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9th Class Physics Chapter 3 Notes - Dynamics

9th ClassBISE LahorePhysicsSyllabus 2026
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<p><em>Parhai — Free Punjab Board Notes</em></p> <h2>Key Concepts</h2> <h3>Newton's First Law of Motion — The Law of Inertia</h3> <p>When no net (resultant) force is applied to an object, it keeps doing what it is already doing: a stationary object stays stationary, and a moving object keeps moving in a straight line at constant speed. This tendency of matter to resist any change in its state of rest or motion is called <strong>inertia</strong>.</p> <p>The mass of a body is the measure of its inertia — a heavier object is harder to set into motion and harder to stop, because it has greater inertia.</p> <p>A familiar example: passengers standing in a bus lurch forward when the driver brakes suddenly. Their lower bodies stop with the bus, but their upper bodies try to keep moving forward because of inertia.</p> <h3>Newton's Second Law of Motion</h3> <p>When a net force F acts on a body of mass m, it produces an acceleration a in the same direction as the force. Mathematically:</p> <p><strong>F = m × a</strong></p> <p>The SI unit of force is the <strong>newton (N)</strong>. One newton is the force that gives a 1 kg mass an acceleration of 1 m/s², so <strong>1 N = 1 kg·m/s²</strong>.</p> <h3>Momentum</h3> <p>Momentum is the product of a body's mass and its velocity: <strong>p = m × v</strong>. Its SI unit is <strong>kg·m/s</strong> (also written as <strong>newton-second, Ns</strong>). A heavier or faster object is harder to stop — that is, it has more momentum.</p> <h3>Force and Change in Momentum</h3> <p>Newton's second law can also be stated in terms of momentum: the force acting on a body equals the rate of change of its momentum.</p> <p><strong>F = (p<sub>f</sub> − p<sub>i</sub>) / t = Δp / t</strong></p> <p>This means a larger force is needed to change momentum quickly (short time), while the same change of momentum needs a smaller force if it happens over a longer time. This is why seatbelts and airbags stretch out the stopping time in an accident — increasing the time reduces the force on the passengers.</p> <h3>Law of Conservation of Momentum</h3> <p>For an <strong>isolated system</strong> (no external force acting), the total momentum of two or more interacting bodies stays constant.</p> <p><strong>m<sub>1</sub>u<sub>1</sub> + m<sub>2</sub>u<sub>2</sub> = m<sub>1</sub>v<sub>1</sub> + m<sub>2</sub>v<sub>2</sub></strong></p> <p>where u<sub>1</sub>, u<sub>2</sub> are velocities before interaction and v<sub>1</sub>, v<sub>2</sub> are velocities after.</p> <p>Examples: when a gun is fired, the bullet moves forward and the gun recoils backward — the forward momentum of the bullet equals the backward momentum of the gun. Rockets work the same way: burning fuel is pushed out the back with huge momentum, so the rocket gains equal momentum in the forward direction.</p> <h3>Newton's Third Law of Motion — Action and Reaction</h3> <p>For every action there is an equal and opposite reaction. These two forces always act on <strong>different bodies</strong> — action on one body, reaction on the other — which is why they never cancel each other out.</p> <p>Examples: a book on a table — its weight presses down on the table (action), and the table pushes back up on the book (reaction). A balloon let loose — air rushes out one way (action of balloon on air), and the escaping air pushes the balloon the opposite way (reaction of air on balloon).</p> <p>If action and reaction are equal and opposite, how does anything move? Because they act on different objects. When you walk, your foot pushes the ground backward (action), and the ground pushes your foot forward (reaction) — and it is this reaction on <em>your</em> body that moves you forward.</p> <h3>Friction</h3> <p>Friction is the force that opposes relative motion between two surfaces in contact. When you push a body that is at rest, friction matches your push — growing with the applied force — up to a maximum value called the <strong>force of limiting friction (F<sub>s</sub>)</strong>. The body only starts moving once the applied force exceeds this limit.</p> <p>The force of limiting friction is proportional to the <strong>normal reaction R</strong> (the pressing force between the surfaces):</p> <p><strong>F<sub>s</sub> = μ × R</strong></p> <p>where <strong>μ (mu)</strong> is the coefficient of friction, a number that depends on the nature of the two surfaces. Smoother surfaces and less pressing force mean less friction; rough surfaces and harder pressing mean more.</p> <h3>Ways to Reduce Friction</h3> <ul> <li>Make the sliding surfaces smooth by polishing.</li> <li>Lubricate the surfaces with oil or grease.</li> <li>Use ball bearings or roller bearings — rolling friction is far smaller than sliding friction.</li> <li>Give fast-moving vehicles a streamlined (fish-like) shape to cut down air resistance.</li> </ul> <h3>Centripetal Force</h3> <p>The force needed to keep a body moving in a circle is called <strong>centripetal force</strong>. It always points toward the center of the circle:</p> <p><strong>F<sub>c</sub> = m × v² / r</strong></p> <p>Applications: when a car turns, friction between the tyres and the road provides the centripetal force — which is why vehicles skid on slippery roads (friction too small → not enough centripetal force). Roads are <strong>banked</strong> (outer edge raised) at turns to help provide it. The spinner of a washing machine uses the same idea in reverse: spinning clothes at very high speed throws water out through the holes because the wet fabric cannot supply enough centripetal force to keep the water moving in the circle. A cream separator spins milk fast to separate fat the same way.</p> <h3>Tension and Acceleration in a String (Atwood Machine)</h3> <p><strong>Case 1 — both bodies hanging vertically over a frictionless pulley (Atwood machine):</strong> For masses m<sub>1</sub> &gt; m<sub>2</sub>:</p> <ul> <li>Acceleration: <strong>a = (m<sub>1</sub> − m<sub>2</sub>) / (m<sub>1</sub> + m<sub>2</sub>) × g</strong></li> <li>Tension in the string: <strong>T = 2m<sub>1</sub>m<sub>2</sub> / (m<sub>1</sub> + m<sub>2</sub>) × g</strong></li> </ul> <p>This arrangement can even be used to determine the value of g experimentally.</p> <p><strong>Case 2 — one body vertical, one on a smooth horizontal table:</strong> For hanging mass m<sub>2</sub> and table mass m<sub>1</sub>:</p> <ul> <li>Acceleration: <strong>a = m<sub>2</sub> / (m<sub>1</sub> + m<sub>2</sub>) × g</strong></li> <li>Tension: <strong>T = m<sub>1</sub>m<sub>2</sub> / (m<sub>1</sub> + m<sub>2</sub>) × g</strong></li> </ul> <h2>Questions & Answers</h2> <h3>1. What is Newton's first law of motion? Explain.</h3> <p>Newton's first law states that a body continues in its state of rest or of uniform motion in a straight line unless a net external force acts on it. A book on a table does not start moving on its own — something must push it. Likewise, a ball rolled on the ground would never stop if no force (like friction) opposed it. Because this law deals with the inertia of matter, it is also called the law of inertia.</p> <h3>2. What is Newton's second law of motion? Explain.</h3> <p>Newton's second law deals with what happens when a net force acts on a body: the force produces an acceleration in the direction of the net force. The acceleration is directly proportional to the net force and inversely proportional to the mass: <strong>a ∝ F</strong> and <strong>a ∝ 1/m</strong>, giving <strong>F = ma</strong>. One newton (1 N) is the force that gives a 1 kg body an acceleration of 1 m/s².</p> <h3>3. What is Newton's third law of motion? Explain.</h3> <p>To every action there is always an equal but opposite reaction. When body A pushes body B (action), body B pushes back on body A with a force of equal magnitude in the opposite direction (reaction). The two forces always act on different bodies, so they cannot cancel each other. A book resting on a table presses the table downward (action), while the table pushes the book upward (reaction).</p> <h3>4. What is the law of conservation of momentum?</h3> <p>The total momentum of an isolated system of two or more interacting bodies remains constant. For example, an inflated balloon released into the air forms a system with the air inside it. Before release the total momentum is zero; once released, the air gains momentum in one direction and the balloon gains equal momentum in the opposite direction, keeping the total unchanged.</p> <h3>5. How is force related to the change in momentum?</h3> <p>A force acting on a body changes its velocity, and therefore its momentum. Dividing the change in momentum (final minus initial) by the time taken gives the rate of change of momentum, which equals the force: <strong>F = (p<sub>f</sub> − p<sub>i</sub>) / t</strong>. The SI unit of momentum (Ns) matches kg·m/s.</p> <h3>6. Define inertia, momentum, force, friction and centripetal force.</h3> <ul> <li><strong>Inertia:</strong> the characteristic of a body by which it resists any change in its state of rest or uniform motion. Mass measures inertia.</li> <li><strong>Momentum:</strong> the product of the mass and velocity of a moving body, <strong>p = mv</strong>.</li> <li><strong>Force:</strong> the agency that changes, or tends to change, the state of rest or uniform motion of a body.</li> <li><strong>Friction:</strong> the force that opposes relative motion between two surfaces in contact.</li> <li><strong>Centripetal force:</strong> the force that keeps a body moving in a circle, always directed toward the center.</li> </ul> <h3>7. Why is it dangerous to travel on the roof of a bus?</h3> <p>Every part of a person sitting inside a moving bus shares the bus's motion. On the roof, a person can easily lose balance — and if the bus brakes or turns suddenly, inertia makes the body keep moving while the bus changes speed, so the person may be thrown off. That is why rooftop travel is extremely dangerous.</p> <h3>8. Why does a passenger move outward when a bus takes a turn?</h3> <p>To follow the curved path, the bus needs a centripetal force, which friction between the tyres and the road supplies. The passenger's body does not receive that force, so inertia carries the body along its original straight-line direction — it feels like being pushed outward.</p> <h3>9. Action and reaction are equal and opposite — so how does anything move?</h3> <p>Action and reaction never act on the same body, so they do not cancel. When a horse pulls a cart, the action (horse pushing the ground backward) and the reaction (ground pushing the horse forward) act on different bodies. The forward reaction on the horse moves the horse, and the horse drags the cart with it.</p> <h3>10. When a gun is fired, why does it recoil?</h3> <p>Firing is an action–reaction pair: the exploding charge pushes the bullet forward (action), and the bullet pushes back on the gun with an equal and opposite force (reaction), making the gun kick backward — this is the recoil.</p> <h3>11. Why is rolling friction less than sliding friction?</h3> <p>When one surface slides over another, tiny contact points (cold welds) form and must be torn apart continuously, which resists motion strongly. A rolling body lifts off each contact point without tearing these welds, so the resistance is far smaller.</p> <h3>12. Describe two situations where friction is needed.</h3> <p>(i) Walking — without friction between shoes and the ground, our feet would slip and we could not move forward. (ii) Vehicles — tyres need friction with the road to move and to stop; brakes work by creating friction. Nails also stay in wood only because of friction.</p> <h2>MCQs</h2> <ol> <li>Newton's first law of motion is valid only in the absence of: (a) force (b) net force (c) friction (d) momentum — <strong>Answer: (b)</strong></li> <li>Inertia depends on: (a) force (b) net force (c) mass (d) velocity — <strong>Answer: (c)</strong></li> <li>When the driver of a moving bus applies brakes suddenly, a passenger falls: (a) toward the moving bus (b) away from the bus (c) in the direction of motion (d) opposite to the direction of motion — <strong>Answer: (c)</strong></li> <li>A string is stretched by two equal and opposite forces of 10 N each. The tension in the string is: (a) zero (b) 5 N (c) 10 N (d) 20 N — <strong>Answer: (c)</strong></li> <li>The mass of a body: (a) decreases when accelerated (b) increases when accelerated (c) decreases when moving fast (d) remains the same — <strong>Answer: (d)</strong></li> <li>Two masses m<sub>1</sub> and m<sub>2</sub> hang from the ends of a string over a frictionless pulley. The acceleration of the bodies is: (a) m<sub>1</sub>m<sub>2</sub>/(m<sub>1</sub>+m<sub>2</sub>) × g (b) (m<sub>1</sub>−m<sub>2</sub>)/(m<sub>1</sub>+m<sub>2</sub>) × g (c) m<sub>1</sub>/(m<sub>1</sub>−m<sub>2</sub>) × g (d) (m<sub>1</sub>+m<sub>2</sub>)/m<sub>2</sub> × g — <strong>Answer: (b)</strong></li> <li>Which of the following is a unit of momentum? (a) Nm (b) kg·m/s (c) Ns (d) Ns² — <strong>Answer: (b) and (c)</strong></li> <li>When a horse pulls a cart, the action is on the: (a) cart (b) earth (c) horse (d) earth and cart — <strong>Answer: (a)</strong></li> <li>Which material lowers friction when pushed between metal plates? (a) water (b) fine marble powder (c) air (d) oil — <strong>Answer: (d)</strong></li> </ol> <h2>Key Formulas</h2> <table border="1" cellpadding="6"> <tr><th>Quantity</th><th>Formula</th></tr> <tr><td>Newton's second law</td><td>F = ma</td></tr> <tr><td>Momentum</td><td>p = mv</td></tr> <tr><td>Force from change in momentum</td><td>F = (p<sub>f</sub> − p<sub>i</sub>) / t</td></tr> <tr><td>Conservation of momentum</td><td>m<sub>1</sub>u<sub>1</sub> + m<sub>2</sub>u<sub>2</sub> = m<sub>1</sub>v<sub>1</sub> + m<sub>2</sub>v<sub>2</sub></td></tr> <tr><td>Limiting friction</td><td>F<sub>s</sub> = μR</td></tr> <tr><td>Centripetal force</td><td>F<sub>c</sub> = mv² / r</td></tr> <tr><td>Atwood machine (both vertical), acceleration</td><td>a = (m<sub>1</sub> − m<sub>2</sub>) / (m<sub>1</sub> + m<sub>2</sub>) × g</td></tr> <tr><td>Atwood machine (both vertical), tension</td><td>T = 2m<sub>1</sub>m<sub>2</sub> / (m<sub>1</sub> + m<sub>2</sub>) × g</td></tr> <tr><td>One vertical, one horizontal, acceleration</td><td>a = m<sub>2</sub> / (m<sub>1</sub> + m<sub>2</sub>) × g</td></tr> <tr><td>One vertical, one horizontal, tension</td><td>T = m<sub>1</sub>m<sub>2</sub> / (m<sub>1</sub> + m<sub>2</sub>) × g</td></tr> <tr><td>Weight</td><td>w = mg</td></tr> <tr><td>Unit of force</td><td>1 N = 1 kg·m/s²</td></tr> </table> <h2>Solved Numericals</h2> <ol> <li><strong>A force of 20 N moves a body with an acceleration of 2 m/s². What is its mass?</strong><br>Given: F = 20 N, a = 2 m/s². Using F = ma → 20 = m × 2 → <strong>m = 10 kg</strong>.</li> <li><strong>The weight of a body is 147 N. What is its mass? (g = 10 m/s²)</strong><br>Using w = mg → 147 = m × 10 → <strong>m = 14.7 kg</strong>.</li> <li><strong>How much force is needed to prevent a body of mass 10 kg from falling?</strong><br>The force must balance the weight: F = w = mg = 10 × 10 → <strong>F = 100 N</strong>.</li> <li><strong>Find the acceleration produced by a force of 100 N in a mass of 50 kg.</strong><br>Using F = ma → 100 = 50 × a → <strong>a = 2 m/s²</strong>.</li> <li><strong>Two masses 52 kg and 48 kg hang from a string over a frictionless pulley. Find the acceleration and the tension.</strong><br>a = (52 − 48) / (52 + 48) × 10 = 4/100 × 10 → <strong>a = 0.4 m/s²</strong>.<br>T = 2 × 52 × 48 / (52 + 48) × 10 = 49920/100 → <strong>T ≈ 499 N</strong>.</li> <li><strong>26 kg lies on a smooth horizontal table, 24 kg hangs vertically, connected by a string over a pulley. Find the acceleration and tension.</strong><br>a = 24 / (26 + 24) × 10 = 240/50 → <strong>a = 4.8 m/s²</strong>.<br>T = 26 × 24 / (26 + 24) × 10 = 6240/50 → <strong>T ≈ 125 N</strong>.</li> <li><strong>How much time is needed to change momentum by 22 Ns with a force of 20 N?</strong><br>F = Δp / t → 20 = 22 / t → <strong>t = 1.1 s</strong>.</li> <li><strong>Find the friction between a 5 kg wooden block and a horizontal marble floor (μ = 0.6).</strong><br>F = μmg = 0.6 × 5 × 10 → <strong>F = 30 N</strong>.</li> <li><strong>What centripetal force is needed to keep a 0.5 kg body moving in a circle of radius 50 cm at 3 m/s?</strong><br>r = 0.5 m. F<sub>c</sub> = mv²/r = 0.5 × 9 / 0.5 → <strong>F<sub>c</sub> = 9 N</strong>.</li> </ol> <p>More free notes at Parhai.</p>

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