12
Question Question 12
A hockey ball of mass 200 g travelling at 10 ms-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms-1. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Given,
Mass (m) = 200 g
Initial velocity (u) = 10 ms-1
Final velocity (v) = -5 ms-1
Initial momentum of the ball = mu = 200 × 10 = 2000 gms-1
Final momentum of the ball = mv = 200 × -5 = -1000 gms-1
∴ The change in momentum (mv - mu) = -1000 - 2000 = -3000 gms-1
Convert gms-1 to kgms-1 :
1000 gms-1 = 1 kgms-1
So, 3000 gms-1 = = -3 kgms-1
Hence, momentum of the ball reduces by 3 kgms-1 after being struck by the hockey stick.
BRIGHT TUTORIALS
BRIGHT TUTORIALS
CBSE Class IX | Academic Year 2026-2027
9403781999
Excellence in Education
Science | Chapter 8: Force and Laws of MotionWeb Content — Quick Reference
Chapter 8: Force and Laws of Motion — Quick Reference
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
BRIGHT TUTORIALS
CBSE Class IX | Academic Year 2026-2027
9403781999
Excellence in Education
Science | Chapter 7: MotionWeb Content — Quick Reference
Chapter 7: Motion — Quick Reference
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