UPCAT Physics — Newton's Laws, Dynamics & MomentumExam Answer Templates

Marks

1

Topic

Newton's Laws of Motion

Difficulty

easy

Template Id

T1

Examiner Tip

The law must be stated exactly - paraphrasing often loses marks in physics

Model Answer

A body at rest remains at rest, and a body in motion continues to move with constant velocity, unless acted upon by an external unbalanced force.

Question Type

very_short_answer

Answer Structure

• Single sentence stating the law completely [1 mark]

Scoring Breakdown

Common Mark Deductions

• Incomplete statement missing 'unbalanced' force
• Using 'net force' instead of 'unbalanced force'

Key Phrases To Include

• body at rest remains at rest
• constant velocity
• external unbalanced force

Marks

2

Topic

Momentum

Difficulty

easy

Template Id

T2

Examiner Tip

Always mention that momentum is a vector quantity for full marks

Model Answer

Momentum is the product of mass and velocity of a moving object. It is a vector quantity. SI unit: kg⋅m/s or N⋅s

Question Type

short_answer

Answer Structure

• Definition of momentum [1 mark]
• SI unit with proper notation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Missing 'vector quantity' specification
• Wrong unit notation

Key Phrases To Include

• product of mass and velocity
• vector quantity
• kg⋅m/s

Marks

2

Topic

Newton's Second Law

Difficulty

easy

Template Id

T3

Examiner Tip

Always show the algebraic rearrangement step (a = F/m) for full method marks

Model Answer

Given: F = 20 N, m = 4 kg Using Newton's Second Law: F = ma Therefore, a = F/m = 20/4 = 5 m/s²

Question Type

numerical

Answer Structure

• State given values clearly [0.5 marks]
• Write correct formula [0.5 marks]
• Substitute and calculate [0.5 marks]
• Final answer with unit [0.5 marks]

Scoring Breakdown

Common Mark Deductions

• Missing 'Given:' section
• No formula stated
• Missing units

Key Phrases To Include

• Given:
• Newton's Second Law
• F = ma
• m/s²

Marks

3

Topic

Newton's Third Law

Difficulty

medium

Template Id

T4

Examiner Tip

Always specify that action and reaction forces act on different objects - this is a common misconception

Model Answer

Newton's Third Law states that for every action, there is an equal and opposite reaction. The action and reaction forces act on different objects and are equal in magnitude but opposite in direction. Example: When walking, we push the ground backward (action), and the ground pushes us forward (reaction) with equal force.

Question Type

short_answer

Answer Structure

• State Newton's Third Law [1 mark]
• Explain key characteristics [1 mark]
• Provide relevant example [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not mentioning forces act on different objects
• Poor example that doesn't show action-reaction clearly

Key Phrases To Include

• equal and opposite reaction
• different objects
• equal in magnitude
• opposite in direction

Marks

5

Topic

Conservation of Momentum

Difficulty

medium

Template Id

T5

Examiner Tip

Always identify whether collision is elastic or inelastic - this shows understanding of the physics involved

Model Answer

Given: m₁ = 1000 kg, u₁ = 20 m/s (initial velocity of car 1) m₂ = 800 kg, u₂ = 0 m/s (initial velocity of car 2) v = common velocity after collision This is an inelastic collision. Using conservation of momentum: Total momentum before collision = Total momentum after collision m₁u₁ + m₂u₂ = (m₁ + m₂)v (1000 × 20) + (800 × 0) = (1000 + 800)v 20,000 + 0 = 1800v v = 20,000/1800 = 11.11 m/s Therefore, the common velocity after collision is 11.11 m/s.

Question Type

numerical

Answer Structure

• State all given values clearly [1 mark]
• Identify type of collision [1 mark]
• Write conservation of momentum equation [1 mark]
• Substitute values correctly [1 mark]
• Calculate final answer with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not identifying collision type
• Arithmetic errors
• Missing final unit
• Wrong momentum formula

Key Phrases To Include

• inelastic collision
• conservation of momentum
• total momentum before = total momentum after

Marks

5

Topic

Impulse and Momentum

Difficulty

hard

Template Id

T6

Examiner Tip

In derivation questions, every algebraic step should be shown - examiners award partial marks for method

Model Answer

Definition: Impulse is the change in momentum of an object when acted upon by a force for a certain time period. Impulse = Force × time = F × t SI unit: N⋅s or kg⋅m/s Derivation of Impulse-Momentum Theorem: From Newton's Second Law: F = ma We know that acceleration a = (v - u)/t, where v = final velocity, u = initial velocity Therefore: F = m(v - u)/t Multiplying both sides by t: F × t = m(v - u) F × t = mv - mu F × t = pf - pi Therefore, Impulse = Change in momentum This is the impulse-momentum theorem.

Question Type

long_answer

Answer Structure

• Define impulse with formula [1 mark]
• State SI unit [1 mark]
• Start derivation from F = ma [1 mark]
• Show algebraic steps clearly [1 mark]
• Conclude with final theorem [1 mark]

Scoring Breakdown

Common Mark Deductions

• Missing definition
• Skipping derivation steps
• Algebraic errors
• No final statement of theorem

Key Phrases To Include

• change in momentum
• F × t
• Newton's Second Law
• impulse-momentum theorem

Marks

3

Topic

Force Analysis

Difficulty

medium

Template Id

T7

Examiner Tip

Always draw force vectors from the center of mass of the object and use proper arrowheads

Model Answer

[Draw inclined plane with block] Forces acting on the block: 1. Weight (mg) acting vertically downward from center of block 2. Normal force (N) acting perpendicular to inclined surface, away from plane 3. Component of weight parallel to plane (mg sin θ) down the slope 4. Component of weight perpendicular to plane (mg cos θ) into the plane

Question Type

diagram_based

Answer Structure

• Draw clear diagram with block on incline [1 mark]
• Show weight vector vertically downward [1 mark]
• Show normal force perpendicular to surface [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forces not from center of mass
• Wrong directions
• Missing labels
• Including friction when specified frictionless

Key Phrases To Include

• vertically downward
• perpendicular to surface
• mg
• N

Marks

3

Topic

Types of Collisions

Difficulty

medium

Template Id

T8

Examiner Tip

Always mention both momentum and energy conservation aspects for complete comparison

Model Answer

Elastic Collision: - Kinetic energy is conserved - Objects separate after collision - Both momentum and kinetic energy remain constant Inelastic Collision: - Kinetic energy is not conserved (some is lost) - Objects may stick together (perfectly inelastic) - Only momentum is conserved

Question Type

short_answer

Answer Structure

• Define elastic collision with key features [1.5 marks]
• Define inelastic collision with key features [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Confusing energy and momentum conservation
• Not mentioning what happens to objects after collision

Key Phrases To Include

• kinetic energy conserved
• kinetic energy not conserved
• momentum conserved
• objects separate
• stick together

Marks

2

Topic

Momentum Applications

Difficulty

easy

Template Id

T9

Examiner Tip

Key insight is recognizing physical condition - at maximum height, velocity is momentarily zero

Model Answer

Given: m = 0.5 kg, initial velocity u = 10 m/s upward At maximum height, velocity v = 0 m/s Momentum p = mv = 0.5 × 0 = 0 kg⋅m/s Therefore, momentum at maximum height is zero.

Question Type

numerical

Answer Structure

• Identify key condition at maximum height [1 mark]
• Apply momentum formula and calculate [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using initial velocity instead of velocity at maximum height
• Missing units

Key Phrases To Include

• maximum height
• velocity = 0
• p = mv

Marks

3

Topic

Conservation Laws

Difficulty

medium

Template Id

T10

Examiner Tip

The key phrase is 'no external forces' - internal forces don't affect total momentum

Model Answer

Law of Conservation of Momentum: The total momentum of a system remains constant when no external force acts on the system. Conditions for validity: 1. No external forces should act on the system 2. The system should be isolated or closed

Question Type

short_answer

Answer Structure

• State the law clearly [1 mark]
• First condition [1 mark]
• Second condition [1 mark]

Scoring Breakdown

Common Mark Deductions

• Vague statement of law
• Incorrect conditions
• Only one condition given

Key Phrases To Include

• total momentum remains constant
• no external force
• isolated system

Marks

3

Topic

Force and Motion

Difficulty

medium

Template Id

T11

Examiner Tip

Always interpret the negative sign in force problems - it indicates direction relative to motion

Model Answer

Given: m = 1200 kg, u = 25 m/s, v = 0 m/s (final velocity), t = 5 s First, find acceleration: a = (v - u)/t = (0 - 25)/5 = -5 m/s² Using Newton's Second Law: F = ma = 1200 × (-5) = -6000 N The magnitude of force required is 6000 N (negative sign indicates force opposes motion).

Question Type

numerical

Answer Structure

• State given values [0.5 marks]
• Calculate acceleration [1 mark]
• Apply F = ma [1 mark]
• Interpret result [0.5 marks]

Scoring Breakdown

Common Mark Deductions

• Wrong acceleration calculation
• Missing negative sign interpretation
• No units

Key Phrases To Include

• acceleration
• Newton's Second Law
• opposes motion

Marks

3

Topic

Impulse Applications

Difficulty

medium

Template Id

T12

Examiner Tip

This is a classic application question - always link physics concepts to real-world observations

Model Answer

When falling, the change in momentum is the same regardless of surface type. From impulse-momentum theorem: Impulse = F × t = Change in momentum On a hard surface: - Contact time (t) is very small - Since F × t is constant, force (F) must be very large On a soft surface: - Contact time (t) is longer - Since F × t is constant, force (F) is smaller Therefore, the large force on hard surface causes more pain and injury.

Question Type

short_answer

Answer Structure

• State that change in momentum is same [0.5 marks]
• Apply impulse-momentum theorem [1 mark]
• Compare contact times and forces [1 mark]
• Conclude about pain/injury [0.5 marks]

Scoring Breakdown

Common Mark Deductions

• Not using impulse concept
• Incorrect relationship between force and time
• No clear conclusion

Key Phrases To Include

• impulse-momentum theorem
• contact time
• F × t = constant
• change in momentum

Marks

2

Topic

Inertia and Newton's Laws

Difficulty

easy

Template Id

T13

Examiner Tip

Inertia is a property of matter, not a force - this is a common misconception

Model Answer

Inertia is the tendency of an object to resist changes in its state of motion. Newton's First Law is also called the Law of Inertia because it describes this property - objects at rest stay at rest and objects in motion continue in motion unless acted upon by an external force.

Question Type

short_answer

Answer Structure

• Define inertia [1 mark]
• Connect to Newton's First Law [1 mark]

Scoring Breakdown

Common Mark Deductions

• Incomplete definition
• No connection to Newton's laws

Key Phrases To Include

• resist changes
• state of motion
• Law of Inertia

Marks

5

Topic

Elastic Collisions

Difficulty

hard

Template Id

T14

Examiner Tip

Elastic collision problems require both conservation laws - don't forget the relative velocity condition

Model Answer

Given: m₁ = 2 kg, u₁ = +4 m/s (taking right as positive) m₂ = 3 kg, u₂ = -2 m/s (opposite direction) For elastic collision, both momentum and kinetic energy are conserved. Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (2)(4) + (3)(-2) = 2v₁ + 3v₂ 8 - 6 = 2v₁ + 3v₂ 2 = 2v₁ + 3v₂ ... (1) For elastic collision, relative velocity reverses: v₂ - v₁ = -(u₂ - u₁) = -(-2 - 4) = 6 v₂ - v₁ = 6 ... (2) Solving equations (1) and (2): From (2): v₂ = v₁ + 6 Substitute in (1): 2 = 2v₁ + 3(v₁ + 6) 2 = 2v₁ + 3v₁ + 18 2 = 5v₁ + 18 v₁ = -3.2 m/s v₂ = -3.2 + 6 = 2.8 m/s Therefore, after collision: Ball 1: velocity = 3.2 m/s to the left Ball 2: velocity = 2.8 m/s to the right

Question Type

numerical

Answer Structure

• State given values with sign convention [1 mark]
• Apply conservation of momentum [1 mark]
• Use elastic collision condition [1 mark]
• Solve simultaneous equations [1 mark]
• State final answers with directions [1 mark]

Scoring Breakdown

Common Mark Deductions

• Wrong sign convention
• Not using elastic collision condition
• Algebraic errors
• Missing direction interpretation

Key Phrases To Include

• sign convention
• conservation of momentum
• elastic collision
• relative velocity reverses