CELE Engineering Mechanics — Dynamics: KineticsExam Answer Templates

Examiner tip

Examiners specifically check that the numerator is separation (after) and denominator is approach (before). Writing it inverted is the single most common error on 1-mark definition questions for this topic.

Model answer

The coefficient of restitution e is the ratio of the relative velocity of separation to the relative velocity of approach of two colliding bodies along the line of impact: e = (v₂ − v₁)/(u₁ − u₂). Its value ranges from 0 (perfectly plastic) to 1 (perfectly elastic).

Answer structure

• Line 1: State the definition with the correct formula and variable identification [1 mark]

Scoring breakdown

Common mark deductions

• Writing e as approach/separation (inverted ratio) — loses the mark
• Stating range incorrectly as e > 1 or e < 0
• Omitting the formula and giving only a verbal description

Key phrases to include

• relative velocity of separation
• relative velocity of approach
• e = (v₂ − v₁)/(u₁ − u₂)
• 0 ≤ e ≤ 1

Examiner tip

The key phrase examiners look for is 'inertia force' and the equation ΣF − ma = 0. A definition without the equation is considered incomplete for this mark level.

Model answer

D'Alembert's principle states that by adding an inertia force (−ma) opposite to the acceleration, the dynamic equation ΣF = ma is rewritten as ΣF − ma = 0, which has the same form as static equilibrium. This allows all forces including the inertia force to be treated as a system in equilibrium.

Answer structure

• Line 1: State the principle and write the pseudo-equilibrium equation ΣF − ma = 0 [1 mark]

Scoring breakdown

Common mark deductions

• Writing the inertia force as +ma instead of −ma
• Failing to mention the equilibrium analogy
• Confusing D'Alembert with the work–energy principle

Key phrases to include

• inertia force −ma
• opposite to acceleration
• ΣF − ma = 0
• pseudo-equilibrium

Examiner tip

Both Impulse–Momentum (ΣFΔt = mΔv) and Newton's law with kinematics (a = Δv/Δt, then ΣF = ma) are accepted. Impulse–Momentum is faster and is the preferred method when time and velocity are given.

Model answer

Given: m = 1 500 kg, v₁ = 0, v₂ = 25 m/s, Δt = 10 s

Applying the Impulse–Momentum Principle:
ΣF · Δt = m(v₂ − v₁)
ΣF (10) = 1 500 (25 − 0)
ΣF = 37 500 / 10

ΣF = 3 750 N

Answer structure

• Line 1: List all given data with units [0.5 mark]
• Line 2: State the principle — Impulse–Momentum: ΣF·Δt = mΔv [0.5 mark]
• Line 3: Substitute and solve for ΣF [1 mark]

Scoring breakdown

Common mark deductions

• Using ΣF = ma with a = Δv/Δt instead of impulse–momentum (acceptable but loses time if misapplied)
• Forgetting units on the final answer
• Dividing by mass instead of time in the last step

Key phrases to include

• Impulse–Momentum Principle
• ΣF · Δt = m(v₂ − v₁)
• 3 750 N

Examiner tip

Always show the FBD step explicitly. Examiners award a method mark for the friction calculation even if the final arithmetic is wrong.

Model answer

Given: m = 50 kg, F = 200 N, μ = 0.20, g = 9.81 m/s²

Step 1 — Normal force:
N = mg = 50 × 9.81 = 490.5 N

Step 2 — Friction force:
F_f = μN = 0.20 × 490.5 = 98.1 N

Step 3 — Applying Newton's 2nd Law (→ positive):
ΣF = ma
200 − 98.1 = 50 a
a = 101.9 / 50

a = 2.04 m/s²

Answer structure

• Line 1: Compute normal force N = mg [0.5 mark]
• Line 2: Compute friction F_f = μN [0.5 mark]
• Line 3: Apply ΣF = ma and solve for a [1 mark]

Scoring breakdown

Common mark deductions

• Using weight (490.5 N) instead of mass (50 kg) in ΣF = ma
• Omitting friction or adding it in the wrong direction
• Computing N = F (applied force) instead of N = mg on horizontal surface

Key phrases to include

• N = mg
• F_f = μN
• ΣF = ma
• 200 − 98.1 = 50a
• 2.04 m/s²

Examiner tip

Board exam 2-mark definition questions expect both the formula and the word statement. A formula alone earns only 1 mark. Write both.

Model answer

Work (U) done by a constant force F over displacement d is defined as:
U = F · d · cosθ
where θ is the angle between the force vector and the displacement vector. Work is a scalar quantity measured in Joules (J = N·m).

The Work–Energy Theorem states that the net work done on a body equals the change in its kinetic energy:
U₁→₂ = ΔKE = ½mv₂² − ½mv₁²

Answer structure

• Line 1: Define work with the formula U = Fd cosθ and identify θ [1 mark]
• Line 2: State the Work–Energy Theorem: U₁→₂ = ½mv₂² − ½mv₁² [1 mark]

Scoring breakdown

Common mark deductions

• Writing U = Fd without the cosθ factor
• Confusing work with power (W = U/t)
• Writing the theorem as U = ½mv² instead of the difference ½mv₂² − ½mv₁²

Key phrases to include

• U = Fd cosθ
• scalar quantity
• Joules (J)
• U₁→₂ = ½mv₂² − ½mv₁²
• net work equals change in kinetic energy

Examiner tip

Examiners award all three marks even if you write work of friction as −F_f × d without expanding N = mg, provided the numerical value (73.6 J) is correct. Show the intermediate N calculation to protect method marks.

Model answer

Given: m = 10 kg, F = 100 N, d = 5 m, μ = 0.15, v₁ = 0, g = 9.81 m/s²

Step 1 — Work done by applied force:
U_F = F · d = 100 × 5 = 500 J

Step 2 — Work done by friction (negative):
N = mg = 10 × 9.81 = 98.1 N
U_f = −μNd = −0.15 × 98.1 × 5 = −73.6 J

Step 3 — Apply Work–Energy Theorem:
U_net = ΔKE
500 − 73.6 = ½(10)v₂² − 0
426.4 = 5 v₂²
v₂ = √(426.4 / 5) = √85.28

v₂ = 9.24 m/s

Answer structure

• Step 1: Calculate work of applied force U_F = Fd [1 mark]
• Step 2: Calculate work of friction U_f = −μNd, with N = mg [1 mark]
• Step 3: Apply U_net = ΔKE and solve for v₂ [1 mark]

Scoring breakdown

Common mark deductions

• Writing friction work as positive (+73.6 J) instead of negative
• Forgetting to set v₁ = 0 (starting from rest), leading to incorrect ΔKE
• Using N = F (applied force) instead of N = mg for horizontal surface
• Not taking the square root at the final step (leaving answer as v² = 85.28)

Key phrases to include

• U_F = Fd = 500 J
• U_f = −μNd
• N = mg = 98.1 N
• U_net = ΔKE = ½mv₂² − ½mv₁²
• v₂ = 9.24 m/s

Examiner tip

On incline problems, draw the FBD with the x-axis along the slope and y-axis perpendicular. Label W sinθ and W cosθ explicitly — this earns the first mark even before any arithmetic.

Model answer

Given: m = 20 kg, θ = 30°, μ = 0.25, g = 9.81 m/s²

FBD — Forces along the incline (positive down the slope):
Driving force: W sinθ = 20 × 9.81 × sin30° = 98.1 N
Normal force: N = W cosθ = 20 × 9.81 × cos30° = 169.9 N
Friction force (opposing motion, up the slope): F_f = μN = 0.25 × 169.9 = 42.5 N

Applying Newton's 2nd Law along the incline:
ΣF = ma
W sinθ − F_f = ma
98.1 − 42.5 = 20 a
55.6 = 20 a

a = 2.78 m/s² (down the incline)

Answer structure

• Step 1: Resolve weight into components along and perpendicular to incline [1 mark]
• Step 2: Compute normal force N = W cosθ and friction F_f = μN [1 mark]
• Step 3: Apply ΣF = ma along incline and solve for a [1 mark]

Scoring breakdown

Common mark deductions

• Using N = mg (full weight) instead of N = mg cosθ on an inclined surface
• Writing friction as mg sinθ (incorrect — that is the driving component)
• Not stating direction of friction (up the incline opposes downward motion)
• Arithmetic error in sin30° or cos30° (use sin30° = 0.5, cos30° = 0.866)

Key phrases to include

• W sinθ = 98.1 N
• N = W cosθ = 169.9 N
• F_f = μN = 42.5 N
• ΣF = ma: W sinθ − F_f = ma
• a = 2.78 m/s²

Examiner tip

The most common 5-mark trap in this type is Part (b) sign error. Always write 'Define upward as positive' before substituting. Examiners specifically award 1 mark for explicit sign convention in impulse problems.

Model answer

Given: m = 5 kg, h = 10 m, v_rebound = 8 m/s (upward), g = 9.81 m/s²

PART (a) — Work–Energy Theorem:
Take ground as datum; initial KE = 0 (dropped from rest)
Work done by gravity = W × h = mgh = 5 × 9.81 × 10 = 490.5 J
U₁→₂ = ΔKE
490.5 = ½mv₂² − 0
v₂ = √(2 × 490.5 / 5) = √196.2

v_impact = 14.0 m/s (downward)

PART (b) — Impulse–Momentum Theorem:
Define upward as positive.
Before impact: v₁ = −14.0 m/s (downward)
After impact: v₂ = +8.0 m/s (upward)

Impulse J = m(v₂ − v₁)
J = 5 × (8.0 − (−14.0))
J = 5 × 22.0

J = 110 N·s (upward)

Answer structure

• Part (a) Step 1: Identify work done by gravity = mgh [1 mark]
• Part (a) Step 2: Apply U = ΔKE and solve for v_impact = 14.0 m/s [1 mark]
• Part (b) Step 1: Define sign convention; identify v₁ = −14.0 m/s, v₂ = +8 m/s [1 mark]
• Part (b) Step 2: Apply Impulse = m(v₂ − v₁) [1 mark]
• Part (b) Step 3: Correct answer J = 110 N·s upward [1 mark]

Scoring breakdown

Common mark deductions

• Part (b): Using same sign for both velocities (not recognizing direction change at impact) — loses 2 marks
• Part (a): Using v² = u² + 2as kinematics instead of energy (acceptable but slower; no mark penalty)
• Forgetting to state direction of impulse (upward) in Part (b)
• Using v_impact = 14.0 m/s as positive in impulse without sign correction

Key phrases to include

• U = mgh = 490.5 J
• U = ΔKE = ½mv₂² − ½mv₁²
• v_impact = 14.0 m/s
• upward positive: v₁ = −14.0 m/s
• J = m(v₂ − v₁) = 110 N·s

Examiner tip

Whenever 'they move together after collision' appears in the problem, immediately write v₁_after = v₂_after = v_c and note e = 0 — this confirms perfectly plastic impact. Writing this observation earns an examiner's recognition mark.

Model answer

Given: m₁ = 2 000 kg, u₁ = 15 m/s; m₂ = 1 000 kg, u₂ = 0; they move together after collision.

PART (a) — Conservation of Momentum (perfectly plastic: v₁ = v₂ = v_c):
m₁u₁ + m₂u₂ = (m₁ + m₂)v_c
2 000(15) + 1 000(0) = (2 000 + 1 000)v_c
30 000 = 3 000 v_c

v_c = 10 m/s

PART (b) — Coefficient of Restitution:
Since the bodies move together after collision, v₁_after = v₂_after = 10 m/s.
e = (v₂_after − v₁_after) / (u₁ − u₂)
e = (10 − 10) / (15 − 0)
e = 0 / 15

e = 0 (perfectly plastic collision)

Verification: Bodies moving together after impact confirms e = 0 (perfectly plastic).

Answer structure

• Part (a) Step 1: Write conservation of momentum equation [1 mark]
• Part (a) Step 2: Substitute and solve v_c = 10 m/s [1 mark]
• Part (b) Step 1: Identify after-collision velocities (both = v_c = 10 m/s) [1 mark]
• Part (b) Step 2: Apply e = (v₂ − v₁)/(u₁ − u₂) [1 mark]
• Part (b) Step 3: e = 0 and state: perfectly plastic collision [1 mark]

Scoring breakdown

Common mark deductions

• Using kinetic energy conservation instead of momentum conservation for Part (a)
• Inverting the e formula: writing (u₁ − u₂)/(v₂ − v₁) instead of the correct ratio
• Not stating that e = 0 corresponds to a perfectly plastic collision
• Arithmetic error: 2000×15 = 30 000, dividing by 2000 instead of 3000

Key phrases to include

• conservation of momentum
• m₁u₁ + m₂u₂ = (m₁ + m₂)v_c
• v_c = 10 m/s
• e = (v₂ − v₁)/(u₁ − u₂)
• e = 0
• perfectly plastic collision

Examiner tip

The shortcut formula μ_min = v²/(rg) can be derived in one line and is worth memorizing. Always verify by checking that μ < 1 for realistic road conditions — if μ > 1, check for arithmetic errors.

Model answer

Given: m = 1 200 kg, r = 50 m, v = 15 m/s, g = 9.81 m/s²

PART (a) — Centripetal (Normal) Acceleration:
a_n = v²/r = (15)²/50 = 225/50

a_n = 4.5 m/s²

PART (b) — Minimum Friction (Normal equation in curvilinear kinetics):
The friction force provides the centripetal force:
F_f = m·a_n = 1 200 × 4.5 = 5 400 N
Normal force on horizontal surface: N = mg = 1 200 × 9.81 = 11 772 N
For impending skid: F_f = μN
μ = F_f / N = 5 400 / 11 772

μ_min = 0.459

Answer structure

• Part (a): Apply a_n = v²/r and compute a_n = 4.5 m/s² [1 mark]
• Part (b) Step 1: Set centripetal force = friction force; compute F_f = ma_n = 5 400 N [1 mark]
• Part (b) Step 2: Compute N = mg and solve μ = F_f/N = 0.459 [1 mark]

Scoring breakdown

Common mark deductions

• Confusing centripetal acceleration with tangential acceleration
• Not recognizing that friction is the only horizontal force providing centripetal acceleration
• Using N = F_applied instead of N = mg on a horizontal road
• Omitting the word 'minimum' in the final answer statement

Key phrases to include

• a_n = v²/r
• ΣF_n = mv²/r
• centripetal force provided by friction
• μ = v²/(rg)
• μ_min = 0.459

Examiner tip

Board exam 3-mark conceptual questions reward structured answers. Use numbered points matching the three principles. An answer written in paragraph form without clear structure risks losing the examiner's reading flow and costs 0.5–1 mark.

Model answer

1. Newton's Second Law (ΣF = ma): Use when forces and accelerations are the primary unknowns at a specific instant. Best for problems where you need to find acceleration or an unknown force given the mass and motion state.
   Example: Finding the tension in a rope supporting an accelerating elevator.
2. Work–Energy Theorem (U₁→₂ = ΔKE): Use when force acts over a distance and the unknown is a speed (not time). Eliminates the need for kinematics equations; most efficient for problems involving height change or spring forces.
   Example: Finding the speed of a block after sliding down a rough incline from rest over a given distance.
3. Impulse–Momentum Theorem (ΣF·Δt = mΔv): Use when force acts over a time interval and the unknown involves velocity change or average force. Ideal for collision and impact problems.
   Example: Finding the average braking force that stops a vehicle in a given time.

Answer structure

• Principle 1: State condition for Newton's Law use + example [1 mark]
• Principle 2: State condition for Work–Energy use + example [1 mark]
• Principle 3: State condition for Impulse–Momentum use + example [1 mark]

Scoring breakdown

Common mark deductions

• Giving the formula without the condition (when to use it)
• Providing the same example for two different principles
• Describing Work–Energy as applicable when time is involved (it is not — that is Impulse–Momentum)

Key phrases to include

• ΣF = ma — forces and acceleration at an instant
• Work–Energy — distance and speed, time not required
• Impulse–Momentum — time interval and velocity change

Examiner tip

For head-on collisions with opposing initial velocities, the sign of u_B is the top source of error. Writing 'Define → as positive; u_A = +6, u_B = −4' on the first line locks in the sign convention and earns an explicit recognition mark from the examiner.

Model answer

Given: m_A = 3 kg, u_A = +6 m/s; m_B = 2 kg, u_B = −4 m/s; e = 0.75
(Define rightward → as positive)

Equation 1 — Conservation of Momentum:
m_A·u_A + m_B·u_B = m_A·v_A + m_B·v_B
3(+6) + 2(−4) = 3v_A + 2v_B
18 − 8 = 3v_A + 2v_B
3v_A + 2v_B = 10 ... (1)

Equation 2 — Coefficient of Restitution:
e = (v_B − v_A) / (u_A − u_B)
0.75 = (v_B − v_A) / (6 − (−4))
0.75 = (v_B − v_A) / 10
v_B − v_A = 7.5 ... (2)

Solving (1) and (2) simultaneously:
From (2): v_B = v_A + 7.5
Substitute into (1): 3v_A + 2(v_A + 7.5) = 10
5v_A + 15 = 10
v_A = −1 m/s
v_B = −1 + 7.5 = +6.5 m/s

v_A = 1 m/s ← (leftward)
v_B = 6.5 m/s → (rightward)

Answer structure

• Step 1: Define positive direction and assign signed values to u_A and u_B [0.5 mark]
• Step 2: Write conservation of momentum equation (1): 3v_A + 2v_B = 10 [1.5 marks]
• Step 3: Write e equation (2): v_B − v_A = 7.5 [1.5 marks]
• Step 4: Solve simultaneously for v_A = −1 m/s and v_B = +6.5 m/s [1 mark]
• Step 5: State directions: v_A leftward, v_B rightward [0.5 mark]

Scoring breakdown

Common mark deductions

• Using u_B = +4 m/s (ignoring the leftward direction) — cascades into wrong answers for both equations
• Writing the e equation as (u_A − u_B)/(v_B − v_A) instead of the correct inversion
• Solving correctly but not stating direction of each velocity in the final answer
• Arithmetic error in the simultaneous equations step

Key phrases to include

• rightward positive
• u_B = −4 m/s
• 3v_A + 2v_B = 10
• v_B − v_A = e(u_A − u_B) = 7.5
• v_A = −1 m/s (leftward)
• v_B = 6.5 m/s (rightward)

Examiner tip

The work done by a spring equals the decrease in spring potential energy: U = ½kx₁² − ½kx₂². On frictionless surfaces, this equals the increase in KE. Always verify: spring releases energy (x decreases) → work is positive → block speeds up.

Model answer

Given: k = 800 N/m, x₁ = 0.3 m (initial compression), x₂ = 0 (natural length), m = 2 kg, frictionless surface

Using the Work–Energy Theorem (conservative forces only):
Work done by spring = ½k·x₁² − ½k·x₂²
= ½(800)(0.3)² − ½(800)(0)²
= ½(800)(0.09) − 0
= 36 J

Applying U_spring = ΔKE:
36 = ½mv₂² − ½mv₁² (block starts at rest: v₁ = 0)
36 = ½(2)v₂²
v₂ = √(36/1) = √36

v₂ = 6 m/s

Answer structure

• Step 1: State work done by spring formula: U_spring = ½k(x₁² − x₂²) [1 mark]
• Step 2: Substitute x₁ = 0.3 m, x₂ = 0 and compute U = 36 J [1 mark]
• Step 3: Apply U = ΔKE with v₁ = 0 and solve v₂ = 6 m/s [1 mark]

Scoring breakdown

Common mark deductions

• Writing U_spring = ½kx₂² − ½kx₁² (wrong order — gives negative work for a spring launching a block)
• Using stored spring energy ½kx² = 36 J directly without stating the work–energy framework
• Confusing spring compression with elongation in the formula
• Not setting v₁ = 0 (block initially at rest)

Key phrases to include

• U_spring = ½k(x₁² − x₂²)
• x₁ = 0.3 m, x₂ = 0
• U = 36 J
• ΔKE = ½mv₂² − ½mv₁²
• v₂ = 6 m/s

Examiner tip

Multi-method verification questions (like this one) are favorite 5-mark board items. Examiners award a full mark specifically for showing both methods and explicitly stating they agree. Never skip the verification statement.

Model answer

Given: m = 1 000 kg, v₁ = 20 m/s, v₂ = 0, Δt = 5 s

PART (a) — Average Braking Force (Impulse–Momentum):
ΣF · Δt = m(v₂ − v₁)
F_brake (5) = 1 000(0 − 20)
F_brake = −20 000 / 5 = −4 000 N

Braking force = 4 000 N (opposing motion)

PART (b) — Braking Distance:
Method 1 — Kinematics:
Deceleration: a = (v₂ − v₁)/Δt = (0 − 20)/5 = −4 m/s²
d = v₁t + ½at² = 20(5) + ½(−4)(5²) = 100 − 50 = 50 m

Method 2 — Work–Energy Verification:
Work done by brake = ΔKE
−F_brake × d = ½m·v₂² − ½m·v₁²
−4 000 × d = 0 − ½(1 000)(20²)
−4 000 × d = −200 000
d = 50 m ✓ Consistent

Braking distance = 50 m

Answer structure

• Part (a): Apply ΣF·Δt = mΔv, solve F_brake = 4 000 N [1 mark]
• Part (b) Method 1: Compute a = Δv/Δt = −4 m/s², apply kinematics for d = 50 m [1.5 marks]
• Part (b) Method 2: Apply Work–Energy: F·d = ΔKE, solve d = 50 m [1.5 marks]
• Consistency check: state both methods give 50 m [1 mark]

Scoring breakdown

Common mark deductions

• Sign error: reporting braking force as positive without stating 'opposing motion'
• Using average velocity formula d = (v₁ + v₂)/2 × t correctly but not verifying with energy method
• Omitting the consistency check statement — loses the final 1 mark
• Using F = ma without first computing a from kinematics

Key phrases to include

• ΣF·Δt = m(v₂ − v₁)
• F_brake = 4 000 N opposing motion
• a = −4 m/s²
• d = 50 m
• Work–Energy: F·d = ΔKE

Examiner tip

PRC board examiners treat 'conservation of energy' and 'work–energy theorem' as distinct principles. Conservation applies only without non-conservative forces; the work–energy theorem is the general case. Distinguishing them earns the second mark.

Model answer

Conservation of Energy states that when only conservative forces (gravity, spring forces) act on a body, the total mechanical energy remains constant:
KE₁ + PE₁ = KE₂ + PE₂
½mv₁² + mgh₁ = ½mv₂² + mgh₂

Conditions for its application in kinetics:
(1) No friction or non-conservative forces are present (or their work is zero).
(2) All forces acting are conservative — gravity and elastic spring forces satisfy this; friction and applied forces do not.
When friction is present, the work–energy theorem must be used instead: U_net = ΔKE (with negative friction work included).

Answer structure

• Line 1: State the principle: KE₁ + PE₁ = KE₂ + PE₂ with the formula [1 mark]
• Line 2: State the condition: only conservative forces; no friction or external non-conservative work [1 mark]

Scoring breakdown

Common mark deductions

• Stating conservation of energy without specifying 'mechanical energy' or 'conservative forces'
• Claiming energy is always conserved (confusing total energy with mechanical energy)
• Not addressing what changes when friction is present

Key phrases to include

• KE₁ + PE₁ = KE₂ + PE₂
• conservative forces only
• no friction
• total mechanical energy is constant
• non-conservative forces

Do

• Write the formula first, then a one-line verbal description
• Include units and the range or limits of the quantity (e.g., 0 ≤ e ≤ 1)
• Use standard engineering notation (ΣF, ΔKE, Δt)

Don't

• Do not derive the formula — state it
• Do not write a full paragraph for a 1-mark answer
• Do not use layman terms instead of engineering terminology

Do

• State the governing principle before substituting numbers
• Show all intermediate steps (N = mg, F_f = μN) even for simple calculations
• Box the final answer and write the unit explicitly

Don't

• Do not skip from given data directly to the final answer — show one intermediate step
• Do not omit direction or sign for vector answers
• Do not round intermediate values prematurely (carry 4 significant figures)

Do

• Number your steps (Step 1, Step 2, Step 3) — makes scoring obvious for the examiner
• Include a labeled FBD sketch (even a simple one) for Newton's Law problems
• Check the sign of friction work — always negative for opposing motion

Don't

• Do not mix steps out of order — setup must precede substitution
• Do not use ΣF = ma when work–energy is 3x faster for the same problem
• Do not omit N = mg computation on horizontal surfaces — it is a separate mark

Do

• Write 'Given' data block at the top with all values and units
• State the principle at the start of each subpart
• Define sign convention explicitly before any impulse–momentum or collision problem
• State the physical interpretation of the answer (e.g., 'negative velocity means leftward motion')
• For collision problems: always write both equations (momentum + e) before solving

Don't

• Do not skip the sign convention definition in head-on collision problems
• Do not attempt Part (b) if you have not solved Part (a) — the chain of marks depends on it
• Do not use KE conservation for collision problems unless the problem states perfectly elastic (e = 1)
• Do not omit direction or physical interpretation for the final answer in vector problems
• Do not erase incorrect work without partial credit note — cross it out and write 'Correction:' beside it