Force and Laws of Motion Class 9th Notes - Free NCERT Class 9 Science Chapter 9 Notes - Download PDF

Force and Laws of Motion Class 9th Notes - Free NCERT Class 9 Science Chapter 9 Notes - Download PDF

Edited By Vishal kumar | Updated on Mar 18, 2024 09:11 AM IST

Force And Laws Of Motion Class 9 CBSE Notes: Chapter 9- Download Free PDF

Force and Laws of Motion is a critical chapter in the Class 9 Physics curriculum, providing the fundamental principles required to understand motion. Careers360's Force and Laws of Motion class 9 notes are written by highly qualified teachers and provide in-depth explanations of key concepts. All students can freely access these ch 9 physics class 9 notes, which are conveniently available in PDF format on the careers360 platform. Force and laws of motion are recognised as fundamental principles in Physics and play an important role in examinations.

This Story also Contains
  1. Force And Laws Of Motion Class 9 CBSE Notes: Chapter 9- Download Free PDF
  2. NCERT class 9 science chapter 9 notes
  3. The first law of motion
  4. The second law of motion
  5. Third law of motion
  6. Conservation of momentum
  7. Significance of NCERT notes for class 9 science chapter 9
  8. NCERT Solutions Of Class 9 Subject-Wise
  9. NCERT Class 9 Exemplar Solutions for Other Subjects:

Students can strengthen their understanding of these concepts by diligently reviewing the class 9 physics chapter 9 notes. careers360 also offers free NCERT Solutions to help students prepare for exams thoroughly. Force is a measurable physical quantity that includes both magnitude and direction. It manifests as a push or pull on an object. Its application can cause changes in the state of motion or rest, alter velocity and direction, and even change the shape of an object. This chapter is a cornerstone for understanding the complex dynamics of motion and lays a solid foundation for future research in Physics.

Also, students can refer,

NCERT class 9 science chapter 9 notes

Force is a push or pull on an object, it can change the velocity of an object, change its direction of motion or change the size or shape of the object.

Balanced and unbalanced forces

When the magnitude of the net force applied to an object is zero, there will be no change in the object’s state of motion, such forces on the object are termed balanced forces. When multiple forces are applied to an object, it starts to move in the direction of the force which is greater in magnitude, these are unbalanced forces, and unbalanced forces cause a change in the state of motion of the object.

To bring a change in the speed or direction of an object, the action of the unbalanced force is required.

The first law of motion

Galileo observed the motion of objects on an inclined plane and concluded that no net force is needed to sustain an object’s uniform motion.

The first law of motion states that until driven to change by an applied force, an object will stay in a condition of rest or uniform motion in a straight path.

The first law of motion states that when an object is subjected to an unbalanced external force, its velocity changes, or the object accelerates. Newton's first law of motion, often known as the law of inertia, emphasises objects' ability to resist change in velocity.

Inertia and mass

Inertia is the property of an object to resist the change in its state of motion, if it’s in motion, it tends to be in motion, if it’s in rest, it’ll tend to remain at rest. An object with greater mass will have more inertia, in other words; mass is a measure of inertia.

The second law of motion

Momentum (p) is the product of an object’s mass (m) and its velocity (v)

p=mv

The direction of momentum is the same as that of velocity, and its SI unit is kg m s-1

Force applied on an object is directly proportional to the change in momentum it caused and, it is inversely proportional to the time taken for this change to take place.

The second law of motion states that an object's rate of change of momentum is proportional to the applied unbalanced force in the force's direction.

The mathematical formulation of the second law of motion

If a force of 'F' is applied to a mass of ‘m' with an initial velocity of 'u'. The force produces a change in its velocity in a time 't,' thus the final velocity is 'v.'

Initial momentum = p1 = mu

Final momentum = p2 = mv

Change in momentum ∝ p2 – p1

mv- mu

∝ m(v-u)

Rate of change of momentum m(v-u)/t

Therefore, F = m(v-u)/t

F = km(v-u)/t

F = ma

The value of k is chosen to be 1, a is the acceleration of the object.

SI unit of force is N (Newton) or kg m s-2

When no force is applied to the object, F=0

F= ma

F= m(v-u)/t

Ft = m(v-u)

Since, F=0, therefore,

v=u

Initial velocity equals final velocity, this is in accordance with Newton’s first law of motion.

Third law of motion

When one item puts a force on another, the second object reacts by exerting a force on the first, according to the third law of motion. These two forces have the same magnitude however operate in opposing directions and on separate objects.

Conservation of momentum

Law of conservation of momentum: The total momentum of two objects is conserved even after the collision.

If there are two objects ‘A’ and ‘B’, let their respective masses and velocities be mA , uA and mB , uB.

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Let uA > uB , so these objects collide with each other.

The collision will last for a time ‘t’, the force exerted by A on B will be FAB , and B exerts a force FBA on A.

Let their final velocities be vA and vB.

Momentum of A before collision = mAuA

Momentum of A before collision = mAvA

Momentum of B before collision = mBuB

Momentum of B before collision = mBvB

FAB = rate of change of momentum of A= mA(vA-uA)/t

FBA = rate of change of momentum of B= mB(vB-uB)/t

According to Newton’s third law, FAB = -FBA

mA(vA-uA)/t = - mB(vB-uB)/t

Simplifying the above equation,

mAuA + mBuB= mAvA +mBvB

Initial sum of momentum = Final sum of momentum

Significance of NCERT notes for class 9 science chapter 9

  • Revision Assistance: Force And Laws Of Motion notes class 9 help students revise the chapter's content and ensure a thorough understanding of the key concepts covered.
  • Simplicity: These CBSE class 9 physics ch 9 notes are intended to be easy to understand, with step-by-step explanations to help students grasp the concepts without difficulty.
  • CBSE Syllabus Coverage: These ch 9 physics class 9 notes are consistent with the CBSE Science syllabus for class 9, ensuring that students cover all relevant topics and concepts.
  • Accessibility: The availability of physics class 9 chapter 9 notes PDF format allows students to access and use them for offline preparation, allowing for greater flexibility in study schedules.
  • Clarity of Concepts: The Force And Laws Of Motion class 9 notes provide clear explanations, allowing students to effectively grasp the principles of laws of motion.
  • Preparation Tool: Using these class 9 physics chapter 9 notes helps students prepare for exams, allowing them to consolidate their knowledge and perform well in assessments.

NCERT Solutions Of Class 9 Subject-Wise

NCERT Class 9 Exemplar Solutions for Other Subjects:

Class 10 Chapter Wise Notes

Frequently Asked Questions (FAQs)

1. In class 9 chapter 9 what is the SI unit of force?

Newton (N) or kg m s^{-2}

2. Define inertia?

Inertia is the property of an object to resist the change in its state of motion, if it’s in motion, it tends to be in motion, if it’s in rest, it’ll tend to remain at rest.

3. What is a balanced force as in NCERT chapter 9 class 9?

When the magnitude of the net force applied on an object is zero, there will be no change in the object’s state of motion, such forces on the object are termed as balanced forces.

4. What is momentum?

According to class 9th science chapter 9 notes Momentum (p) is the product of an object’s mass (m) and its velocity (v)

p=mv. The direction of momentum is the same as that of velocity, and its SI unit is kg m/s

5. State the law of conservation of momentum mentioned in cbse class 9 science chapter 9 notes?

The total momentum of the two objects is unchanged or conserved by the collision. These topics can also be downloaded from Class 9 Science chapter 9 notes pdf download.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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