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)1.08 × 10−3 N
V = 4/3 πr3 = 1.41 × 10−8 m3 ✓
W = ρVg = 1.08 × 10−3 N ✓
2
(b)8.2 m s−2
buoyancy = 1.75 × 10−4 N ✓
a = (W − B)/m = 8.2 m s−2 ✓
2
(c)drag increases with speed until forces balance
viscous drag increases as the speed increases ✓
when drag + buoyancy = weight the resultant force is zero ✓
2
(d)0.023 m s−1
6πηrv = W − B ✓
v = 2r2g(ρs − ρf)/9η ✓
= 0.023 m s−1 ✓
3
Question 4Structured · 10 marks

Glider A, of mass 0.3 kg, moves at 0.8 m s−1 along a horizontal air track towards a stationary glider B of mass 0.6 kg. After the collision A rebounds at 0.1 m s−1.

(a)

State the condition for the total momentum of the two gliders to be conserved.

[1]
(b)

Calculate the velocity of B after the collision.

[2]
(c)

Determine whether the collision is elastic.

[3]
(d)

The force on B during the collision varies with time as a triangle, lasting 0.08 s. Determine the maximum force on B.

[2]
(e)

Explain, with reference to Newton’s third law, why the change of momentum of A is equal and opposite to that of B.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)no external force
no resultant external force acts on the system ✓
1
(b)0.450 m s−1
0.3(0.8) = 0.3(-0.1) + 0.6v ✓
v = 0.450 m s−1 ✓
2
(c)inelastic
initial KE = 0.0960 J ✓
final KE = 0.0623 J ✓
kinetic energy is not conserved, so inelastic ✓
3
(d)6.75 N
impulse = Δp = 0.270 N s = area = ½Fmaxt ✓
Fmax = 6.75 N ✓
2
(e)equal and opposite forces for same time
the force of A on B is equal and opposite to the force of B on A ✓
they act for the same time, so the impulses are equal and opposite ✓
2
Question 5Structured · 7 marks

A car of mass 1200 kg travelling at 25 m s−1 brakes and stops in a distance of 40 m.

(a)

Calculate the deceleration of the car.

[2]
(b)

Calculate the average braking force.

[1]
(c)

Determine the minimum coefficient of friction between the tyres and the road.

[2]
(d)

Explain, with reference to momentum, why a seat belt that stretches reduces injury in a collision.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)7.8 m s−2
v2 = u2 + 2as ✓
a = 625/80 = 7.81 m s−2 ✓
2
(b)9.4 kN
F = 1200 × 7.81 = 9.4 × 103 N ✓
1
(c)0.80
μmg = ma ✓
μ = 7.81/9.8 = 0.80 ✓
2
(d)longer time, smaller force
the change of momentum of the passenger is the same ✓
it occurs over a longer time so the force is smaller ✓
2
Question 6Structured · 7 marksA* challenge

A box of mass 12 kg is pulled along a horizontal floor by a rope at 30° above the horizontal. The tension is 60 N and the box accelerates at 1.5 m s−2.

(a)

Calculate the horizontal component of the tension.

[1]
(b)

Calculate the frictional force on the box.

[2]
(c)

Calculate the normal reaction on the box.

[2]
(d)

Hence determine the coefficient of dynamic friction.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)52 N
60 cos 30° = 52 N ✓
1
(b)34 N
Fnet = ma = 18 N ✓
friction = 52 − 18 = 34 N ✓
2
(c)88 N
N = mg − 60 sin 30° ✓
118 − 30 = 88 N ✓
2
(d)0.39
μ = F/N ✓
0.39 ✓
2

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in