AL MAWAKEB AL KHAWANEEJ

Final Exam
Revision Session

Grade 12 Physics — 2025–2026
June 2, 2026
Have ready: paper, pen, calculator, formula sheet.
This is not a lecture. You are solving problems for 60 minutes.
AMK PHYSICS Your Exam on Tuesday
Standard Points
Forces & Newton's 2nd Law 20
Work, Energy & Conservation 30
Momentum, Impulse & Conservation 50

Half your exam is momentum. That's where we start.

AMK PHYSICS Block 1 of 4
Block 1

Momentum, Impulse &
Conservation

50 points — 5 rounds
AMK PHYSICS Momentum — Warm-up
Warm-up Chat Flood 60 sec
A 3 kg object moves to the right at +4 m/s.

Calculate the momentum p of the object.

Type your answer in chat — don't press Enter until I say GO.
Solution
p = m · v = 3 × (+4) = +12 kg·m/s (to the right)
Watch for
If you wrote 12 without the + sign — momentum has direction. Always include the sign.
AMK PHYSICS Momentum — Round 1
Round 1 Chat Flood 90 sec
A 2 kg ball moves to the right at +8 m/s.
It then moves to the left at −8 m/s (same speed, opposite direction).

Calculate the change in momentum Δp.
Solution
p₁ = m · v₁ = 2 × (+8) = +16 kg·m/s
p₂ = m · v₂ = 2 × (−8) = −16 kg·m/s

Δp = p₂ − p₁ = (−16) − (+16) = −32 kg·m/s
Trap
Common wrong answer: 0 (from 8 − 8 = 0).
Δp = pf − pi — subtract the vectors with signs, not the magnitudes.
AMK PHYSICS Momentum — Round 2
Round 2 Chat Flood 2 min
A 0.15 kg baseball approaches a bat at +30 m/s.
After being struck, it moves at −20 m/s (opposite direction).

a) Calculate Δp of the ball.
b) If contact time Δt = 0.02 s, calculate the force F exerted by the bat.
Solution
pi = 0.15 × (+30) = +4.5 kg·m/s
pf = 0.15 × (−20) = −3.0 kg·m/s

a) Δp = pf − pi = (−3.0) − (+4.5) = −7.5 kg·m/s

b) F = Δp / Δt = (−7.5) / (0.02) = −375 N
Trap
The mistake: doing 30 − 20 = 10, then 0.15 × 10 = 1.5.
No. Compute pf and pi SEPARATELY, then subtract. Signs matter.
AMK PHYSICS Momentum — Round 3
Round 3 Chat Flood 90 sec
A 60 kg skater and a 40 kg skater stand at rest on frictionless ice.
They push off each other. The 60 kg skater moves left at −2 m/s.

Find the velocity of the 40 kg skater after the push.
Solution
Both at rest → ptotal,i = 0

0 = m₁·v₁f + m₂·v₂f
0 = (60)(−2) + (40)(v₂f)
0 = −120 + 40·v₂f
v₂f = 120 / 40 = +3 m/s (to the right)
Trap
The equation starts with 0 because they were at rest.
If you forget that total initial momentum = 0, you can't set up the equation.
AMK PHYSICS Momentum — Round 4
Round 4 Chat Flood 90 sec
Car A (1200 kg) travels east at +25 m/s.
It collides with stationary Car B (1800 kg).
The two cars lock together after impact.

Find the common velocity vf after the collision.
Solution
ptotal,i = (1200)(+25) + (1800)(0) = +30,000 kg·m/s

(mA + mB)·vf = 30,000
(3000)·vf = 30,000
vf = +10 m/s (east)
Trap
"Lock together" = perfectly inelastic → divide by m₁ + m₂ = 3000, not by 1200 alone.
Common wrong answer: 25 m/s (divided by one mass only).
AMK PHYSICS Momentum — Quick Check
Raise Hand
Raise your hand if this is True:

"Momentum is a scalar quantity."
FALSE

Momentum is a vector — it has magnitude AND direction.
This is on your T/F section. Don't lose this mark.
AMK PHYSICS Block 2 of 4
Block 2

Work, Energy &
Conservation

30 points — 3 rounds
AMK PHYSICS Energy — Round 5
Round 5 Chat Flood 90 sec
A worker pulls a toolbox 20 m with a constant horizontal pull of 50 N.
Friction force = 15 N opposing the motion.

a) Calculate Wpull
b) Calculate Wfriction
c) Calculate Wnet
Solution
a) Wpull = 50 × 20 × cos(0°) = +1000 J
b) Wfriction = −f · d = −15 × 20 = −300 J
c) Wnet = (+1000) + (−300) = +700 J
Trap
Friction always does negative work when the object moves.
If you wrote Wfriction = +300 J, that's marks gone.
AMK PHYSICS Energy — Round 6
Round 6 Chat Flood 90 sec
A 3 kg object is initially moving at 4 m/s.
A net force does +30 J of net work on it.

a) Calculate the initial kinetic energy KEi
b) Using Wnet = ΔKE, find the final kinetic energy KEf
Solution
a) KEi = ½ × 3 × (4)² = ½ × 3 × 16 = 24 J

b) Wnet = KEf − KEi
KEf = Wnet + KEi = 30 + 24 = 54 J
Trap
Don't forget to square the velocity before multiplying.
½ × 3 × 4 = 6 is WRONG. It's ½ × 3 × 16 = 24.
AMK PHYSICS Energy — Round 7
Round 7 Chat Flood 90 sec
A 0.5 kg ball is released from rest at a height of 5 m.
Air resistance is negligible. Use g = 10 m/s². Ground = h = 0.

a) Calculate PEg and KE at the release point.
b) Calculate the total mechanical energy ME at release.
Solution
a) PEg = m·g·h = 0.5 × 10 × 5 = 25 J
   KE = ½ × 0.5 × 0² = 0 J (released from rest)

b) ME = KE + PEg = 0 + 25 = 25 J
Trap
ME = KE + PEg, not KE − PEg.
This exact trap appears twice on your exam (MCQ Q12 and T/F Q7). Free marks if you remember.
AMK PHYSICS Energy — Quick Check
Raise Hand
Raise your hand if this is True:

"Work done by a force is positive when the force
opposes the displacement."
FALSE

Positive work = force and displacement in the same direction.
Opposing = 180° = cos(180°) = −1 → negative work.
AMK PHYSICS Block 3 of 4
Block 3

Forces &
Newton's 2nd Law

20 points — 2 rounds
AMK PHYSICS Forces — Round 8
Round 8 Chat Flood 60 sec
Two horizontal forces act on a 4 kg block on a frictionless surface.
F₁ = 18 N to the right, F₂ = 6 N to the left.

a) Find Fnet
b) Find the acceleration
c) Balanced or unbalanced? Justify.
Solution
a) Fnet = (+18) + (−6) = +12 N (right)
b) a = Fnet / m = 12 / 4 = +3 m/s² (right)
c) Unbalanced — Fnet ≠ 0, so the block accelerates.
Trap
Use signs: right = positive, left = negative.
Don't add magnitudes: 18 + 6 = 24 is wrong.
AMK PHYSICS Forces — Round 9
Round 9 Chat Flood 60 sec
A 25 kg suitcase rests on the floor of an elevator.
The elevator is at rest. Use g = 10 m/s².

a) Find the weight Fg
b) Find the normal force Fn
Solution
a) Fg = m · g = 25 × 10 = 250 N (downward)

b) Elevator at rest → a = 0 → Fnet = 0
Fn − Fg = 0 → Fn = Fg = 250 N (upward)
Trap
On a flat surface at rest: Fn = Fg. It is never 2Fg.
Also: equilibrium means Fnet = 0, not "velocity = 0."
AMK PHYSICS Block 4 of 4
Block 4

The 10 Traps That
Cost You Marks

Screenshot this next slide
AMK PHYSICS Screenshot This
  1. 1 ME = KE + PEg — not minus
  2. 2 Momentum is a vector — direction matters, it is not scalar
  3. 3 Δp = pf − pi — subtract vectors with signs, not magnitudes
  4. 4 Friction work is always negative when the object moves
  5. 5 Fn = Fg on a flat surface — never 2Fg
  6. 6 Equilibrium = Fnet = 0 — does not mean velocity = 0
  7. 7 Perfectly inelastic = stick together → divide by combined mass
  8. 8 Recoil from rest → total initial p = 0
  9. 9 J = Δp = F·Δt — they're the same thing, pick whichever fits
  10. 10 KE = ½mv² — don't forget to square v first
AMK PHYSICS Formula Sheet

Forces & Motion

Fnet = m · a
Fg = m · g
Fn = Fg = m · g  (flat, no extra force)
Equilibrium: Fnet = 0

Work

W = F · d · cos θ
W = F · d  (same direction)
W = 0  (perpendicular)
Wf = −f · d

Energy

KE = ½ m v²
PEg = m · g · h
ME = KE + PEg
Wnet = ΔKE = KEf − KEi

Momentum & Impulse

p = m · v
Δp = pf − pi
J = F · Δt = Δp

Conservation of Momentum

m₁v₁i + m₂v₂i = m₁v₁f + m₂v₂f  (general)
m₁v₁i + m₂v₂i = (m₁ + m₂) · Vf  (perfectly inelastic)
0 = m₁v₁f + m₂v₂f  (recoil from rest)

AMK PHYSICS

Tonight

Solve 3 momentum problems from your revision sheet.

Focus on recoil, perfectly inelastic, and impulse with force — momentum is the biggest section on your exam.

Formula sheet is the last page of your exam.
Know where each formula is.

Good luck tomorrow.

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