Physics — Motion and Gravitation

Overview

Motion and Gravitation forms a foundational chapter in upper-primary science, appearing consistently in KAR TET Paper II. This topic tests your understanding of how objects move, why they move, and what governs their motion under gravitational influence. Questions typically involve numerical problems using equations of motion, conceptual questions on Newton's laws, and application-based problems on free fall and gravitational force.

Mastering this topic requires a clear grasp of definitions (distance vs displacement, speed vs velocity), the three equations of motion, Newton's three laws, and the universal law of gravitation. Expect 3–5 questions combining direct formula application with conceptual reasoning about everyday phenomena like falling objects, moving vehicles, and planetary motion.

Key Concepts

• **Rest and Motion are relative**: An object is at rest or in motion only with respect to a reference point. A passenger in a moving bus is at rest relative to the bus but in motion relative to the road.

• **Distance vs Displacement**: Distance is the total path length (scalar, always positive); displacement is the shortest straight-line distance from initial to final position (vector, can be zero or negative).

• **Speed vs Velocity**: Speed = distance/time (scalar); velocity = displacement/time (vector). An object moving in a circle at constant speed has changing velocity because direction changes.

• **Uniform vs Non-uniform Motion**: Uniform motion has constant velocity (zero acceleration); non-uniform motion has changing velocity (non-zero acceleration).

• **Acceleration**: Rate of change of velocity. Positive acceleration means speeding up in the direction of motion; negative acceleration (retardation/deceleration) means slowing down.

• **Newton's First Law (Inertia)**: An object remains at rest or in uniform motion unless acted upon by an external unbalanced force. Explains why passengers lurch forward when a bus stops suddenly.

• **Newton's Second Law**: Force = mass × acceleration (F = ma). Greater mass requires greater force for the same acceleration.

• **Newton's Third Law**: Every action has an equal and opposite reaction. The forces act on different bodies — a swimmer pushes water backward, water pushes swimmer forward.

• **Universal Law of Gravitation**: Every object attracts every other object with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.

Formulas / Key Facts

**Equations of Motion (for uniformly accelerated motion in a straight line):**

| Equation | Formula | Use | |----------|---------|-----| | First | v = u + at | Find final velocity when time is known | | Second | s = ut + ½at² | Find displacement when time is known | | Third | v² = u² + 2as | Find velocity when displacement is known (time not given) |

Where: u = initial velocity, v = final velocity, a = acceleration, t = time, s = displacement

**Newton's Second Law:**

• F = ma (Force in Newtons, mass in kg, acceleration in m/s²)
• 1 Newton = force needed to accelerate 1 kg by 1 m/s²

**Universal Law of Gravitation:**

• F = G(m₁m₂)/d²
• G = 6.67 × 10⁻¹¹ Nm²/kg² (universal gravitational constant)
• m₁, m₂ = masses of two objects; d = distance between their centres

**Free Fall:**

• Acceleration due to gravity (g) = 9.8 m/s² (approximately 10 m/s² for calculations)
• For a freely falling body: u = 0, a = g = 9.8 m/s²
• For an object thrown upward: a = –g (takes g as negative)

**Weight vs Mass:**

• Mass is constant everywhere (measured in kg)
• Weight = mg (force due to gravity, measured in Newtons)
• Weight varies with location (less on Moon, more on Jupiter)

Worked Examples

**Example 1: Using equations of motion**

A car starts from rest and accelerates uniformly at 2 m/s² for 10 seconds. Find (a) final velocity, (b) distance covered.

Solution:

• Given: u = 0, a = 2 m/s², t = 10 s
• (a) v = u + at = 0 + 2 × 10 = 20 m/s
• (b) s = ut + ½at² = 0 + ½ × 2 × 10² = 100 m

**Example 2: Free fall problem**

A stone is dropped from a height of 80 m. How long does it take to reach the ground? (Take g = 10 m/s²)

Solution:

• Given: u = 0 (dropped, not thrown), s = 80 m, a = g = 10 m/s²
• Using s = ut + ½at²
• 80 = 0 + ½ × 10 × t²
• 80 = 5t²
• t² = 16
• t = 4 seconds

**Example 3: Newton's Second Law**

A force of 20 N acts on a body of mass 4 kg at rest. Find the acceleration and velocity after 5 seconds.

Solution:

• F = ma → 20 = 4 × a → a = 5 m/s²
• v = u + at = 0 + 5 × 5 = 25 m/s

Common Mistakes

• **Confusing distance with displacement** → Distance is always positive and can be longer than displacement. A person walking 3 m east then 3 m west covers 6 m distance but has 0 displacement.

• **Forgetting to assign correct sign to acceleration** → When an object slows down, acceleration is negative (opposite to velocity direction). Students often use positive values and get wrong answers for braking problems.

• **Using wrong units** → Always convert km/h to m/s before applying equations. Multiply km/h by 5/18 to get m/s. Example: 36 km/h = 36 × 5/18 = 10 m/s.

• **Applying F = ma to weight problems incorrectly** → Weight is mg, not m. A 50 kg person's weight is 50 × 10 = 500 N, not 50 N.

• **Misunderstanding Newton's Third Law** → Action and reaction act on different bodies, not the same body. They don't cancel out because they don't act on the same object.

• **Assuming g is same everywhere** → Value of g varies: 9.8 m/s² on Earth's surface, about 1.6 m/s² on Moon. Questions may specify different values.

Quick Reference

• v = u + at; s = ut + ½at²; v² = u² + 2as — memorise all three equations
• F = ma — force equals mass times acceleration
• F = Gm₁m₂/d² — gravitational force formula; G = 6.67 × 10⁻¹¹ Nm²/kg²
• Free fall: u = 0, a = g = 9.8 m/s² (use 10 for quick calculation)
• 1 km/h = 5/18 m/s; 1 m/s = 18/5 km/h
• Weight = mg (in Newtons); mass is in kg and stays constant