Class 9 — Force and Laws of Motion

Chapter Overview & Weightage

Force and Laws of Motion is one of the foundation chapters in CBSE Class 9 Physics. It introduces Newton’s three laws — the conceptual backbone for ALL of mechanics in Class 11 and beyond. JEE/NEET concepts (momentum conservation, friction, free body diagrams) literally start here.

In CBSE Class 9 board exams, this chapter contributes about 6-8 marks out of 80 total Physics marks. That’s roughly 8-10% — significant, and entirely scoring with consistent practice.

Topics tested: definitions of force types (balanced/unbalanced), Newton’s three laws (statement + example), inertia and mass, momentum, conservation of momentum, simple numericals on $F = ma$ and momentum.

Key Concepts You Must Know

Prioritized by exam frequency:

Newton’s First Law (Inertia) — every object continues at rest or uniform motion unless an unbalanced force acts. Statement + example carries 2-3 marks every year.
Newton’s Second Law — $F = ma$ derivation from $F = dp/dt$ . Numerical applications.
Newton’s Third Law — action and reaction. Often asked as “explain why a swimmer pushes water backward to move forward.”
Conservation of Momentum — derivation from third law + numerical (gun-bullet recoil, collisions).
Inertia of Rest, Motion, and Direction — three types with everyday examples.

Important Formulas

$F = ma = \frac{dp}{dt}$

Use when: any problem giving force and mass, or momentum change.

$p = mv$

Use when: comparing or combining motion of objects with different masses or velocities.

$F \cdot \Delta t = \Delta p = m(v - u)$

Use when: a force acts over a known time and we want velocity change, or vice versa.

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

Use when: collisions, recoil, explosions — anywhere external force is zero or negligible.

Solved Previous Year Questions

PYQ 1 (CBSE Class 9, 2023)

A force of $5 \, \text{N}$ acts on a body of mass $2 \, \text{kg}$ for $4$ seconds. Find the velocity acquired by the body if it starts from rest.

Solution: $a = F/m = 5/2 = 2.5 \, \text{m/s}^2$ . Then $v = u + at = 0 + 2.5 \times 4 = 10 \, \text{m/s}$ .

PYQ 2 (CBSE Class 9, 2022)

A bullet of mass $20 \, \text{g}$ is fired from a gun of mass $4 \, \text{kg}$ with a velocity of $400 \, \text{m/s}$ . Calculate the recoil velocity of the gun.

Solution: Conservation of momentum: $0 = m_b v_b + m_g v_g$ . So $v_g = -m_b v_b / m_g = -(0.02)(400)/4 = -2 \, \text{m/s}$ . Recoil velocity is $2 \, \text{m/s}$ in the opposite direction.

PYQ 3 (CBSE Class 9, 2021)

State Newton’s second law of motion. Derive $F = ma$ from it.

Solution: Statement: rate of change of momentum is directly proportional to applied force, and acts in the direction of force. Derivation: $F \propto dp/dt = d(mv)/dt = m(dv/dt) = ma$ (for constant mass). Choosing units to make the proportionality constant 1 gives $F = ma$ .

Difficulty Distribution

For this chapter in CBSE Class 9 boards:

Easy (definitions, statements): ~3 marks
Medium (numerical on $F = ma$ , momentum): ~3 marks
Hard (multi-step momentum conservation): ~2 marks

Almost all questions are 1-mark MCQs, 2-mark short answers, or 3-mark numericals. No 5-mark long answer typically.

Expert Strategy

Strategy 1: Master statements word-perfect. CBSE awards full marks only for textbook-exact statements of Newton’s laws. Sloppy paraphrasing loses 1 mark.

Strategy 2: Always show units. A numerical answer without units gets half the marks. Write " $10 \, \text{m/s}$ " not just ” $10$ .”

Strategy 3: Diagram every force problem. Even simple ones. CBSE markers explicitly look for FBDs in 3+ mark questions and award 1 mark for a correct diagram.

Strategy 4: For momentum questions, write the conservation equation clearly with sign convention. Lose marks for sign confusion in recoil/collision problems.

Common Traps

Trap 1: Confusing mass with weight. Mass is constant; weight is $mg$ and changes with location. CBSE often asks “what is the difference?” as a 2-mark question.

Trap 2: Forgetting to convert grams to kilograms. A bullet mass given as " $20 \, \text{g}$ " must be converted to $0.02 \, \text{kg}$ before plugging into $F = ma$ .

Trap 3: Mistaking the third law as “equal action and reaction on the same body.” The pair acts on DIFFERENT bodies — that’s why they don’t cancel.

Trap 4: Treating “constant velocity” as zero force. Constant velocity means zero NET force — there can still be balanced forces present.

In CBSE 2024, a tricky question asked: “A horse pulls a cart. By Newton’s third law, the cart pulls the horse with equal force. So how does the cart move?” The answer requires explaining that the horse-cart pair experiences external force from the ground (friction), which moves them together. Master this conceptual question — it appears in some form every alternate year.