CBSE Weightage:

CBSE Class 9 Science — Force and Laws of Motion

CBSE Class 9 Science — Force and Laws of Motion — chapter overview, key concepts, solved examples, and exam strategy.

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Chapter Overview & Weightage

Force and Laws of Motion is Chapter 9 in CBSE Class 9 Science. It’s one of the foundational chapters for physics and carries approximately 8–10 marks in the annual exam. Questions range from one-mark definitions to 5-mark numerical and explanation questions.

Newton’s three laws are tested in every year’s Class 9 exam. Expect: (1) one statement question about a specific law, (2) one numerical on F=maF = ma, and (3) one conservation of momentum problem. These alone account for 6–8 marks reliably.

TopicExpected marks
Newton’s First Law (inertia)1–2 marks
Newton’s Second Law (F=maF = ma)2–3 marks
Newton’s Third Law (action-reaction)1–2 marks
Conservation of momentum2–3 marks

Key Concepts You Must Know

  • Force: A push or pull on an object that can change its state of rest or motion. SI unit: Newton (N).
  • Balanced forces: Forces that cancel out — no change in motion.
  • Unbalanced forces: Net force ≠ 0 — causes acceleration.
  • Inertia: The tendency of an object to resist changes to its state of motion. Greater mass = greater inertia.
  • Momentum (p): p=mvp = mv. A vector quantity. SI unit: kg·m/s.
  • Newton’s First Law: An object at rest stays at rest, and an object in motion stays in motion with the same velocity, unless acted upon by an unbalanced external force.
  • Newton’s Second Law: F=maF = ma. The rate of change of momentum is proportional to the applied force.
  • Newton’s Third Law: For every action, there is an equal and opposite reaction. Forces always act in pairs on different objects.
  • Law of Conservation of Momentum: Total momentum before collision = Total momentum after collision (when no external forces act).

Important Formulas

F=ma=ΔpΔt=mvmutF = ma = \frac{\Delta p}{\Delta t} = \frac{mv - mu}{t}

FF in Newton (N), mm in kg, aa in m/s²

1 Newton = force that gives 1 kg mass an acceleration of 1 m/s²

m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

Before collision (left side) = After collision (right side).

For a stationary second object (u2=0u_2 = 0): m1u1=m1v1+m2v2m_1 u_1 = m_1 v_1 + m_2 v_2

Solved Previous Year Questions

PYQ 1 — First Law and inertia (2 marks)

Q: Give one example each of inertia of rest and inertia of motion.

Answer:

Inertia of rest: When a bus suddenly starts moving, passengers are thrown backward. Their bodies were at rest and tend to stay at rest.

Inertia of motion: When a moving bus brakes suddenly, passengers lean forward. Their bodies were in motion and tend to continue moving forward.

PYQ 2 — Numerical: F = ma (3 marks)

Q: A force of 20 N acts on a body of mass 4 kg for 5 seconds. If the body starts from rest, find: (a) acceleration, (b) velocity after 5 seconds, (c) distance covered.

Solution:

(a) a=F/m=20/4=5 m/s2a = F/m = 20/4 = 5 \text{ m/s}^2

(b) Using v=u+at=0+5×5=25 m/sv = u + at = 0 + 5 \times 5 = 25 \text{ m/s}

(c) Using s=ut+12at2=0+12×5×25=62.5 ms = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 5 \times 25 = 62.5 \text{ m}

PYQ 3 — Conservation of momentum (4 marks)

Q: A rifle of mass 4 kg fires a bullet of mass 50 g with a velocity of 200 m/s. Find the recoil velocity of the rifle.

Solution:

Initial momentum of system = 0 (both at rest).

By conservation of momentum: m1v1+m2v2=0m_1 v_1 + m_2 v_2 = 0

Let rifle recoil velocity = vv (backward).

0.05×200+4×v=00.05 \times 200 + 4 \times v = 0

10+4v=010 + 4v = 0

v=2.5 m/sv = -2.5 \text{ m/s}

The rifle recoils at 2.5 m/s in the direction opposite to the bullet.

Difficulty Distribution

LevelMarksType
Easy1 markStatement of law, definition
Medium2–3 marksExplanation with example, basic numerical
Hard4–5 marksMulti-step numerical, conservation of momentum

Expert Strategy

For 5-mark questions on conservation of momentum, always write: (1) state the law, (2) write the equation m1u1+m2u2=m1v1+m2v2m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2, (3) substitute values, (4) solve and state the direction. Students who write all four parts get full marks even if there’s an arithmetic error.

For Newton’s Second Law numericals, always identify what’s given and what’s unknown before picking the right kinematic equation.

Know the real-life examples of each law cold — examiners use them in 1-mark fill-in-the-blank and 2-mark short answer questions:

  • First Law: passengers lurching in a bus, karate board break (inertia broken by brief impulse)
  • Third Law: rocket propulsion, swimming (push water back → move forward), recoil of a gun
  • Conservation: explosion, collision, gun-bullet system

Common Traps

Trap 1: Third Law confusion. “Action and reaction are equal and opposite — so they cancel.” They DON’T cancel because they act on different objects. A horse pulls a cart (action on cart), the cart pulls the horse back (reaction on horse). Two different objects, so they don’t cancel each other.

Trap 2: Units. Momentum is in kg·m/s, not kg·m/s². The extra per-second comes only when you’re talking about force (F=dp/dtF = dp/dt).

Trap 3: Forgetting direction in momentum problems. Momentum is a vector. Taking one direction as positive and applying signs consistently is essential. Missing the sign gives the wrong recoil direction.

Trap 4: Newton’s first law applies only when net force = 0. If asked “what force is needed to keep an object moving at constant velocity on a frictionless surface?” — the answer is zero, not any specific value. On a surface with friction, you need a force equal to friction to maintain constant velocity.