Grade 12 Physical Sciences · Paper 1 (Physics)

Mechanics Grade 12

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Question 4.4 · May/June 2025 · Paper 1 · National (DBE) · 2 marks

Trolley A of mass

1, 5

kg is at rest on a frictionless horizontal surface. A second trolley B, of mass

2

kg, travelling horizontally at a constant speed v, collides with trolley A. The trolleys stick together and move at a constant velocity to the right, covering a distance of

0, 8

m in

2

s. See the diagram below.

Ignore ALL frictional and rotational effects. Diagram for this question

4.4 During another collision, trolley B exerts a greater force on trolley A and the change in momentum of trolley A is the same as before. How is the time for the collision affected?

Choose from INCREASES, DECREASES or REMAINS THE SAME.

Write down a relevant equation that supports the answer.(2)

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

What's in Mechanics Grade 12

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

Momentum and Impulse42 past paper questions

Vertical Projectile Motion48 past paper questions

Work, Energy and Power89 past paper questions

More Mechanics Grade 12 questions and answers

Try each one on paper before you open the memo.

1. Question 4.2.2 · November 2025 · Paper 1 · National (DBE) · 4 marks
Cricketers at an academy conduct an experiment to determine the relationship between the contact time and the final momentum of a ball for a constant average net force.

A cricket ball of mass $150$ g is thrown horizontally with a certain initial velocity and is struck by a bat so that the ball moves in the opposite direction, as shown in the diagram below. The average net force acting on the ball is horizontal.

The experiment is repeated using the same ball. The average net force and the initial velocity of the ball are kept constant. The contact time between the bat and the ball is changed EACH time. The results obtained are shown in the sketch graph below.
4.2 Calculate the:
4.2.2 Magnitude of the initial velocity of the ball(4)
Show the memo (4 ticks)Memo
1. ✓
  METHOD 1: FormulaAny one:
  $F_{n e t} Δ t = Δ p$
  $F_{n e t} Δ t = p_{f} - p_{i}$
  $F_{n e t} Δ t = m v_{f} - m v_{i}$
  1 mark · Use the impulse-momentum theorem
2. ✓
  METHOD 1: Substitution of impulse $(225) (0, 03) = \dots$
  1 mark · Use the impulse-momentum theorem
3. ✓
  METHOD 1: Substitution of momentum $(225) (0, 03) = 4, 5 - (0, 15) v_{i}$
  1 mark · Use the impulse-momentum theorem
4. ✓
  METHOD 1: Answer $v_{i} = - 15$ m·s $^{- 1}$
  $v_{i} = 15$ m·s $^{- 1}$
  1 mark · Use the impulse-momentum theorem
2. Question 3.3.1 · May/June 2024 · Paper 1 · National (DBE) · 3 marks
A hot-air balloon moves vertically downwards at a constant velocity of $3, 4$ m·s $^{- 1}$ . When the balloon is $15$ m above the ground, a small ball is dropped from the balloon. Refer to the diagram below.

The ball strikes the ground and bounces vertically upwards. The hot-air balloon continues to move downwards at the same constant velocity.

Ignore the effects of air friction acting on the ball.
3.3 Use only EQUATIONS OF MOTION to calculate:
3.3.1 The value of $t_{1}$ indicated on the graph(3)
Show the memo (3 ticks)Memo
1. ✓
  Formula with $Δ t$ Downwards as positive:
  $Δ y = v_{i} Δ t + \frac{1}{2} a Δ t^{2}$
  1 mark · Use equations of motion for projectiles
2. ✓
  Substitution $15 = 3, 4Δ t + \frac{1}{2} (9, 8) Δ t^{2}$
  1 mark · Use equations of motion for projectiles
3. ✓
  Final answer $Δ t = 1, 44$ s, so $t_{1} = 1, 44$ s
  1 mark · Use equations of motion for projectiles
3. Question 5.3 · November 2024 · Paper 1 · National (DBE) · 4 marks
A constant force F is applied at an angle of $30°$ to the horizontal on a crate of mass $6$ kg that is initially at rest, as shown in the diagram below.

A constant frictional force of $10$ N acts on the crate as it moves from rest at point A along a horizontal surface to point B. The distance between point A and point B is $1, 5$ m. The speed of the crate at point B is $2$ m·s $^{- 1}$ .
5.3 Using ENERGY PRINCIPLES ONLY, calculate the magnitude of force F.(4)
Show the memo (4 ticks)Memo
1. ✓
  METHOD 1: Formula (any one) $W_{n c} = Δ E_{k} + Δ E_{p}$
  $W_{F} + W_{f} = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) + m g (h_{f} - h_{i})$
  $F Δ x cos θ + f Δ x cos θ = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) + m g (h_{f} - h_{i})$
  1 mark · Solve energy problems with friction
2. ✓
  METHOD 1: Work done by the forces $F (1, 5) cos 30° + 10 (1, 5) cos 180°$
  1 mark · Solve energy problems with friction
3. ✓
  METHOD 1: Energy changes $= \frac{1}{2} (6) (2^{2} - 0^{2}) + 0$
  1 mark · Solve energy problems with friction
4. ✓
  METHOD 1: Answer $F = 20, 78$ N
  1 mark · Solve energy problems with friction
4. Question 4.2 · May/June 2024 · Paper 1 · National (DBE) · 5 marks
Two trolleys A and B of mass $3, 2$ kg and $2, 6$ kg respectively are held at rest on a straight horizontal, frictionless track, with a compressed spring between them, as shown in the diagram below.

After the trolleys are released, the spring extends to its natural length and then falls onto the track. Trolley A now moves with a constant velocity of $0, 4$ m·s $^{- 1}$ to the left, while trolley B moves with a constant unknown velocity to the right. Trolley B reaches the end of the track after $1, 3$ s.
4.2 Calculate the distance travelled by trolley B in $1, 3$ s.(5)
Show the memo (5 ticks)Memo
1. ✓
  Momentum formulaRight as positive:
  $\sum p_{i} = \sum p_{f}$
  
  Any one: $m_{A} v_{i A} + m_{B} v_{i B} = m_{A} v_{f A} + m_{B} v_{f B}$
  1 mark · Apply conservation of momentum
2. ✓
  Initial momentum is zero $0 = \dots$
  1 mark · Apply conservation of momentum
3. ✓
  Substitution of final momenta $0 = (3, 2) (- 0, 4) + (2, 6) v_{f}$
  $v_{f} = 0, 49$ m·s $^{- 1}$
  1 mark · Apply conservation of momentum
4. ✓
  Substitution into equation of motion $v = \frac{Δ x}{Δ t}$
  $0, 49 = \frac{Δ x}{1 , 3}$
  
  OR $Δ x = (\frac{v _{i} + v _{f}}{2}) Δ t = (\frac{0 , 49 + 0 , 49}{2}) 1, 3$
  1 mark · Calculate distance, speed and velocity
5. ✓
  Final answer $Δ x_{B} = 0, 64$ m
  1 mark · Calculate distance, speed and velocity
5. Question 3.1.1 · November 2024 · Paper 1 · National (DBE) · 5 marks
Ball A is thrown vertically upwards at $12$ m·s $^{- 1}$ from the top of a building. Two seconds after ball A was thrown upwards, ball B is thrown vertically downwards at $5, 4$ m·s $^{- 1}$ from the top of the same building. Both balls, A and B, strike the ground at time t seconds.

Ignore the effects of air friction.

The position-time sketch graphs for both balls are shown below.
3.1 Using EQUATIONS OF MOTION ONLY, calculate the value of EACH of the following, as shown in the graphs:
3.1.1 t(5)
Show the memo (5 ticks)Memo
1. ✓
  FormulaUpwards as positive:
  Ball A: $Δ y = v_{i} Δ t + \frac{1}{2} a Δ t^{2}$
  1 mark · Use equations of motion for projectiles
2. ✓
  Substitution for ball A $Δ y = 12 t + \frac{1}{2} (- 9, 8) t^{2}$ ...... Eqn (1)
  1 mark · Use equations of motion for projectiles
3. ✓
  Substitution for ball B $Δ y = - 5, 4 (t - 2) + \frac{1}{2} (- 9, 8) (t - 2)^{2}$ ...... Eqn (2)
  1 mark · Use equations of motion for projectiles
4. ✓
  Equate displacements $Δ y_{A} = Δ y_{B}$ :
  $12 t + \frac{1}{2} (- 9, 8) t^{2} = - 5, 4 (t - 2) + \frac{1}{2} (- 9, 8) (t - 2)^{2}$
  1 mark · Use equations of motion for projectiles
5. ✓
  Answer $12 t = - 5, 4 t + 10, 8 - 19, 6 + 19, 6 t$
  $t = 4$ (s)
  1 mark · Use equations of motion for projectiles

Mechanics Grade 12 videos

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

Equations of Motion Introduction Grade 10 PhysicsMiss Martins Maths and Science

Mechanical Energy: Kinetic energy and Gravitational Potential energy IntroductionMiss Martins Maths and Science

Principle of Conservation of linear momentum Grade 12Miss Martins Maths and Science

2)Change momentum gr 12 | IntroKevinmathscience

6) Impulse gr 12 | Intro part 1Kevinmathscience

4) Linear Momentum Conservation | IntroKevinmathscience

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