UPCAT Physics — Work, Energy & ImpulseExam Answer Templates

Marks

1

Topic

Work

Difficulty

easy

Template Id

T1

Examiner Tip

Always include the formula - definitions without formulas rarely get full marks

Model Answer

Work done by a force is the product of force and displacement in the direction of force. W = F × s × cos θ

Question Type

very_short_answer

Answer Structure

• Line 1: State the definition with formula [1 mark]

Scoring Breakdown

Common Mark Deductions

• Missing formula
• Incorrect definition
• Not mentioning direction

Key Phrases To Include

• product of force and displacement
• direction of force
• W = F × s × cos θ

Marks

2

Topic

Work-Energy Theorem

Difficulty

medium

Template Id

T2

Examiner Tip

Start with 'Statement:' and 'Explanation:' headings for clarity

Model Answer

Statement: The work done by all forces acting on a body is equal to the change in kinetic energy of the body. Explanation: W = ΔKE = KE_final - KE_initial = ½mv² - ½mu² This theorem shows that work done results in change of motion of the body.

Question Type

short_answer

Answer Structure

• Line 1: State the theorem clearly [1 mark]
• Line 2: Write the formula and brief explanation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Incomplete statement
• Missing formula
• No explanation

Key Phrases To Include

• work done by all forces
• change in kinetic energy
• W = ΔKE

Marks

3

Topic

Work Calculation

Difficulty

medium

Template Id

T3

Examiner Tip

Always show the value of trigonometric ratios used in calculations

Model Answer

Given: Force, F = 20 N Displacement, s = 5 m Angle, θ = 60° To find: Work done, W = ? Formula: W = F × s × cos θ Substituting values: W = 20 × 5 × cos 60° W = 20 × 5 × 0.5 W = 50 J Therefore, work done = 50 J

Question Type

numerical

Answer Structure

• Lines 1-3: Write given data clearly [1 mark]
• Line 4: Write formula [1 mark]
• Lines 5-7: Substitute and calculate [1 mark]

Scoring Breakdown

Common Mark Deductions

• Missing units
• Wrong formula
• Calculation errors
• Not showing cos 60° value

Key Phrases To Include

• Given:
• To find:
• W = F × s × cos θ
• cos 60° = 0.5

Marks

3

Topic

Kinetic Energy Derivation

Difficulty

hard

Template Id

T4

Examiner Tip

Start derivations with clear assumptions and use well-known laws step by step

Model Answer

Consider a body of mass 'm' initially at rest. Let a constant force 'F' act on it. From Newton's second law: F = ma Let the body acquire velocity 'v' after moving distance 's'. From equation of motion: v² = u² + 2as Since u = 0: v² = 2as, therefore a = v²/2s Work done by force = F × s = ma × s = m × (v²/2s) × s = mv²/2 By work-energy theorem, this work done equals kinetic energy gained. Therefore, KE = ½mv²

Question Type

long_answer

Answer Structure

• Lines 1-2: State initial conditions and Newton's law [1 mark]
• Lines 3-4: Apply kinematic equation [1 mark]
• Lines 5-6: Calculate work done and apply work-energy theorem [1 mark]

Scoring Breakdown

Common Mark Deductions

• Skipping steps
• Wrong equations
• Not stating assumptions
• Missing final expression

Key Phrases To Include

• Newton's second law
• equation of motion
• work-energy theorem
• KE = ½mv²

Marks

2

Topic

Potential Energy

Difficulty

easy

Template Id

T5

Examiner Tip

Examples from daily life like water in dams, stretched springs get full marks

Model Answer

Potential energy is the stored energy possessed by a body due to its position or configuration. PE = mgh (for gravitational potential energy) Example: Water stored in a dam at height has gravitational potential energy.

Question Type

short_answer

Answer Structure

• Line 1: Define potential energy with formula [1 mark]
• Line 2: Give a relevant example [1 mark]

Scoring Breakdown

Common Mark Deductions

• Vague definition
• No formula
• Inappropriate example

Key Phrases To Include

• stored energy
• position or configuration
• PE = mgh

Marks

5

Topic

Energy Calculations

Difficulty

medium

Template Id

T6

Examiner Tip

Remember at highest point velocity is zero, so KE = 0, and use energy conservation concepts

Model Answer

Given: Mass, m = 2 kg Initial velocity, u = 10 m/s Acceleration due to gravity, g = 10 m/s² At highest point, final velocity v = 0 To find: KE and PE at highest point Step 1: Find height at highest point Using v² = u² - 2gh (negative because motion is against gravity) 0² = 10² - 2 × 10 × h 0 = 100 - 20h h = 5 m Step 2: Find KE at highest point KE = ½mv² = ½ × 2 × 0² = 0 J Step 3: Find PE at highest point PE = mgh = 2 × 10 × 5 = 100 J Therefore, KE = 0 J and PE = 100 J at the highest point.

Question Type

numerical

Answer Structure

• Lines 1-5: Write given data and what to find [1 mark]
• Lines 6-9: Calculate maximum height using kinematics [2 marks]
• Line 10: Calculate KE at highest point [1 mark]
• Line 11: Calculate PE at highest point [1 mark]

Scoring Breakdown

Common Mark Deductions

• Wrong kinematic equation
• Forgetting v = 0 at top
• Unit errors
• Calculation mistakes

Key Phrases To Include

• v = 0 at highest point
• v² = u² - 2gh
• KE = ½mv²
• PE = mgh

Marks

1

Topic

Energy Conservation

Difficulty

easy

Template Id

T7

Examiner Tip

Mention 'conservative forces' - this shows deeper understanding

Model Answer

The total mechanical energy (KE + PE) of a system remains constant when only conservative forces act on it.

Question Type

very_short_answer

Answer Structure

• Line 1: State the law clearly [1 mark]

Scoring Breakdown

Common Mark Deductions

• Incomplete statement
• Not mentioning conservative forces

Key Phrases To Include

• total mechanical energy
• KE + PE
• remains constant
• conservative forces

Marks

2

Topic

Impulse

Difficulty

easy

Template Id

T8

Examiner Tip

Give both units (N·s and kg·m/s) to show they are equivalent

Model Answer

Impulse is the product of force and time for which the force acts. It equals the change in momentum. J = F × t = Δp = mv - mu Unit: N·s or kg·m/s

Question Type

short_answer

Answer Structure

• Line 1: Define impulse [1 mark]
• Line 2: Give formula and unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Missing formula
• Wrong unit
• Incomplete definition

Key Phrases To Include

• product of force and time
• change in momentum
• J = F × t
• N·s

Marks

3

Topic

Impulse Calculation

Difficulty

medium

Template Id

T9

Examiner Tip

Always mention impulse-momentum theorem when asked for both quantities

Model Answer

Given: Force, F = 50 N Time, t = 0.2 s To find: Impulse (J) and change in momentum (Δp) Formula: J = F × t Calculation: J = 50 × 0.2 = 10 N·s By impulse-momentum theorem: J = Δp Therefore, change in momentum = 10 kg·m/s Impulse = 10 N·s and change in momentum = 10 kg·m/s

Question Type

numerical

Answer Structure

• Lines 1-2: Write given data [1 mark]
• Lines 3-4: Apply impulse formula [1 mark]
• Lines 5-6: Apply impulse-momentum theorem [1 mark]

Scoring Breakdown

Common Mark Deductions

• Wrong formula
• Missing units
• Not connecting impulse to momentum change

Key Phrases To Include

• J = F × t
• impulse-momentum theorem
• J = Δp

Marks

3

Topic

Impulse-Momentum Relationship

Difficulty

medium

Template Id

T10

Examiner Tip

Sports examples like cricket catching or football kicking are always well-received

Model Answer

Impulse-momentum theorem states that impulse equals change in momentum. J = F × t = Δp = m(v - u) This means the effect of force depends on both magnitude and duration. Example: In cricket, a fielder pulls his hands backward while catching a ball to increase the time of contact, reducing the force experienced and preventing injury.

Question Type

short_answer

Answer Structure

• Line 1: State the theorem [1 mark]
• Line 2: Write the relationship formula [1 mark]
• Lines 3-4: Give practical example with explanation [1 mark]

Scoring Breakdown

Common Mark Deductions

• No theorem statement
• Poor example
• Missing formula

Key Phrases To Include

• impulse-momentum theorem
• J = Δp
• increase time
• reduce force

Marks

5

Topic

Impulse-Momentum Problems

Difficulty

hard

Template Id

T11

Examiner Tip

Pay attention to direction - negative values indicate force opposite to motion

Model Answer

Given: Mass, m = 5 kg Initial velocity, u = 8 m/s Final velocity, v = 0 m/s (brought to rest) Time, t = 4 s To find: Impulse (J) and Force (F) Step 1: Calculate change in momentum Δp = m(v - u) = 5(0 - 8) = -40 kg·m/s Step 2: Calculate impulse J = Δp = -40 N·s (Magnitude of impulse = 40 N·s) Step 3: Calculate force From J = F × t F = J/t = -40/4 = -10 N (Magnitude of force = 10 N) Therefore, impulse = 40 N·s and force = 10 N (in opposite direction to motion).

Question Type

numerical

Answer Structure

• Lines 1-5: Write given data clearly [1 mark]
• Lines 6-7: Calculate change in momentum [1 mark]
• Lines 8-9: Find impulse using J = Δp [1 mark]
• Lines 10-11: Calculate force using J = F × t [1 mark]
• Line 12: State final answer with direction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Ignoring negative sign
• Wrong momentum calculation
• Missing direction
• Unit errors

Key Phrases To Include

• brought to rest
• v = 0
• Δp = m(v - u)
• J = Δp
• opposite direction

Marks

2

Topic

Work and Power

Difficulty

easy

Template Id

T12

Examiner Tip

Structure as 'Work: definition, formula, unit' and 'Power: definition, formula, unit'

Model Answer

Work: Energy transferred when force acts through displacement. W = F × s. Unit: Joule (J). Power: Rate of doing work or energy transfer per unit time. P = W/t. Unit: Watt (W).

Question Type

short_answer

Answer Structure

• Line 1: Define work with formula and unit [1 mark]
• Line 2: Define power with formula and unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Missing formulas
• Wrong units
• Incomplete definitions

Key Phrases To Include

• energy transferred
• force through displacement
• rate of doing work
• W = F × s
• P = W/t

Marks

3

Topic

Power Calculations

Difficulty

medium

Template Id

T13

Examiner Tip

Show both cases clearly - examiners look for systematic approach

Model Answer

Given: Work done, W = 500 J Time₁ = 10 s, Time₂ = 5 s To find: Power in both cases Formula: P = W/t Case 1: P₁ = W/t₁ = 500/10 = 50 W Case 2: P₂ = W/t₂ = 500/5 = 100 W Therefore, initial power = 50 W and new power = 100 W

Question Type

numerical

Answer Structure

• Lines 1-2: Write given data [1 mark]
• Line 3: Apply power formula for first case [1 mark]
• Line 4: Calculate power for second case [1 mark]

Scoring Breakdown

Common Mark Deductions

• Wrong formula
• Calculation errors
• Missing units

Key Phrases To Include

• P = W/t
• same work
• different time

Marks

2

Topic

Energy Graphs

Difficulty

medium

Template Id

T14

Examiner Tip

Always label axes and mention the mathematical relationship

Model Answer

The graph shows a parabolic curve starting from origin. [Graph description: X-axis: Velocity (v), Y-axis: Kinetic Energy (KE), Curve: Parabola passing through origin with equation KE = ½mv²] Shape: Parabolic curve because KE ∝ v² Starting point: Origin (0,0) because when v = 0, KE = 0

Question Type

diagram_based

Answer Structure

• Line 1: Draw properly labeled graph [1 mark]
• Line 2: Explain the shape and relationship [1 mark]

Scoring Breakdown

Common Mark Deductions

• Wrong curve shape
• Missing labels
• No explanation

Key Phrases To Include

• parabolic curve
• KE ∝ v²
• passes through origin

Marks

3

Topic

Applications of Work-Energy Theorem

Difficulty

hard

Template Id

T15

Examiner Tip

Connect physics concepts to real-life situations for better marks

Model Answer

According to work-energy theorem, work done by braking force equals change in kinetic energy. W = F × s = ΔKE = ½mv² - 0 = ½mv² For constant braking force F: s = ½mv²/F Since s ∝ v², stopping distance increases with square of velocity. Example: If speed doubles, stopping distance becomes four times because KE increases by factor of four.

Question Type

long_answer

Answer Structure

• Line 1: State work-energy theorem [1 mark]
• Line 2: Derive relationship between stopping distance and velocity [1 mark]
• Line 3: Explain with numerical example [1 mark]

Scoring Breakdown

Common Mark Deductions

• No theorem reference
• Missing mathematical relationship
• Poor explanation

Key Phrases To Include

• work-energy theorem
• s ∝ v²
• kinetic energy increases