Given,

Mass of the bullet (m) = 10 g

Convert g into kg

1000 g = 1 kg

So, 10 g = $\dfrac{10}{1000}$ = 0.01 kg

Initial velocity of the bullet (u) = 150 ms^-1

Terminal velocity of the bullet (v) = 0

Time period (t) = 0.03 s

To find the distance of penetration, the acceleration of the bullet must be calculated

Let the distance of penetration be s

According to the first equation of motion,

v = u + at

or

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

Substituting we get,

a = $\dfrac{0-150}{0.03} = \dfrac{150}{0.03} =\dfrac{-15000}{3}$ = -5000 ms^-2

∴ Acceleration = -5000 ms^-2.

According to the third equation of motion,

2as = v² - u²

or

a = $\dfrac{\text{v}^2 - \text{u}^2 }{2\text{s}}$

Substituting we get,

s = $\dfrac{0 - (150)^2}{2 \times -5000} = \dfrac{-22500}{-10000} =$ = 2.25 m.

Hence, the total distance travelled is 2.25 m

Now, as Force = Mass x Acceleration

Substituting we get,

F = 0.01 × -5000 = -50 N

Negative sign indicates direction of recoil of bullet is opposite to the direction of bullet.

Hence, magnitude of the force = 50 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