202110243-TRIUMPH-STUDENT-WORKBOOK-PHYSICAL_SCIENCE-G09-PART1

AS2-Asking questions and making hypothesis Short Answer Type Questions 12. Answer the following questions in 3-4 sentences. (i) [(Session 2.2)] In a school debate, Arun said, “A car can have a constant speed and a changing velocity” Amit said, “A car can have a constant velocity and a changing speed.” Are these statements correct? What questions will you ask your friend to check whether these hypotheses are correct or not? (ii) [(Session 2.2)] a. The object in motion may or may not possess acceleration. b. The velocity of a body moving with constant speed can be changed. What questions would you ask your teacher to clarify your doubts regarding both the statements given above? CHAPTER 2. MOTION 49

Long Answer Type Questions 13. Answer the following questions in 6-8 sentences. (i) [(Session 2.2)] In this chapter you came across terms like velocity, average velocity, instantaneous velocity. Is it necessary to have different terms to describe the same physical quantity ? Give reasons. AS3-Experimentation and field investigation Long Answer Type Questions 14. Answer the following questions in 6-8 sentences. (i) [(Session 2.2)] Write an activity to show a situation where the speed changes but the direction of motion remains constant. CHAPTER 2. MOTION 50

AS4-Information skills and projects Very Short Answer Type Questions 15. Answer the following questions in one sentence. [Refer to Session 2.2 ] (i) Complete the table. v (m/s) u (m/s) a (m/s2 ) t(s) 0 30 20 3 0 6 CHAPTER 2. MOTION 51

AS5-Communication through drawing and model making Long Answer Type Questions 16. Answer the following questions. (i) [(Session 2.3)] Draw displacement v/s time graphs to show: a. Body which is stationary b. Body which is moving in a uniform velocity c. Body with variable velocity which is accelerating d. Body with variable velocity which is decelerating CHAPTER 2. MOTION 52

(ii) [(Session 2.3)] Draw velocity–time graph to show: a) Zero acceleration b) Uniform acceleration c) Variable acceleration CHAPTER 2. MOTION 53

17. Answer the following question. (i) [(Session 2.2)] Draw a neat diagram to show an ant moving on the surface of a ball. AS7-Application to daily life, concern to bio diversity Short Answer Type Questions 18. Answer the following questions in 3-4 sentences. (i) [(Session 2.2)] What is the acceleration of a vehicle moving in a straight line that changes its velocity from 100 km/h to a dead stop in 10 s? CHAPTER 2. MOTION 54

Long Answer Type Questions 19. Answer the following questions in 6-8 sentences. (i) [(Session 2.3)] Write real life examples to show : a. When the speed becomes zero for an instant. b. When the acceleration of an object is zero. c. When the body is accelerating. d. When the body is in negative acceleration. CHAPTER 2. MOTION 55

Objective Questions AS1-Conceptual Understanding 20. Choose the correct answer. (i) Average velocity is (B) Displacement + Time taken (A) Displacement+ 1/2 Time taken (C)Displacement /Total time taken (D)Displacement –1/2 Time taken (ii) The change of position of an object with respect to time is called (A) harmony (B) constant velocity (C) motion (D)constant acceleration (iii) Distance and displacement of an object are both the same when (A) the path of the object is circular (B) path of the object is random (C)path of the object is uniform (D)path of the object is linear (iv) If a car travels in such a way that the initial and the final position of the car is the same then (A) the distance travelled and displacement of car both are same (B) the distance travelled by the car is more than its displacement (C)the displacement of the car is zero (D)both B & C . (v) The motion in which the velocity is not constant is called as CHAPTER 2. MOTION 56

(A) non–uniform motion (B) uniform motion (C)linear motion (D)random motion (vi) For uniform motion, the velocity time graph is always (A) a curve (B) a straight line (C)not a straight line (D)none of the above (vii) The rate of change of velocity is called as . (A) force (B) speed (C) acceleration (D) momentum (viii) Considering the speed of the object remains constant, the velocity of the object can change in (A) non linear motion (B) circular motion (C) linear motion (D)both A and B AS2-Asking questions and making hypothesis 21. Choose the correct answer. (i) While whirling a stone tied to the end of a string, which of the statements is true? (A) The velocity of the stone changes everytime even though the speed is constant. (B) The velocity of the stone does not change. (C)The speed is constant and the velocity is also constant. (D)None of the above AS4-Information skills and projects 22. Choose the correct answer. (i) Speed is a scalar quantity as CHAPTER 2. MOTION 57

(A) it gives direction and magnitude both. (B) it gives direction only. (C)it gives magnitude only. (D)it does not give magnitude and direction. CHAPTER 2. MOTION 58

3. LAWS OF MOTION SESSION 1 INTRODUCTION AND FIRST LAW OF MOTION 1.1 Mind Map 1.2 Terminology i. Uniform Motion – Motion in which an object covers equal distances in equal intervals of time. ii. Inertia – The tendency to resist the change of state of motion. iii. Mass –The quantity of matter present in a body is known as its mass. SESSION 1. INTRODUCTION AND FIRST LAW OF MOTION 59

iv. Force – The push or pull acting on an object. v. Isaac Newton – Scientist who gave the Laws of motion and gravitation. vi. Galileo – Scientist who invented the concept of relative motion and Telescope. 1.3 Solved Examples Q1. A body of mass ‘m’ is kept on the horizontal floor and it is pushed in the horizontal direction with a force of 10N continuously, so that it moves steadily. [Refer to TB page 36] a) Draw FBD (a diagram showing all the forces acting on the body at a point of time) b) What is the value of friction? A. a) b) Given that the body is moving steadily, hence the net force on the body is zero both in horizontal and vertical directions. Force acting on it along horizontal directions are force of friction(f), force of push(F). We know that Fnet.x = 0 F+ (-f) = 0 F=f Hence the value of force of friction is 10N SESSION 1. INTRODUCTION AND FIRST LAW OF MOTION 60

1.4 Key Concepts i. Isaac Newton formulated the laws of motion. ii. The property of matter that resists changes in its state of motion or rest is called inertia. iii. Inertia depends on the mass of the object or mass is considered as the measure of inertia. iv. Greater the mass of the object, the more it resists its change of state. v. Newton’s first law of motion: “A body continues its state of rest or of uniform motion unless a net force acts on it.” 1.5 Reflection on Concepts Q1. [AS1] Explain the reasons for the following. [Refer to TB page 46 Q1] i. Why dust comes out of a carpet when it is is beaten with a stick? ii. Luggage kept on the roof of a bus is tied with a rope. iii. Why a pace bowler in cricket runs a long distance before he bowls? A. a) When a carpet is beaten with a stick dust comes out of it. When a carpet is beaten with a stick, the particles of material of the carpet are set into motion whereas the particles of dust in the carpet are at rest due to inertia. Thus the particles of dust are separated from the carpet and the dust comes out of it. b) Luggage kept on the roof of a bus is being tied with a rope. As the bus moves, the luggage kept on the roof which is at rest is set to motion along with the bus. When brakes are applied the bus comes to rest but the luggage will be still under motion due to the property of inertia. In such condition, the luggage will fall down from the roof of the bus. So the luggage is tied down with a rope to prevent its loss or breakage. c) A pace bowler in cricket runs a long distance before he bowls. The cricket ball in the hands of the pace bowler is at rest and continues to be at rest due to the property of inertia. But as the bowler runs towards the crease at bowler’s end, the ball also is under motion and when he releases the ball, it continues to move forward and travels towards the batsman though the bowler stops running. SESSION 1. INTRODUCTION AND FIRST LAW OF MOTION 61

By running the bowler gets some momentum. That momentum will be shifted to ball when he releases the ball. Momentum, before and after releasing will be same. As the mass of the ball is less, velocity increases. Q2. [AS1] Illustrate an example for first law of motion. [Refer to TB page 46 Q2] A. Example of first law of motion: When we are travelling in a bus, we tend to remain at rest with respect to the seat until the driver applies the brakes to stop the bus. With the application of the brakes, the bus stops but our body tends to continue to move forward because of its inertia. Thus our body experiences a jerk and we have to apply force to prevent our body falling forward. 1.6 Application of Concepts Q1. [AS1] Two objects having masses 8 kg and 25 kg. Which one has more inertia? Why? [Refer to TB page 46 Q1] A. As the inertia is a property of matter that depends on the mass of the object, the object of mass 25 kg has more inertia than the object of mass 8 kg. Since more the mass greater is the inertia. 1.7 Suggested Experiments Q1. [AS3] Conduct an experiment to prove Newton’s first law of motion and write a report. [Refer to TB page 48 Q1] A. If we kick a ball, it will fly in the direction you kicked, with certain speed, until a force slows it down or stops it. If the ball goes high, the force of gravity slows it down. If the ball rolls on the ground, the force of friction makes the ball slow down and stop. If the net force acting on an object is zero, the object which is at rest remains at rest or if the object is already moving with a certain velocity it continues to move with the same velocity. Thus we can represent the first law of motion as: If Fnet = 0 then the velocity of an object is either zero or constant. Thus when the net force acting on a body is zero, we say that the body is in equilibrium. Newton’s first Law of motion is also known as the Law of Inertia. SESSION 1. INTRODUCTION AND FIRST LAW OF MOTION 62

1.8 Suggested Projects Q1. [AS4] Observe some daily life examples for Newton’s first law of motion and explain the situations.Write a report on your observations. [Refer to TB page 48 Q1] A. Students’ activity. Objective Questions (1) The scientist who said \"An object in motion will remain in the same motion as long as no external force is applied ,” is (Pg 47; TB Q1) (A) Aristotle (B) Galilleo (C) Newton (D) Dalton Correct Answer: B (2) If the net force acting on an object is zero, then the body is said to be in the state of ( Pg 47; TB Q 2) (A) Equilibrium (B) Motion (C)Inertia of motion (D)Uniform motion Correct Answer: A (3) Inertia of a body depends on (Pg 47: TB Q 3) (A) Shape (B) Volume (C) Mass (D) Area Correct Answer: C (4) Newton used the word ‘mass in motion’ to represent (Pg 47; TB Q 4) (A) Linear momentum (B) Inertia of motion (C)Third law of motion (D)Inertia at rest Correct Answer: A SESSION 1. INTRODUCTION AND FIRST LAW OF MOTION 63

SESSION 2 SECOND LAW OF MOTION 2.1 Mind Map SESSION 2. SECOND LAW OF MOTION 64

2.2 Terminology i. Linear momentum – Product of velocity and mass. ii. Acceleration – Rate of change of velocity. iii. Newton – SI unit of force. 2.3 Solved Examples Q1. A mat of mass 1 kg and length 1m is placed on the floor. One end of the mat is pulled with a constant speed of 1m/s towards the other end till the other end comes to motion (till the mat is reversed). How much force is required to do this? [Refer to TB page 38] SESSION 2. SECOND LAW OF MOTION 65

A. A mat is being pulled with a constant speed of v = 1 m/s, so that the mass of the part of the mat is continuously increasing. Hence, here the mass is a variable. The time required for bringing the entire mat in motion is given by ∆t = distance covered by the end = 2 = 2s s peed 1 (Distance covered by the end = 1m + 1m = 2m) From Newton’s second law of motion, Fnet = ∆p = ∆(mv) ∆t ∆t Here v is constant, so we get Fnet = v ∆m ∆t Where, ∆m is the change of mass in time ∆t . The change of mass in 2 s is equal to entire mass of mat. Fnet = (1m/s)×(1kg) 2 = 1 N 2 In the horizontal direction only one force is acting. Hence the required force is 1 N. 2 SESSION 2. SECOND LAW OF MOTION 66

Q2. An Atwood machine consists of two loads of masses m1 and m2 attached to the ends of a limp of inextensible string as shown in the figure. The string runs over a pulley. Find the acceleration of each load and tension in the string m1>m2. [Refer to TB page 39] A. We know that tension of string always tries to pull the bodies up. From the FBD of the mass m1, there exist two forces on the load of mass m1, one is tension of the string acting in upward direction and weight of the load (m1 g) acting in downward direction. The net force on m1, Fnet = m1 a Thus, m1 g – T = m1 a . . . . . . . . . . . . . . . . . . .. (1) Thus, the net force (Fnet ) acting on mass m1 produces an acceleration ‘a’ in it. When m1 moves down, m2 moves up. So, the magnitudes of acceleration are same. SESSION 2. SECOND LAW OF MOTION 67

From the FBD of mass m2 Fnet = T–m2 g = m2 a . . . . . . . . . . . . . . . . . . . . . . . . (2) From (1) and (2), 2.4 Key Concepts i. Newton’s second law of motion: “The rate of change of momentum of a body is directly proportional to the force acting on it and it takes place in the direction of the net force.” ii. Mathematical form of the second law: F = m × a, (where m = Mass of the body, a = acceleration of the body). iii. The quantity of matter contained in a body is called mass. iv. The rate of change of velocity is called acceleration. v. Linear momentum of a body is given by the product of its mass and velocity, i.e., p = mv. vi. The force acting on a body of mass 1 kg and producing an acceleration of 1 m/s2 is called one “Newton”. The S.I. unit of force is 1 Newton (N). vii. One Newton = 1 kg x 1 m/s2 = 1 kgm/s2 SESSION 2. SECOND LAW OF MOTION 68

2.5 Reflection on Concepts Q1. [AS1] Illustrate the second law of motion with an example. [Refer to TB page 46 Q2] A. Second Law of Motion: The rate of change of momentum of a body is di- rectly proportional to the net force acting on it and it takes place in the direction of net force. Example–The harder you hit the ball, the more quickly it moves away from you, because you impart a greater acceleration to it. 2.6 Application of Concepts Q1. [AS1] A force acts for 0.2 s on an object having mass 1.4 kg initially at rest. The force stops to act but the object moves through 4 m in the next 2 s, find the magnitude of the force. [Refer to TB page 46 Q4] A. Since it is at rest initially, initial velocity, u = 0 Let the force that acts on it be F. Then, F = ma, where m = mass of the body = 1.4 kg, and a = acceleration of the body Velocity after 0.2s, V=u + at = 0 + 0.2a = 0.2a . . . ..(1) After 0.2s, the body moves with uniform velocity, acceleration is zero because force is removed. Therefore, velocity, v = s/t = 4/2 = 2 m/s . . . . . . . . . . . . .(2) From (1) and (2) v = 0.2a ⇒ 2 = 0.2a ⇒ a=2/0.2 = 10 m/s2 Therefore, force applied F=ma = 1.4kg × 10 m/s2 = 14 N SESSION 2. SECOND LAW OF MOTION 69

Q2. [AS1] An object of mass 5 kg is moving with a velocity of 10ms−1. A force is applied so that in 20 s, it attains a velocity of 25ms−1. What is the amount of force applied on the object? [Refer to TB page 46 Q5] A. Initial velocity, u= 10 m/s Final velocity, v = 25 m/s Time = 20 s Mass, m = 5 kg Acceleration is given by a = v−u = 25−10 = 15 = 0.75 ms−2 t 20 20 Force, F = ma = 5 × 0.75 = 3.75N Q3. [AS1] A hammer of mass 400 g moving at 30ms−1 strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer? [Refer to TB page 46 Q6] A. Mass of the hammer (m) = 400g = 0.4kg Velocity of the hammer, v= 30 m/s Momentum, ∆p= 30 × 0.4 N–s The nail stops the hammer with in a time 0.01s Therefore, ∆t = 0.01s The stopping force of the nail on the hammer, Fnet = ∆p/∆t = 30 × 0.4 / 0.01 = 1200N SESSION 2. SECOND LAW OF MOTION 70

Q4. [AS1] Two people push a car for 3 s with a combined net force of 200 N. i. Calculate the impulse provided to the car. ii. If the car has a mass of 1200 kg, what will be the change in velocity? [Refer to TB page 46 Q3] A. i. Impulse = Force × time = 200 N × 3 s = 600 Ns. ii. Acceleration = Force ÷ mass = 200 N/ 1200 kg = 1/6 ms-2 Change in velocity = acceleration × time = 0.5 ms-1 Q5. [AS7] A man of mass 30 kg uses a rope to climb which bears only 450 N. What is the maximum acceleration with which he can climb safely. [Refer to TB page 46 Q7] A. Let us consider that the mass of rope is zero. Then the tension throughout the rope is uniform. So, tension of the rope = 450 N Mass of the man = 30 kg From the formula, F=ma We have a = F/m=450/30=15 Acceleration with which the man can climb safely = 15 ms-2 2.7 Higher Order Thinking Skills Q1. [AS1] Three identical blocks, each of mass 10 kg, are pulled as shown on the horizontal frictionless surface. If the tension (F) in the rope is 30 N, what is the acceleration of each block? And what are the tensions in the other ropes? (Neglect the masses of the ropes.) [Refer to TB page 47 Q3] A. Recall: F = ma⇒ a = F m SESSION 2. SECOND LAW OF MOTION 71

Force acting on the system, F = 30 N Total mass on which the force acts, m = 10 kg × 3= 30 kg ∴ a = 30N = 1m/s2 30kg ∴ Acceleration of each block = 1 m/s2 Again, T1 = Force acting on first block T1 = 30N = 10N 3 T2 = Tension acting on second block T2 = 30N×2 = 20N 3 Q2. [AS7] A vehicle has a mass of 1500 kg. What must be the force between the vehicle and the road if the vehicle is to be stopped with a negative acceleration of 1.7ms−2 ? [Refer to TB page 47 Q1] A. Force applied, F = ma m, mass of automobile = 1500 kg negative acceleration, a = −1.7ms−2 Force between vehicle and road F = ? F, f orce = 1500 × −1.7N = −2550N (Note: The negative sign indicates the direction of force is opposite to the direction of motion.) Objective Questions (1) The SI unit of momentum is (Pg 48; TB 5) (A) m/sec (B) kg.m (C) kg.m/sec (D) kg.m/sec2 Correct Answer: C SESSION 2. SECOND LAW OF MOTION 72

SESSION 3 THIRD LAW OF MOTION 3.1 Mind Map 3.2 Terminology i. Action and reaction pair - Equal and opposite forces that act on different objects. 3.3 Key Concepts i. Two opposing forces are called the action and reaction pair. ii. Newton’s third law of motion: If one object exerts a force on another object, the second object exerts a force on the first one with equal magnitude but in opposite direction. SESSION 3. THIRD LAW OF MOTION 73

3.4 Reflection on Concepts Q1. [AS1] Illustrate an example of the third Law of motion. [Refer to TB page 46 Q2] A. The third law of motion is observed when a passenger jumps out of a rowing boat. As the passenger jumps forward, the force on the boat moves it backwards. 3.5 Higher Order Thinking Skills Q1. [AS2] If a fly collides with the windshield of a fast moving bus: [Refer to TB page 47 Q4] (a) Is the impact force experienced same for the fly and the bus? Why? (b) Is the same acceleration experienced by the fly and the bus? Why? A. (a) The impact force experienced by the fly and the bus is the same. According to Newton’s third law of motion, the total momentum before impact and after impact remains unchanged and as the masses of the fly and car are both constant, the forces are equal but opposite in direction. (b) The accelerations experienced by the fly and the bus are different since the masses of the two bodies are different but the forces are the same. 3.6 Suggested Experiments Q1. [AS3] Conduct an experiment to show the action and reaction forces acting on two dif- ferent objects. [Refer to TB page 48 Q2] A. Aim: To show the action and reaction forces acting on two different objects. Materials required: Test tube, rubber cork cap, Bunsen burner, laboratory stand and thread. Procedure : • Take a test tube of good quality glass and put small amount of water in it. • Place a cork cap at its mouth to close it. • Now suspend the test tube horizontally with the help of two strings as shown in the given figure. • Heat the test tube with a Bunsen burner until water vaporizes and cork cap blows out. SESSION 3. THIRD LAW OF MOTION 74

Observation : The test tube recoils, while the cork cap blows out. Conclusion: When the cork cap blows off, the test tube hanging on the strings recoils. This is in accordance with Newton’s third law of motion. 3.7 Suggested Projects Q1. [AS4] Write a report on the action and reaction in the systems that you have observed in your daily life which are the evident of Newton’s third law of motion. [Refer to TB page 48 Q2] A. Students’ activity. SESSION 3. THIRD LAW OF MOTION 75

SESSION 4 CONSERVATION OF MOMENTUM AND IMPULSE 4.1 Terminology i. Momentum – Product of mass and velocity. ii. Conservation of momentum – In an isolated system in absence of a net external force, the total momentum is conserved. iii. Impulse – Product of force and time. iv. Impulsive forces – Forces that act for a very short time. 4.2 Solved Examples Q1. A cannon of mass m1 = 12000 kg located on a smooth horizontal platform fires a shell of mass m2 = 300 kg in horizontal direction with a velocity v2 = 400m/s. Find the velocity of the cannon after it is shot. [Refer to TB page 44] A. Since the pressure of the powder gases in the bore of the cannon is an internal force so, the net external force acting on cannon during the firing is zero. Let v1 be the velocity of the cannon after shot. The initial momentum of system is zero. The final momentum of the system = m1v1 + m2v2 From the conservation of linear momentum, we get m1v1 + m2v2 = 0 m1v1 = −m2v2 or v1 = –m2 v2 / m1 Substituting the given values in the above equation, we get SESSION 4. CONSERVATION OF MOMENTUM AND IMPULSE 76

v1 = (300kg)× (400m/s) = −10m/s 12000kg Here ’–’ sign indicates that the canon moves in a direction opposite to the direction of bullet. 4.3 Key Concepts i. The product of mass and velocity is called as momentum. ii. When two bodies collide, the total momentum is unchanged before and after the colli- sion.The total quantity of momentum remains constant in a system until it is subjected to external influence. This is known as law of conservation of momentum. iii. An indefinitely large force acting for a very short time but producing a finite change of momentum is called impulse. iv. Product of net force and interaction time is called impulse. v. Impulse is equivalent to rate of change of momentum. vi. Force exerted over a limited time is called impulsive force. 4.4 Reflection on Concepts Q1. [AS1] Explain the following –[Refer to TB page 46 Q3] a) Static Inertia b) Inertia of motion c) Momentum d) Impulse e) Impulsive force A. a) Static Inertia – The object at rest will try to remain at rest until we apply an external force there will be no change in the position of the object. This is known as static inertia. b) Inertia of motion – Inertia is the resistance of any physical object to any change in its state of motion. c) Momentum – The momentum (p) of a body is simply defined as the product of its mass (m) and its velocity (v): i.e. Momentum = (mass) x (velocity) d) Impulse – Impulse is equivalent to the change in momentum that an object experi- ences during an interaction. e) Impulsive force – Forces exerted over a limited time are called impulsive forces. SESSION 4. CONSERVATION OF MOMENTUM AND IMPULSE 77

4.5 Application of Concepts Q1. [AS1] What is the momentum of a 6.0 kg bowling ball with a velocity of 2.2 m/s? [Refer to TB page 46 Q2] A. Momentum of a body p = mv Mass of bowling ball, m = 6.0 kg Velocity of the ball, v =2.2 m/s Momentum of bowling ball = 6.0 x 2.2 kg –m/s = 13.2 kg –m/s 4.6 Higher Order Thinking Skills Q1. [AS1] Two ice skaters initially at rest, push each other. If one skater whose mass is 60 kg has a velocity of 2 m/s, what is the velocity of other skater whose mass is 40 kg? [Refer to TB page 47 Q2] A. From the conservation of linear momentum, we have m1v1 + m2v2 = 0 ⇒ m1v1 = −m2v2.................(1) m1 = mass of first skater = 60kg v1 = velocity of first skater = 2m/s m2 = mass of second skater = 40kg v1 = velocity of second skater =? Substituting the above values in equation (1) 60 × 2 = - 40 × v2 ⇒ v2 = - 60×2 = - 3 m/s 40 ∴ Velocity of second skater = - 3 m/s SESSION 4. CONSERVATION OF MOMENTUM AND IMPULSE 78

—— CCE Based Practice Questions —— AS1-Conceptual Understanding Very Short Answer Type Questions 1. State true or false. [Refer to Session 3.2 ] (i) Force = mass of the body x acceleration due to gravity. (ii) Quantity of matter contained in a body is called weight. [ ] (iii) One Newton = 1kg x 1 m/s2. [ ] [ ] 2. Fill in the blanks. is the acceleration. [Refer to Session 3.2 ] and . (i) Larger is the mass or . (ii) Momentum depends on is the acceleration. (iii) SI unit of momentum is (iv) Larger the net force 3. Answer the following questions in one sentence. [Refer to Session 3.3 ] (i) A vehicle has a mass of 1500 kg. What must be the force between the vehicles and the road if the vehicle is to be stopped with a negative acceleration of 1.7 m/s2 ? CHAPTER 3. LAWS OF MOTION 79

4. Fill in the blanks. [Refer to Session 3.3 ] (i) For every action there is and reaction. (ii) The two opposing force is known as and . (iii) Newton’s third law is applicable to interaction between bodies. (iv) The reaction force is indicated by sign. (v) The force exerted between two bodies is in magnitude and opposite in . 5. Answer the following questions in one sentence. [Refer to Session 3.1 ] (i) Who formulated laws of motion? (ii) Do all bodies have same inertia? (iii) Define inertia. CHAPTER 3. LAWS OF MOTION 80

(iv) Is it possible to have a world with no friction as imagined by Galileo? (v) How many laws of motion were proposed by Newton? 6. Fill in the blanks. of inertia. [Refer to Session 3.1 ] 81 (i) Mass is considered as the CHAPTER 3. LAWS OF MOTION

(ii) If the mass of the object is greater, the more it resists its ___________________. (iii) Inertia depends on the of the object. 7. State true or false. [Refer to Session 3.4 ] (i) Law of conservation of momentum states in the absence of a net external force on the system, the momentum of the system remains changed. [] (ii) A system is said to be isolated when net external force acting on it is zero. ] [ 8. Fill in the blanks. [Refer to Session 3.4 ] (i) A system is said to be isolated when net external force acting on it is . (ii) When the resultant force acting on a system is zero, the total momentum of the system . This is called . (iii) The velocity with which the gun moves in backward direction is called as velocity. (iv) According to second law of motion Fnet = . (v) When the lift is stationary or moves with uniform velocity the acceleration is zero and the net force is . (vi) The change in momentum over a longer time requires force. (vii) Forces exerted over a limited time are called . CHAPTER 3. LAWS OF MOTION 82

Short Answer Type Questions 9. Answer the following questions in 3-4 sentences. (i) [(Session 3.2)] State Newton’s second law of motion. (ii) [(Session 3.2)] What is the momentum of a 6 kg bowling ball with velocity 2.2 m/s? (iii) [(Session 3.2)] Write the equation to show how force and acceleration are related. 10. Answer the following questions in 3-4 sentences. 83 (i) [(Session 3.3)] State Newton’s third law of motion. CHAPTER 3. LAWS OF MOTION

(ii) [(Session 3.3)] Explain the motion of a rocket. 11. Answer the following questions in 3-4 sentences. (i) [(Session 3.1)] State the first law of motion. (ii) [(Session 3.1)] State the relationship between inertia and mass. (iii) [(Session 3.1)] What happens if you kick a ball high? 84 CHAPTER 3. LAWS OF MOTION

(iv) [(Session 3.1)] Give an example to show the relationship between inertia and mass. 12. Answer the following questions in 3-4 sentences. (i) [(Session 3.4)] State the law of conservation of momentum. (ii) [(Session 3.4)] Define impulse and impulsive force. CHAPTER 3. LAWS OF MOTION 85

(iii) [(Session 3.4)] Why are air bags used in cars? (iv) [(Session 3.4)] Derive an equation to say impulse is equal to change in momentum. Long Answer Type Questions 13. Answer the following questions in 6-8 sentences. (i) [(Session 3.2)] A force of 500 dynes acts on a mass of 0.05 kg over a distance of 200 m. Assuming that the mass is initially at rest, find the final velocity and time for which the force acts. CHAPTER 3. LAWS OF MOTION 86

14. Answer the following questions in 6-8 sentences. (i) [(Session 3.3)] Identify the action and reaction process involved in the act of walking. CHAPTER 3. LAWS OF MOTION 87

15. Answer the following questions in 6-8 sentences. (i) [(Session 3.1)] Explain the reason – a. A rider falls forward, when a galloping horse suddenly stops. b. Before taking a long jump, a boy runs a certain distance. c. While jumping out of a bus, it’s important to run along the same direction of the moving bus. CHAPTER 3. LAWS OF MOTION 88

(ii) [(Session 3.1)] Two objects have masses 8 kg and 2500 g. Which one has more inertia? Why? CHAPTER 3. LAWS OF MOTION 89

16. Answer the following questions in 6-8 sentences. (i) [(Session 3.4)] An electron of mass of 9 × 10-31 kg is moving with a linear velocity of 6 × 107 ms-1. Calculate the linear momentum of the electron. (ii) [(Session 3.4)] A body has a linear momentum of 5 Ns. If the velocity of the body is 200 ms-1, find the mass of the body. CHAPTER 3. LAWS OF MOTION 90

AS2-Asking questions and making hypothesis Short Answer Type Questions 17. Answer the following questions in 3-4 sentences. (i) [(Session 3.1)] When driver applies the brake of the bus, we move forward. Why do you think this happens? AS3-Experimentation and field investigation Long Answer Type Questions 18. Answer the following questions in 6-8 sentences. (i) [(Session 3.3)] Write the experiment to show the action and reaction forces acting on two different objects. CHAPTER 3. LAWS OF MOTION 91

CHAPTER 3. LAWS OF MOTION 92

AS4-Information skills and projects Very Short Answer Type Questions 19. Answer the following questions in one sentence. [Refer to Session 3.2 ] (i) On what factors does momentum depend? Long Answer Type Questions 20. Answer the following questions in 6-8 sentences. (i) [(Session 3.1)] Complete the table. Quantity Formula Unit newton Linear momentum Force Fnet = k ∆p/∆t CHAPTER 3. LAWS OF MOTION 93

AS5-Communication through drawing and model making Short Answer Type Questions 21. Answer the following questions in 3-4 sentences. (i) [(Session 3.3)] Observe the given diagram carefully. What are the momenta of mar- bles before and after collision? AS6-Appreciation and aesthetic sense, Values 94 Long Answer Type Questions CHAPTER 3. LAWS OF MOTION

22. Answer the following questions in 6-8 sentences. (i) [(Session 3.1)] Why do you appreciate Galileo Galilei? AS7-Application to daily life, concern to bio diversity Short Answer Type Questions 23. Answer the following questions in 3-4 sentences. (i) [(Session 3.2)] Solve the following problem – Two ice skaters initially at rest push each other. If one skater whose mass is 60 kg has a velocity of 2 m/s. What is the velocity of other skater whose mass is 40 kg? CHAPTER 3. LAWS OF MOTION 95

(ii) [(Session 3.2)] Calculate the momentum of a bullet having mass of 25 g is thrown using hand with a velocity of 0.1 m/s. Long Answer Type Questions 24. Answer the following questions in 6-8 sentences. (i) [(Session 3.2)] A car of mass 1000 kg develops a force of 500 N over a distance of 49 m. If initially the car is at rest find a. Final velocity b. Time for which it accelerates CHAPTER 3. LAWS OF MOTION 96

25. Answer the following questions in 6-8 sentences. (i) [(Session 3.4)] Which has more momentum between a metal sphere of mass 10 g moving with a velocity of 400 m/s and a cricket ball of mass 400 g thrown with a speed of 90 km/h? CHAPTER 3. LAWS OF MOTION 97

Objective Questions AS1-Conceptual Understanding 26. Choose the correct answer. (i) An object will remain in same motion as long as (A) it is pulled (B) no external force is applied to it (C)it’s acceleration is constant (D)external force is applied to it (ii) The property of matter that resists change in its state of motion or rest is called (A) acceleration (B) force (C) inertia (D)centripetal force (iii) Greater the mass of the body (A) more it will resist the change in state (B) lesser it will resist the change in state (C)mass does not affect the motion (D)none of the above (iv) Larger the force the acceleration. (A) lesser (B) greater (C) same (D) zero (v) The following represents Newton’s second law of motion best: (A) p = mv (B) m = fa (C)m= a/f (D)a = fm CHAPTER 3. LAWS OF MOTION 98