Grade 11 Physical Sciences · Paper 1 (Physics)

Mechanics Grade 11

Real past paper questions and answers from 2024 to 2025, with each memo split into the ticks examiners give. Try one, mark yourself, and see exactly which skill cost you marks.

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Question 1.1 · November 2025 · Paper 1 · National (DBE) · 2 marks

Various options are provided as possible answers to the following questions. Each question has only ONE correct answer. Choose the answer and write only the letter (A–D) next to the question numbers (1.1 to 1.10) in the ANSWER BOOK, e.g. 1.11 E.

1.1 The net force acting on an object on a horizontal surface is always ...(2)

A in the direction of motion of the object.
B in the direction of acceleration of the object.
C zero when the object moves at constant acceleration.
D increasing when the object moves at constant acceleration.

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

What's in Mechanics Grade 11

Each skill is one thing a memo gives a tick for. Tap a skill for its own questions, notes and videos.

Vectors in Two Dimensions17 past paper questions

Newton's Laws50 past paper questions

Forces36 past paper questions

Force Diagrams4 past paper questions

Draw a free-body or force diagramDraw every force acting on an object as a labelled arrow, for example weight, normal force, friction and applied force on a box.20 past paper questions · 3 videos

Application of Newton's Laws66 past paper questions

More Mechanics Grade 11 questions and answers

Try each one on paper before you open the memo.

1. Question 2.3.1 · November 2025 · Paper 1 · National (DBE) · 2 marks
A block of mass $5$ kg is at rest on a rough horizontal surface. When a constant force F of magnitude $18$ N acts on the block at an angle $θ$ to the horizontal, the block experiences maximum static friction. See the diagram below.
2.3 The horizontal component of force F is $15$ N.

Calculate:
2.3.1 $θ$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Substitution $F_{x} = F cos θ$
  $15 = 18 cos θ$
  1 mark · Resolve a vector into components
2. ✓
  Answer $θ = 33, 56°$
  1 mark · Resolve a vector into components
2. Question 2.3.2 · November 2024 · Paper 1 · National (DBE) · 5 marks
In an experiment, a crate of mass $8, 5$ kg, lying stationary on a rough horizontal table, is connected to a mass hanger by means of a light inextensible string passing over a frictionless pulley, as shown in the diagram below. Mass pieces are added to the mass hanger and the acceleration of the crate is measured. The experiment is repeated several times by adding different masses to increase the hanging mass each time.

Ignore the effects of air friction.

The results obtained were used to draw the sketch graph below.
2.3 Calculate the:
2.3.2 Magnitude of the acceleration represented by Y on the graph if the coefficient of kinetic friction between the crate and the table is $0, 40$ (5)
Show the memo (5 ticks)Memo
1. ✓
  Newton's second law for the crate (any one)For the crate: $F_{n e t} = ma$
  $T - f_{k} = ma$
  (OR $f_{k} - T = ma$ )
  $T - μ_{k} m g = ma$
  1 mark · Solve problems with connected bodies
2. ✓
  Friction term for the crate $T - 0, 4 (8, 5) (9, 8) = 8, 5 a$ ...... Eqn (1)
  
  (The tick is for $T - 0, 4 (8, 5) (9, 8)$ .)
  1 mark · Solve problems with connected bodies
3. ✓
  Forces on the hanging massFor the hanging mass: $F_{n e t} = ma$
  $m g - T = ma$
  $7, 4 (9, 8) - T = 7, 4 a$ ...... Eqn (2)
  
  (The tick is for $7, 4 (9, 8) - T$ .)
  1 mark · Solve problems with connected bodies
4. ✓
  $ma$ side of the equations (any one) $= 8, 5 a$ in Eqn (1) OR $= 7, 4 a$ in Eqn (2)
  1 mark · Solve problems with connected bodies
5. ✓
  AnswerAdding Eqn (1) and Eqn (2): $72, 52 - 33, 32 = 15, 9 a$
  $Y = a = 2, 47$ (m·s $^{- 2}$ )
  1 mark · Solve problems with connected bodies
3. Question 1.1 · May/June 2025 · Paper 1 · National (DBE) · 2 marks
Various options are provided as possible answers to the following questions. Each question has only ONE correct answer. Choose the answer and write only the letter (A–D) next to the question numbers (1.1 to 1.10) in the ANSWER BOOK, e.g. 1.11 E.
1.1 Three forces act on an object so that the resultant force is zero.

Which ONE of the following vector diagrams is the CORRECT representation of the three forces?(2)
1. A
2. B
3. C
4. D
Show the memo (1 tick)Memo
1. ✓
  Correct optionWhen the resultant of three forces is zero, the force vectors drawn head-to-tail form a CLOSED triangle: every arrow starts where the previous one ends, all the way round.
  
  In D the vertical arrow points up, the slanted arrow runs from the top down to the right, and the horizontal arrow runs back to the starting point. In A, B and C two arrowheads meet at the same corner, so the diagram does not close and the resultant is not zero.
  
  Answer: D
  2 marks · Draw a closed vector diagram
4. Question 1.1 · November 2024 · Paper 1 · National (DBE) · 2 marks
1.1 Several forces are acting on a moving object.

Which ONE of the following statements is CORRECT when these forces are in equilibrium?(2)
1. A The velocity of the object is increasing.
2. B The object is moving at a constant velocity.
3. C The kinetic energy of the object is decreasing.
4. D The object has a non-zero acceleration.
Show the memo (1 tick)Memo
1. ✓
  Correct optionForces in equilibrium have a resultant (net) force of zero. By Newton's second law, $F_{n e t} = ma = 0$ , so $a = 0$ . A moving object with zero acceleration keeps moving at constant velocity (Newton's first law).
  
  A and D need a non-zero acceleration, and C needs the speed to change, which also needs a net force.
  
  Answer: B
  2 marks · Explain using Newton's laws
5. Question 2.3.1 · May/June 2024 · Paper 1 · National (DBE) · 3 marks
Block A of mass $4, 1$ kg is connected to block B of mass $2, 3$ kg by a light inextensible string passing over a frictionless pulley. Block A is at rest on a rough horizontal table and block B hangs vertically, as shown in the diagram below.

A force F of magnitude $49$ N is applied on block A at an angle of $50°$ to the horizontal, causing block A to accelerate TO THE LEFT from rest along the table.

The coefficient of kinetic friction between the surface of the table and block A is $0, 35$ .
2.3 Calculate the magnitude of the:
2.3.1 Kinetic frictional force exerted on block A(3)
Show the memo (3 ticks)Memo
1. ✓
  Formula $f_{k} = μ_{k} F_{N}$
  
  Any one: $f_{k} = μ_{k} (F_{g} - F_{y})$ or $f_{k} = μ_{k} (m g - F sin θ)$
  1 mark · Calculate and explain frictional force
2. ✓
  Substitution $f_{k} = (0, 35) [(4, 1) (9, 8) - 49 sin 50°]$
  1 mark · Calculate and explain frictional force
3. ✓
  Answer with unit $f_{k} = 0, 93$ N
  1 mark · Calculate and explain frictional force

Mechanics Grade 11 videos

Short lessons from South African teachers first. Videos load only when you tap play, to save your data.

Vectors and Scalars PhysicsMiss Martins Maths and Science

Grade 11 Resultant vector and Vector DirectionsMiss Martins Maths and Science

Vectors in 2D grade 11: Resolving a vector into perpendicular and parallel componentsMiss Martins Maths and Science

Grade 11 Physics Closed Vector Diagrams: Triangle of forcesMiss Martins Maths and Science

Grade 11 Newton Laws: Newtons 1stKevinmathscience

Grade 11 Newton Laws: Fnet = maKevinmathscience

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