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Case Study Questions Class 9 Science Force and Laws of Motion

Case study questions class 9 science chapter 9 force and laws of motion.

CBSE Class 9 Case Study Questions Science Force and Laws of Motion. Important Case Study Questions for Class 9 Exam. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Force and Laws of Motion.

At Case Study Questions there will given a Paragraph. In where some Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks or 4 marks.

CBSE Case Study Questions Class 9 Science – Force and Laws of Motion

(1 ) Newton’s first law of motion states that a body at rest will remain at rest position only and a body which is in motion continues to be in motion unless otherwise they are acted upon by an external force. In other words, all objects resist a changein their state of motion. In a qualitative way, the tendency of undisturbed objects to stayat rest or to keep moving with the same velocity is called inertia. This is why, the firstlaw of motion is also known as the law of inertia. Answer the following questions.

(i) The first law of motion is also known as

(a)law of inertia

(b)law of thermodynamics

(c)both a and b

(d)none of these

(ii) If no external force acts on object which is at rest. it will

(a)remain at rest

(b)start to move

(c)both a and b can possible

(iii) If no external force acts on moving object. it will

(a)stop moving

(b)continue to move with same speed in same direction

(c)changes its direction of motion

(iv) State Newton’s first law of motion.

(v) why Newton’s first law of motion is called law of inertia

Answer key -1

(iv) Newton’s first law of motion states that a body at rest will remain at rest position only and a body which is in motion continues to be in motion unless otherwise they are acted upon by an external force.

(v) All objects resist a change in their state of motion. In a qualitative way, the tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia. This is why, the first law of motion is also known as the law of inertia.

(2) Two strings X and Y are tied to the two opposite faces of the block as shown in figure. If we apply a force by pulling the string X, the block begins to move to the right. Similarly, if we pull the string Y, the block moves to the left. But, if the block is pulled from both the sides with equal forces, the block will not move. Such forces are called balanced forces and do not change the state of rest or of motion of an object. Now, let us consider a situation in which two opposite forces of different magnitudes pull the block. In this case, the block would begin to move in the direction ofthe greater force. Thus, the two forces are not balanced and the unbalanced force acts in the direction the block moves. This suggests that an unbalanced force acting on an object brings it in motion. Force is push or pull.

(i) Force is nothing but

(c)both push or pulls

(d)none of the above

(ii) When balanced forces acting on moving object then

(a) Object continue  to move with same speed

(b) Object will change its direction of motion

(c) Object will come to rest

(d) None of the above

(iii) When unbalanced force acts on moving object opposite to direction of motion then

(b) Object will come to rest

(c) Both can be possible

(iv) Differentiate between balanced and unbalanced force. give 3 points each .

(v) From above diagram if one person pull from Y rope with 10N force and another person pull from X rope with 5N force. In which direction box will move? Is this a case of unbalanced force or balance force?

Answer key -2

(iv) Difference between balanced and unbalanced force

(3) The second law of motion is quantitative expression of force and it states that the rate of change of momentum of an object isproportional to the applied unbalanced force in the direction of force. Mathematically, F = ma, the unit of force is kg-m/s 2 or Newton,which has the symbol N. The second law ofmotion gives us a method to measure the force acting on an object as a product of its mass and acceleration.Answer the following questions.

(i) SI unit of force is

(b) Kg-m/s 2

(c) Both a and b

(ii) The quantitative expression of force is given by

(a) First law of motion

(b) Second law of motion

(c) Third law of motion

(iii) Force is directly proportional to

(a) Acceleration of object

(b) Time for which force acts on object

(iv) State second law of motion. State whether it is scalar or vector quantity

(v) Differentiate between first law and second law of motion.(give 3 points)

Answer key -3

(iv) The second law of motion is quantitative expression of force and it states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force. Mathematically, F = ma, the unit of force is kg-m/s 2 or Newton.

(v) Difference between first law and second law of motion is given by

(4) The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and neveron the same object. It is important to note that even though the action and reaction forces are always equal in magnitude; these forces may not produce accelerations of equal magnitudes, this is because each force acts on a different object that may have a different mass.The two opposing forces are also known as action and reaction forces. Answer the following questions.

(i) Action reaction forces are always

(a) Equal and in same direction

(b) Equal and in opposite direction

(c) Unequal and in same direction

(ii) Which of the following is correct about action reaction forces?

(a) They act on different objects

(b) They are equal in magnitude and opposite in direction

(c) Both forces acted on different object simultaneously

(d) All the above

(iii) State third law of motion

(iv) Give 5 examples of third law of motion

(v) Even though the action and reaction forces are always equal in magnitude; these forces may not produce accelerations of equal magnitudes. Give your justification on this statement

Answer key -4

(iii) The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and neveron the same object.

(iv) Examples of third law of  motion are

• Swimming or rowing a boat. • Static friction while pushing an object. • Walking. • Standing on the ground or sitting on a chair. • The upward thrust of a rocket. • Resting against a wall or tree.

(v) Even though the action and reaction forces are always equal in magnitude; these forces may not produce accelerations of equal magnitudes because these action reaction forces are acting on two different objects having different masses that’s why they are acceleration with different magnitude.

(5) The sum of momentum of the two objects before collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision. Law of conservation of momentum is applicable to system of particle. Answer the following questions.

(i)Law of conservation of momentum is applicable to

(a) A system of particles

(b) Only for 2 particles

(c) Only for 1 particle

(ii) Law of conservation of momentum holds good provided that

(a) There should be external unbalanced force acting on particles

(b) There should not be any external unbalanced force acting on particles

(c) No internal forces acting on particles

(iii)The total momentum of the two objects when collision occurs is

(a) Changed

(b) Remains conserved

(c) Become zero

(iv) State law of conservation of momentum.

(v) If action and Reaction are equal in magnitude and opposite in direction then why they do not cancel each other?

Answer key -5

(iv) The sum of momentum of the two objects before collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision.

(v) Action and Reaction are equal in magnitude and opposite in direction but they do not cancel each other because they are not action on sane object. As these forces are acting on different object hence produces different acceleration and does not cancel each other.

This makes us understand easily And we can cover all the topics from this chapter, I think this helps us to score good marks in exam. Thankyou for providing this.

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Case Study Questions of Chapter 9 Force and Laws of Motion PDF Download

Case study Questions on Class 9 Science Chapter 9 are very important to solve for your exam. Class 9 Science Chapter 9 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Science Chapter 9 Force and Laws of Motion

In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Force and Laws of Motion Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 9 Science  Chapter 9 Force and Laws of Motion

Case Study/Passage-Based Questions

Question 1:

The sum of the momentum of the two objects before the collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision. The Law of conservation of momentum is applicable to the system of particles. Answer the following questions.

(i)Law of conservation of momentum is applicable to

(a) A system of particles

(b) Only for 2 particles

(c) Only for 1 particle

(d) None of the above

Answer: (a) A system of particles

(ii) Law of conservation of momentum holds good provided that

(a) There should be external unbalanced force acting on particles

(b) There should not be any external unbalanced force acting on particles

(c) No internal forces acting on particles

Answer: (b) There should not be any external unbalanced force acting on particles

(iii)The total momentum of the two objects when collision occurs is

(a) Changed

(b) Remains conserved

(c) Become zero

Answer: (b) Remains conserved

(iv) State law of conservation of momentum.

Answer: The sum of momentum of the two objects before collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision.

(v) If action and Reaction are equal in magnitude and opposite in direction then why they do not cancel each other?

Answer: Action and Reaction are equal in magnitude and opposite in direction but they do not cancel each other because they are not action on sane object. As these forces are acting on different object hence produces different acceleration and does not cancel each other.

Question 2:

The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and never on the same object. It is important to note that even though the action and reaction forces are always equal in magnitude; these forces may not produce accelerations of equal magnitudes, this is because each force acts on a different object that may have a different mass. The two opposing forces are also known as action and reaction forces. Answer the following questions.

(i) Action reaction forces are always

(a) Equal and in the same direction

(b) Equal and in the opposite direction

(c) Unequal and in the same direction

Answer: (b) Equal and in the opposite direction

(ii) Which of the following is correct about action reaction forces?

(a) They act on different objects

(b) They are equal in magnitude and opposite in direction

(c) Both forces acted on different object simultaneously

(d) All the above

Answer: (d) All the above

(iii) State third law of motion

Answer: The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and neveron the same object.

(iv) Give 5 examples of third law of motion

Answer: Examples of third law of motion are Swimming or rowing a boat. •Static friction while pushing an object. •Walking. •Standing on the ground or sitting on a chair. •The upward thrust of a rocket. •Resting against a wall or tree.

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Science Chapter 9 Force and Laws of Motion with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 9 Science Force and Laws of Motion Case Study and passage-based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible

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Class 9th Science - Force and Laws of Motion Case Study Questions and Answers 2022 - 2023

By QB365 on 09 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 9th Science Subject - Force and Laws of Motion, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

QB365 - Question Bank Software

Force and laws of motion case study questions with answer key.

9th Standard CBSE

Final Semester - June 2015

(iii) If the above coin is replaced by a heavy five rupee coin, what will be your observation. Give reason. (a) Heavy coin will possess more inertia so it will not fall in tumbler. (b) Heavy coin will possess less inertia so it will fall in tumbler. (c) Heavy coin will possess more inertia so it will fall in tumbler. (d) Heavy coin will possess less inertia so it will not fall in tumbler. (iv) Name the law which provides the definition of force.

(v) State Newton's first law of motion. (a) Energy can neither be created nor be destroyed, it can be converted from one form to another, total amount of energy always remains constant. (b) A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless it is acted upon by an external unbalanced force. (c) For every action in nature there is an equal and opposite reaction. (d) The acceleration in an object is directly related to the net force and inversely related to its mass.

(ii) What is the total momentum of car A and car B before collision?

(iii) What is the momentum of the car A after collision?

(iv) What is the total momentum of car A and car B after collision?

(v) What is the velocity of car B after the collision?

*****************************************

Force and laws of motion case study questions with answer key answer keys.

(i) (a) The coin possesses inertia of rest, it resists the change and hence falls in the glass. (ii) (b) Newton's first law of motion. (iii) (c) Heavy coin will possess more inertia so it will fall in tumbler. (iv) (d) Newton's second law. (v) (b) A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless it is acted upon by an external unbalanced force.

(i) (b) 37500 kg. m/s Momentum of car A (before collision) = Mass of car A × Velocity of car A = 1500 × 25 = 37500 kg.m/s (ii) (c) 15000 kg. m/s Momentum of car B (before collision) = Mass of car B × Velocity of car B = 1000 × 15 = 15000 kg.m/s Total momentum of car A and car B (before collision) = 37500 + 15000 = 52500 kg.m/s (iii) (a) 30000 kg. m/s After collision, the velocity of car A of mass 1500 kg becomes 20 m/s. So, Momentum of car A (after collision) = 1500 × 20 = 30000 kg.m/s (iv) (d) 52500 kg. m/s According to the law of conservation of momentum, Total momentum before collision = Total momentum after collision ∴ Total momentum after collision = 52500 kg. m/s (v) (a) 22 m/s Momentum of car B (after collision) = 1000 × v = 1000 v kg.m/s Total momentum of car A and car B (after collision) = 30000 + 1000 v According to the law of conservation of momentum, Total momentum before collision = Total momentum after collision = 52500 kg. m/s ∴ 52500 = 30000 + 1000 v ⇒1000 v = 52500 – 30000 ⇒ 1000 v = 22500 ⇒ v = 22500/1000 ⇒ v = 22.5 m/s

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Chapter 8 Class 9 - Motion

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In this chapter, we will learn

What is Motion and What are the Different Types of Motion

What is meaning of Distance and Displacement

What is Uniform Motion and Non Uniform Motion - with graphs

What is Speed

How do we calculate Average Speed

What is Velocity

How do we calculate Average Velocity

What is Acceleration  

First Equation of Motion - Explanation and Derivation

Second Equation of Motion - Explanation and Derivation

Third Equation of Motion - Explanation and Derivation

What is Uniform Circular Motion

Distance Time Graph - and finding velocity from it

Speed Time Graph or Velocity Time Graph - and how to find acceleration and distance from it

Graphical Derivation of Equations of Motion

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• NCERT Solutions
• NCERT Class 9
• NCERT 9 Science
• Chapter 8: Motion

NCERT Solutions for Class 9 Science Chapter 8: Motion

Ncert solutions class 9 science chapter 8 – free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 7.

NCERT Solutions for Class 9 Science Chapter 8 Motion is designed with the intention of clarifying doubts and concepts easily. Class 9 NCERT Solutions in Science is a beneficial reference and guide that helps students clear doubts instantly in an effective way. 

Download Exclusively Curated Chapter Notes for Class 9 Science Chapter – 8 Motion

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NCERT Solutions for Class 9 Science approaches students in a student-friendly way and is loaded with questions, activities, and exercises that are CBSE exam and competitive exam-oriented. NCERT Solutions for Class 9 Science is the contribution of our faculty, having vast teaching experience. It is developed keeping in mind the concept-based approach along with the precise answering method for CBSE examinations. Refer to NCERT Solutions for Class 9 for best scores in CBSE and competitive exams. It is a detailed and well-structured solution for a solid grip on the concept-based learning experience. NCERT for Class 9 Science Solutions is made available in both web and PDF format for ease of access.

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Access Answers of Science NCERT class 9 Chapter 8: Motion  (All intext and exercise questions solved)

Intext Questions – 1   Page: 100

1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Yes, an object which has moved through a distance can have zero displacement if it comes back to its initial position.

Example: If a person jogs in a circular park which is circular and completes one round. His initial and final position is the same.

Hence, his displacement is zero.

2. A farmer moves along the boundary of a square field of side 10m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Side of the given square field = 10m

Hence, the perimeter of a square = 40 m

Time taken by the farmer  to cover the boundary of 40 m = 40 s

So, in 1 s, the farmer covers a distance of 1 m

Distance covered by the farmer in 2 min 20 sec = 1 x 140 = 140 m

The total number of rotations taken by the farmer to cover a distance of 140 m = total distance/perimeter

At this point, let us say the farmer is at point B from the origin O

Therefore,  from Pythagoras theorem, the displacement s = √(10 2 +10 2 )

s =  10 √ 2

s = 14.14 m

3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object.

Neither of the statements is true.

(a) Given statement is false because the displacement of an object which travels a certain distance and comes back to its initial position is zero.

(b) Given statement is false because the displacement of an object can be equal to, but never greater than the distance travelled.

Intext Questions – 2   Page: 102

1. Distinguish between speed and velocity.

2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Since average speed is the total distance travelled in a time frame and velocity is the total displacement in the time frame, the magnitude of average velocity and average speed will be the same when the total distance travelled is equal to the displacement.

3. What does the odometer of an automobile measure?

An odometer, or odograph, is a device that measures the distance travelled by an automobile based on the perimeter of the wheel as the wheel rotates.

4. What does the path of an object look like when it is in uniform motion?

The path of an object in uniform motion is a straight line.

5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 10 8 m/s.

Given that the signal travels in a straight line, the distance between the spaceship and the ground station is equal to the total distance travelled by the signal.

5 minutes = 5*60 seconds = 300 seconds.

Speed of the signal = 3 × 10 8 m/s.

Therefore, total distance = (3 × 10 8 m/s) * 300s

= 9*10 10 meters.

Intext Questions – 3   Page: 103

1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

Uniform Acceleration:  When an object is travelling in a straight line with an increase in velocity at equal intervals of time, then the object is said to be in uniform acceleration.

The free-falling of an object is an example of uniform acceleration.

Non-Uniform Acceleration:  When an object is travelling with an increase in velocity but not at equal intervals of time is known as non-uniform acceleration.

Bus moving or leaving from the bus stop is an example of non-uniform acceleration.

2. A bus decreases its speed from 80 km h –1 to 60 km h –1 in 5 s. Find the acceleration of the bus.

Given, the initial velocity (u) = 80km/hour = 80000m/3600s= 22.22 m.s -1

The final velocity (v) = 60km/hour = 60000m/3600s= 16.66 m.s -1

Time frame, t = 5 seconds.

Therefore, acceleration (a) =(v-u)/t = (16.66 m.s -1 – 22.22 m.s -1 )/5s

= -1.112 m.s -2

Therefore, the total acceleration of the bus is -1.112m.s -2 . It can be noted that the negative sign indicates that the velocity of the bus is decreasing.

3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h –1 in 10 minutes. Find its acceleration.

Given parameters

Initial velocity (u) = 0

Final velocity (v) = 40 km/h

v = 40 × (5/18)

v = 11.1111 m/s

Time (t) = 10 minute

t = 60 x 10

Acceleration (a) =?

Consider the formula

11.11 = 0 + a × 600

11,11 = 600 a

a = 11.11/600

a = 0.0185 ms -2

Intext Questions – 4   Page: 107

1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

For uniform motion, the distance-time graph is a straight line. On the other hand, the distance-time graph of an object in non-uniform motion is a curve.

The first graph describes the uniform motion and the second one describes the non-uniform motion.

2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

The distance-time graph can be plotted as follows.

When the slope of the distance-time graph is a straight line parallel to the time axis, the object is at the same position as time passes. That means the object is at rest.

3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

The speed-time graph can be plotted as follows.

Since there is no change in the velocity of the object (Y-Axis value) at any point of time (X-axis value), the object is said to be in uniform motion.

4. What is the quantity which is measured by the area occupied below the velocity-time graph?

Considering an object in uniform motion, its velocity-time graph can be represented as follows.

Now, the area below the velocity-time graph is the area of the rectangle OABC, which is given by OA*OC. But OA is the velocity of the object and OC represents time. Therefore, the shaded area can be represented as:

Area under the velocity-time graph = velocity*time.

Substituting the value of velocity as displacement/time in the previous equation, it is found that the area under the velocity-time graph represents the total displacement of the object.

Intext Questions – 5 Page: 109,110

1. A bus starting from rest moves with a uniform acceleration of 0.1 m s -2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

(a) Given, the bus starts from rest. Therefore, initial velocity (u) = 0 m/s

Acceleration (a) = 0.1 m.s -2

Time = 2 minutes = 120 s

Acceleration is given by the equation a=(v-u)/t

Therefore, terminal velocity (v) = (at)+u

= (0.1 m.s -2 * 120 s) + 0 m.s -1

= 12 m.s -1 + 0 m.s -1

Therefore, terminal velocity (v) = 12 m/s

(b) As per the third motion equation, 2as = v 2 – u 2

Since a = 0.1 m.s -2 , v = 12 m.s -1 , u = 0 m.s -1 , and t = 120 s, the following value for s (distance) can be obtained.

Distance, s =(v 2 – u 2 )/2a

=(12 2 – 0 2 )/2(0.1)

Therefore, s = 720 m.

The speed acquired is 12 m.s -1 and the total distance travelled is 720 m.

2. A train is travelling at a speed of 90 km h –1 . Brakes are applied so as to produce a uniform acceleration of –0.5 m s -2 . Find how far the train will go before it is brought to rest.

Given, initial velocity (u) = 90 km/hour = 25 m.s -1

Terminal velocity (v) = 0 m.s -1

Acceleration (a) = -0.5 m.s -2

As per the third motion equation, v 2 -u 2 =2as

Therefore, distance traveled by the train (s) =(v 2 -u 2 )/2a

s = (0 2 -25 2 )/2(-0.5) meters = 625 meters

The train must travel 625 meters at an acceleration of -0.5 ms -2 before it reaches the rest position.

3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s -2 . What will be its velocity 3 s after the start?

Given, initial velocity (u) = 0 (the trolley begins from the rest position)

Acceleration (a) = 0.02 ms -2

Time (t) = 3s

As per the first motion equation, v=u+at

Therefore, terminal velocity of the trolley (v) = 0 + (0.02 ms -2 )(3s)= 0.06 ms -1

Therefore, the velocity of the trolley after 3 seconds will be 6 cm.s -1

4. A racing car has a uniform acceleration of 4 m s -2 . What distance will it cover in 10 s after start?

Given, the car is initially at rest; initial velocity (u) = 0 ms -1

Acceleration (a) = 4 ms -2

Time period (t) = 10 s

As per the second motion equation, s = ut+1/2 at 2

Therefore, the total distance covered by the car (s) = 0 * 10m + 1/2 (4ms -2 )(10s) 2

= 200 meters

Therefore, the car will cover a distance of 200 meters after 10 seconds.

5. A stone is thrown in a vertically upward direction with a velocity of 5 m s -1 . If the acceleration of the stone during its motion is 10 m s –2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Given, initial velocity (u) = 5 m/s

Terminal velocity (v) = 0 m/s (since the stone will reach a position of rest at the point of maximum height)

Acceleration = 10 ms -2 in the direction opposite to the trajectory of the stone = -10 ms -2

As per the third motion equation, v 2 – u 2 = 2as

Therefore, the distance travelled by the stone (s) = (0 2 – 5 2 )/ 2(10)

Distance (s) = 1.25 meters

As per the first motion equation, v = u + at

Therefore, time taken by the stone to reach a position of rest (maximum height) = (v – u) /a

=(0-5)/-10 s

Time taken = 0.5 seconds

Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.

Exercises Page: 112,113

1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Given, diameter of the track (d) = 200m

Therefore, the circumference of the track (π*d) = 200π meters.

Distance covered in 40 seconds = 200π meters

Distance covered in 1 second = 200π/40

Distance covered in 2minutes and 20 seconds (140 seconds) = 140 * 200π/40 meters

= (140*200*22)/(40* 7) meters = 2200 meters

Number of rounds completed by the athlete in 140 seconds = 140/40 = 3.5

Therefore, the final position of the athlete (with respect to the initial position) is at the opposite end of the circular track. Therefore, the net displacement will be equal to the diameter of the track, which is 200m.

Therefore, the net distance covered by the athlete is 2200 meters and the total displacement of the athlete is 200m.

2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Given, distance covered from point A to point B = 300 meters

Distance covered from point A to point C = 300m + 100m = 400 meters

Time taken to travel from point A to point B = 2 minutes and 30 seconds = 150 seconds

Time taken to travel from point A to point C = 2 min 30 secs + 1 min = 210 seconds

Displacement from A to B = 300 meters

Displacement from A to C = 300m – 100m = 200 meters

Average speed = total distance travelled/ total time taken

Average velocity = total displacement/ total time taken

Therefore, the average speed while traveling from A to B = 300/150 ms -1 = 2 m/s

Average speed while traveling from A to C = 400/210 ms -1 = 1.9 m/s

Average velocity while traveling from A to B =300/150 ms -1 = 2 m/s

Average velocity while traveling from A to C =200/210 ms -1 = 0.95 m/s

3. Abdul, while driving to school, computes the average speed for his trip to be 20 km.h –1 . On his return trip along the same route, there is less traffic and the average speed is 30 km.h –1 . What is the average speed for Abdul’s trip?

Distance travelled to reach the school = distance travelled to reach home = d (say)

Time taken to reach school = t 1

Time taken to reach home = t 2

therefore, average speed while going to school = total distance travelled/ total time taken = d/t 1 = 20 kmph

Average speed while going home = total distance travelled/ total time taken = d/t 2 = 30 kmph

Therefore, t 1 = d/20 and t 2 = d/30

Now, the average speed for the entire trip is given by total distance travelled/ total time taken

= (d+d)/(t 1 +t 2 )kmph = 2d/(d/20+d/30)kmph

= 2/[(3 + 2)/60]

= 120/5 kmh -1 = 24 kmh -1

Therefore, Abdul’s average speed for the entire trip is 24 kilometers per hour.

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s –2 for 8.0 s. How far does the boat travel during this time?

Given, initial velocity of the boat = 0 m/s

Acceleration of the boat = 3 ms -2

Time period = 8s

As per the second motion equation, s = ut + 1/2 at 2

Therefore, the total distance travelled by boat in 8 seconds = 0 + 1/2 (3)(8) 2

= 96 meters

Therefore, the motorboat travels a distance of 96 meters in a time frame of 8 seconds.

5. A driver of a car travelling at 52 km h –1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h –1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

The speed v/s time graphs for the two cars can be plotted as follows.

The total displacement of each car can be obtained by calculating the area beneath the speed-time graph.

Therefore, displacement of the first car = area of triangle AOB

= (1/2)*(OB)*(OA)

But OB = 5 seconds and OA = 52 km.h -1 = 14.44 m/s

Therefore, the area of the triangle AOB is given by: (1/2)*(5s)*(14.44ms -1 ) = 36 meters

Now, the displacement of the second car is given by the area of the triangle COD

= (1/2)*(OD)*(OC)

But OC = 10 seconds and OC = 3km.h -1 = 0.83 m/s

Therefore, area of triangle COD = (1/2)*(10s)*(0.83ms -1 ) = 4.15 meters

Therefore, the first car is displaced by 36 meters whereas the second car is displaced by 4.15 meters. Therefore, the first car (which was traveling at 52 kmph) travelled farther post the application of brakes.

6. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?

(a) Since the slope of line B is the greatest, B is travelling at the fastest speed.

(b) Since the three lines do not intersect at a single point, the three objects never meet at the same point on the road.

(c) Since there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals 4/7 km.

Since the initial point of an object C is 4 graph units away from the origin, Its initial distance from the origin is 4*(4/7)km = 16/7 km

When B passes A, the distance between the origin and C is 8km

Therefore, total distance travelled by C in this time = 8 – (16/7) km = 5.71 km

(d) The distance that object B has covered at the point where it passes C is equal to 9 graph units.

Therefore, total distance travelled by B when it crosses C = 9*(4/7) = 5.14 km

7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s -2 , with what velocity will it strike the ground? After what time will it strike the ground?

Given, initial velocity of the ball (u) = 0 (since it began at the rest position)

Distance travelled by the ball (s) = 20m

Acceleration (a) = 10 ms -2

As per the third motion equation,

v 2 – u 2 = 2as

= 2*(10ms -2 )*(20m) + 0

v 2 = 400m 2 s -2

Therefore, v= 20ms -1

The ball hits the ground with a velocity of 20 meters per second.

As per the first motion equation,

Therefore, t = (v-u)/a

= (20-0)ms -1 / 10ms -2

= 2 seconds

Therefore, the ball reaches the ground after 2 seconds.

8. The speed-time graph for a car is shown in Fig. 8.12

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car?

The shaded area represents the displacement of the car over a time period of 4 seconds. It can be calculated as:

(1/2)*4*6 = 12 meters. Therefore the car travels a total of 12 meters in the first four seconds.

(b) Since the speed of the car does not change from the points (x=6) and (x=10), the car is said to be in uniform motion from the 6 th to the 10 th second.

9. State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity (b) an object moving with an acceleration but with uniform speed. (c) an object moving in a certain direction with an acceleration in the perpendicular direction.

(a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) it is possible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time

Circular motion is an example of an object moving with acceleration but with uniform speed.

An object moving in a circular path with uniform speed is still under acceleration because the velocity changes due to continuous changes in the direction of motion.

(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Given, the radius of the orbit = 42250 km

Therefore, circumference of the orbit = 2*π*42250km = 265571.42 km

Time is taken for the orbit = 24 hours

Therefore, speed of the satellite = 11065.4 km.h -1

The satellite orbits the Earth at a speed of 11065.4 kilometers per hour.

NCERT Class 9 Science Chapter 8 explains the concept of motion, and types of motion with relevant examples for a clear understanding of the concept. It explains the causes of phenomena like sunrise, sunset, and changing of the seasons. It helps students understand uniform and non-uniform motion. Distance-time graphs and velocity-time graphs, which are considered important concepts for examination, are explained in an easy way in NCERT Solutions . It describes how the acceleration of an object is the change in velocity per unit time. NCERT Class 9 Science Chapter 8 is covered under Unit III: Motion, Force and Work and can get you maximum marks.

• NCERT Solutions for Class 9 explains motion in terms of distance moved or displacement.
• Uniform and non-uniform motions of objects are explained through the graph and examples.
• Uniform circular motion concept is made understandable in a simple way.
• Problems on acceleration, velocity, and average velocity are also solved.

Key Features of NCERT Solutions for Class 9 Science Chapter 8: Motion

• A simple and easily understandable approach is followed in NCERT Solutions to make students aware of topics.
• Provides complete solutions to all the questions present in the respective NCERT textbooks.
• NCERT Solutions offers detailed answers to all the questions to help students in their preparations.
• These solutions will be useful for CBSE exams, Science Olympiads, and other competitive exams.

Disclaimer:

Dropped Topics –  8.5 Equations of motion by graphical method, 8.5.1 Equation for Velocity–Time Relation, 8.5.2 Equation for Position–Time relation and 8.5.3 Equation for Position– Velocity.

Frequently Asked Questions on NCERT Solutions for Class 9 Science Chapter 8

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• Class 9 Science Case...

Class 9 Science Case Study Questions

Table of Contents

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

If you are wondering how to solve class 9 science case study questions, then myCBSEguide is the best platform to choose. With the help of our well-trained and experienced faculty, we provide solved examples and detailed explanations for the recently added Class 9 Science case study questions.

You can find a wide range of solved case studies on myCBSEguide, covering various topics and concepts. Class 9 Science case studies are designed to help you understand the application of various concepts in real-life situations.

The rationale behind Science

Science is crucial for Class 9 students’ cognitive, emotional, and psychomotor development. It encourages curiosity, inventiveness, objectivity, and aesthetic sense.

In the upper primary stage, students should be given a variety of opportunities to engage with scientific processes such as observing, recording observations, drawing, tabulating, plotting graphs, and so on, whereas in the secondary stage, abstraction and quantitative reasoning should take a more prominent role in science teaching and learning. As a result, the concept of atoms and molecules as matter’s building units, as well as Newton’s law of gravitation, emerges.

Science is important because it allows Class 9 Science students to understand the world around us. It helps to find out how things work and to find solutions to problems at the Class 9 Science level. Science is also a source of enjoyment for many people. It can be a hobby, a career, or a source of intellectual stimulation.

Case study questions in Class 9 Science

The inclusion of case study questions in Class 9 science CBSE is a great way to engage students in critical thinking and problem-solving. By working through real-world scenarios, Class 9 Science students will be better prepared to tackle challenges they may face in their future studies and careers. Class 9 Science Case study questions also promote higher-order thinking skills, such as analysis and synthesis. In addition, case study questions can help to foster creativity and innovation in students. As per the recent pattern of the Class 9 Science examination, a few questions based on case studies/passages will be included in the CBSE Class 9 Science Paper. There will be a paragraph presented, followed by questions based on it.

Examples of Class 9 science class case study questions

Class 9 science case study questions have been prepared by myCBSEguide’s qualified teachers. Class 9 case study questions are meant to evaluate students’ knowledge and comprehension of the material. They are not intended to be difficult, but they will require you to think critically about the material. We hope you find Class 9 science case study questions beneficial and that they assist you in your exam preparation.

The following are a few examples of Class 9 science case study questions.

Class 9 science case study question 1

• due to its high compressibility
• large volumes of a gas can be compressed into a small cylinder
• transported easily
• all of these
• shape, volume
• volume, shape
• shape, size
• size, shape
• the presence of dissolved carbon dioxide in water
• the presence of dissolved oxygen in the water
• the presence of dissolved Nitrogen in the water
• liquid particles move freely
• liquid have greater space between each other
• both (a) and (b)
• none of these
• Only gases behave like fluids
• Gases and solids behave like fluids
• Gases and liquids behave like fluids
• Only liquids are fluids

Answer Key:

• (d) all of these
• (a) shape, volume
• (b) the presence of dissolved oxygen in the water
• (c) both (a) and (b)
• (c) Gases and liquids behave like fluids

Class 9 science case study question 2

• 12/32 times
• 18 g of O 2
• 18 g of CO 2
• 18 g of CH 4
• 1 g of CO 2
• 1 g of CH 4 CH 4
• 2 moles of H2O
• 20 moles of water
• 6.022  ×  1023 molecules of water
• 1.2044  ×  1025 molecules of water
• (I) and (IV)
• (II) and (III)
• (II) and (IV)
• Sulphate molecule
• Ozone molecule
• Phosphorus molecule
• Methane molecule
• (c) 8/3 times
• (d) 18g of CH ​​​​​4
• (c) 1g of H ​​​​​​2
• (d) (II) and (IV)
• (c) phosphorus molecule

Class 9 science case study question 3

• collenchyma
• chlorenchyma
• It performs photosynthesis
• It helps the aquatic plant to float
• It provides mechanical support
• Sclerenchyma
• Collenchyma
• Epithelial tissue
• Parenchyma tissues have intercellular spaces.
• Collenchymatous tissues are irregularly thickened at corners.
• Apical and intercalary meristems are permanent tissues.
• Meristematic tissues, in its early stage, lack vacuoles, muscles
• (I) and (II)
• (III) and (I)
• Transpiration
• Provides mechanical support
• Provides strength to the plant parts
• None of these
• (a) Collenchyma
• (b) help aquatic plant to float
• (b) Sclerenchyma
• (d) Only (III)
• (c) provide strength to plant parts

Cracking Class 9 Science Case Study Questions

There is no one definitive answer to Class 9 Science case study questions. Every case study is unique and will necessitate a unique strategy. There are, nevertheless, certain general guidelines to follow while answering case study questions.

• To begin, double-check that you understand the Class 9 science case study questions. Make sure you understand what is being asked by reading it carefully. If you’re unclear, seek clarification from your teacher or tutor.
• It’s critical to read the Class 9 Science case study material thoroughly once you’ve grasped the question. This will provide you with a thorough understanding of the problem as well as the various potential solutions.
• Brainstorming potential solutions with classmates or other students might also be beneficial. This might provide you with multiple viewpoints on the situation and assist you in determining the best solution.
• Finally, make sure your answer is presented simply and concisely. Make sure you clarify your rationale and back up your claim with evidence.

A look at the Class 9 Science Syllabus

The CBSE class 9 science syllabus provides a strong foundation for students who want to pursue a career in science. The topics are chosen in such a way that they build on the concepts learned in the previous classes and provide a strong foundation for further studies in science. The table below lists the topics covered in the Class 9 Science syllabus of the Central Board of Secondary Education (CBSE). As can be seen, the Class 9 science syllabus is divided into three sections: Physics, Chemistry and Biology. Each section contains a number of topics that Class 9 science students must study during the course.

CBSE Class 9 Science (Code No. 086)

Theme: Materials Unit I: Matter-Nature and Behaviour Definition of matter; solid, liquid and gas; characteristics – shape, volume, density; change of state-melting (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation. Nature of matter:  Elements, compounds and mixtures. Heterogeneous and homogenous mixtures, colloids and suspensions. Particle nature and their basic units:  Atoms and molecules, Law of constant proportions, Atomic and molecular masses. Mole concept: Relationship of mole to mass of the particles and numbers. Structure of atoms:  Electrons, protons and neutrons, valency, the chemical formula of common compounds. Isotopes and Isobars.

Theme: The World of the Living Unit II: Organization in the Living World Cell – Basic Unit of life:  Cell as a basic unit of life; prokaryotic and eukaryotic cells, multicellular organisms; cell membrane and cell wall, cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus; nucleus, chromosomes – basic structure, number. Tissues, Organs, Organ System, Organism: Structure and functions of animal and plant tissues (only four types of tissues in animals; Meristematic and Permanent tissues in plants).

Theme: Moving Things, People and Ideas Unit III: Motion, Force and Work Motion:  Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, derivation of equations of motion by graphical method; elementary idea of uniform circular motion. Force and Newton’s laws:  Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration. Elementary idea of conservation of Momentum. Gravitation:  Gravitation; Universal Law of Gravitation, Force of Gravitation of the earth (gravity), Acceleration due to Gravity; Mass and Weight; Free fall. Floatation:  Thrust and Pressure. Archimedes’ Principle; Buoyancy. Work, energy and power:  Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy. Sound:  Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.

Theme: Food Unit IV: Food Production Plant and animal breeding and selection for quality improvement and management; Use of fertilizers and manures; Protection from pests and diseases; Organic farming.

PRESCRIBED BOOKS:

• Science-Textbook for class IX-NCERT Publication
• Assessment of Practical Skills in Science-Class IX – CBSE Publication
• Laboratory Manual-Science-Class IX, NCERT Publication
• Exemplar Problems Class IX – NCERT Publication

myCBSEguide: A true helper

There are numerous advantages to using myCBSEguide to achieve the highest results in Class 9 Science.

• myCBSEguide offers high-quality study materials that cover all of the topics in the Class 9 Science curriculum.
• myCBSEguide provides practice questions and mock examinations to assist students in the best possible preparation for their exams.
• On our myCBSEguide app, you’ll find a variety of solved Class 9 Science case study questions covering a variety of topics and concepts. These case studies are intended to help you understand how certain principles are applied in real-world settings
• myCBSEguide is that the study material and practice problems are developed by a team of specialists who are always accessible to assist students with any questions they may have. As a result, students may be confident that they will receive the finest possible assistance and support when studying for their exams.

So, if you’re seeking the most effective strategy to study for your Class 9 Science examinations, myCBSEguide is the place to go!

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• Important Questions for CBSE Class 9 Science Chapter 9 - Force and Laws of Motion

CBSE Class 9 Science Chapter-9 Important Questions - Free PDF Download

Vedantu now offers Chapter 9's crucial science questions for Class 9 in a downloadable PDF format. Students can download the PDF and study anytime and anywhere in offline mode. Force and Laws of Motion important questions along with their detailed answers are prepared by our subject experts to help students get a clear idea of the concepts covered in this chapter. The chapter Forces and Laws of Motion explains the change in the state of an object in motion or at rest. 

Vedantu provides all the necessary notes and questions on this chapter, including the revision notes , NCERT solutions with step-by-step explanations, etc. to make it easier for students to understand the concepts. Enroll for Class 9 Science tutoring at Vedantu.com to enhance your exam scores. Vedantu offers free CBSE Solutions (NCERT) and additional study resources. Students seeking improved Math solutions can access Class 9 Maths NCERT Solutions, aiding comprehensive syllabus revision and higher exam scores.

Download CBSE Class 9 Science Important Questions 2024-25 PDF

Also, check CBSE Class 9 Science Important Questions for other chapters:

Study Important Questions for Class 9 Science Chapter 9 - Forces and Laws of Motion

Very short answer questions                             (1 mark).

1. A batsman hits a cricket ball which then rolls on level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because – 

The batsman did not hit the ball hard enough.

Velocity is proportional to the force exerted on the ball.

There is a force on the ball opposing the motion.

There is no unbalanced force on the ball, so the ball would want to come to rest.

Ans: (c) There is a force on the ball opposing the motion.

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

\[{{(mv)}^{2}}\]

\[m{{v}^{2}}\]

$\dfrac{1}{2}m{{v}^{2}}$

Ans: (d) $mv$

3. Using a horizontal force of $200N$, 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: In order to move the cabinet across the floor at a constant velocity, the net force experienced by it must be zero. Thus a frictional force of $200N$ must be exerted on the cabinet to move it across the floor at a constant velocity, against the horizontal force of $200N$.

4. What is the S.I. unit of momentum?

$\dfrac{ms}{kg}$

$kgm{{s}^{-1}}$

$\dfrac{kg}{ms}$

Ans: (c) $kgm{{s}^{-1}}$

5. What is the numerical formula for force?

$\mathsf{F=ma}$

$\mathsf{F=}\dfrac{\mathsf{m}}{\mathsf{a}}$

$\mathsf{F=m}{{\mathsf{a}}^{\mathsf{2}}}$

$F={{m}^{2}}a$

Ans: (a) $\mathrm{F=ma}$

6. If the initial velocity is zero then the force acting is 

Acceleration

Ans: (a) Acceleration

7. What is the S.I. unit of force?

$\dfrac{kgm}{{{s}^{2}}}$

$\dfrac{kgm}{s}$

$\dfrac{kg{{m}^{2}}}{{{s}^{2}}}$

$kg{{m}^{2}}{{s}^{2}}$

Ans: (a) $\dfrac{\mathrm{kgm}}{{{\mathrm{s}}^{\mathrm{2}}}}$

8. Newton’s first law of motion is also called

Law of Inertia

Law of Momentum

Law of Action & Reaction

None of these

Ans: (a) Law of Inertia

9. Which law explains swimming?

Newton’s first law

Newton’s second law

Newton’s third law

All of these

Ans: (c) Newton’s third law

10. The S.I. unit of weight is:

$\dfrac{N}{s}$

$\dfrac{Nm}{s}$

Ans: (a) $\mathrm{N}$

11. Which equation defines Newton’s Second law of motion?

$\mathsf{F=ma=}\dfrac{\mathsf{dp}}{\mathsf{dt}}$

$\mathsf{F=m}\dfrac{\mathsf{da}}{\mathsf{dt}}\mathsf{=P}$

$\dfrac{\mathsf{dF}}{\mathsf{dt}}\mathsf{=ma=P}$

$\mathsf{F=ma=P}$

Ans: (d) $\mathrm{F=ma=P}$

12. The people in the bus are pushed backwards when the bus starts suddenly due to

Inertia due to Rest

Inertia due to Motion

Inertia due to Direction

Ans: (a) Inertia due to Rest

13. If the force acting on the body is zero. Its momentum is

Ans: (b) constant

14. The inability of the body to change its state of rest or motion is

Acceleration.

Ans: (c) Inertia

Short Answer Questions                                          (2 Marks)

1. An object experiences a net zero external unbalanced force. Is it possible for the object to be traveling 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 for an object to be traveling with a non-zero velocity if it experiences a net zero external unbalanced force. This is based on Newton’s First law of motion, which states that an object will change its state of motion if and only if the net force acting on it is non-zero.  

Thus, in order to change its motion or to bring about motion, some external unbalanced force is required. 

In this case, the object experiences a net zero unbalanced force, which will cause it to move with some non-zero velocity provided that the object was previously moving with a constant velocity.

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

Ans: When a carpet is beaten with a stick, dust comes out of it because of Newton’s First Law of Motion, the law of Inertia. Initially, the dust particles and the carpet are in a state of rest. When the carpet is beaten with a stick, it causes the carpet to move, while the dust particles, due to inertia of rest, will resist the change in motion. Thus, the carpet’s forward motion will exert a backward force on the dust particles, which makes them move in the opposite direction. Therefore the dust comes out of the carpet.

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

It is advised to tie any luggage kept on the roof of a bus with a rope because of Newton’s First Law of Motion, the law of Inertia. 

When the bus moves, the luggage will be in inertia of motion and say the bus suddenly stops, then the luggage tends to resist this change in motion, causing it to move forward and fall off, if not tied up by a rope. 

Similarly, when the bus decelerates or changes its direction while turning, the inertia of motion of the luggage will try to resist this change in motion, causing the luggage to move oppositely and fall off, if not tied up by a rope.

4. A stone of $1kg$ is thrown with a velocity of $20m{{s}^{-1}}$ across the frozen surface of a lake and comes to rest after traveling a distance of $50m$. What is the force of friction between the stone and the ice?

Ans: Given:

Mass of stone: $\mathrm{m=1kg}$

Initial velocity of stone: $\mathrm{u=20m}{{\mathrm{s}}^{\mathrm{-1}}}$

Final velocity of stone:  $\mathrm{v=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it comes to rest)

Distance traveled on ice: $\mathrm{s=50m}$

To find: Force of friction between stone and ice.

First, we need to find the deceleration:

It is known that – ${{\mathrm{v}}^{\mathrm{2}}}\mathrm{=}{{\mathrm{u}}^{\mathrm{2}}}\mathrm{+2as}$

Thus, ${{\mathrm{0}}^{\mathrm{2}}}\mathrm{=(20}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{+2a(50)}$

$\Rightarrow \mathrm{0=400+100a}$

$\Rightarrow -400=100a$

$\Rightarrow a=-4m{{s}^{-2}}$

The negative sign implies deceleration.

Next, finding the frictional force:

$\mathrm{F=ma}$

$\Rightarrow \mathrm{F=(1)(}-4)$

$\Rightarrow \mathrm{F=}-4N$

Thus, the force of friction between stone and ice is $-4N$ .

5. An automobile vehicle has a mass of $1500kg$. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of $1.7m{{s}^{-2}}$?

Mass of vehicle:$\mathrm{m=1500kg}$

Negative acceleration:$\mathrm{a=}-\mathrm{1}\mathrm{.7m}{{\mathrm{s}}^{\mathrm{-2}}}$

To find: Force of friction between road and vehicle.

It is known that - $\mathrm{F=ma}$

$\Rightarrow \mathrm{F=(1500)(}-1.7)$

$\Rightarrow \mathrm{F=}-2550N$

Thus the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of $1.7m{{s}^{-2}}$is $-2550N$.

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

Mass of object:$\mathrm{m=100kg}$

Initial velocity of object:$\mathrm{u=5m}{{\mathrm{s}}^{\mathrm{-1}}}$

Final velocity of object: $\mathrm{v=8m}{{\mathrm{s}}^{\mathrm{-1}}}$

Time duration of acceleration:$t=6s$

To find: 

Initial momentum

Final momentum

Force exerted on the object

It is known that – $momentum=\operatorname{ma}ss\times velocity$

$Initial\_momentum=\operatorname{ma}ss\times initial\_velocity$

$\Rightarrow Initial\_momentum=100\times 5$

$\Rightarrow Initial\_momentum=500kgm{{s}^{-1}}$

$Final\_momentum=\operatorname{ma}ss\times final\_velocity$

$\Rightarrow Final\_momentum=100\times 8$

$\Rightarrow Final\_momentum=800kgm{{s}^{-1}}$

Now, the force – $\mathrm{F=ma}$

$\mathrm{F=m(}\dfrac{v-u}{t})$

$\Rightarrow \mathrm{F=100(}\dfrac{8-5}{6})$

$\Rightarrow \mathrm{F=100(}\dfrac{3}{6})$

$\Rightarrow \mathrm{F=50N}$

Initial momentum:$500kgm{{s}^{-1}}$

Final momentum: $800kgm{{s}^{-1}}$

Force exerted on object: $50N$

7. 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.

Kiran’s statement – 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). 

Based on the law of conservation of momentum, the change of momentum experienced by both the insect and car should be equal. The change in velocity of the insect will be greater, due to its small mass, while the change in velocity of the car is insignificant, due to its larger mass. But the change in momentum before and after collision would be the same. Thus, Kiran’s statement is false.

Akhtar’s statement – since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result, the insect died. 

According to Newton’s third law of motion, for every action, there is an equal and opposite reaction. Thus both the car and insect would experience the same force. So, we can say that Akhtar’s statement is also false.

Rahul’s statement – both the motorcar and the insect experienced the same force and a change in their momentum.

Inferring from the law of conservation of momentum and Newton’s third law of motion, we can say that Rahul’s statement is true.

8. State Newton’s second law of motion?

Ans: Newton’s Second law of motion states that when an unbalanced external force acts on an object, its velocity changes, that is, the object gets accelerated. It can also be stated as the time rate of change of the momentum of a body is equal in both magnitude and direction to the force applied on it.

Mathematically – $\mathrm{F=ma}$, where ‘F’ is the force, ‘m’ is the mass of the object, and ‘a’ is the acceleration.

9. What is the momentum of a body of mass $200g$ moving with a velocity of $15m{{s}^{-1}}$?

Ans: Given: 

Mass of body:$\mathrm{m=200g=0}\mathrm{.2kg}$

Velocity of body: $\mathrm{v=15m}{{\mathrm{s}}^{\mathrm{-1}}}$

To find: Momentum of the body.

$\Rightarrow momentum=0.2\times 15$

$\Rightarrow momentum=3kgm{{s}^{-1}}$

Thus, the momentum of the body is $3kgm{{s}^{-1}}$.

10. Define force and what are the various types of forces?

Ans: Force is defined as the push or pulls on an object that produces a change in the state or shape of the object. It can also cause a change in the speed and/or direction of motion of the object.

The various types of force are:

Mechanical force

Gravitational force

Frictional force

Electrostatic force

Electromagnetic force

Nuclear force

11. A force of $25N$ acts on a mass of $500g$ resting on a frictionless surface. What is the acceleration produced?

Mass:$\mathrm{m=500g=0}\mathrm{.5kg}$

Force exerted: $\mathrm{F=25N}$

To find: Acceleration.

It is known that – $\mathrm{F=ma}$

$\Rightarrow a=\dfrac{F}{m}$

$\Rightarrow a=\dfrac{25}{0.5}$

$\Rightarrow a=50m{{s}^{-2}}$

Thus, the acceleration produced is \[50m{{s}^{-2}}\].

12. State Newton’s first law of Motion?

Ans: Newton’s first law of motion is also called the law of Inertia. It states that an object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force. Or, an object rest will continue to be at rest and an object in motion will continue to be in motion until and unless it is acted upon by an external force.

13. A body of mass $5kg$ starts and rolls down $32m$of an inclined plane in $4s$. Find the force acting on the body?

Mass of body:$\mathrm{m=5kg}$

Initial velocity of body: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starting to roll)

Distance travelled on inclined plane: $\mathrm{s=32m}$

Time duration of rolling:$t=4s$

To find: Force acting on the body.

First we need to find the acceleration:

It is known that – $\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

Thus, $\mathrm{32=(0}\times \mathrm{4)+}\dfrac{1}{2}(a\times {{4}^{2}})$

$\Rightarrow \mathrm{32=}\dfrac{1}{2}(a\times 16)$

$\Rightarrow \mathrm{32=}(a\times 8)$

$\Rightarrow a=4m{{s}^{-2}}$

Next, finding the force:

$\Rightarrow \mathrm{F=(5}\times 4)$

$\Rightarrow \mathrm{F=20}N$

Thus, the force acting on the body is $20N$ .

14. On a certain planet, a small stone tossed up at $15m{{s}^{-1}}$ vertically upwards takes $7.5s$ to return to the ground. What is the acceleration due to gravity on the planet?

Initial velocity of stone:$\mathrm{u=15m}{{\mathrm{s}}^{\mathrm{-1}}}$

Final velocity of stone: $\mathrm{v=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it becomes zero at the highest point)

Total time duration of flight (tossed up and falling down to the ground):$t=7.5s$

To find: Acceleration due to gravity of the planet.

It is known that – $\mathrm{v=u+at}$

Thus, $\mathrm{0=15+(at)}$

$\mathrm{t=}\dfrac{-15}{a}$  this denotes the time for one-half of the entire flight.

Thus the total duration of the flight is twice this duration.

i.e. $7.5s=2t$

$\Rightarrow 7.5=2(\dfrac{-15}{a})$

Now, the acceleration due to gravity is –

$a=\dfrac{2\times (-15)}{7.5}$

\[\Rightarrow a=-4m{{s}^{-2}}\]

Thus, the acceleration due to gravity of the planet is $-4m{{s}^{-2}}$.

15. Why is the weight of the object more at the poles than at the equator?

The weight of the object is more at the poles than at the equator because the acceleration due to gravity is slightly greater at the poles than at the equator. This is because - $g=\dfrac{GM}{{{r}^{2}}}$, meaning acceleration due to gravity is inversely proportional to the square of the radius. Since the radius of the earth at the equator is greater than at the poles, the acceleration due to gravity is slightly less at the equator, than at the poles.

Also, we know that weight is directly proportional to acceleration due to gravity $(\because w=m\times g\Rightarrow w\alpha g)$.

Using these two implications, we can say that at the equator, where the radius is larger, the acceleration due to gravity is smaller, the weight is lower. And at the poles, where the radius is smaller, the acceleration due to gravity is greater, the weight is higher.

16. Why does the passenger sitting in a moving bus are pushed in the forward direction when the bus stops suddenly?

Ans: The passengers sitting in the moving bus are pushed in the forward direction when the bus stops suddenly due to inertia because the passengers' upper body continues to be in a state of motion, while the lower part of the body that is in contact with the seat remains at rest. As a result, the passenger’s upper body is pushed in the forward direction, in the direction in which the bus was moving before coming to a halt.

17. Why does the boat move backward when the sailor jumps in the forward direction?

Ans: The boat moves backward when the sailor jumps in the forward direction because of Newton’s third law of motion which states that for every action, there is an equal and opposite reaction. Thus, when the sailor jumps in the forward direction he is causing an action force due to which the boat moves backward. In response, the boat exerts an equal and opposite force (the reaction force) on the sailor due to which he is pushed in the forward direction.

18. Derive the law of conservation of momentum from Newton’s third law?

Ans: Newton’s third law of motion states that for every action, there is an equal and opposite reaction, that acts on different bodies.

Say we have two objects A and B of masses ${{m}_{A}}$ and ${{m}_{B}}$ are traveling in the same direction along a straight line at different velocities ${{u}_{A}}$ and ${{u}_{B}}$, respectively. 

Consider that there are no other external unbalanced forces acting on them. 

Let ${{u}_{A}}>{{u}_{B}}$ and the two objects collide with each other.

During collision which lasts for a time $t$, A exerts a force ${{F}_{AB}}$ on B and B exerts a force ${{F}_{BA}}$ on  A. 

Say, ${{v}_{A}}$ and ${{v}_{B}}$ are the velocities of the two A and B after the collision, respectively.

Momentum of A before collision: \[{{m}_{A}}\times {{u}_{A}}\]

Momentum of A after collision: \[{{m}_{A}}\times {{v}_{A}}\]

Momentum of B before collision: \[{{m}_{B}}\times {{u}_{B}}\]

Momentum of B after collision: \[{{m}_{B}}\times {{v}_{B}}\]

It is also known that force can also be defined as the rate of change of momentum, i.e. $F=ma=m(\dfrac{v-u}{t})=\dfrac{mv-mu}{t}$

Now, the rate of change of momentum of A during collision is $\dfrac{{{m}_{A}}{{v}_{A}}-{{m}_{A}}{{u}_{A}}}{t}$, which is the force ${{F}_{AB}}$.

And the rate of change of momentum of B during collision is \[\dfrac{{{m}_{B}}{{v}_{B}}-{{m}_{B}}{{u}_{B}}}{t}\], which is the force ${{F}_{BA}}$.

Based on Newton’s third law of motion, the force ${{F}_{AB}}$  exerted by Aon B and force ${{F}_{BA}}$ exerted by B on A are equal in magnitude but opposite in direction.

i.e.  ${{F}_{AB}}=-{{F}_{BA}}$

Using the formulae – 

${{F}_{AB}}=-{{F}_{BA}}$

\[\Rightarrow \dfrac{{{m}_{A}}{{v}_{A}}-{{m}_{A}}{{u}_{A}}}{t}=-(\dfrac{{{m}_{B}}{{v}_{B}}-{{m}_{B}}{{u}_{B}}}{t})\]

Simplifying,

\[\Rightarrow {{m}_{A}}{{v}_{A}}-{{m}_{A}}{{u}_{A}}=-({{m}_{B}}{{v}_{B}}-{{m}_{B}}{{u}_{B}})\]

\[\Rightarrow {{m}_{A}}{{v}_{A}}-{{m}_{A}}{{u}_{A}}=-{{m}_{B}}{{v}_{B}}+{{m}_{B}}{{u}_{B}}\]

Rearranging,

\[{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}\]

Here, \[{{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}\]is the total sum of momentum of A and B before collision and \[{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}\] is the total sum of momentum of A and B after collision.

This equation implies that the final momentum of the two objects after the collision is equal to the initial momentum of the two objects before the collision.

Thus the law of conservation of momentum, for an isolated system, is that the total initial momentum for an event is equal to total initial momentum.

19. An astronaut has $80kg$ mass on earth.

i) What is his weight on earth?

Mass of astronaut: $\mathrm{m=80kg}$

To find his weight on earth.

It is known that,

Acceleration due to gravity on earth: ${{g}_{e}}=10m{{s}^{-2}}$

Acceleration due to gravity on mars:${{g}_{m}}=3.7m{{s}^{-2}}$

Weight: $w=m\times g$

Weight on earth: ${{w}_{e}}=m\times {{g}_{e}}$

$\Rightarrow {{w}_{e}}=80\times 10$

${{w}_{e}}=800N$

ii) What will be his mass and weight on mars with ${{g}_{m}}=3.7m{{s}^{-2}}$?

Mass of astronaut:$\mathrm{m=80kg}$

To find his mass and weight on mars.

Weight on mars: ${{w}_{m}}=m\times {{g}_{m}}$

$\Rightarrow {{w}_{m}}=80\times 3.7$

${{w}_{m}}=296N$

The mass of astronauts remains the same on mars because it is a constant value. 

Thus, mass on mars is $\mathrm{m=80kg}$.

Short Answer Questions                                     (3 Marks)

1. Which of the following has more inertia:

a. A rubber ball and a stone of the same size?

Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a rubber ball and a stone of the same size, it is clear that the stone will have greater inertia than the ball. It is because, despite being the same size, the stone weighs more than the rubber ball.

b. A bicycle and a train?

Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a bicycle and a train, it is clear that the train will have greater inertia than the bicycle because the train weighs more than the bicycle.

c. A five rupees coin and a one-rupee coin?

Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a five rupees coin and a one-rupee coin, the five rupees coin will have greater inertia than the one-rupee coin because five rupees coin weighs more than a one-rupee coin.

2. In the following example, try to identify the number of times the velocity of the ball changes:

“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”.

Also, identify the agent supplying the force in each case.

Ans: In the example, the number of times the velocity of football changes is four.

(i) The velocity of the football changes first when a player kicks the ball towards another player on his team. Here, the agent supplying the force is the foot of the football player who is kicking the ball.

(ii) The velocity of the football changes second when that another player kicks the ball towards the goal. Here, the agent supplying the force is the foot of the other player who is now kicking the ball towards the goal.

(iii) The velocity of the football changes for the third time when the goalkeeper of the opposite team stops the football by collecting it. Here, the agent supplying the force are the hands of the goalkeeper who collects the ball.

(iv) The velocity of the football changes for the fourth time when the goalkeeper kicks it towards a player of his team. Here, the agent supplying the force is the foot of the goalkeeper who is now kicking the ball towards his teammate.

3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Ans: Some of the leaves may get detached from a tree if we vigorously shake its branch because the branches of the tree will come into motion while the leaves tend to continue in their state of rest. This is due to the inertia of rest of the leaves. The force of shaking will act on the leaves with the change in direction rapidly, which results in the leaves detaching and falling off from the tree.

4. Why do you fall in the forward direction when a moving bus breaks to a stop and fall back when it accelerates from rest?

Similarly, the passengers sitting in the bus are pushed in the backward direction when the bus accelerates from rest due to inertia, because the passengers’ upper body continues to be in a state of rest, while the lower part of the body that is in contact with the seat is set in motion. As a result, the passenger’s upper body is pushed in the backward direction, in the opposite to which the bus starts to move.

5. If action is always equal to the reaction, explain how a horse can pull a cart.

In this case, we are dealing with unbalanced forces. It is true that the horse exerts an action force on the cart and experiences a reaction force from the cart. But also, the horse creates an action force on the ground over which it is walking, and experiences a reaction force from the ground.

In pulling the cart, the action force of the horse pulling the cart is greater than the reaction force of the cart, resisting the pull. Thus the cart moves in the direction of the pull of the horse.

In stepping on the ground, the horse creates an action force on the ground in the backward direction during each step, while the ground creates a reaction force pushing the horse forward.

In this was a horse can pull a cart.

6. Explain why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

Ans: It is difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity because of Newton’s third law of motion. In this case, the water being ejected out in the forward direction with great force (action) will create a backward force that results in the backward movement (reaction) of the hose pipe. As a result of this backward force and movement, it will be difficult for the fireman to hold the hose properly with stability.

7. From a rifle of mass $4kg$, a bullet of mass $50g$ is fired with an initial velocity of $35m{{s}^{-1}}$. Calculate the initial recoil velocity of the rifle.

Mass of rifle: ${{m}_{1}}=4kg$

Mass of bullet: ${{m}_{2}}=50g=0.05kg$ 

Initial velocity of rifle: ${{u}_{1}}=0m{{s}^{-1}}$ (it is stationary during firing)

Initial velocity of bullet:${{u}_{2}}=0m{{s}^{-1}}$(it starts from rest, inside the barrel of the rifle)

Fired velocity of bullet:${{v}_{2}}=35m{{s}^{-1}}$

To find: Recoil velocity of rifle:${{v}_{1}}$

By using the law of conservation of momentum, for an isolated system, is that the total initial momentum for an event is equal to total initial momentum, we get – 

\[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

Here, \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]is the total sum of momentum of rifle and bullet before firing and \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] is the total sum of momentum of rifle and bullet after firing.

Substituting the values in – \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

\[\Rightarrow (4\times {{v}_{1}})+(0.05\times 35)=(4\times 0)+(0.05\times 0)\]

\[\Rightarrow (4\times {{v}_{1}})+(17.5)=0\]

\[\Rightarrow (4\times {{v}_{1}})=-17.5\]

\[\Rightarrow {{v}_{1}}=-4.375m{{s}^{-1}}\](The negative sign indicates the backward direction in which the rifle moves when it recoils)

Thus, the recoil velocity of the rifle is \[4.375m{{s}^{-1}}\].

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

(a) The net accelerating force

Force exerted by engine on wagons: $F=40000N$

Frictional exerted on wagons:$f=5000N$

Mass of engine:${{m}_{e}}=8000kg$

Mass of each wagon:${{m}_{w}}=2000kg$

Mass of all five wagons:${{m}_{W}}=5\times {{m}_{w}}=5\times 2000=10000kg$

Mass of entire train: ${{m}_{T}}={{m}_{e}}+{{m}_{W}}=8000+10000=18000kg$

To find accelerating force.

Net accelerating force can be found by subtracting the frictional force from the force exerted by the engine on the wagons.

Thus, $NetAcceleratingForce=ForceOfEngine-FrictionalForce$

$\Rightarrow NetAcceleratingForce=F-f$

$\Rightarrow NetAcceleratingForce=40000-5000$

$\Rightarrow NetAcceleratingForce=35000N$

(b) The acceleration of the train

Ans:  Given:

To find the acceleration of the train.

$\Rightarrow a=\dfrac{NetAcceleratingForce}{MassOfTrain}$

$\Rightarrow a=\dfrac{35000}{18000}$

$\Rightarrow a=1.944m{{s}^{-2}}$

(c) The force of wagon $1$ on wagon $2$.

To find the force exerted by wagon $1$ on wagon $2$

Here, wagon $1$ exerts a pulling force on the remaining $4$ wagons

$\Rightarrow {{\mathrm{F}}_{21}}\mathrm{=(4}{{\mathrm{m}}_{w}}\mathrm{)}\times \mathrm{a}$

$\Rightarrow {{\mathrm{F}}_{21}}\mathrm{=(4}\times 2000\mathrm{)}\times 1.944$

$\Rightarrow {{\mathrm{F}}_{21}}\mathrm{=15552}N$

9. Two objects, each of mass $1.5kg$, are moving in the same straight line but in opposite directions. The velocity of each object is $2.5m{{s}^{-1}}$ before the collision during which they stick together. What will be the velocity of the combined object after collision?

Mass of object 1: ${{m}_{1}}=1.5kg$

Mass of object 2: ${{m}_{2}}=1.5kg$ 

Initial velocity of object 1: ${{u}_{1}}=2.5m{{s}^{-1}}$

Initial velocity of object 2:${{u}_{2}}=-2.5m{{s}^{-1}}$(negative sign because it is moving in the opposite direction)

Mass of combined object after collision: $m={{m}_{1}}+{{m}_{2}}=1.5+1.5=3kg$

To find: Final velocity of the combined object after collision:$v$

\[mv={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

Here, \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]is the total sum of the momentum of objects before the collision and \[mv\] is the total momentum of the combined objects after the collision.

Substituting the values in – \[mv={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

\[\Rightarrow (3\times v)=(1.5\times 2.5)+(1.5\times -2.5)\]

\[\Rightarrow (3\times v)=(3.75)+(-3.75s)\]

\[\Rightarrow (3\times v)=0\]

$\Rightarrow v=0m{{s}^{-1}}$

Thus, the velocity of the combined object after collision is \[0m{{s}^{-1}}\].

10. 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.

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. This pair of forces is called the action-reaction pair. 

In the case of the massive truck parked alongside the road, the action-reaction pair is the weight of the truck exerting a force on the road in the downward direction (action), and the static friction of the road in the upward direction (reaction), which keeps the truck at rest. These two equal and opposite forces cancel out each other, which is why the truck will not move.

For it to move, we need to apply additional external force to overcome the static friction of the road. 

Thus, as the student explained, the truck does not move because the two opposite and equal forces of the truck and road cancel out each other is valid.

11. A hockey ball of mass $200g$ traveling at $10m{{s}^{-1}}$ is struck by a hockey stick so as to return it along its original path with a velocity at $5m{{s}^{-1}}$. Calculate the change of momentum that occurred in the motion of the hockey ball by the force applied by the hockey stick.

Mass of hockey ball: $m=200g=0.2kg$

Initial velocity of hockey ball: $u=10m{{s}^{-1}}$

Final velocity of hockey ball:$v=-5m{{s}^{-1}}$ (because it moves back in its original direction)

To find: Change in momentum of hockey ball due to the force of hockey stick

\[ChangeOfMomentum=mv-mu\]

Here, \[mu\]is the initial momentum of the hockey ball and \[mv\] is the final momentum of the hockey ball.

Substituting the values in –\[ChangeOfMomentum=mv-mu\]

\[\Rightarrow ChangeOfMomentum=(0.2\times -5)-(0.2\times 10)\]

\[\Rightarrow ChangeOfMomentum=(-1)-(2)\]

\[\Rightarrow ChangeOfMomentum=-3kgm{{s}^{-1}}\]

Thus, the change in momentum of hockey ball due to the force of hockey stick is \[-3kgm{{s}^{-1}}\].

12. A bullet of mass $10g$ traveling horizontally with a velocity of $150m{{s}^{-1}}$ strikes a stationary wooden block and comes to rest in $0.03s$. 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.

Mass of bullet: $m=10g=0.01kg$

Initial velocity of bullet: $u=150m{{s}^{-1}}$

Final velocity of bullet: $\mathrm{v=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it comes to rest after penetration)

Time duration of bullet travel:$t=0.03s$

Distance of penetration of bullet into the block

Force exerted by the block on the bullet

Distance of penetration:

Thus, $\mathrm{0=150+(a}\times 0.03\mathrm{)}$

$\Rightarrow \mathrm{(a}\times 0.03\mathrm{)=}-\mathrm{150}$

$\Rightarrow \mathrm{a=}-5000m{{s}^{-2}}$

$\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

$\Rightarrow \mathrm{s=(150}\times 0.03\mathrm{)+}\dfrac{1}{2}(-5000\times {{0.03}^{2}})$

$\Rightarrow \mathrm{s=(}4.5\mathrm{)+}(-2.25)$

$\Rightarrow \mathrm{s=}2.25m$

$\Rightarrow \mathrm{F=(0}\mathrm{.01}\times -5000)$

$\Rightarrow \mathrm{F=}-5\mathrm{0}N$

Distance of penetration of bullet into the block is $2.25m$

Force exerted by the block on the bullet is $-50N$

13. An object of mass $1kg$ traveling in a straight line with a velocity of $10m{{s}^{-1}}$ collides with and sticks to, a stationary wooden block of mass $5kg$. 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.

Mass of object 1: ${{m}_{1}}=1kg$

Mass of object 2: ${{m}_{2}}=5kg$ 

Initial velocity of object 1: ${{u}_{1}}=10m{{s}^{-1}}$

Initial velocity of object 2:${{u}_{2}}=0m{{s}^{-1}}$(because it is stationary)

Mass of combined object after collision: $m={{m}_{1}}+{{m}_{2}}=1+5=6kg$

Momentum before impact

Momentum after impact

The final velocity of the combined object after collision:$v$

Momentum before impact is the Initial momentum:

\[InitialMomentum={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

\[\Rightarrow InitialMomentum=(1\times 10)+(5\times 0)\]

\[\Rightarrow InitialMomentum=10kgm{{s}^{-1}}\]

Momentum after impact is the Final momentum:

\[FinalMomentum=mv\]

Thus we get – \[FinalMomentum=mv={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}=10kgm{{s}^{-1}}\]

\[\Rightarrow FinalMomentum=10kgm{{s}^{-1}}\]

Finding the final velocity of the combined object:

\[FinalMomentum=10kgm{{s}^{-1}}\]

\[\Rightarrow mv=10\]

\[\Rightarrow (6\times v)=10\]

\[\Rightarrow v=1.67m{{s}^{-1}}\]

Momentum before impact is \[10kgm{{s}^{-1}}\]

Momentum after impact is \[10kgm{{s}^{-1}}\]

Final velocity of the combined object after collision is \[1.67m{{s}^{-1}}\]

14. How much momentum will a dumb-bell of mass $10kg$ transfer to the floor if it falls from a height of $80cm$? Take its downward acceleration to be $10m{{s}^{-1}}$.

Mass of dumbbell: $m=10kg$

Initial velocity of dumbbell: $u=0m{{s}^{-1}}$(as it starts from rest)

Height of fall of dumbbell: $\mathrm{h=80cm=0}\mathrm{.8m}$ 

Acceleration due to gravity: $g=10m{{s}^{-2}}$

To find: Momentum transferred to the ground by dumbbell.

It is known that – ${{v}^{2}}={{u}^{2}}+2gh$

$\Rightarrow {{v}^{2}}={{(0)}^{2}}+(2\times 10\times 0.8)$

$\Rightarrow {{v}^{2}}=16$

$\Rightarrow v=4m{{s}^{-1}}$

Now, \[Momentum=mv\]

\[\Rightarrow Momentum=(10\times 4)\]

\[\Rightarrow Momentum=40kgm{{s}^{-1}}\]

Thus, the momentum transferred to the ground by dumbbell is \[40kgm{{s}^{-1}}\]

15. A force of $15N$acts for $5s$on a body of mass $5kg$ which is initially at rest. Calculate.

(a) Final velocity of the body

Mass of body: $m=5kg$

Initial velocity of body: $u=0m{{s}^{-1}}$(as it starts from rest)

Force acting on the body: $F=15N$

Time: $t=5s$

To find the final velocity of the body.

First we need to find the acceleration produced.

$\Rightarrow \mathrm{a=}\dfrac{F}{m}$

$\Rightarrow \mathrm{a=}\dfrac{15}{5}$

$\Rightarrow a=3m{{s}^{-2}}$

Thus, $\mathrm{v=0+(3}\times 5\mathrm{)}$

$\Rightarrow v\mathrm{=15}m{{s}^{-1}}$

(b) The displacement of the body

To find the displacement of the body.

Next, the distance of penetration:

$\Rightarrow \mathrm{s=(0}\times 5)\mathrm{+}\dfrac{1}{2}(3\times {{5}^{2}})$

$\Rightarrow \mathrm{s=(0)+}(37.5)$

$\Rightarrow \mathrm{s=37}.5m$

16. Differentiate between mass and weight?

Ans: The difference between mass and weight is given below,

17. A scooter is moving with a velocity of $20m{{s}^{-1}}$when brakes are applied. The mass of the scooter and the rider is $180kg$. The constant force applied by the brakes is $500N$.

(a) How long should the brakes be applied to make the scooter comes to a halt?

Mass of scooter and rider: $m=180kg$

Initial velocity of scooter: $u=20m{{s}^{-1}}$

Final velocity of scooter: $v=0m{{s}^{-1}}$(as it halts after applying the brake)

Force of the brake: $F=-500N$(as it opposes the motion)

To find the time duration over which brake should be applied to stop the scooter.

$\Rightarrow -500\mathrm{=180(}\dfrac{0-20}{t})$

$\Rightarrow t\mathrm{=(}\dfrac{180\times (-20)}{-500})$

$\Rightarrow t\mathrm{=(}\dfrac{-3600}{-500})$

$\Rightarrow t\mathrm{=}7.2s$

(b) How far does the scooter travel before it comes to rest?

To find distance travelled by scooter before coming to halt.

Acceleration – $\mathrm{a=(}\dfrac{v-u}{t})$

$\Rightarrow \mathrm{a=(}\dfrac{0-20}{7.2})$

$\Rightarrow \mathrm{a=}-2.78m{{s}^{-2}}$

Acceleration is negative because it is retarding the motion of the scooter.

$\Rightarrow \mathrm{s=(20}\times 7.2)\mathrm{+}\dfrac{1}{2}(-2.78\times {{7.2}^{2}})$

$\Rightarrow \mathrm{s=(144)+}(-72.1)$

$\Rightarrow \mathrm{s=}71.9m$

18. State Newton’s third law of motion and how does it explain the walking of man on the ground?

Ans: Newton’s third law of motion states that for every action, there is an equal and opposite reaction acting on different bodies. This implies the existence of the action-reaction force pair. That is, for every action force created an equal and opposite reaction force will be created.

The walking of a man on the ground can be explained with Newton’s third law of motion. During walking on the ground, the man creates an active force on the ground in the backward direction during each step, while the ground creates a reaction force pushing the man forward enabling him to walk.

19. With what speed must a ball be thrown vertically up in order to rise to a maximum height of $45m$? And for how long will it be in the air?

Final velocity of stone: $v=0m{{s}^{-1}}$(as it attains zero velocity at maximum height)

Height to which stone is to be thrown: $h=45m$

Acceleration due to gravity: $g=-10m{{s}^{-2}}$(it is negative because the stone is thrown against gravity upwards)

The initial velocity with which the stone is to be thrown: $u$

Time duration over which the stone stays in the air

Initial velocity:

$\Rightarrow {{0}^{2}}={{u}^{2}}+(2\times -10\times 45)$

\[\Rightarrow 0={{u}^{2}}+(-900)\]

\[\Rightarrow {{u}^{2}}=900\]

$\Rightarrow u=30m{{s}^{-1}}$

It is known that – $\mathrm{v=u+gt}$

Thus, $\mathrm{0=30+(}-10\times t\mathrm{)}$

$\Rightarrow t=3s$

It takes $3s$ to go up and another $3s$ to come down. So we can say that the total time the stone is air bound is $3s+3s=6s$

The initial velocity with which the stone is to be thrown is $30m{{s}^{-1}}$

Time duration over which the stone stays in the air is $6s$

20. State Newton’s second law of motion and derive it mathematically?

Ans: 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 the force.

Mathematical derivation:

Say we have an object of mass $m$that is moving along a straight line with an initial velocity, $u$. 

It is then uniformly accelerated to velocity, $v$ in time, $t$ by the application of a constant force, $F$ throughout the time, $t$. 

Initial Momentum of object: \[{{p}_{1}}=m\times u\]

Final Momentum of object: \[{{p}_{2}}=m\times v\]

Now, the change of momentum is the Final momentum subtracted by the Initial momentum

Thus, $\Delta p={{p}_{2}}-{{p}_{1}}=mv-mu=m(v-u)$

$\Rightarrow \Delta p=m(v-u)$

The rate of change of momentum is $\dfrac{\Delta p}{t}$

i.e. $\dfrac{\Delta p}{t}=\dfrac{m(v-u)}{t}$

We know that the applied force is proportional to the rate of change of momentum of the object.

$F=\dfrac{\Delta p}{t}$

$\Rightarrow F=\dfrac{m(v-u)}{t}$

But, acceleration $a=\dfrac{v-u}{t}$

Using these, we get

$\Rightarrow F=ma$

The SI unit of force is Newton ($Kgm{{s}^{-2}}$)

The second law of motion gives a method to measure the force acting on an object as a product of its mass and acceleration.

21. A bullet traveling at $360m{{s}^{-1}}$ strikes a block of softwood. The mass of the bullet is $2.0g$. Does the bullet come to rest after penetrating $10cm$ into the wood?

Find the average deceleration force exerted by the wood.

Mass of bullet: $m=2.0g=0.002kg$

Initial velocity of bullet: $u=360m{{s}^{-1}}$

Distance travelled by the bullet into the block:$s=10cm=0.1m$

To find the average deceleration force exerted by the wood block.

It is known that – ${{v}^{2}}={{u}^{2}}+2as$

$\Rightarrow {{0}^{2}}={{360}^{2}}+(2\times a\times 0.1)$

\[\Rightarrow 0=129600+(0.2a)\]

\[\Rightarrow 0.2a=-129600\]

$\Rightarrow a=-648000m{{s}^{-2}}$

The acceleration is negative because it opposes the motion of bullet.

Next, force

$\Rightarrow \mathrm{F=(0}\mathrm{.002}\times -648000)$

$\Rightarrow \mathrm{F=}-1296N$

(b) Find the time taken by the bullet to come to rest.

Distance travelled by the bullet into the block: $s=10cm=0.1m$

To find the time taken by the bullet to come to rest.

Thus, $\mathrm{0=360+(}-648000\times t\mathrm{)}$

$\Rightarrow -648000t\mathrm{=}-36\mathrm{0}$

$\Rightarrow t=5.56\times {{10}^{-4}}s$

22. Two objects A and B are dropped from a height. The object B being dropped $1s$ after A was dropped. How long after A was dropped will A and B be $10m$apart?

Ans: Given: Object B is dropped one second after object A.

To find: Time at which A and B will be $10m$ apart

We can say that the initial velocity of both A and B as ${{\mathrm{u}}_{A}}={{\mathrm{u}}_{B}}\mathrm{=0m}{{\mathrm{s}}^{\mathrm{-1}}}$, since they are dropped from rest.

For object A – ${{\mathrm{s}}_{A}}\mathrm{=}{{\mathrm{u}}_{A}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{A}}^{\mathrm{2}}$

For object B –${{\mathrm{s}}_{B}}\mathrm{=}{{\mathrm{u}}_{B}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{B}}^{\mathrm{2}}$

We take acceleration to be acceleration due to gravity because it is being dropped from a height downwards to the earth.

Since we need to find the time at which A and B will be $10m$ apart

Let’s say - ${{\mathrm{s}}_{A}}-{{\mathrm{s}}_{B}}\mathrm{=10m}$

Also, since object B is dropped one second after object A, we can say that ${{t}_{B}}={{t}_{A}}-1$

Substituting in – ${{\mathrm{s}}_{A}}-{{\mathrm{s}}_{B}}\mathrm{=10m}$

$\Rightarrow {{\mathrm{s}}_{A}}-{{\mathrm{s}}_{B}}\mathrm{=}{{\mathrm{u}}_{A}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{A}}^{\mathrm{2}}-{{\mathrm{u}}_{B}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{B}}^{\mathrm{2}}$

$\Rightarrow 10\mathrm{=}{{\mathrm{u}}_{A}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{A}}^{\mathrm{2}}-\left[ {{\mathrm{u}}_{B}}\mathrm{t+}\dfrac{1}{2}\mathrm{g(}{{\mathrm{t}}_{A}}-1{{)}^{\mathrm{2}}} \right]$

$\Rightarrow 10\mathrm{=(0}\times \mathrm{t)+}\dfrac{1}{2}\mathrm{(10}\times {{\mathrm{t}}_{A}}^{\mathrm{2}})-\left[ \mathrm{(0}\times \mathrm{t)+}\dfrac{1}{2}\mathrm{(10}\times {{\mathrm{(}{{\mathrm{t}}_{A}}-1)}^{\mathrm{2}}}) \right]$

$\Rightarrow 10\mathrm{=(5}\times {{\mathrm{t}}_{A}}^{\mathrm{2}})-\left[ 5\times \mathrm{(}{{\mathrm{t}}_{A}}^{\mathrm{2}}-2{{\mathrm{t}}_{A}}+1) \right]$

$\Rightarrow 10\mathrm{=5}{{\mathrm{t}}_{A}}^{\mathrm{2}}-\mathrm{5}{{\mathrm{t}}_{A}}^{\mathrm{2}}+10{{\mathrm{t}}_{A}}-5$

$\Rightarrow 10\mathrm{=}10{{\mathrm{t}}_{A}}-5$

\[\Rightarrow 10{{\mathrm{t}}_{A}}=15\]

$\Rightarrow {{\mathrm{t}}_{A}}=1.5s$

Thus, the time at which A and B will be $10m$ apart is $1.5s$

23. A boy throws a stone up with a velocity of $60m{{s}^{-1}}$.

(a) How long will it take to reach the maximum height? $(g=-10m{{s}^{-2}})$

Initial velocity of stone: $\mathrm{u=60m}{{\mathrm{s}}^{\mathrm{-1}}}$ 

To find time to reach maximum height.

$\Rightarrow \mathrm{v=u+gt}$

$\Rightarrow 0\mathrm{=60+(}-10\mathrm{t)}$

$\Rightarrow -10\mathrm{t=}-60$

$\Rightarrow t=6s$

(b) What will be the maximum height reached by the stone?

To find maximum height.

$\Rightarrow \mathrm{h=ut+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}^{\mathrm{2}}}$

$\Rightarrow \mathrm{h=(60}\times \mathrm{6)+}\dfrac{1}{2}\mathrm{(}-10\times {{6}^{\mathrm{2}}})$

$\Rightarrow \mathrm{h=(360)+(}-180)$

$\Rightarrow \mathrm{h=180m}$

(c) What will be its velocity when it reaches the ground?

To find velocity when it reaches the ground.

Velocity when reaching the ground:

For this, we consider the initial velocity (from its maximum attained height) is zero. And the acceleration due to gravity becomes positive because it is falling down.

i.e. Initial velocity of stone: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ 

Now,  ${{v}^{2}}={{u}^{2}}+2as$

$\Rightarrow {{v}^{2}}={{u}^{2}}+2gs$

$\Rightarrow {{v}^{2}}={{0}^{2}}+(2\times 10\times 180)$

\[\Rightarrow {{v}^{2}}=3600\]

\[\Rightarrow v=\sqrt{3600}\]

$\Rightarrow v=60m{{s}^{-1}}$

24. A certain particle has a weight of $30N$at a place where the acceleration due to gravity is $9.8m{{s}^{-2}}$.

(a) What are its mass and weight at a place where the acceleration due to gravity is $3.5m{{s}^{-2}}$?

Weight of particle:$\mathrm{w=30N}$

Acceleration due to gravity on that planet: ${{g}_{1}}=9.8m{{s}^{-2}}$

Mass of particle: $m$

To find mass and weight of particle on planet with${{g}_{2}}=3.5m{{s}^{-2}}$.

Mass of particle: $m=\dfrac{w}{g}$

$\Rightarrow m=\dfrac{30}{9.8}$

$\Rightarrow m=3.06kg$

Mass and Weight on planet with${{g}_{2}}=3.5m{{s}^{-2}}$

Mass is a constant quantity irrespective of place. 

So, $\Rightarrow m=3.06kg$

Weight: $w=m\times {{g}_{2}}$

$\Rightarrow w=3.06\times 3.5$

(b) What will be its mass and weight at a place where the acceleration due to gravity is zero?

To find mass and weight of particle on planet with${{g}_{3}}=0m{{s}^{-2}}$ s.

Mass and Weight on planet with${{g}_{3}}=0m{{s}^{-2}}$

Weight: $w=m\times {{g}_{3}}$

$\Rightarrow w=3.06\times 0$

25. Why does a person while firing a bullet holds the gun tightly to his shoulders?

Ans: A person while firing a bullet holds the gun tightly to his shoulder because of the recoil of the gun when the bullet is fired. This is in accordance with Newton’s third law of motion, which states that for every action, there is an equal and opposite reaction, that acts on different bodies. 

When a bullet is fired, the forward motion of the bullet (action) creates a recoil or backward motion of the gun (reaction). The action force being much greater will create an equivalent recoil force in the backward direction. 

If the person who holds the gun does not hold it properly to his shoulders that can result in injury. This is because the shoulder absorbs most of the force during recoil that enables the shooter to take a steady shot.

Thus, if not held tightly to his shoulders, the shot will not be precise and this can also cause the gun to fly away from his hands.

26. A car is moving with a velocity of $16m{{s}^{-1}}$ when brakes are applied. The force applied by the brakes is $1000N$. The mass of the car its passengers is $1200kg$.

How long should the brakes be applied to make the car come to a halt?

Mass of car and passengers: $m=1200kg$

Initial velocity of car: $u=16m{{s}^{-1}}$

Final velocity of car: $v=0m{{s}^{-1}}$(as it halts after applying the brake)

Force of the brake: $F=-1000N$(as it opposes the motion)

To find time duration over which brake should be applied to stop the car.

$\Rightarrow -1000\mathrm{=1200(}\dfrac{0-16}{t})$

$\Rightarrow t\mathrm{=(}\dfrac{1200\times (-16)}{-1000})$

$\Rightarrow t\mathrm{=(}\dfrac{-19200}{-1000})$

$\Rightarrow t\mathrm{=19}\mathrm{.2}s$

How far does the car travel before it comes to rest?

To find distance travelled by car before coming to halt.

$\Rightarrow \mathrm{a=(}\dfrac{0-16}{19.2})$

$\Rightarrow \mathrm{a=}-0.83m{{s}^{-2}}$

$\Rightarrow \mathrm{s=(16}\times 19.2)\mathrm{+}\dfrac{1}{2}(-0.83\times {{19.2}^{2}})$

$\Rightarrow \mathrm{s=(307}\mathrm{.2)+}(-152.98)$

$\Rightarrow \mathrm{s=154}\mathrm{.2}m$

Long Answer Questions                       (5 Marks)

1. Two objects of masses $100g$ and $200g$ are moving along the same line and direction with velocities of $2m{{s}^{-1}}$ and $1m{{s}^{-1}}$ respectively. They collide and after the collision, the first object moves at a velocity of $1.67m{{s}^{-1}}$. Determine the velocity of the second object.

Mass of object 1: ${{m}_{1}}=100g=0.1kg$

Mass of object 2: ${{m}_{2}}=200g=0.2kg$ 

Initial velocity of object 1: ${{u}_{1}}=2m{{s}^{-1}}$

Initial velocity of object 2:${{u}_{2}}=1m{{s}^{-1}}$

Final velocity of object 1:${{v}_{1}}=1.67m{{s}^{-1}}$

To find: Final velocity of object 2:${{v}_{2}}$

Here, \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]is the total sum of momentum of objects before collision and \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] is the total sum of momentum of objects after collision.

\[\Rightarrow (0.1\times 1.67)+(0.2\times {{v}_{2}})=(0.1\times 2)+(0.2\times 1)\]

\[\Rightarrow (0.167)+(0.2\times {{v}_{2}})=(0.2)+(0.2)\]

\[\Rightarrow (0.167)+(0.2\times {{v}_{2}})=0.4\]

\[\Rightarrow 0.2\times {{v}_{2}}=0.4-0.167\]

\[\Rightarrow 0.2\times {{v}_{2}}=0.233\]

$\Rightarrow {{v}_{2}}=1.165m{{s}^{-1}}$

Thus, the velocity of the second object is \[1.165m{{s}^{-1}}\].

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

Mass of truck:$\mathrm{m=7metricTonne=7000kg}$

Initial velocity of truck: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is stars from rest)

Distance travelled: $\mathrm{s=400m}$

Time duration of travel:$t=20s$

Acceleration of the truck

Force acting on the truck

Thus, $\mathrm{400=(0}\times 20\mathrm{)+}\dfrac{1}{2}(a\times {{20}^{2}})$

$\Rightarrow 400\mathrm{=}\dfrac{1}{2}(a\times 400)$

$\Rightarrow 800\mathrm{=}(a\times 400)$

$\Rightarrow a=2m{{s}^{-2}}$

$\Rightarrow \mathrm{F=(7000}\times 2)$

$\Rightarrow \mathrm{F=14000}N$

Thus, the acceleration of the truck is $2m{{s}^{-2}}$, and the force acting on the truck is $14000N$ .

3. A stone is dropped from a $100m$high tower. How long does it take to fall?

The first $50m$

Initial velocity of stone: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starts from rest, before being dropped)

Height of the tower: $\mathrm{s=100m}$ 

Distance travelled in case A: ${{\mathrm{s}}_{1}}\mathrm{=50m}$ (first half distance)

Distance travelled in case B: ${{\mathrm{s}}_{2}}\mathrm{=50m}$ (next half distance)

To find: Time duration of travel:$t$ during the first $50m$.

Since the stone is dropped from a height, we can consider its acceleration to be equal to the acceleration due to gravity.

Acceleration of stone: Acceleration due to gravity  $\Rightarrow a=g=10m{{s}^{-2}}$

${{\mathrm{s}}_{1}}\mathrm{=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

$\Rightarrow \mathrm{50=(0}\times t\mathrm{)+}\dfrac{1}{2}(10\times {{t}^{2}})$

$\Rightarrow \mathrm{50=5}\times {{t}^{2}}$

$\Rightarrow {{t}^{2}}=10$

$\Rightarrow t=\sqrt{10}=3.16s$

$\Rightarrow {{t}_{1}}=3.16s$

(b) The second $50m$

To find: Time duration of travel:$t$ during the second $50m$.

Time duration for the next $50m$can be found by subtracting time for the first half distance from the time for the total distance of travel.

$\Rightarrow 10\mathrm{0=(0}\times t\mathrm{)+}\dfrac{1}{2}(10\times {{t}^{2}})$

$\Rightarrow 100\mathrm{=5}\times {{t}^{2}}$

$\Rightarrow {{t}^{2}}=20$

$\Rightarrow t=\sqrt{20}=4.47s$

$\Rightarrow t=4.47s$

Thus the time for the second half is – ${{t}_{2}}=t-{{t}_{1}}$

$\Rightarrow {{t}_{2}}=4.47-3.16$

$\Rightarrow {{t}_{2}}=1.31s$

4. A body of mass $10kg$ starts from rest and rolls down an inclined plane. It rolls down $10m$ in $2s$.

What is the acceleration attained by the body?

Mass of body:$\mathrm{m=10kg}$

Distance travelled on inclined plane: $\mathrm{s=10m}$

Time duration of rolling:$t=2s$

To find acceleration attained by the body.

Thus, $\mathrm{10=(0}\times 2\mathrm{)+}\dfrac{1}{2}(a\times {{2}^{2}})$

$\Rightarrow 10\mathrm{=}\dfrac{1}{2}(a\times 4)$

$\Rightarrow 10\mathrm{=}(a\times 2)$

$\Rightarrow a=5m{{s}^{-2}}$

What is the velocity of the body at $2s$?

To find velocity of body at $t=2s$.

Thus, $\mathrm{v=0+(5}\times 2\mathrm{)}$

$\Rightarrow v=10m{{s}^{-1}}$

What is the force acting on the body?

To find force acting on the body.

$\Rightarrow \mathrm{F=(10}\times 5)$

$\Rightarrow \mathrm{F=50}N$

5. A body of mass $2kg$ is at rest at the origin of a frame of reference. A force of $5N$acts on it at $t=0s$. The force acts for $4s$ and then stops.

(a) What is the acceleration produced by the force on the body?

Mass of body:$\mathrm{m=2kg}$

Initial velocity of body: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starting from rest)

Force acting on the body: $F=5N$

Time duration for which force is exerted:$t=4s$

To find acceleration produced by the force on the body.

$\Rightarrow a=\dfrac{5}{2}$

$\Rightarrow a=2.5m{{s}^{-2}}$

(b) What is the velocity at $t=4s$?

To find velocity of body at $t=4s$.

Thus, $\mathrm{v=0+(2}\mathrm{.5}\times 4\mathrm{)}$

(c) Draw the v-t graph for the period $t=0s$ to $t=6s$.

To plot the v-t graph.

Plotting the v-t graph

Using the formula – $\mathrm{v=u+at}$

At $t=0s\Rightarrow v=0+(2.5\times 0)=0m{{s}^{-1}}$

At $t=1s\Rightarrow v=0+(2.5\times 1)=2.5m{{s}^{-1}}$

At $t=2s\Rightarrow v=0+(2.5\times 2)=5m{{s}^{-1}}$

At $t=3s\Rightarrow v=0+(2.5\times 3)=7.5m{{s}^{-1}}$

At $t=4s\Rightarrow v=0+(2.5\times 4)=10m{{s}^{-1}}$

At $t=5s\Rightarrow v=0+(2.5\times 5)=12.5m{{s}^{-1}}$

At $t=6s\Rightarrow v=0+(2.5\times 6)=15m{{s}^{-1}}$

(Image will be Uploaded Soon)

(d) Find the distance traveled in $6s$.

To find distance travelled in $t=6s$.

This can be found by calculating the area under the v-t graph.

This is a triangle with base as $6$ and height as $15$

Area of triangle: $\dfrac{1}{2}\times b\times h=\dfrac{1}{2}\times 6\times 15=45$

Thus, the distance travelled in $t=6s$ is $45m$.

Why is Chapter 9 the Forces and Laws of Motion Important for Students?

Forces and Laws of Motion is a crucial chapter. In this chapter, students learn about the Laws of Motion as given by Sir Issac Newton, the various terminologies used for denoting forces of nature and motion of objects, the relation between objects, force, motion, etc. It gives students valuable knowledge about the working of the universe. 

Force and Laws of Motion Class 9 Important Questions with Answers 

Force and Laws of Motion important questions will ensure that students are prepared for all types of questions that might appear in exams. It will help them understand how well they have prepared, how good their time management is, and how they can improve. 

We suggest that students keep practicing the class 9 science chapter 9 important questions to familiarize themselves with the various types of questions, save them precious time in exams, and score good marks. 

The Relevance of Chapter 9 Forces and Laws of Motion

Chapter 9 of Class 9 Physics deals with Forces and Laws of Motion. The chapter talks about the multiple forces in nature, how they act upon objects, how things react, and most importantly, the fundamental Laws of Motion as given by Sir Issac Newton. 

Students need to study this chapter thoroughly and from a young age, so remember them later.  

How to Study for Chapter 9 Forces and Laws of Motion

Given below is the proforma we suggest our students follow for the best possible preparation- 

Study the given class and study notes carefully, and make their notes for better understanding.

Consult online classes on our website.

Watch videos of inertia, forces of motion, and simple examples to understand the concept better.

Practice the Force and Laws of Motion Class 9 Important Questions so that they get to know all the possible questions that might appear.

Practice the questions repeatedly so that they get the best preparation possible and attain good marks in exams. 

Introduction to Forces and Laws of Motion

A force refers to the effort to change an object's state at rest or even at motion. It might also change the object's velocity and direction. The shape of an object can also be changed by force. 

Forces can be Two Types- 

Balanced Forces: Balanced forces do not result in any changes in motion. When it is applied to an object, there will be no such force acting upon the object.

Unbalanced Forces: Unbalanced Forces move in the direction of the force in the highest magnitude. It acts upon an object and can change its speed and direction of motion. 

Types of Forces 

There are different types of forces acting around us. 

Gravitational Force: One of the most commonly known forces, gravitational force, refers to the force that exists due to the attraction between two bodies due to their masses. It is denoted by 

F= G (m1m2/r2) 

Here, G refers to the universal constant while m1 and m2 are the masses of the bodies and r is the distance between them. 

Electromagnetic Force- Electromagnetic Force refers to the force exerted by two charged particles at each other. Two common examples of electromagnetic forces are Friction and Tension. 

Nuclear Force- There are protons and neutrons in every atom. The nuclear force helps to bind neutrons and protons and holds them together in an atom. It is also known as Strong Force or Nuclear Interactions. This force is larger in magnitude than any other force. However, it has a concise range of influence, so in that respect, the other forces are more dominating. 

Weak Force- Weak Forces are responsible for phenomenons known as beta decay. Sometimes a neutron changes itself into a proton, emits an electron, and a particle known as antineutrino. This process is known as Beta Decay. 

The weak force is a force of attraction that works at a concise range of 0.1% of a proton's diameter. These forces differ from gravitational, electromagnetic, and nuclear forces and are known as Weak Forces.

Three Main Laws of Motion

The three primary Laws of Motion are Newton's Laws of Motion. They are as follows-

Newton's First Law of Motion states that an object remains at rest unless an external force acts upon it. Similarly, an object in motion stays in motion in the same direction unless acted upon by an external force. 

Newton's Second Law of Motion states that the force acting on a body will be directly proportional to the rate of change in its momentum.

Newton's Third Law of Motion states that "For every action, there is an equal and opposite reaction." 

The Terminology Used in this Chapter

Net Force: Net force acting on a body refers to the fact that when multiple forces work on one body, it tends to change into one component. 

Frictional Force: The force present between two surfaces in contact and opposes relative motion is known as Frictional Force. 

Inertia: Inertia refers to all bodies' tendency to resist changes in a state of rest or motion. However, all bodies do not have the same inertia. The inertia of a body is directly proportional to its mass.

Momentum: The momentum of an object refers to the product of its mass and velocity. p=mv The impacts that an object or body produces depend on its mass and speed, and this vector quantity has direction and magnitude. 

Inertial and Non-Inertial Frames: Inertial Frame refers to the frame where Newton's Laws hold. On the other hand, Non-Inertial Frames refer to the reference frame where Newton's Law of Motion does not fit. A non-inertial frame undergoes acceleration with respect to an inertial frame. In a non-inertial frame, an accelerometer will detect a non-zero acceleration. 

Significance of Key Questions in CBSE Class 9 Science Chapter 9 - Force and Laws of Motion

The significance of Key Questions for CBSE Class 9 Science Chapter 9 - "Force and Laws of Motion" provided by Vedantu is substantial and cannot be emphasized enough. These curated questions play a crucial role in a student's exam preparation journey. They serve as a targeted resource for revising key concepts and assessing one's understanding of the subject matter. Vedantu's focus on quality education is evident in these questions, which are designed to align with the CBSE curriculum and examination pattern. By practicing these questions, students gain confidence in their problem-solving abilities and enhance their grasp of the fundamental principles of force and motion. These questions are not just aids for academic excellence but also tools for fostering a deeper understanding and appreciation of the laws that govern the physical world.

Why Choose Vedantu to Study?

Vedantu is a one-stop platform for all students who require class notes and solved questions to study. After careful assessment and research, we have provided the best possible class notes, revision notes, and class 9 science chapter 9 important questions for our students. These notes and questions are written by our experts who have immense knowledge of the respective subjects. They carefully go through all the syllabus, notes, and guidelines given by the board and the previous years' question papers before writing the letters and questions available on our website. 

Vedantu also conducts online classes on its website, which can help students understand the study material better. These sessions are also recorded and available on our website if anyone misses the classes and wishes to go through the classes later.

Conclusion 

To conclude, we can say that motion and motion laws are a significant chapter for students in Class 9. Not only does it hold importance in school exams, but it also has considerable weightage in board exams and competitive exams later on. So, we at Vedantu make sure that our students get all the study notes, revision notes, and essential questions of force and motion laws in one place. They are also available in downloadable PDF format so that students can study anytime and anywhere they want. We suggest students practice the critical questions repeatedly for the best preparation possible and fetch good marks. 

Important Related Links for CBSE Class 9 

FAQs on Important Questions for CBSE Class 9 Science Chapter 9 - Force and Laws of Motion

1. When a Carpet is Beaten with a Stick, Dust Comes off it. Explain why?

Answer: When a carpet is beaten with a stick, the carpet is set to motion and it moves back and forth. However, the dust particles on the carpet are not in motion due to the inertia of rest, and they come off the carpet fibers. It is due to this same reason, fruits fall off from the tree on moving the branches vigorously.

2. What are the three Laws of Motion?

Answer: Newton’s first law of motion states that an object at rest remains to be in the state of rest unless acted upon by an unbalanced external force, and an object in motion remains to be in the state of motion in the same direction unless acted upon by an unbalanced external force.

Newton’s second law of motion states that the force acting on a body is directly proportional to the rate of change of its momentum.

Newton’s third law of motion states that for every action there is an equal and opposite reaction.

3. What are the Forces Acting upon a Book Lying on a Table?

Answer: The net force acting upon a book lying on a table is zero. The weight of the book (which is a force) acts in the downward direction. The normal reaction force on the book, which is equal to the weight of the book, acts in the opposite direction to its weight.

4. Are the Important Questions for CBSE Class 9 Science Chapter 9- Force and Laws of Motion Helpful for the Exam Preparation?

Answer: Yes, the important questions for CBSE Class 9 Science Chapter 9- Force and Laws of Motion are definitely helpful for the exam preparation. These important questions and answers from Force and Laws are written and compiled by the subject-matter experts at Vedantu, in a very easy to understand manner. Students can download the important questions for CBSE Class 9 Science Chapter 9- Force and Laws of Motion PDF for free of cost from Vedantu. Since questions from all the essential topics are covered in this PDF, students can rely upon it for their exam preparation and practice purposes. 

5. Which chapter is important for Class 9 Science?

Ans: Class 9 Science textbook contains 15 chapters. Competition and the struggle to fare better than others is increasing and will continue to increase in future. Science can be challenging to study if you don’t have a strong grasp of the basics. You will be forced to cram. You should read all the chapters. You should not start your preparation with the thought of skipping certain chapters. However, you can give more time to study topics such as Motion, Matter, Atoms and Molecules, Work, Energy and Power and Tissue and Natural Resources.

6. What is force according to Chapter 9 of Class 9 Science?

Ans: Force is any movement of push or pull. It is an interaction between two objects that are in contact with each other. If there is no interaction, there will be no force between objects. Depending on the movement and relative force, either the speed of the object increases or it stops moving. Force can also change the shape of an object. Newton (N) is the SI unit of force. We apply force daily when we open a door or when we walk.

7. Why do we fall ill according to Chapter 9 of Class 9 Science?

Ans: Health and disease are complicated terms. Health is a broad area that encompasses mental, physical and social well being. Our body’s immune system protects us from various pathogens that may invade our body. We become ill when we consume contaminated food, water or air. We can even become ill by being in contact with someone who has diseases that can be transmitted. Vectors such as mosquitoes also act as agents carrying pathogens from an infected person to another. To find more about this chapter refer to the study materials provided by Vedantu that can be downloaded absolutely free of cost.

8. Why do we need safety belts when we apply brakes suddenly? Explain scientifically from the information obtained from Chapter 9 of Class 9 Science.

Ans: Newton’s first law of motion is the law of inertia. The law states that an object that is at rest will stay at rest unless force is applied to it. When we are sitting in a car, the car is moving but our body is at rest. When a car stops we move out of the inertia. If the car stops suddenly, our bodies too will move suddenly and abruptly. This can cause injury. Seat belts are a type of safety mechanism that stops our body from moving forward and protects us.

9. What do Newton’s laws of motion say according to Chapter 9 of Class 9 Science?

Ans: There are three laws of motion given by Newton. It helps us understand why an object is at rest, how it interacts with other objects at rest and when it moves. The first law of motion describes the state of inertia. It tells us why we need to apply force to change the state, speed or direction of an object. The second law states that force is equal to the product of mass and acceleration. The third law explains the way two bodies interact. It states that when two bodies interact they apply force on each other which is equal in magnitude but opposite in direction.

CBSE Class 9 Science Important Questions

Cbse study materials.

Class 9 Science Case Study Questions Chapter 9 Force and Laws of Motion

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Case study Questions in Class 9 Science Chapter 9 are very important to solve for your exam. Class 9 Science Chapter 9 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving  Class 9 Science Case Study Questions  Chapter 9 Force and Laws of Motion

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In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Force and Laws of Motion Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Science  Chapter 9 Force and Laws of Motion

Case Study/Passage-Based Questions

Case Study 1: The sum of the momentum of the two objects before the collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision. The Law of conservation of momentum is applicable to the system of particles. Answer the following questions.

(i)Law of conservation of momentum is applicable to

(a) A system of particles

(b) Only for 2 particles

(c) Only for 1 particle

(d) None of the above

Answer: (a) A system of particles

(ii) Law of conservation of momentum holds good provided that

(a) There should be external unbalanced force acting on particles

(b) There should not be any external unbalanced force acting on particles

(c) No internal forces acting on particles

Answer: (b) There should not be any external unbalanced force acting on particles

(iii)The total momentum of the two objects when collision occurs is

(a) Changed

(b) Remains conserved

(c) Become zero

Answer: (b) Remains conserved

(iv) State law of conservation of momentum.

Answer: The sum of momentum of the two objects before collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision.

(v) If action and Reaction are equal in magnitude and opposite in direction then why they do not cancel each other?

Answer: Action and Reaction are equal in magnitude and opposite in direction but they do not cancel each other because they are not action on sane object. As these forces are acting on different object hence produces different acceleration and does not cancel each other.

Case Study 2: The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and never on the same object. It is important to note that even though the action and reaction forces are always equal in magnitude; these forces may not produce accelerations of equal magnitudes, this is because each force acts on a different object that may have a different mass. The two opposing forces are also known as action and reaction forces. Answer the following questions.

(i) Action reaction forces are always

(a) Equal and in the same direction

(b) Equal and in the opposite direction

(c) Unequal and in the same direction

Answer: (b) Equal and in the opposite direction

(ii) Which of the following is correct about action reaction forces?

(a) They act on different objects

(b) They are equal in magnitude and opposite in direction

(c) Both forces acted on different object simultaneously

(d) All the above

Answer: (d) All the above

(iii) State third law of motion

Answer: The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and neveron the same object.

(iv) Give 5 examples of third law of motion

Answer: Examples of third law of motion are Swimming or rowing a boat. •Static friction while pushing an object. •Walking. •Standing on the ground or sitting on a chair. •The upward thrust of a rocket. •Resting against a wall or tree.

Case Study 3:

Force is a push or pull that can change the state of motion of an object. According to Newton’s first law of motion, an object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force. This is known as the law of inertia. Newton’s second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this can be expressed as F = ma, where F is the force, m is the mass of the object, and a is the acceleration produced. Newton’s third law of motion states that for every action, there is an equal and opposite reaction. This means that whenever an object exerts a force on another object, the second object exerts an equal and opposite force on the first object. Understanding the concepts of force and the laws of motion helps us explain the behavior of objects and the factors that influence their motion.

What is force? a) A change in the state of motion of an object b) A push or pull that can change the state of motion of an object c) The mass of an object d) The velocity of an object Answer: b) A push or pull that can change the state of motion of an object

What does Newton’s first law of motion state? a) An object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force. b) The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. c) For every action, there is an equal and opposite reaction. d) The force exerted by an object is equal to its mass multiplied by its acceleration. Answer: a) An object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force.

What is Newton’s second law of motion? a) An object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force. b) The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. c) For every action, there is an equal and opposite reaction. d) The force exerted by an object is equal to its mass multiplied by its acceleration. Answer: b) The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

What is Newton’s third law of motion? a) An object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force. b) The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. c) For every action, there is an equal and opposite reaction. d) The force exerted by an object is equal to its mass multiplied by its acceleration. Answer: c) For every action, there is an equal and opposite reaction.

How do the concepts of force and the laws of motion help us? a) Explain the behavior of objects and the factors that influence their motion. b) Calculate the speed of objects. c) Classify objects into different categories. d) Determine the position of objects. Answer: a) Explain the behavior of objects and the factors that influence their motion.

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Extra Questions for Class 9 Science Chapter 8 Motion

Extra questions for Class 9 Science Chapter 8 Motion with answers is given below. Our subject expert prepared these solutions as per the latest NCERT textbook. These questions will be helpful to revise the all topics and concepts. CBSE Class 9 extra questions are the most simple and conceptual questions that are prepared by subject experts for the students to study well for the final exams. By solving these extra questions, students can be very efficient in their exam preparations.

Motion Class 9 Science Extra Questions and Answers

Very short answer questions.

1: The phenomenon of motion was placed on a sound scientific footing by two scientists. Write their names.  Answer:  Galileo Galilei and Isaac Newton.

2: Are rest and motion absolute or relative terms?  Answer:  They are relative terms.

3: Suppose a ball is thrown vertically upwards from a position P above the ground. It rises to the highest point Q and returns to the same point P. What is the net displacement and distance travelled by the ball?  Answer:  Displacement is zero. Distance is twice the distance between position P and Q.

4: Which speed is greater: 30 m/s or 30 km/h? Answer:  30 m/s

5: What do you mean by 2 m/s 2 ? Answer:  The velocity of the body increases by 2 m/s after every second.

6: Can uniform linear motion be accelerated?  Answer:  No

7: Define one radian.  Answer:  It is the angle which is subtended at the centre by an arc having a length equal to the radius of the circle.

8: What is the relation between linear velocity and angular velocity Answer:  Linear velocity = Angular velocity x Radius of circular path.

Short Answer Type Questions

1: Give an example of a body which may appear to be moving for one person and stationary for the other. 

Answer:  The passengers in a moving bus observe that the trees, buildings as well as the people on the roadside appear to be moving backwards. Similarly, a person standing on the roadside observes that the bus (along with its passengers) is moving in forward direction. But, at the same time, each passenger in a moving bus or train observes, his fellow passengers sitting and not moving. Thus, we can tell that motion is relative.

2: How can we describe the location of an object? 

Answer:  To describe the position of an object we need to specify a reference point called the origin. For example, suppose that a library in a city is 2 km north of the railway station. We have specified the position of the library with respect to the railway station i.e., in this case, the railway station acts as the reference point.

3: What do you mean by average speed? What are its units? 

Answer:  Average speed is defined as the average distance travelled per unit time and is obtained by dividing the total distance travelled by the total time taken. The unit of average speed is the same as that of the speed, that is, ms -1 .

4: What is the difference between uniform velocity and non-uniform velocity?

Answer: Uniform velocity: An object with uniform velocity covers equal distances in equal intervals of time in a specified direction, e.g., an object moving with speed of 40km h -1 towards west has uniform velocity.

Non-uniform velocity: When an object covers unequal distances in equal intervals of time in a specified direction, or if the direction of motion changes, it is said to be moving with a non-uniform or variable velocity, e.g., revolving fan at a constant speed has variable velocity.

5: Differentiate between distance and displacement. Answer: 

6: What are the uses of a distance-time graph?

Answer:  The various uses of a distance-time graph are as follows: (a) It tells us about the position of the body at any instant of time. (b) From the graph, we can see the distance covered by the body during a particular interval of time. (c) It also gives us information about the velocity of the body at any instant of time.

7: A cyclist cycles for t second at a speed of 3 m/s and then for the same time at a speed of 5 m/s along a straight road due north. What is the average speed of the cyclist?

Answer: You may compute arithmetic mean to compute average speed. Alternately, Total Time taken = 2t seconds Total Distance covered = 3t + 5t = 8t m

8: When will you say a body is in (i) uniform acceleration? (ii) nonuniform acceleration?

Answer:  (i) A body is said to be in uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time.

(ii) A body is said to be in nonuniform acceleration if the rate of change of its velocity is not constant in equal intervals of time. 9: A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

10: A train starting from a railway station and moving with uniform acceleration attains a speed 40 km/h in 10 minutes. Find its acceleration.

11: A bus starting from rest moves with a uniform acceleration of 0.1 ms –2   for 2 minutes.  Find (a) the speed acquired, (b) the distance travelled.

Answer: Given,  Initial speed (u) = 0 m/s  Acceleration of bus (a) = 0.1 ms –2    Time taken (t) = 2 min = 2 × 60 = 120 s  Final velocity = ?  Distance Covered (S) = ?

∴  v = u + at    = 0 + 0.1 × 120 = 12 m/s

∴ S = ut + ½at 2   = 0 + (0.1 × 120 2  )/2 = 0.1 × 14400 /2 = 720m  

12: A train is travelling at a speed of 90 kmh –1 . Brakes are applied so as to produce a uniform acceleration of – 0.5 ms –2 . Find how far the train will go before it is brought to rest.

Answer: Given,

13: A trolley, while going down an inclined plane, has an acceleration of 2 cm s –2 . What will be its velocity 3 s after the start?

Answer: Given, Initial velocity (u) = 0 cm s –1 . acceleration (a) = 2 cm s –2 . time taken (t) = 3s

 ∴ v = u + at ⇒ v = 0 + 2 × 3 ⇒ v = 6 cm s –1          

14: A racing car has a uniform acceleration of 4 m s –2 . What distance will it cover in 10 s after start?

Answer: Given, Initial velocity of car (u) = 0 m s –1 . acceleration (a) = 4 m s –2 . time taken (t) = 10 s Distance covered (S) = ?

15: A train is running at a speed of 72 km/h. It crosses a bridge of length half kilometer in 1 minute. Calculate the length of the train.

16: What is the nature of the distance-time graphs for uniform and non-uniform motion of an object? 

Answer: (a) When an object moves with uniform (constant) speed, it covers equal distances in equal intervals of time. It will make a straight line on distance-time graph making an angle with time (x-axis) axis.

(b) For an object moving in a non-uniform motion will cover unequal distances in equal intervals of time. It will be a curve on a distance-time graph.

Long Answer Type Questions

1: With the help of a graph, derive the relation v = u + at. 

Answer:  Consider the velocity-time graph of an object that moves under uniform acceleration as shown in the figure (u≠0).

From this graph, we can see that initial velocity of the object (at point A) is u and then it increases to v (at point B) in time t. The velocity changes at uniform rate a. As shown in the figure, the lines BC and BE are drawn from point B on the time and the velocity axes respectively. The initial velocity is represented by OA. The final velocity is represented by BC. The time interval t is represented by OC. BD = BC – CD, represents the change in velocity in time interval t. If we draw AD parallel to OC, we observe that BC = BD + DC = BD + OA

Substituting, BC with v and OA with u, we get v = BD + u or, BD = v – u    …(i)

Thus, from the given velocity-time graph, the acceleration of the object is given by Change in velocity a = (Change in velocity)/(Time Taken) = BD/AD = BD/OC

Substituting OC with t, we get a = BD/t ⇒ BD = at   …(ii)

From equations (i) and (ii), we have v-u = at or, v =u + at

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Assertion and Reason Questions for Class 9 Science Chapter 8 Motion

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Here we are providing assertion reason questions for class 9 science. This article covers assertion reason questions based on Class 9 Science Chapter 8 Motion . To check the answer, click on ‘Answer’ given below each question. After clicking it will expand.

Directions: In each of the following questions, a statement of Assertion is given and a corresponding statement of Reason is given just below it. Of the statements, given below, mark the correct answer as: (a) Both assertion and reason are true and reason is the correct explanation of assertion. (b) Both assertion and reason are true but reason is not the correct explanation of assertion. (c) Assertion is true but reason is false. (d) Both Assertion and Reason are false.

Q.1. Assertion : An object may acquire acceleration even if it is moving at a constant speed. Reason : With change in the direction of motion, an object can acquire acceleration.

Q.2. Assertion : Displacement of an object may be zero even if the distance covered by it is not zero. Reason : Displacement is the shortest distance between the initial and final position.

Q.3. Assertion : The graph between two physical quantities P and Q is straight line, when P/Q is constant. Reason : The straight line graph means that P is proportional to Q or P is equal to constant multiplied by Q.

Q.4. Assertion : Velocity versus time graph of a particle in uniform motion along a straight path is a line parallel to the time axis. Reason : In uniform motion the velocity of a particle increases as the square of the time elapsed.

Q.5. Assertion : the speedometer of a car measures the instantaneous speed of the car. Reason : Average speed is equal to the total distance covered by an object divided by the total time taken.

Q.6. Assertion : An object may have acceleration even if it is moving with uniform speed. Reason : An object may be moving with uniform speed but it may be changing its direction of motion.

Q.7. Assertion : Motion with uniform velocity is always along a straight line path. Reason : In uniform velocity, speed is the magnitude of the velocity and is equal to the instantaneous velocity.

Q.8. Assertion : If a particle is moving with constant velocity, then average velocity for any time interval is equal to instantaneous velocity. Reason : If average velocity of a particle moving on a straight line is zero for a given time interval, then instantaneous velocity at some instant within this interval must be zero.

Q.9. Assertion : The displacement of an object can be either positive, negative or zero. Reason : Displacement has both the magnitude and direction.

Q.10. Assertion : When the displacement of a body is directly proportional to the square of the time. Then the body is moving with uniform acceleration. Reason : The slope of velocity-time graph with time axis gives acceleration.

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Test: Motion- Case Based Type Questions- 2 - Class 9 MCQ

10 questions mcq test - test: motion- case based type questions- 2, direction: the distance moved by a student at different intervals of time, while walking to school, is given in the table below. a distance time graph for the motion of the student is also shown. study the table and the graph and choose correct options to answer any four questions given below: q. what will be the speed if the body travels a distance of 15 metres in 10 seconds.

Direction: The distance moved by a student at different intervals of time, while walking to school, is given in the table below. A distance time graph for the motion of the student is also shown. Study the table and the graph and choose correct options to answer any four questions given below: Q. Name the physical quantities denoted by the slope of the distance–time graph.

Displacement

All of these

Speed is the physical quantity denoted by slope of distance-time graph.

Direction: The distance moved by a student at different intervals of time, while walking to school, is given in the table below. A distance time graph for the motion of the student is also shown. Study the table and the graph and choose correct options to answer any four questions given below: Q. What does the shape of the graph suggest about the type of motion ?

The graph indicates circular motion.

The graph indicates non-uniform motion.

The graph indicates uniform motion.

The graph indicates uniform circular motion.

As from point A to point C, the student moves with constant speed and from point C to D he is reducing the speed and from D to E again constant speed acceleration.

Direction: The distance moved by a student at different intervals of time, while walking to school, is given in the table below.

A distance time graph for the motion of the student is also shown. Study the table and the graph and choose correct options to answer any four questions given below:

Q. State a condition under which a body moves in such a way that the magnitude of its average velocity is equal to its average speed.

• A. When a body moves along a circular path.
• B. When a body moves along a straight line.
• C. When a body is in a uniform circular motion.
• D. None of the above

Q. An object is said to be moving with ___________ if it covers equal distances in equal intervals of time.

• A. non-uniform speed
• B. uniform speed
• C. average speed
• D. uniform displacement

Direction: The graph shows the distance-time graph of three objects A, B and C. Study the graph and answer any four questions.

Q. How far has B travelled by the time it passes C

Q. All the three objects A,B & C __________

• A. once met at a single point.
• B. were at the same point at the beginning.
• C. will meet at the same point at the end.
• D. were never at the same point on the road.

Q. Which of the three is travelling the fastest?

All of them are at rest.

Speed = Distance/Time

From the graph, we see that,

A travels from 6 to 12 km in 2 hours.

B travels from 0 to 12 km in 1.4 hours.

C travels from 2 to 12 km in 1.6 hours.

Since, B travels the most distance, in the smallest time, it is travelling the fastest.

Q. How far has C travelled when B passes A?

The distance travelled by A is 8 km.

Q. The distance-time graph is a ______________ when the object is at rest.

• A. straight line passing through origin
• B. straight line parallel to time axis
• D. straight line intersecting y-axis

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