Given,

Mass of the object (m₁) = 1 kg

Mass of the block (m₂) = 5 kg

Initial velocity of the object (u₁) = 10 ms^-1

Initial velocity of the block (u₂) = 0

Mass of the resulting object = m₁ + m₂ = 6 kg

Velocity of the resulting object (v) = ?

Total momentum before the collision = m₁u₁ + m₂u₂ = (1 × 10) + 0 = 10 kgms^-1

According to the law of conservation of momentum,

total momentum before the collision = total momentum after the collision.

∴ the total momentum post the collision is also 10 kgms^-1

Hence, total momentum just before the impact = 10 kgms^-1 and just after the impact = 10 kgms^-1

Now, (m₁ + m₂) × v = 10 kgms^-1

6 x v = 10

v = $\dfrac{10}{6}$ = $\dfrac{5}{3}$ ms^-1

Hence, velocity of the combined object = $\dfrac{5}{3}$ ms^-1

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