Panhellenic Exams · 90 minutes · 40 marks

Physics · Panhellenic Exams: Collisions and oscillations

Collisions and relative motion · Oscillations

Original pilot material — human educator review pending Written specifically for YourFavTeacher with AI assistance, without copying past-paper prompts. Technical and numerical checks do not replace review by a human educator. It is not an official paper or endorsed by an examination body.

Before you start

A complete original revision assessment of the two listed units only — not a full Panhellenic mock paper or whole-year assessment.
Timing is indicative. Show method, calculation and checking. In physics, include units and direction where required.
The indicative rubric accepts equivalent correct methods. Do not penalise the same arithmetic error again when subsequent working is consistent. This is not a diagnostic instrument or official grade.

Materials: Pencil, paper and ruler. Leave π exact where needed. · The numerical data do not require a calculator. This is not an official examination rule.

Collisions and relative motion Oscillations

Go to questions

Questions and marks

Work on paper first. Solutions stay closed until you choose to reveal them. Marks are for self-assessment, not automated or official grading.

1. Question 1

5 marks

In an isolated horizontal system, a 1 kg body at v=14 m/s sticks to a stationary 1 kg body. Find their common velocity.

Worked solution and marking guide

$Total momentum is conserved; kinetic energy need not be.$
$1\cdot14+1\cdot0=(1+1)V \Rightarrow V=7 m/s$
$The common velocity is in the initial positive direction.$

7 m/s

Mark allocation

Total momentum is conserved; kinetic energy need not be. — 1 marks
1·14+1·0=(1+1)V ⇒ V=7 m/s — 2 marks
The common velocity is in the initial positive direction. — 2 marks

2. Question 2

5 marks

Equal masses collide head-on elastically. Initially v₁=7 m/s and v₂=-6 m/s. Find final velocities.

Worked solution and marking guide

$Equal masses in a head-on elastic collision exchange velocities.$
$υ₁'=-6 m/s; υ₂'=7 m/s$
$Both total momentum and kinetic energy are preserved.$

υ₁'=-6 m/s; υ₂'=7 m/s

Mark allocation

Equal masses in a head-on elastic collision exchange velocities. — 1 marks
υ₁′=−6 m/s; υ₂′=7 m/s — 2 marks
Both total momentum and kinetic energy are preserved. — 2 marks

3. Question 3

5 marks

A 1 kg ball moves at v=7 m/s and rebounds elastically, normally from a fixed wall. Find its momentum change along the initially positive axis.

Worked solution and marking guide

$Speed is unchanged, but the direction reverses.$
$Δp=1\cdot(-7)-1\cdot7=-14 kg\cdotm/s$
$The ball alone does not conserve momentum: the wall exerts an external impulse.$

-14 kg\cdotm/s

Mark allocation

Speed is unchanged, but the direction reverses. — 1 marks
Δp=1·(−7)−1·7=−14 kg·m/s — 2 marks
The ball alone does not conserve momentum: the wall exerts an external impulse. — 2 marks

4. Question 4

5 marks

A 1 kg body's velocity changes from -6 to 7 m/s. Find its momentum change.

Worked solution and marking guide

$Calculate final minus initial momentum with signs.$
$Δp=1\cdot[7-(-6)]=13 kg\cdotm/s$
$Subtracting a negative initial value becomes addition.$

13 kg\cdotm/s

Mark allocation

Calculate final minus initial momentum with signs. — 1 marks
Δp=1·[7−(−6)]=13 kg·m/s — 2 marks
Subtracting a negative initial value becomes addition. — 2 marks

5. Question 5

5 marks

An ideal horizontal spring system has m=1 kg and k=49 N/m. Find ω and T.

Worked solution and marking guide

$For an ideal spring, ω=\sqrt(k/m) and T=2π/ω.$
$ω=\sqrt49=7 rad/s; T=2π/7 s$
$Period is independent of amplitude in the ideal model.$

ω=7 rad/s; T=2π/7 s

Mark allocation

For an ideal spring, ω=√(k/m) and T=2π/ω. — 1 marks
ω=√49=7 rad/s; T=2π/7 s — 2 marks
Period is independent of amplitude in the ideal model. — 2 marks

6. Question 6

5 marks

A spring with k=100 N/m oscillates with amplitude A=0.06 m. Find total energy.

Worked solution and marking guide

$In the ideal system E=kA²/2.$
$E=0.5\cdot100\cdot(0.06)²=0.18 J$
$Amplitude is in metres and is squared.$

0.18 J

Mark allocation

In the ideal system E=kA²/2. — 1 marks
E=0.5·100·(0.06)²=0.18 J — 2 marks
Amplitude is in metres and is squared. — 2 marks

7. Question 7

5 marks

For SHM with ω=7 rad/s and x=0.1 m, find acceleration.

Worked solution and marking guide

$Acceleration has the opposite sign to displacement.$
$a=-ω²x=-7²\cdot0.1=-4.9 m/s²$
$The negative sign points towards equilibrium.$

-4.9 m/s²

Mark allocation

Acceleration has the opposite sign to displacement. — 1 marks
a=−ω²x=−7²·0.1=−4.9 m/s² — 2 marks
The negative sign points towards equilibrium. — 2 marks

8. Question 8

5 marks

At an oscillation endpoint, velocity is momentarily zero. Is the restoring force also zero?

Worked solution and marking guide

$F=-kx$
$At an endpoint |x|=A, so |F|=kA.$
$The force towards equilibrium reverses the motion.$

No; the force magnitude is maximal.

Mark allocation

F=−kx — 1 marks
At an endpoint |x|=A, so |F|=kA. — 2 marks
The force towards equilibrium reverses the motion. — 2 marks

Provenance and scope

Original authorship for YourFavTeacher with AI assistance.

Official sources were checked only for topic and level reference. No specific past-paper prompts, diagrams or solutions were copied or adapted. No external endorsement or licence to republish third-party papers is claimed.

Edition: 2026-09-06

Ministry of Education — 2027 examinable material (decision, July 2026) (new tab)

Record mistakes, revisit the relevant chapter and ask a tutor for an explanation if you need support.

Find a tutor in this subject All assessments

Panhellenic Exams · 90 minutes · 40 marks

Physics · Panhellenic Exams: Collisions and oscillations

Collisions and relative motion · Oscillations

Before you start

A complete original revision assessment of the two listed units only — not a full Panhellenic mock paper or whole-year assessment.
Timing is indicative. Show method, calculation and checking. In physics, include units and direction where required.
The indicative rubric accepts equivalent correct methods. Do not penalise the same arithmetic error again when subsequent working is consistent. This is not a diagnostic instrument or official grade.

Materials: Pencil, paper and ruler. Leave π exact where needed. · The numerical data do not require a calculator. This is not an official examination rule.

Collisions and relative motion Oscillations

Go to questions

Questions and marks

Work on paper first. Solutions stay closed until you choose to reveal them. Marks are for self-assessment, not automated or official grading.

1. Question 1

5 marks

In an isolated horizontal system, a 1 kg body at v=14 m/s sticks to a stationary 1 kg body. Find their common velocity.

Worked solution and marking guide

$Total momentum is conserved; kinetic energy need not be.$
$1\cdot14+1\cdot0=(1+1)V \Rightarrow V=7 m/s$
$The common velocity is in the initial positive direction.$

7 m/s

Mark allocation

Total momentum is conserved; kinetic energy need not be. — 1 marks
1·14+1·0=(1+1)V ⇒ V=7 m/s — 2 marks
The common velocity is in the initial positive direction. — 2 marks

2. Question 2

5 marks

Equal masses collide head-on elastically. Initially v₁=7 m/s and v₂=-6 m/s. Find final velocities.

Worked solution and marking guide

$Equal masses in a head-on elastic collision exchange velocities.$
$υ₁'=-6 m/s; υ₂'=7 m/s$
$Both total momentum and kinetic energy are preserved.$

υ₁'=-6 m/s; υ₂'=7 m/s

Mark allocation

Equal masses in a head-on elastic collision exchange velocities. — 1 marks
υ₁′=−6 m/s; υ₂′=7 m/s — 2 marks
Both total momentum and kinetic energy are preserved. — 2 marks

3. Question 3

5 marks

A 1 kg ball moves at v=7 m/s and rebounds elastically, normally from a fixed wall. Find its momentum change along the initially positive axis.

Worked solution and marking guide

$Speed is unchanged, but the direction reverses.$
$Δp=1\cdot(-7)-1\cdot7=-14 kg\cdotm/s$
$The ball alone does not conserve momentum: the wall exerts an external impulse.$

-14 kg\cdotm/s

Mark allocation

Speed is unchanged, but the direction reverses. — 1 marks
Δp=1·(−7)−1·7=−14 kg·m/s — 2 marks
The ball alone does not conserve momentum: the wall exerts an external impulse. — 2 marks

4. Question 4

5 marks

A 1 kg body's velocity changes from -6 to 7 m/s. Find its momentum change.

Worked solution and marking guide

$Calculate final minus initial momentum with signs.$
$Δp=1\cdot[7-(-6)]=13 kg\cdotm/s$
$Subtracting a negative initial value becomes addition.$

13 kg\cdotm/s

Mark allocation

Calculate final minus initial momentum with signs. — 1 marks
Δp=1·[7−(−6)]=13 kg·m/s — 2 marks
Subtracting a negative initial value becomes addition. — 2 marks

5. Question 5

5 marks

An ideal horizontal spring system has m=1 kg and k=49 N/m. Find ω and T.

Worked solution and marking guide

$For an ideal spring, ω=\sqrt(k/m) and T=2π/ω.$
$ω=\sqrt49=7 rad/s; T=2π/7 s$
$Period is independent of amplitude in the ideal model.$

ω=7 rad/s; T=2π/7 s

Mark allocation

For an ideal spring, ω=√(k/m) and T=2π/ω. — 1 marks
ω=√49=7 rad/s; T=2π/7 s — 2 marks
Period is independent of amplitude in the ideal model. — 2 marks

6. Question 6

5 marks

A spring with k=100 N/m oscillates with amplitude A=0.06 m. Find total energy.

Worked solution and marking guide

$In the ideal system E=kA²/2.$
$E=0.5\cdot100\cdot(0.06)²=0.18 J$
$Amplitude is in metres and is squared.$

0.18 J

Mark allocation

In the ideal system E=kA²/2. — 1 marks
E=0.5·100·(0.06)²=0.18 J — 2 marks
Amplitude is in metres and is squared. — 2 marks

7. Question 7

5 marks

For SHM with ω=7 rad/s and x=0.1 m, find acceleration.

Worked solution and marking guide

$Acceleration has the opposite sign to displacement.$
$a=-ω²x=-7²\cdot0.1=-4.9 m/s²$
$The negative sign points towards equilibrium.$

-4.9 m/s²

Mark allocation

Acceleration has the opposite sign to displacement. — 1 marks
a=−ω²x=−7²·0.1=−4.9 m/s² — 2 marks
The negative sign points towards equilibrium. — 2 marks

8. Question 8

5 marks

At an oscillation endpoint, velocity is momentarily zero. Is the restoring force also zero?

Worked solution and marking guide

$F=-kx$
$At an endpoint |x|=A, so |F|=kA.$
$The force towards equilibrium reverses the motion.$

No; the force magnitude is maximal.

Mark allocation

F=−kx — 1 marks
At an endpoint |x|=A, so |F|=kA. — 2 marks
The force towards equilibrium reverses the motion. — 2 marks

Provenance and scope

Original authorship for YourFavTeacher with AI assistance.

Edition: 2026-09-06

Ministry of Education — 2027 examinable material (decision, July 2026) (new tab)

Record mistakes, revisit the relevant chapter and ask a tutor for an explanation if you need support.

Find a tutor in this subject All assessments