# NCERT Solutions Class 9 Science Chapter 9

## NCERT Solutions for Class 9 Science Chapter 9: Forces and Laws of Motion

Science is an interesting subject, and it help students to dwell into the world of knowing and learning new things. This subject also opens numerous opportunities for students with a particular inclination toward Science in a myriad of ways. However, the difficulty level keeps increasing class by class. Therefore, students of class 9 will have to upgrade their learning skills and focus on scoring better in this subject. Class 9 is also an important class for many reasons as it prepares them for the upcoming board classes.

Class 9 Science NCERT Solutions Chapter 9: Forces and Laws of Motion introduces the correlations between the forces acting on a body. Students will get to learn about constant velocity and static equilibrium. It is crucial to understand the basic concepts related to the chapter, such as the structure of the atom and gravitation. Students will get a gist of all the previous chapters and gain knowledge of Newton's theory. It also discusses essential facts to help students throughout the NCERT Solutions for Class 9 Science Chapter 9.

NCERT Solutions for Class 9 Science Chapter 9 are available for students. The solutions are helpful for the students as they help them score high marks in the examination. It contains detailed step-by-step explanations of all the problems under chapter 9. The Class 9 Science Chapter 9 NCERT Solutions are developed keeping in mind the concept-based approach; along with this, they have a precise answering method for examinations. NCERT Solutions for Class 9 Science Chapter 9 is a detailed and well-structured concept-based learning solution.

Extramarks is one of the leading online learning platforms in India. Lakhs of students have faith in our study materials and teaching methods. Students pursuing primary and secondary education can start with a trial account on our website and can explore other study materials and revision notes. NCERT Solutions for Class 9 Science Chapter 9 is also available for the students on our website.

Students can visit our website for the latest updates and news related to NCERT. Further, students can also refer to other class solutions such as NCERT solutions class 10, NCERT solutions class 11, and NCERT solutions class 12.

## Key Topics Covered in NCERT Solutions for Class 9 Science Chapter 9

NCERT solutions class 9 Science chapter 9 explains motion and causes of motion . Further, students will learn 1st laws of motion, 2nd law of motion and 3rd law of motion. It is presented in a detailed manner with pictorial presentation and mathematical formulations. The solutions will be helpful for CBSE term-wise exams, Science Olympiads, and other competitive exams.

Some of the key topics featured in NCERT Solutions for Class 9 Science Chapter 9 are

### Balanced and Unbalanced Forces

Forces are all around us. Everything involves a force, from lifting a pen to kicking a ball. For a force to be applied, you need to shift, rotate, turn, move, stop, catch, or drag, as well as turn, stop, catch, drag, or drop.  Force is that causes a stationary object or object to move. It can also cause an object moving to slow down, stop, or change its motion direction.

There are two types of forces explained in NCERT Solutions for Class 9 Science Chapter 9, these are:

1. Balanced forces are created when two opposing forces work together on a surface to create balance. Balanced Force describes forces that act on a body in opposite directions but are equal in magnitude. A balanced force, in simple terms, is when two equal forces work on a person in the opposite direction.
2. Unbalanced forces: When two forces act in opposing directions on a surface that is not equal in size or magnitude, these forces are called unbalanced forces. Unbalanced forces are applied to a body and cause a change in its state of motion. The force's magnitude is more significant—a body's state of motion changes when an unbalanced force is applied.

### Newton’s First Law of Motion

Newton's First Law governs motion. A body doesn't change its state of rest, uniform motion or motion, unless an unbalanced force acts on it. Therefore, the body will continue moving in the same direction or state before a force intervenes. It means that the body will remain in the same state of rest and uniform motion along a straight line unless it is forced to do so by an applied unbalanced force. Students can refer to NCERT Solutions for Class 9 Science Chapter 9 to understand the formation of the Law of Motion and its formula.

### Inertia and Mass

Inertia refers to any property resistant to change in its current state of motion or rest. In other words, inertia refers to the property inherent in a body that attempts to maintain its current form. For example, a body at rest will remain at rest unless disturbed by an external force.

The greater the inertia, the larger the object's mass, and vice versa. A greater object's mass will result in a greater force required to alter its state. A heavier thing will resist changes in its state of motion or rest.

### Newton’s Second Law of motion

Newton’s Second Law of Motion states that  the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force.

Newton's  Second law is also called the Law of momentum.

To understand how the Second Law of motion acts on various objects, students can refer to NCERT Solutions for Class 9 Science Chapter 9.

### Newton’s Third Law of motion

This law states that when two bodies interact, their forces are equal in magnitude but opposite in direction. Newton's First Law of Motion defines force. Newton's Second Law of Motion measures force. Newton's Third Law of Motion links forces between two related bodies.

Newton's Third Law of Motion explains the nature of force and what happens when one body exerts a force on another. It also reveals the directions of forces during the interaction between two bodies. What will be the relative magnitude of these forces? To find the answer to these questions, students can refer to NCERT Solutions for Class 9 Science Chapter 9.

### Conservation of Momentum

Conservation of Momentum is when two bodies interact without an external force. The interaction does not affect the total momentum of the bodies involved. Two balls, for example, are moving towards one another and collide. An external force did not cause the collision of the balls; they collided as they moved toward each other.

## NCERT Solutions for Class 9 Science Chapter 9: Exercise & Answer Solutions

NCERT Solutions for Class 9 Science Chapter 9, exercises, and answer solutions are available on the Extramarks website. It will help students understand the concepts quickly and help them score excellently on their exams. In addition, it helps them attain a clear understanding of Science and strengthens their subject knowledge. Students can clear their conceptual doubts by referring to our NCERT Solutions for Class 9 Science Chapter 9.

Chapter 9 is about Force, Types of Forces, and Laws of motion proposed by Sir Isaac Newton. Students will learn inertia, mass,  Laws of motion, and mathematical formulation. Furthermore, the concepts are explained in a simple language combined with diagrams, activities involved and an explanation of the numerical problems. Our subject matter experts have efficiently prepared the solutions. It helps students  build a strong foundation. Students can register on our website to get a trial account.

Students can click on the links below for specific questions and solutions:

• Chapter 9: Exercise 9.1 Solutions 4 Questions
• Chapter 9: Exercise 9.2 Solutions 4 Questions
• Chapter 9: Exercise 9.3 Solutions 18 Questions

Along with this, students can also refer to other solutions for primary and secondary classes:

• NCERT Solutions Class 1
• NCERT Solutions Class 2
• NCERT Solutions Class 3
• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions Class 9
• NCERT Solutions Class 10
• NCERT Solutions Class 11
• NCERT Solutions Class 12

## NCERT Exemplar for Class 9 Science:

Science is a thought-provoking subject. Most students share a curiosity about this subject, but students face difficulty in remembering concepts and complex theories. Therefore, tackling this is beneficial for them to solve different questions at a higher difficulty level. In addition, it helps them  remember the key concepts and definitions.

The NCERT Exemplar offers different types of questions such as multiple-choice questions, fill in the blanks, match the pair and true or false. The questions have a higher level of difficulty, which helps students get a gist of the topic and understand how to approach complex types of questions. The example is beneficial for every student in their exams. Each question asked in CBSE annual exams is wholly based on NCERT books. Therefore, students must have a good knowledge of the NCERT Exemplar. It helps crack the annual exam and clear the toughest competitive exams.

Few questions are taken from NCERT Solutions for Class 9 Science Chapter 9. However, it helps students understand how to calculate the inertia depending on the object's mass.

## Key Features of NCERT Solutions for Class 9 Science Chapter 9:

NCERT Solutions for Class 9 Science Chapter 9 is an excellent way for students to have a firm grip on the chapter's topics. The solutions help to build concepts and guidelines for students to excel in exams. Students who are referred to our Extramarks NCERT solutions are at an advantage. They get to increase their knowledge and improve overall scores in the exams.

The key concepts from NCERT Solutions for Class 9 Science Chapter 9 are listed here:

• The solutions help one understand all the related concepts in Science that they may have been taught to date at the school level.
• NCERT Solutions for Class 9 Science Chapter 9 are presented clearly and concisely, and every topic is explained in detail in the NCERT textbooks used by the author.
• The solutions also help improve students' overall performance by assisting them in understanding all the concepts required for analysis and practice.
• Students will be able to understand core concepts and increase their knowledge.
• Since the solutions are revised and updated, every chapter is explained in the light of current topics and the main concepts.
• The best part about the Extramarks NCERT Solutions for Class 9 Science Chapter 9 is that they provide a perfect learning platform for students. The student doesn't have to go through lengthy calculations and try to find out the answers.

Q.1 An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

Ans-

Yes. It is possible that an object is moving with a non-zero velocity even when the object experiences a net zero external unbalanced force. It is possible only when the object has been moving with a constant velocity in a certain direction. In order to change the state of motion, a net non-zero external unbalanced force must be applied on the object.

Q.2 When a carpet is beaten with a stick, dust comes out of it. Explain.

Ans-

Due to inertia, an object opposes any change in its state of rest or state of motion. As soon as the carpet is beaten with a stick, the carpet comes into motion while the dust particles oppose their state of rest. As per Newton’s first law of motion, the dust particles remain stationary, while the carpet moves. Therefore, the dust particles come out of the carpet.

Q.3 Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Ans-

When the bus starts accelerating in forward direction, it comes in the state of motion but the luggage kept on the roof remains in the state of rest. Therefore, with the forward movement of the bus, the luggage remains at its original position and finally falls from the roof of the bus. Hence, it is advised to tie any luggage kept on the roof of a bus.

Q.4 A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.

Ans-

The correct option is (c)

Explanation: Here, the ball comes to rest due to frictional force which opposes its motion. Frictional force always acts in the direction opposite to the direction of motion. Therefore, this force tries to stop the ball.

Q.5 A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)

Ans-

$Given, intial velocity of the truck, u = 0 distance travelled, s = 400 m time taken, t = 20 s acceleration, a = ? On using the relation s = ut + 1 2 at 2 400 m = 0(20 s) + 1 2 a(20 s) 2 or, a = 2 ms -2 Now, 1 metric tonne = 1000 kg ∴ 7 metric tonnes = 7×1000 kg = 7000 kg By using Newton’s second law, F = ma = (7000 kg) (2 ms -2 ) = 14000 N Thus, the acceleration of the truck is 2 ms -2 and the force acting on it is 14000 N.$

Q.6 A stone of 1 kg is thrown with a velocity of 20 ms−1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

Ans-

$Given, u = 20 ms -1 v = 0 s = 50 m mass of the stone, m = 1 kg On using the third equation of motion, v 2 = u 2 + 2as (a = acceleration of the stone) ∴ 0 2 = (20 ms -1 ) 2 +2×a×50 m or, a = – 400 m 2 s −2 100 m = – 4 ms -2 Here, negative sign shows that acceleration is acting against the motion of the stone. By using Newton’s second law of motion, Force, F = ma or, = 1 kg ×(-4 ms -2 ) = -4 N Therefore, the force of friction between the stone and the ice is – 4 N.$

Q.7 A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:

(a) the net accelerating force and

(b) the acceleration of the train.

$\begin{array}{l}\left(a\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}Given,}\text{force exerted by the engine, F = 40000 N}\\ \text{frictional force due to tracks, F’ = 5000 N}\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{\hspace{0.17em}}\therefore {\text{Net force, F}}_{\text{net}}=40000-5000\text{= 35000 N}\text{}\\ \text{(b)}\text{\hspace{0.17em}}\text{Net force on the wagons},\text{}{F}_{a}=\text{35}000\text{N}\\ \text{Mass of the 5 wagons},\text{}m=\text{2}000×\text{5}=\text{1}0000\text{kg}\\ \text{Mass of the engine},\text{}m\prime =\text{8}000\text{kg}\\ \therefore \text{Total mass},\text{M}=m+m\prime =\text{18}000\text{kg}\\ \text{Using Newton}’\text{s second law of motion,}\\ {\text{F}}_{a}=Ma\\ \text{or, a}\text{=}\frac{{\text{F}}_{a}}{M}\text{=}\frac{35000}{18000}=\text{1}{\text{.944 ms}}^{\text{-2}}\\ \text{Hence, the acceleration of the wagons and the train is}\\ 1.944{\text{ms}}^{\text{-2}}.\\ \end{array}$

Q.8 An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms−2?

Ans-

$Given, m = 1500 kg v = 0 a = -1 .7 ms -2 On using Newton’s second law of motion, Force, F = m × a =1500 kg × ( -1 .7 ms -2 )= -2550 N Therefore, the force between the automobile and the road is 2550 N, in the direction opposite to the motion of the automobile.$

Q.9 What is the momentum of an object of mass m, moving with a velocity v?

1. (mv)2
2. mv2
3. ½ mv2
4. mv

Ans-

The correct option is (d).
Explanation:

$\begin{array}{l}\text{Given, mass of the object = m}\\ \text{velocity of the object = v}\\ \text{Momentum, P = mass}×\text{velocity}\\ \text{= mv}\end{array}$

Q.10 Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Ans-

When the force of 200 N is applied on wooden cabinet, an equal and opposite force acts on it. This opposite force is the frictional force. Therefore, a frictional force of 200 N is exerted on the cabinet.

Q.11 Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?

Ans-

$Given, masses of the objects, m 1 =m 2 = 1.5 kg velocities of the objects, v 1 = 2 .5 ms -1 v 2 =- 2 .5 ms -1 Here, velocity of m 2 is considered negative as it is moving in opposite direction before collision. Both the objects stick together after collision. Hence, total mass M = m 1 +m 2 = 1.5 kg + 1.5 kg = 3 kg Let, velocity of the combined mass after collision = V As per conservation law of linear momentum, m 1 v 1 + m 2 v 2 = MV ∴ (1.5 kg × 2 .5 ms -1 )+(1.5 kg × -2 .5 ms -1 ) = 3V or, V = 0 Thus, the velocity of the combined object after collision will be zero.$

Q.12 According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

Ans-

The static friction acting between the truck and the road is very large as the truck is very massive. Thus, truck does not move when someone pushes it. Therefore, it can be concluded that the applied force in one direction is cancelled out by the frictional force of equal amount acting in the opposite direction.

Q.13 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.

Ans-

$Given, mass of the hockey ball, m = 200 g = 0.2 kg velocity of the ball, v 1 = 10 ms -1 Initial momentum, P 1 = mv 1 Now, hockey ball travels in the opposite direction with velocity, v 2 = – 5 ms -1 Final momentum, P 2 = mv 2 ∴ Change in momentum, ΔP = P 1 -P 2 = m[ v 1 – v 2 ] = 0.2 kg[ 10 ms -1 – (-5 ms -1 ) ] = 3 kg ms -1 . The change in momentum comes out to be 3 kgms -1 and the negative sign indicates that the ball started moving in opposite direction.$

Q.14 A bullet of mass 10 g travelling horizontally with a velocity of 150 ms−1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Ans-

$Given, inital velocity of the bullet, u = 150 ms -1 final velocity of the bullet, v = 0 (as the bullet comes to rest) time taken to come to rest for bullet, t = 0.03 s mass of the bullet, m = 0.01 kg On using the first equation of motion, v = u +at or, 0 = 150 + a×0.03 or, a = -5000 ms -2 (Here, negative sign shows that the velocity of the bullet is decreasing) On using the third equation of motion, v 2 = u 2 +2as or, 0 2 = (150) 2 +2×(−5000)×s or, s = 22500 10000 =2.25 m Hence, the distance of penentration of the bullet into the block is 2.25 m. On using Newton’s second law of motion, Force exerted by the wooden block on the bullet, F = ma = 0.01×5000=50 N. 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Q.15 An object of mass 1 kg travelling in a straight line with a velocity of 10 ms-1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Ans-

$\begin{array}{l}\mathrm{Given},\text{mass of the object, m = 1 kg}\\ \text{mass of the wooden block, M = 5 kg}\\ \mathrm{velocity}{\text{of the object before collision, u}}_{\text{1}}{\text{= 10 ms}}^{\text{-1}}\\ \mathrm{velocity}{\text{of the wooden block befor collision, u}}_{\text{2}}\text{= 0}\\ \\ \mathrm{Total}\text{momentum before collision,}\\ {\text{P}}_{\text{inital}}={\text{mu}}_{\text{1}}{\text{+Mu}}_{\text{2}}\text{= 1}×\text{10}+\text{5}×0=10{\text{kgms}}^{\text{-1}}\\ \mathrm{After}\text{collision, both the blocks stick together, hence}\\ \mathrm{the}\text{the combined mass = M+m = 5+1 = 6 kg}\\ \text{Let, the velocity of the combined mass is V}\\ \text{Then, as per conservation law of momentum,}\\ {\text{P}}_{\text{inital}}{\text{= P}}_{\text{final}}\\ \text{10 = 6}×\text{V}\\ \text{or V =}\frac{10}{6}{\text{= 1.67 ms}}^{\text{-1}}\\ \text{Hence, the velocity of the combined mass after}\\ {\text{collision would be 1.67 ms}}^{\text{-1}}\text{.}\\ {\text{Now, final momentum, P}}_{\text{final}}\text{= (M+m)V =6}×\text{1.67}\\ {\text{=10 kgms}}^{\text{-1}}\text{.}\end{array}$

Q.16 An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms-1 to 8 ms-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Ans-

$\begin{array}{l}\mathrm{Given},{\text{inital velocity, u = 5 ms}}^{\text{-1}}\\ {\text{final velocity, v = 8 ms}}^{\text{-1}}\\ \text{mass , m = 100 kg}\\ \text{time taken by the object to accelerate, t = 6 s}\\ \\ \therefore \text{Intial momentum = mu = 100}×{\text{5 = 500 kgms}}^{\text{-1}}\\ \text{Final momentum = mv = 100}×8{\text{= 800 kgms}}^{\text{-1}}\\ \mathrm{Force}\text{exerted on the object, F =}\frac{\mathrm{mv}-\mathrm{mu}}{\mathrm{t}}\\ \text{=}\frac{800-500}{6}\text{=}\frac{300}{6}\text{= 50 N.}\end{array}$

Q.17 Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.

Ans-

As per the conservation of momentum,

(momentum of motorcar + momentum of insect)before

= (momentum of motorcar + momentum of insect)after

∴ Change in in momentum = 0

The direction of the insect gets reversed when it stuck on the windscreen and the velocity of the insect changes to a great extend. The motorcar continues moving with a constant velocity in the forward direction.

As, Kiran suggested that the insect suffers a greater change in momentum as compared to the motorcar hence, it is correct. The momentum of the insect after collision becomes very high as the motorcar is moving at a very high speed. Therefore, the momentum gained by the insect is equal to the momentum lost by the motorcar.

Secondly, Akhtar is also correct as the mass of the motorcar is very large as compared to the mass of the insect and the speed of the motorcar is also very high.

Now, Rahul’s explanation is correct as both the car and the insect experienced equal forces caused by the Newton’s action-reaction law. But Rahul’s statement becomes incorrect when he said that the system suffers a change in momentum because the momentum before the collision is equal to the momentum after the collision.

Q.18 How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s−2.

Ans-

$Given, mass of the dumbell, m = 10 kg distance covered by the dumbell, s = 80 cm = 0.8 m acceleration in the downward direction, a = 10 ms -2 initial velocity of the dumbell, u = 0 final velocity of the dumbell, v = ? On using the third equation of motion, v 2 = u 2 +2as On putting the given values, v 2 = 0 2 + 2(10 ms -2 )(0.8 m) = 16 m 2 s −2 or, v = 16 m 2 s −2 = 4 ms -1 Therefore, the momentum with which the dumbell hits the floor is = mv= 10 kg × 4 ms -1 = 40 kgms -1 .$