Given,

Mass of the hammer (m) = 500 g

Convert g into kg

1000 g = 1 kg

500 g = $\dfrac{500}{1000}$ = 0.5 kg

Initial velocity of the hammer (u) = 50 ms^-1

Terminal velocity of the hammer (v) = 0

Time period (t) = 0.01 s

As per the first equation of motion,

a = $\dfrac{\text{v - u}}{\text{t}}$

Substituting we get,

a = $\dfrac{0-50}{0.01}$ = -5000 ms^-2

We know, Force = mass x acceleration

Substituting we get,

F = 0.5 x 5000 = -2500 N

According to the third law of motion, the nail exerts an equal and opposite force on the hammer and as the force exerted on the nail by the hammer is -2500 N, hence, the force exerted on the hammer by the nail will be 2500 N.

BRIGHT TUTORIALS

CBSE Class IX | Academic Year 2026-2027

9403781999

Excellence in Education

Chapter 8: Force and Laws of Motion — Quick Reference

Newton laws inertia momentum F=ma conservation of momentum action reaction

Quick Revision Points

• Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
• First Law (Inertia): object stays at rest/uniform motion unless external force acts
• Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
• 1 N = force giving 1 kg an acceleration of 1 m/s²
• Third Law: every action has equal and opposite reaction (on different bodies)
• Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8

• Read the detailed chapter notes for complete coverage of all NCERT topics.
• Practice all NCERT in-text and back exercise questions — they are frequently asked in exams.
• Focus on comparison tables, diagrams, and definitions — these are high-scoring areas.
• For numericals (if applicable), practice at least 20 problems of varying difficulty.
• Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources

• Detailed Notes: ch08-force-and-laws-of-motion.html
• Practice Questions: 100+ questions with answers in 05-practice-questions/
• Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
• Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALS

CBSE Class IX | Academic Year 2026-2027

9403781999

Excellence in Education

Chapter 7: Motion — Quick Reference

distance displacement speed velocity acceleration equations of motion graphs circular motion

Quick Revision Points

• Distance (scalar, total path) vs Displacement (vector, shortest path)
• Speed = distance/time (scalar); Velocity = displacement/time (vector)
• Acceleration a = (v−u)/t; unit: m/s²
• Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
• d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
• 1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
• Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7

• Read the detailed chapter notes for complete coverage of all NCERT topics.
• Practice all NCERT in-text and back exercise questions — they are frequently asked in exams.
• Focus on comparison tables, diagrams, and definitions — these are high-scoring areas.
• For numericals (if applicable), practice at least 20 problems of varying difficulty.
• Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources

• Detailed Notes: ch07-motion.html
• Practice Questions: 100+ questions with answers in 05-practice-questions/
• Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
• Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html