Class 9th Physics Chapter 3 Exercise Solved Numericals

Physics isn’t just about theories; it is the science that brings clarity to the natural phenomena around us. In Class 9th Physics Chapter 3 Exercise with solved numericals, we dive deep into solving exercise numericals, focusing on essential concepts like force, motion, and momentum.

With step-by-step solutions, these problems help solidify foundational knowledge, making students confident in approaching real-life applications and competitive exams.

Topics Covered in These Numericals

From the uploaded notes, the exercise numericals in Chapter 3 revolve around Newton’s laws, motion dynamics, and momentum conservation. These include problems that enhance understanding of:

1. Force Calculations: Application of F=maF = maF=ma.
2. Tension in Strings: Understanding Atwood’s Machine cases.
3. Momentum and Collisions: Solving equations involving conservation principles.

Class 9th Physics Chapter 3 with solved numericals

Selected Numerical Problems with Solutions

Example 1: Acceleration and Force Calculation

Problem: Find the acceleration produced by a force of 20 N acting on a body of 8 kg.

Solution:

1. Given:
   Force (FFF) = 20 N, Mass (mmm) = 8 kg.
2. Formula: F=maF = maF=ma.
3. Rearranged: a=Fma = \frac{F}{m}a=mF​.
4. Calculation: a=208=2.5 ms−2a = \frac{20}{8} = 2.5 \, \text{ms}^{-2}a=820​=2.5ms−2

Result: Acceleration = 2.5 ms−22.5 \, \text{ms}^{-2}2.5ms−2.

Example 2: Calculating Tension in Strings

Problem: Two masses of 5.2 kg and 4.8 kg are attached to a string over a frictionless pulley. Find acceleration and tension.

Solution:

1. Given:
   m1=5.2 kgm_1 = 5.2 \, \text{kg}m1​=5.2kg, m2=4.8 kgm_2 = 4.8 \, \text{kg}m2​=4.8kg, g=10 ms−2g = 10 \, \text{ms}^{-2}g=10ms−2.
   
2. Formulas:
   Acceleration (aaa): a=(m1−m2)gm1+m2a = \frac{(m_1 – m_2)g}{m_1 + m_2}a=m1​+m2​(m1​−m2​)g​ Tension (TTT): T=2m1m2gm1+m2T = \frac{2m_1m_2g}{m_1 + m_2}T=m1​+m2​2m1​m2​g​
   
3. Calculations:
   Acceleration: a=(5.2−4.8)105.2+4.8=410=0.4 ms−2a = \frac{(5.2 – 4.8)10}{5.2 + 4.8} = \frac{4}{10} = 0.4 \, \text{ms}^{-2}a=5.2+4.8(5.2−4.8)10​=104​=0.4ms−2 Tension: T=2(5.2)(4.8)(10)5.2+4.8=499.210=50 NT = \frac{2(5.2)(4.8)(10)}{5.2 + 4.8} = \frac{499.2}{10} = 50 \, \text{N}T=5.2+4.82(5.2)(4.8)(10)​=10499.2​=50N

Results: Acceleration = 0.4 ms−20.4 \, \text{ms}^{-2}0.4ms−2, Tension = 50 N50 \, \text{N}50N.

Concept Highlights from Examples

1. Acceleration depends inversely on mass when force is constant.
   
2. In Atwood’s Machine, larger mass differences result in higher acceleration.
   
3. Tension remains uniform across the string in frictionless systems.

Why These Problems Are Vital

Understanding these problems lays a strong foundation for future topics in mechanics and beyond. They teach the importance of equations in real-world applications like pulley systems and collision analysis.

Example 3: Combined Motion in Atwood’s Machine

Problem: Two masses, 4 kg and 6 kg, are attached to the ends of a string over a pulley. The 6 kg mass moves horizontally on a smooth surface, while the 4 kg mass moves vertically. Find the system’s acceleration and string tension.

Solution:

1. Given:
   m1=4 kgm_1 = 4 \, \text{kg}m1​=4kg, m2=6 kgm_2 = 6 \, \text{kg}m2​=6kg, g=10 ms−2g = 10 \, \text{ms}^{-2}g=10ms−2.
   
2. Formulas:
   Acceleration (aaa): a=m1gm1+m2a = \frac{m_1g}{m_1 + m_2}a=m1​+m2​m1​g​ Tension (TTT): T=m1m2gm1+m2T = \frac{m_1m_2g}{m_1 + m_2}T=m1​+m2​m1​m2​g​
   
3. Calculations:
   Acceleration: a=(4)(10)4+6=4010=4 ms−2a = \frac{(4)(10)}{4 + 6} = \frac{40}{10} = 4 \, \text{ms}^{-2}a=4+6(4)(10)​=1040​=4ms−2 Tension: T=(4)(6)(10)4+6=24010=24 NT = \frac{(4)(6)(10)}{4 + 6} = \frac{240}{10} = 24 \, \text{N}T=4+6(4)(6)(10)​=10240​=24N

Results: Acceleration = 4 ms−24 \, \text{ms}^{-2}4ms−2, Tension = 24 N24 \, \text{N}24N.

Example 4: Force Required to Stop a Moving Body

Problem: A 5 kg body moves at 10 ms−110 \, \text{ms}^{-1}10ms−1. Calculate the force required to stop it in 2 seconds.

Solution:

1. Given:
   m=5 kgm = 5 \, \text{kg}m=5kg, vi=10 ms−1v_i = 10 \, \text{ms}^{-1}vi​=10ms−1, vf=0 ms−1v_f = 0 \, \text{ms}^{-1}vf​=0ms−1, t=2 st = 2 \, \text{s}t=2s.
   
2. Formula:
   F=m(vf−vi)tF = m \frac{(v_f – v_i)}{t}F=mt(vf​−vi​)​.
   
3. Calculation: F=5×(0−10)2=5×(−5)=−25 NF = 5 \times \frac{(0 – 10)}{2} = 5 \times (-5) = -25 \, \text{N}F=5×2(0−10)​=5×(−5)=−25N

Result: The negative sign indicates the force acts opposite to motion. Force = 25 N25 \, \text{N}25N.

Why Practice Solved Numericals?

1. Clarity of Concepts: These problems reinforce the fundamental principles of motion, force, and tension.
   
2. Exam Preparation: Familiarity with such questions enhances problem-solving efficiency during exams.
   
3. Real-World Applications: The concepts are directly applicable in engineering, mechanics, and everyday physics.

Key Tips for Students

1. Always write given data and equations before solving.
   
2. Check units to avoid calculation errors.
   
3. Understand the physics behind each equation rather than memorizing it.

Chapter 3 Class 9th Notes

1. Class 9th Physics Chapter 3 Solved Exercise ( MCQs, Short & Long Questions)
2. Class 9th Physics Chapter 3 Notes (MCQs, Short & Long Questions)

Conclusion

Class 9th Physics Chapter 3 numericals are a crucial part of mastering dynamics. From calculating forces to understanding string tension, each solved example bridges theoretical knowledge with practical application.

Practicing these problems not only ensures better grades but also cultivates critical thinking skills essential for higher studies.

FAQs

Other Chapter Class 9th Physics Notes

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2. Class 9th Physics Chapter 1 Notes (MCQs, Short & Long Questions)
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4. Class 9th Physics Chapter 1 Solved Exercise ( MCQs, Short & Long Questions)
5. Class 9th Physics Chapter 2 Notes (MCQs, Short & Long Questions)
6. Class 9th Physics Chapter 2 Exercise Solved Numericals