What is third law & conservation of momentum in Class 9 Science (CBSE/NCERT)?

Third Law & Conservation of Momentum is a key topic in NCERT Class 9 Science Chapter 6, covering action-reaction pairs, momentum, impulse and the law of conservation of momentum with applications like gun recoil and rocket propulsion.

What is third law & conservation of momentum in Class 9 Science (CBSE/NCERT)?

Third Law & Conservation of Momentum is a key topic in NCERT Class 9 Science Chapter 6 u2014 How Forces Affect Motion. It explains Newton's third law of motion and the law of conservation of momentum with applications and numerical problems.

Why is Newton's third law important in NCERT Class 9 Science?

Newton's third law is important because it forms the foundation for understanding third law & conservation of momentum in Chapter 6. Without a clear idea of Newton's third law, students cannot answer higher-order CBSE questions involving action and reaction, momentum, conservation of momentum.

How is third law & conservation of momentum tested in the Class 9 Science CBSE exam?

CBSE tests this through 1-mark MCQs, 2-mark short answers, 3-mark explanations, 5-mark descriptive questions and 4-mark competency-based questions on Newton's third law, action and reaction and momentum drawn from NCERT activities.

What are the key terms to remember for third law & conservation of momentum in Class 9 Science?

Key terms: Newton's third law, action and reaction, momentum, conservation of momentum, recoil of gun, rocket propulsion. Each carries exam weightage and regularly appears in the CBSE Class 9 paper.

Is Third Law & Conservation of Momentum included in the Class 9 Science syllabus for 2025u201326 CBSE?

Yes, Third Law & Conservation of Momentum is part of the NCERT Class 9 Science syllabus (2025u201326) under the new NCERT Exploration textbook. It falls under Chapter 6 and is examined in the annual paper.

How should I prepare third law & conservation of momentum for the CBSE Class 9 Science exam?

Prepare in three steps. First, read this NCERT part carefully, highlighting definitions and diagrams of Newton's third law, action and reaction and momentum. Second, solve every in-text question and end-of-chapter exercise. Third, practise competency-based and assertion-reason questions.

What is third law & conservation of momentum in Class 9 Science (CBSE/NCERT)?

Third Law & Conservation of Momentum is a key topic in NCERT Class 9 Science Chapter 6 — How Forces Affect Motion. It explains newton's third law of motion and the law of conservation of momentum with applications and numerical problems. Core ideas covered include Newton's third law, action and reaction, momentum, conservation of momentum. Mastering this subtopic is essential for scoring well in the CBSE Class 9 Science exam and for building a strong foundation for the Class 10 board exam, because these concepts repeatedly appear in MCQs, short answers and long-answer questions. This part gives a complete, exam-ready explanation with activities, diagrams and competency-based practice aligned to NCERT.

Why is Newton's third law important in NCERT Class 9 Science?

Newton's third law is important in NCERT Class 9 Science because it forms the foundation for understanding third law & conservation of momentum in Chapter 6 — How Forces Affect Motion. Without a clear idea of Newton's third law, students cannot answer higher-order CBSE questions involving action and reaction, momentum, conservation of momentum. School and competitive papers regularly include 2-mark and 3-mark questions on this concept, and competency-based questions often link Newton's third law to real-life situations. Building clarity here pays off directly in marks at Class 9 and again in the Class 10 board exam.

How is third law & conservation of momentum tested in the Class 9 Science CBSE exam?

The CBSE Class 9 Science exam tests third law & conservation of momentum through a mix of 1-mark MCQs, 2-mark short answers, 3-mark explanations with examples, 5-mark descriptive questions (often with diagrams or derivations) and 4-mark competency-based questions. Expect direct questions on Newton's third law, action and reaction, momentum and application-based questions drawn from NCERT activities. Students who follow the NCERT Exploration textbook thoroughly and practise this chapter's questions consistently score in the 90%+ range.

What are the key terms to remember for third law & conservation of momentum in Class 9 Science?

The key terms to remember for third law & conservation of momentum in NCERT Class 9 Science Chapter 6 are: Newton's third law, action and reaction, momentum, conservation of momentum, recoil of gun, rocket propulsion. Each of these concepts carries exam weightage and regularly appears in the CBSE Class 9 paper. Write clear one-line definitions of every term in your revision notes and revisit them before the exam. Linking these terms visually through a flowchart or concept map makes recall easier during the Class 9 Science exam.

Is Third Law & Conservation of Momentum included in the Class 9 Science syllabus for 2025–26 CBSE?

Yes, Third Law & Conservation of Momentum is part of the NCERT Class 9 Science syllabus (2025–26) prescribed by CBSE under the new NCERT Exploration textbook. It falls under Chapter 6 — How Forces Affect Motion — and is examined in the annual paper. The current syllabus retains the full treatment of Newton's third law, action and reaction, momentum as per the NCERT textbook. Because CBSE bases every Class 9 question on NCERT, studying this part thoroughly ensures complete syllabus coverage and guarantees marks from this chapter.

How should I prepare third law & conservation of momentum for the CBSE Class 9 Science exam?

Prepare third law & conservation of momentum for the CBSE Class 9 Science exam in three steps. First, read this NCERT part carefully, highlighting definitions and diagrams of Newton's third law, action and reaction, momentum. Second, solve every in-text question and end-of-chapter exercise — CBSE questions often come directly from NCERT. Third, practise competency-based and assertion-reason questions to sharpen reasoning. Write answers in the exam-style format (point-wise with diagrams) and time yourself. This method delivers confidence and full marks in the exam.

TOPIC 21 OF 50

Newton’s Third Law, Momentum and Conservation of Momentum

🎓 Class 9 Science CBSE Theory Ch 6 — How Forces Affect Motion ⏱ ~13 min

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6.8 Newton's Third Law of Motion

If you have ever stepped off a small boat onto a jetty, you may have felt the boat slide backward as you stepped forward. Why does that happen? You exerted a force on the boat through your foot — and the boat exerted an equal and opposite force on you. This mutual interaction is the heart of Newton's Third Law.

Newton's Third Law of Motion: To every action, there is an equal and opposite reaction. Forces always occur in pairs — if body A exerts a force on body B, body B simultaneously exerts an equal force on body A in the opposite direction.

Two key features of action–reaction pairs:

They are equal in magnitude and opposite in direction.
They act on two different bodies — never on the same body. (That is why they do not cancel out the way balanced forces do.)

Examples of action–reaction pairs

Walking: Your foot pushes the ground backward (action). The ground pushes your foot forward with an equal force (reaction). That is what propels you ahead.
Swimmer: A swimmer pushes the water backward with the hands; the water pushes the swimmer forward.
Rocket launch: Hot gases are expelled downward at high speed (action). The gases push the rocket upward (reaction).
Recoil of a gun: When a gun fires, the gun pushes the bullet forward, and the bullet pushes the gun back. The gun is felt as a "kick" on the shoulder.
Jumping off a boat: The person pushes the boat backward; the boat pushes the person forward toward the shore.

🚶🚀 Action–Reaction Pairs — Compare two scenes L4 Analyse

Newton's Third Law lives in both an everyday walk and a giant rocket launch. Click each scene and analyse: which body is the action acting on, and which body is the reaction acting on?

Fig 6.6: Action–reaction pairs in walking and rocket launch.

Click either the walker or the rocket above to analyse the action–reaction pair.

6.9 Momentum

If a fast-moving cricket ball hits you, it hurts more than a slow-moving one of the same mass. A heavy lorry at 5 m/s causes more damage in a collision than a bicycle at 5 m/s. Both mass and velocity matter when describing the "quantity of motion" of a body. Newton combined the two into a single quantity called momentum.

Momentum: \(p = m v\). Momentum is a vector quantity — its direction is the same as that of the velocity. Its SI unit is kg·m/s.

Change in momentum

If a body's velocity changes from \(u\) to \(v\) in a time \(t\), the change in momentum is \(\Delta p = mv - mu = m(v-u)\). This is closely linked to the Second Law:

\(F = m a = m \dfrac{v-u}{t} = \dfrac{m v - m u}{t} = \dfrac{\Delta p}{t}\)

So Newton's Second Law can also be stated as: the rate of change of momentum of a body is equal to the net force acting on it.

Impulse

Multiplying both sides of \(F = \Delta p/t\) by \(t\) gives:

Impulse: \(F \cdot t = \Delta p\)

Impulse measures the total effect of a force acting for a time \(t\). The same change in momentum can be produced by a large force for a short time or a small force for a long time. That is why we bend our knees on landing from a jump (longer \(t\) → smaller \(F\)) and why a cricketer pulls the hands back while catching a fast ball.

6.10 Law of Conservation of Momentum

Consider two bodies that interact only with each other (an "isolated system" — no external forces). Newton's Third Law guarantees that whatever force one exerts on the other is exactly cancelled by the equal and opposite force on itself. The result is one of the deepest laws in physics:

Law of Conservation of Momentum: The total momentum of an isolated system of bodies remains constant if no external unbalanced force acts on it.

If two bodies of masses \(m_1\) and \(m_2\) have initial velocities \(u_1\) and \(u_2\), and after they interact the velocities become \(v_1\) and \(v_2\), then:

\(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\)

Recoil of a gun — a textbook application

Example 1 — Recoil velocity of a gun

Q. A gun of mass 4 kg fires a bullet of mass 20 g with a velocity of 200 m/s. Find the recoil velocity of the gun.

Solution. Mass of bullet \(m_1 = 0.020\) kg, velocity \(v_1 = 200\) m/s. Mass of gun \(m_2 = 4\) kg, recoil velocity \(v_2 = ?\). Initially both are at rest, so total initial momentum is 0.

By conservation of momentum: \(m_1 v_1 + m_2 v_2 = 0\)

\(v_2 = -\dfrac{m_1 v_1}{m_2} = -\dfrac{0.020 \times 200}{4} = -1\) m/s.

Answer: Gun recoils at 1 m/s in the direction opposite to the bullet.

Fig 6.7: Recoil of a gun — bullet and gun acquire equal and opposite momenta.

Activity 6.3 — Balloon RocketL3 Apply

Predict first: A balloon is inflated and released without tying its mouth. Why does it shoot across the room? Which two of Newton's laws does it illustrate?

Inflate a long balloon. Pinch the mouth shut so the air does not escape.
Tape a drinking straw along the upper side of the balloon.
Pass a long thread through the straw and stretch the thread tight across the room.
Release the balloon's mouth and watch it shoot along the thread.

Observation: The balloon zooms forward as the air rushes backward.

Explanation: The balloon pushes air backward (action) and the air pushes the balloon forward (reaction) — Newton's Third Law. Also, the total momentum of (balloon + air) is conserved: the air gains backward momentum, the balloon gains forward momentum so that the sum is zero. This is the same idea behind rockets and jet engines.

More worked numericals

Example 2 — Momentum of a moving body

Q. Find the momentum of a 2 kg ball moving at 3 m/s.

Solution. \(p = mv = 2 \times 3 = 6\) kg·m/s in the direction of motion.

Example 3 — Force from change in momentum

Q. A cricket ball of mass 150 g moving at 20 m/s is brought to rest by a fielder in 0.1 s. Find the force exerted on the ball.

Solution. \(\Delta p = m(v-u) = 0.15(0-20) = -3\) kg·m/s. \(F = \Delta p/t = -3/0.1 = -30\) N. Magnitude of force = 30 N (opposite to motion).

Example 4 — Two-body collision (one body initially at rest)

Q. A truck of mass 2000 kg moving at 5 m/s collides with a stationary car of mass 1000 kg and they move together. Find their common velocity.

Solution. Conservation of momentum: \(m_1 u_1 + m_2 u_2 = (m_1+m_2)v\).

\(2000 \times 5 + 1000 \times 0 = 3000 \times v\) ⇒ \(v = 10000/3000 \approx 3.33\) m/s.

Example 5 — Why bending knees helps

Q. A 60 kg person jumping from a height lands at 4 m/s. Compare the average force on the knees if (i) the person stops in 0.05 s by landing stiff, (ii) stops in 0.5 s by bending the knees.

Solution. \(\Delta p = 60 \times 4 = 240\) kg·m/s.

(i) \(F = 240/0.05 = 4800\) N. (ii) \(F = 240/0.5 = 480\) N.

Bending the knees reduces the force ten-fold by stretching the stopping time — the same lesson behind air-bags and crumple zones.

Example 6 — Two trolleys recoiling

Q. Two boys of mass 40 kg each stand on a frictionless trolley (total mass 100 kg) at rest. One boy jumps off horizontally at 2 m/s. Find the velocity of the trolley + remaining boy.

Solution. Initial momentum = 0. After jump: jumper momentum = 40 × 2 = 80 kg·m/s. Remaining mass = 60 (trolley) + 40 (boy) = 100 kg. So 100 v + 80 = 0 ⇒ v = −0.8 m/s. The trolley moves at 0.8 m/s opposite to the jumper.

Quick Recap

Concept	Formula / Idea
Newton's Third Law	Forces occur in equal-and-opposite pairs on different bodies.
Momentum	\(p = mv\); unit kg·m/s; vector.
2nd Law (momentum form)	\(F = \Delta p / t\)
Impulse	\(F t = \Delta p\)
Conservation of momentum	\(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\) (isolated system)

Competency-Based Questions

In a friendly experiment, two students stand on identical skateboards facing each other on a smooth horizontal floor. Student A (mass 50 kg) throws a 2 kg medicine ball to Student B (mass 40 kg) at 6 m/s relative to the ground.

Q1. With what velocity does Student A move backward after throwing the ball? L3

Initial momentum = 0. After throw: ball = 2 × 6 = 12 kg·m/s forward. Hence A's momentum = −12. \(v_A = -12/50 = -0.24\) m/s. Student A moves backward at 0.24 m/s.

Q2. With what common velocity do Student B and the ball move after B catches it? L3

Before catch: ball momentum = 12, B at rest. After catch: combined mass = 42 kg. \(v = 12/42 \approx 0.286\) m/s in the original direction.

Q3. Which law explains why Student A moves backward? L2

(a) Law of inertia
(b) Conservation of momentum / Newton's Third Law
(c) Newton's Second Law alone
(d) Law of friction

(b) The push on the ball is matched by an equal and opposite push on Student A; total momentum stays zero.

Q4. State whether True or False: "When you catch a fast cricket ball you instinctively pull your hands back to reduce the force." L1

True. Pulling the hands back lengthens the stopping time \(t\); since \(F = \Delta p/t\), a larger \(t\) means a smaller force on the hands.

Q5. A 1500 kg car changes its velocity from 10 m/s to 25 m/s in 5 s. Find the magnitude of the force acting on it. L3

\(\Delta p = 1500(25-10) = 22500\) kg·m/s. \(F = 22500/5 = 4500\) N.

Assertion–Reason Questions

Options: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.

A: A rocket can move forward in outer space even though there is no air to push against.

R: The rocket exhausts hot gases backward; by Newton's Third Law the gases push the rocket forward.

(A) Both true and R explains A correctly. Rocket motion is a textbook example of the Third Law.

A: The action–reaction forces on a book lying on a table cancel each other out.

R: Action and reaction forces always act on the same body.

(D) Assertion is false (action–reaction forces act on different bodies and so cannot cancel). Reason is also false. Choose (D) only if your option list reads R-true-A-false; here both are false, but standard NCERT answer keys mark such pairs as "Both A and R are false". Final answer: both A and R are false.

A: The total momentum of an isolated system stays constant.

R: An isolated system experiences no net external force.

(A) Both true and R correctly explains A — that is precisely the basis of the conservation law.

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