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# 202110207-APEX-STUDENT-WORKBOOK-PHYSICAL_SCIENCE-G09-PART1

## Description: 202110207-APEX-STUDENT-WORKBOOK-PHYSICAL_SCIENCE-G09-PART1

Physical Science Workbook_9_P_1.pdf 1 18-10-2019 14:09:23 Name: ___________________________________ Section: ________________ Roll No.: _________ School: __________________________________

Table of Contents 1 32 1 MOTION 63 2 LAWS OF MOTION 88 3 IS MATTER PURE? 113 4 ATOMS AND MOLECULES 133 5 WHAT IS INSIDE THE ATOM? 151 6 CHEMICAL REACTIONS AND EQUATIONS 152 PROJECT BASED QUESTIONS ADDITIONAL AS-BASED PRACTICE QUESTIONS TABLE OF CONTENTS

SESSION 1 1. MOTION 1.1 Mind Map MOTION, DISTANCE AND DISPLACEMENT SESSION 1. MOTION, DISTANCE AND DISPLACEMENT 1

1.2 Terminology i. Distance – The length of the actual path traversed by an object. ii. Displacement – The length of the straight line path between the initial and ﬁnal posi- tions of the object. iii. Vector – Quantities that are expressed with both magnitude and direction. iv. Scalar – Quantities that can be expressed only using magnitude. v. Relative – Considered in relation or in proportion to something else. 1.3 Key Concepts i. The change of position of an object with respect to time is called motion. ii. Motion is relative. It depends on the observer. iii. It can be described in terms of distance travelled or on the displacement. 1.4 Asking Questions and Making Hypothesis Q1. Suppose that the three balls shown in ﬁgure start simultaneously from the top of the hills. Which one reaches the bottom ﬁrst? Explain. [Refer to TB page 19 Q6, Try These] A. Among all the three cases, in the second and third hill the ball takes a curved path and covers a large distance. Since, the distance in the ﬁrst hill is the least, the ball in the ﬁrst case will reach the bottom ﬁrst. SESSION 1. MOTION, DISTANCE AND DISPLACEMENT 2

1.5 Communication Through Drawing and Model Making Q1. As shown in ﬁgure, a point traverses along the curved path. Draw the displacement vector from given points A and B. [Refer to TB page 18 Q8] A. The displacement vector for the above point can be shown as below: SESSION 1. MOTION, DISTANCE AND DISPLACEMENT 3

SESSION 2 SPEED AND VELOCITY 2.1 Mind Map 2.2 Terminology i. Average speed –The ratio of total distance covered to total time taken. ii. Average velocity – The ratio of total displacement covered to total time taken. iii. Instantaneous speed – Speed of any moving body at any instance of time. iv. Velocity – Displacement per unit time. v. Acceleration – Velocity per unit time. 2.3 Key Concepts i. Average speed is distance covered in unit time and average velocity is displacement covered in unit time in a particular direction. ii. Velocity has both magnitude and direction and so it is a vector. iii. Speed at a particular instant is called instantaneous speed which gives an idea of how fast the position of the body is changing. SESSION 2. SPEED AND VELOCITY 4

2.4 Conceptual Understanding Q1. In the ﬁgure given below distance vs. time graphs showing motion of two cars A and B are given. Which car moves faster? [Refer to TB page 17 Q2] A. 1. If we draw perpendiculars to X and Y axes from A and B respectively, we can observe that A covers large distance (s1) within a short time (t1). 2. Find the slopes of the lines OA and OB at any instant. Slope of OA is high. Hence Car A moves faster. Q2. Distinguish between speed and velocity. [Refer to TB page 18 Q3] SESSION 2. SPEED AND VELOCITY 5

A. Speed Velocity 1. The distance 1. Velocity is the displacement covered per unit per unit time. time is called speed of an object. 2. Speed has 2. Velocity has both magnitude magnitude but no and direction. direction. 3. Speed is a scalar. 3. Velocity is a vector. 4. Total Distance = 4. Displacement = Average velocity Average speed × × Time taken Time Taken Q3. A train of length 50 m is moving with a constant speed of 10 m/s. Calculate the time taken by the train to cross an electric pole and a bridge of length 250 m. [Refer to TB page 19 Q7, Try These] A. Distance = Speed × T ime Note: A train has to travel its own length in order to cross an electric pole, railway signal, a tree, a kilometre stone or stationary man. Distance to be travelled = Length of the train = 50 m Speed of train = 10 m/s Time taken to cross the electric pole = 50 m/10 = 5 s To cross a bridge/ platform/ another train, a train has to travel the total length of both the bridge as well as the train itself. Distance to be travelled by the train = length of train + length of bridge SESSION 2. SPEED AND VELOCITY 6

Length of train = 50 m Length of bridge = 250 m Distance to be travelled = 50 m + 250 m = 300 m Distance = 300m Time taken by train to cross the bridge = Distance = 300 = 30s Speed 10 Q4. Consider a train that can accelerate with an acceleration of 20 cm/s2 and slow down with deceleration of 100 cm/s2 . Find the minimum time for the train to travel between the stations 2.7 km apart. [Refer to TB page 19 Q4, Try These] A. The train would travel between the stations with minimum time if for the beginning part of the journey it accelerates and for the rest it decelerates. Let us say, the acceleration part is covered in time and deceleration part in time t2 . Initial velocity, u1 = 0 Acceleration, a1 = 20 cm/s2 Final velocity, v1 = u1 + at1 = 20t1 Now, this velocity would be the initial velocity for the decelerating part of journey. Initial velocity, u2 = 20t1 Acceleration, a2 = −100 cm/s2 Final velocity,v2 = 0 t1 t2 v2 = u2 + a2t2 ⇒ 0 = 20t1 + at2 ⇒ 20t1 = 100t2 ⇒ = 5 s1 = u1t1 + 1 a1t12 and s2 = u2t2 + 1 a2t22 , where s1 and s2 represent the displace- 2 2 ments in accelerating and decelerating part respectively. Adding s1 and s2 , s1 + s2 = 2.7km = u1t1 + 1 a1t12 + u2t2 + 1 a2t22 2 2 ⇒ 270000 = 0 + 1 (20)t12 + 20t1t2 + 1 (−100)t22 2 2 ⇒ 270000 = 10t12 + 20t1t2 − 50t22 ⇒ 270000 = 250t22 + 100t22 − 50t22 SESSION 2. SPEED AND VELOCITY 7

⇒ 270000 = 200t22 + 100t22 ⇒ 300t22 = 270000 ⇒ t22 = 900 ⇒ t2 = 30 t1 = 5t2 ⇒ t1 = 150 We get, t1 + t2 = 180 s . Therefore, total time = 180 s 2.5 Asking Questions and Making Hypothesis Q1. When the velocity is constant, can the average velocity over any time interval differ from instantaneous velocity at any instant? If so, give an example; if not, explain why? [Refer to TB page 18 Q6] A. When the velocity is constant, the average velocity over any time interval does not differ from instantaneous velocity at any instant. Example: If we draw the velocity–time graph for a car moving at an average velocity of 50 kmph, we get a straight line graph parallel to x–axis and the height of the straight line will not change with time. It will be a straight line parallel to the x–axis. Q2. Can the direction of velocity of an object reverse when its acceleration is constant? If so give an example; if not, explain why? [Refer to TB page 18 Q7] A. Yes, it can. If the initial velocity vector of an object was in opposite direction to its constant acceleration. Example: Anything you toss with your hand has constant acceleration after you toss it, the acceleration due to gravity directed downward. If you toss upward, it starts out with upward velocity, which reverses and eventually becomes downward velocity. SESSION 2. SPEED AND VELOCITY 8

Q3. Correct your friend who says, ‘The car rounded the curve at a constant velocity of 70 km/h’. [Refer to TB page 18 Q5] A. As in circular motion, the displacement is zero and hence, the velocity is also zero. So, the correct statement is–‘The car rounded the curve at a constant speed of 70 km/h’. 2.6 Communication Through Drawing and Model Making Q1. You may have heard the story of the race between the rabbit and tortoise. They started from same point simultaneously with constant speeds. During the journey, rabbit took rest somewhere along the way for a while. But the tortoise moved steadily with lesser speed and reached the ﬁnishing point before the rabbit. The rabbit awoke and ran, but the tortoise reached the goal even before the rabbit ﬁnished the race and won the race. Draw distance vs time graph for this story. [Refer to TB page 18 Q10] A. SESSION 2. SPEED AND VELOCITY 9

2.7 Application to Daily Life, Concern to Bio Diversity Q1. What is the average speed of a cheetah that sprints 100 m in 4 s? How about if it sprints 50 m in 2 s? [Refer to TB page 18 Q11] A. Total distance = 100 m Time taken = 4 s Average speed of cheetah = 100 / 4 = 25 m/s If it sprints at 50 m in 2 s Total distance = 50 m, Average speed = Total Distance/ Time Taken Time taken = 2 s Average speed of cheetah = 50 / 2 = 25 m/s Q2. Two trains, each having a speed of 30 km/h are headed at each other on the same track. A bird ﬂies off one train to another with a constant speed of 60 km/h, when they are 60 km apart till before they crash. Find the distance covered by the bird and how many trips the bird can make from one train to other before they crash? [Refer to TB page 18 Q12] A. Speed of each train = 30 km/h. Relative speed of the trains when they travel in the opposite direction = sum of speeds of the two trains. Relative speed of trains = 30 km/h + 30 km/h = 60 km/h. So, time taken to crash = distance to be travelled/ relative speed = 60 / 60 = 1 h. Bird ﬂight time = 1 h. Bird ﬂight distance = 1 h × 60 km/ h = 60 km. The bird can make number of trips( inﬁnity) before the trains crash. SESSION 2. SPEED AND VELOCITY 10

Q3. A car travels at a speed of 80 km/h during the ﬁrst half of its running time and at 40 km/h during the other half. Find the average speed of the car. [Refer to TB page 19 Q8, Try These] A. Let us say that the car has travelled for t hours. Distance travelled in ﬁrst half of time = velocity in ﬁrst half x t = 80t km 2 2 Distance travelled in second half of time= velocity in second half × t = 40t km 2 2 Therefore, average speed = Total distance travelled = 80t + 40t = 60 km/h Total time taken 2 2 t Q4. A car covers half the distance at a speed of 50 km/h and the other half at 40 km/h. Find the average speed of the car. [Refer to TB page 19 Q9, Try These]] A. Let us say that the car has travelled for s km. s Time taken for travelling ﬁrst half of distance = Distance traveled /Speed = 2 50 Distance traveled s Speed Time taken for travelling second half of distance = = 2 40 Therefore, average speed = Total distance travelled = s = 44.44 km/h Total time taken ss 2 + 2 50 40 SESSION 2. SPEED AND VELOCITY 11

SESSION 3 UNIFORM AND NON-UNIFORM MOTION 3.1 Mind Map SESSION 3. UNIFORM AND NON-UNIFORM MOTION 12

3.2 Terminology i. Acceleration – Rate of change of velocity. ii. Rectilinear motion – Motion along a straight line. 3.3 Key Concepts i. When the velocity is constant we say the motion is uniform. ii. If the velocity of a moving body changes, the body is said to have acceleration. iii. Rate of change of velocity is called acceleration. iv. If the acceleration is constant, the motion is said to have uniform acceleration. v. Negative acceleration is called deceleration. vi. Equations of motion are (i) v = u + at, (ii) s = ut+1/2 at2 and (iii) v2 – u2 = 2as 3.4 Conceptual Understanding Q1. “She moves at a constant speed in a constant direction.” Rephrase the same sentence in a fewer words. [Refer to TB page 17 Q1] A. The above sentence can be rewritten in fewer words as, “She moves at a constant velocity”. SESSION 3. UNIFORM AND NON-UNIFORM MOTION 13

Q2. Derive the equation for uniform accelerated motion for the displacement covered in its nth second of its motion. [Refer to TB page 19 Q1, Try These]] A. Let Sn be the displacement of the body in n seconds, i.e. t = n. Similarly let Sn-1 be the displacement of the body in (n–1) seconds, i.e. t = (n–1). Let initial velocity at time t = 0 seconds be u and acceleration be a. S = ut + 1 at2 is the kinematic equation to be used. 2 S n = un + 1 an2 2 S n−1 = u(n − 1) + 1 a(n − 1)2 2 So, distance travelled in second, Sn= S(n sec) – S((n-1)sec) = (un + ½ an2 ) – [ u(n–1)+ ½ a(n-1)2 ] = un + ½ an2 – un + u – ½ an2 + ½ a.2n – ½ a = u + an – ½ a = u +a (n–½) Sn = u + a (n–½) Displacement in nth second = S n − S n−1 = u + 1 a(2n − 1) 2 Q3. A body leaving a certain point “O” moves with a constant acceleration. At the end of the 5th second its velocity is 1.5m/s. At the end of the sixth second the body stops and then begins to move backwards. Find the distance traversed by the body before it stops. Determine the velocity with which the body returns to point “O”. [Refer to TB page 19 Q2, Try These] A. Let v = 1.5 m/s; t1 = 5 s; t2 = 6 s From v = u + at1 , we have 1.5 = u + a(5) SESSION 3. UNIFORM AND NON-UNIFORM MOTION 14

For the sixth second, using v = u + at, we have 0 = 1.5 + a(1) or, a = –1.5 m/s2 Therefore, u = 1.5 –5( –1.5) = 1.5 + 7.5 = 9 m/s Now, distance from O to stop, i.e. 0 to 6 seconds, S1 = v2−u2 = 0−81 = 81 = 27m 2a 2(−1.5) 3 For motion in reverse with the same acceleration, the velocity of the body when it reaches O at the same acceleration, is the same as u, i.e. 9 m/s, in the opposite di- rection. Q4. What do you mean by constant acceleration? [Refer to TB page 18 Q4] A. Constant acceleration means that the object accelerats uniformly over equal intervals of time however small the time intervals may be. Q5. A point mass starts moving in a straight line with a constant acceleration ‘a’. At a time ‘t’ after the beginning of motion, the acceleration changes sign, without changing in magnitude. Determine the time from the beginning of the motion in which the point mass returns to the initial position. [Refer to TB page 19 Q3, Try These] A. Velocity at time t is, v = at At time, t, acceleration changes sign. After time t1, the body comes to a stop and starts moving in the opposite direction. Then it takes time t2 to come back to its initial position. Therefore, t0 = t + t1 + t2 Let velocity be v1 after time t1 and v2 after time t2 . Then, v1 = 0 = v–at1; and v2 = at2 v1 = 0 = v–at1 = at–at1 = a(t–t1) Therefore, t1 = t v2 = at2 SESSION 3. UNIFORM AND NON-UNIFORM MOTION 15

We know that the distance between the starting point and the end point remains same. So, in the ﬁrst case, s1 = 1 at2 + vt1 − 1 at12 2 2 But, t1 = t Therefore, s1 = 1 at2 + vt − 1 at2 = vt 2 2 In the second case, S2 = 1 at22 2 Equating the two, we get, vt = 1 at22 2 (at)t = 1 at22 2 2at2 = at22 √ or, t2 = 2t √√ Therefore, t0 = t + t1 + t2 = t + t + 2t = (2 + 2)t Q6. A stone dropped from top of a well reaches the surface of the water in 2 seconds, find the velocity of stone while it touches the surface of water and what is the depth of the water surface from the top of the well (g = 10 m/s2). [Refer TB page 18 Q13] A. t= 2 s U = 0 m/s a = 10 m/s2 The velocity of stone while it touches the surface of water at time t, V =? The depth of the water surface from the top of the well, S =? V = U + at V = 0 + 10 × 2 = 20 m/s S = Ut + ½ at2 S = 0 + ½ 10 × 22 = 20 m Q7. An object moving with 6m per second execute an acceleration 2 m/s2 in next 3 seconds. How much distance did it cover? [Refer TB page 18 Q14] SESSION 3. UNIFORM AND NON-UNIFORM MOTION 16

A. t= 3 s U = 6 m/s a = 10 m/s2 S = Ut + ½ at2 S = 6 × 3 + ½ × 2 × 32 = 27 m Q8. A car stopped after travelling distance 8m due to applying brakes at the speed of 40 m/s. Find the acceleration and retardation of car in that period. [Refer TB page 18 Q15] A. V = 0 U = 40 m/s S= 8 m a =? V2 - U2 = 2aS 0 - 402 = 2 × a × 8 - 1600 = 16 a a = - 1600/16 = - 100 m/s2 3.5 Communication Through Drawing and Model Making Q1. Draw the distance v/s time graph when the speed of the body increases uniformly. [Refer to TB page 19 Q7, Try These] A. Q2. Draw the distance–time graph when its speed decreases uniformly. [Refer to TB page 18 Q9] SESSION 3. UNIFORM AND NON-UNIFORM MOTION 17

A. 3.6 Application to Daily Life, Concern to Bio Diversity Q1. A particle covers 10 m in ﬁrst 5 s and 10 m in next 3 s. Assuming constant acceleration ﬁnd initial speed, acceleration and distance covered in next 2 s. [Refer to TB page 19 Q10, Try These] A. The particle covered 10 m in ﬁrst 5 s. That is s5 = 10 m; t = 5 s The particle covered 10 m in the next 3 s. That is s – s = 10 m 85 Let initial velocity be ‘u’ m/s, and acceleration be ‘a’ m/s2 Then, s 5 = u(5) + ½ a (5)2 i.e. 10 = 5u + (25/2) a Similarly, s = u(8) + ½ a (8)2 8 s = 8u + 32a 8 Therefore, s – s5 = 8u + 32a – 10 = 10 8 i.e. 8u + 32a = 20 or, 2u + 8a = 5 or, u = 5−8a 2 SESSION 3. UNIFORM AND NON-UNIFORM MOTION 18

Substituting this in the equation for s5, we get 10 = 5 5−8a + 25 a 2 2 10 = 5 5−8a + 25 a 2 2 10 = 25 − 20a + 25 a 2 2 10 − 25 = − 15 a 2 2 − 5 = − 15 a 2 2 ⇒ a = 1 m/s2 3 Therefore, u = 5−8a = 5−8 1 = 7 m/s 2 2 3 6 Distance covered in next two seconds, s = s – s 10 8 s= 7 10 + 1 1 102 − 20 6 2 3 s = 70 + 100 − 20 = 170 − 20 = 50 = 8.33m 6 6 6 6 Q2. A man is 48 m behind a bus which is at rest. The bus starts accelerating at the rate of 1m/s2. At the same time the man starts running with uniform velocity of 10 m/s. What is the minimum time in which the man catches the bus? [Refer to TB page 19 Q12, Try These] A. a = 1 m/s2 19 SESSION 3. UNIFORM AND NON-UNIFORM MOTION

s = ut + 1at2 2 n = 0(t) + 1(1)t2 2 ∴ n = t2 2 48 + n = 10t 48 + t2 = 10t 2 96 + t2 = 20t t2 − 20t + 96 = 0 or, t = 8s, 12s Thus, minimum time = 8 seconds Q3. A car starts from rest and travels with uniform acceleration ‘α’ for some time and then with uniform retardation ‘B‘ and comes to rest. The time of motion is ‘t’. Find the maximum velocity attained by it. [Refer to TB page 19 Q11, Try These] A. Acceleration a = α m/s2 Initial speed u = 0 m/sec Let the time be t1 sec From equation v = u + at => v = 0 + αt1 ∴ t1 = v sec α Retardation a = –B m/s2 Initial speed ‘u’ is equal to the ﬁnal velocity with acceleration ‘α ’ =>u = αt1 m/s Final velocity v = 0 m/s Let the time be t2 sec From equation, v = u + at SESSION 3. UNIFORM AND NON-UNIFORM MOTION 20

⇒ 0 = αt1 + (−β)t2 ∴ αt1 = βt2 (∵ v = αt1) v = βt2 ⇒ t2 = v β Total time of motion t= t1 + t2 t= v + v α β =>t = v( 1 + 1 ) α β =>t = v[ (α+β) ] αβ =>v = αβt (α+β) The maximum velocity attained by the car = αβt (α+β) SESSION 3. UNIFORM AND NON-UNIFORM MOTION 21

—— CCE Based Practice Questions —— AS1-Conceptual Understanding Very Short Answer Type Questions 1. Answer the following questions in one sentence. [Refer to Session 1.3 ] (i) State which of the following is uniform motion and which is a non–uniform motion. a. A car going on a straight level road with steady speed– b. A cooling fan running– c. A roller coaster ride– d. A motorist riding on a hill– e. Planets revolving around sun– 2. Fill in the blanks. . [Refer to Session 1.3 ] (i) The motion is uniform when the velocity is (ii) The body has when the velocity of body changes. accelerated motion if acceleration is (iii) The motion is said to be constant. CHAPTER 1. MOTION 22

(iv) In a circular path changes but remains constant. (v) SI unit of acceleration is . 3. State true or false. [Refer to Session 1.2 ] (i) The distance covered and displacement are time dependent quantities. ] [ (ii) In rectilinear motion the distance and magnitude of displacement become unequal. [] (iii) Motion is relative and it is dependent on the observer. [] 4. Fill in the blanks. [Refer to Session 1.2 ] (i) Distance travelled by a body in unit time is known as . (ii) Change in velocity of a body per unit time is known as ________________ . (iii) Speed at any instant is . (iv) Displacement of body per unit time is . (v) Equation for average speed is . or CHAPTER 1. MOTION 23

5. Match the following. [(Session 1.1)] Column A [] Column B i. Displacement a. Motion b. Displacement/time taken ii. A body’s position is changing [ ] c. Vector d. Scalar iii. Distance [] e. Total distance/time taken iv. Average speed [] v. Average velocity [] 6. Fill in the blanks. [Refer to Session 1.1 ] (i) Motion of an object depends on . (ii) A vector can be represented as a directed segment. (iii) The SI unit of distance and displacement is . (iv) Velocity is speed in speciﬁed . (v) Physical quantity which does not require any direction is . Short Answer Type Questions 7. Answer the following questions in 3-4 sentences. (i) [(Session 1.2)] Which device(s) on a car can be used to change: a. Its speed? b. Its velocity but not its speed? (ii) [(Session 1.2)] Can the direction of velocity of an object reverse when its acceleration is constant? If so give an example; if not, explain why? CHAPTER 1. MOTION 24

8. Answer the following questions in 3-4 sentences. (i) [(Session 1.1)] Deﬁne the terms scalar and vector. (ii) [(Session 1.1)] Deﬁne motion, average speed and average velocity. Give equations for each of them. Long Answer Type Questions 9. Answer the following questions in 6-8 sentences. (i) [(Session 1.3)] Write and derive the three equations of motion. (ii) [(Session 1.3)] A car, initially at rest, picks up a velocity of 72 kmh-1 in 1/4th minutes. Calculate (a) Acceleration (b) Distance covered by the car. 10. Answer the following questions in 6-8 sentences. (i) [(Session 1.2)] From the diagram, calculate: (a) De–acceleration in region AB 25 (b) Acceleration in the region BC (c) Total distance travelled in the region ABC (d) Average velocity between 10s and 30s CHAPTER 1. MOTION

(ii) [(Session 1.2)] The velocity of a car changes from 18 kmh -1 to 72 kmh -1 in 30 s. Calculate (A) Change in velocity in ms -1 (B) Acceleration in (a) km h -2 (b) ms-2 11. Answer the following questions in 6-8 sentences. (i) [(Session 1.1)] An aeroplane touches the ground at a speed of 225 m/s and stops after 2 minutes. Calculate – (a) Acceleration (b) Length of runway (ii) [(Session 1.1)] From the displacement–time graph given in the ﬁgure calculate: (A) Velocity between 0 and 4s (B) Velocity between 4s and 6s (C) Velocity between 6s and 9s (D) Average velocity from: (a) 0 – 4s (b) 0 – 6s (c) 0 – 9s CHAPTER 1. MOTION 26

AS4-Information skills and projects Short Answer Type Questions 15. Answer the following questions in 3-4 sentences. (i) [(Session 1.2)] Complete the table. v (m/s) u (m/s) a (m/s2 ) t(s) 0 30 3 0 20 6 AS5-Communication through drawing and model making Long Answer Type Questions 16. Answer the following questions. (i) [(Session 1.3)] Draw displacement vs 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 (ii) [(Session 1.3)] Draw velocity–time graph to show: a) Zero acceleration b) Uniform acceleration c) Variable acceleration CHAPTER 1. MOTION 28

17. Answer the following question. (i) [(Session 1.2)] 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 1.2)] 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? Long Answer Type Questions 19. Answer the following questions in 6-8 sentences. (i) [(Session 1.3)] Write real life examples to show: 1. When the speed becomes zero for an instant. 2. When the acceleration of an object is zero. 3. When the body is accelerating. 4. When the body is in negative acceleration. CHAPTER 1. MOTION 29

Objective Questions AS1-Conceptual Understanding 20. Choose the correct answer. (i) Average velocity is (B) Displacement + Time taken (A) Displacement+ 1/2 Time taken (D)Displacement - 1/2 Time taken (C)Displacement / Total 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 (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 CHAPTER 1. MOTION 30

(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 (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 1. MOTION 31

2. 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 32

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 Q.1 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 25) 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 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 state of uniform motion unless a net force acts on it.” SESSION 1. INTRODUCTION AND FIRST LAW OF MOTION 33

1.5 Conceptual Understanding Q1. Explain the reasons for the following. [Refer to TB page 35 Q1] a) When a carpet is beaten with a stick, dust comes out of it. b) Luggage kept on the roof of a bus is tied with a rope. c) 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 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. By running the bowler gets some momentum. That momentum will be shifted to ball when he releases the ball. Momentum, before and after releasing the ball will be same. As the mass of the ball is less, velocity increases. SESSION 1. INTRODUCTION AND FIRST LAW OF MOTION 34

Q2. Two objects have masses 8 kg and 25 kg. Which one has more inertia? Why? [Refer to TB page 35 Q2] 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. Q3. Illustrate an example for ﬁrst law of motion. [Refer to TB page 35 Q7] 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 Asking Questions and Making Hypothesis Q1. Keep a small rectangular piece of paper on the edge a table and place an old ﬁve rupee coin on its surface vertically as shown in the ﬁgure. Now give a push (quick speed) to the paper with your ﬁnger. How do you explain inertia with this experiment? [Refer to TB page 37 Q9, Try Thess] A. In this experiment, both the ﬁve rupee coin and the paper are at rest. Now, when you give a quick push to the paper, the paper is set to motion. But the coin is still at rest since no external force is acting on it. As per the law of inertia the coin sticks on to its position. So, it remains in its state of rest and is not disturbed. SESSION 1. INTRODUCTION AND FIRST LAW OF MOTION 35

1.7 Appreciation and Aesthetic Sense, Values Q1. How do you appreciate Galileo’s thought of “any moving body continues in its state of motion only until some external force acts on it” which is a contradiction to Aristotle’s belief of “any moving body naturally comes to rest”? [Refer to TB page 35 Q10] A. Galileo’s thought of “any moving body continues in the state of motion only until some external force acts on it” which is a contradiction to the Aristotle’s belief of “any moving body naturally comes to rest” is true. It is based on several experiments and experi- ences. For example: i) A spinning top on ﬂoor continues to spin if the frictional force of the ﬂoor does not resist its motion. ii) The planets continue to revolve around the sun for ages together. iii) A mountain rock continues to lie stationary in its state of rest unless it is re- moved from its place. Aristotle’s believe proved to be wrong only by the experiments conducted by Galileo. Aristotle and Galileo’s contradictory thought led Newton to propose most popular laws of motion. 1.8 Application to Daily Life, Concern to Bio Diversity Q1. A passenger in moving train tosses a coin which falls behind him. It means that the motion of the train is [Refer to TB page 35 Q13] a) Accelerated b) Uniform c) Retarded d) Circular motion A. (a) Accelerated SESSION 1. INTRODUCTION AND FIRST LAW OF MOTION 36

SESSION 2 SECOND LAW OF MOTION 2.1 Mind Map SESSION 2. SECOND LAW OF MOTION 37

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 27] 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. SESSION 2. SECOND LAW OF MOTION 38

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 Q2. 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 28] A. We know that tension of string always tries to pull the bodies up. 39 SESSION 2. SECOND LAW OF MOTION

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 = m1a Thus, m1g–T = m1a . . . . . . . . . . . . . . . . . . ..(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. From the FBD of mass m2 40 Fnet = T–m2 g = m2 a . . . . . . . . . . . . . . . . . . . . . . . . (2) From (1) and (2), SESSION 2. SECOND LAW OF MOTION

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 1m/s2 is called one “Newton”. The S.I. unit of force is 1 Newton (N). vii. One Newton = 1kg × 1m/s2 2.5 Conceptual Understanding Q1. What force is required to produce an acceleration of 3 m/s2 in an object of mass 0.7 kg? [Refer to TB page 35 Q5] A. Force, F = ma Mass of object, m = 0.7 kg; Acceleration, a = 3m/s2 Force required = 0.7 × 3N = 2.1N SESSION 2. SECOND LAW OF MOTION 41

Q2. 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 36 Q1, Try These] 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 Q3. An object of mass 5 kg is moving with a velocity of 10 ms−1. A force is applied so that in 20 s, it attains a velocity of 25 ms−1. What is the amount of force applied on the object? [Refer to TB page 35 Q6] A. Initial velocity, u= 10 m/s Final velocity, v = 25 m/s Time = 15 s Mass, m = 5 kg Acceleration is given by a = v−u = 25−10 = 15 = 1 ms−2 t 15 15 SESSION 2. SECOND LAW OF MOTION 42

Force, F = ma = 5 × 1 = 5N Q4. Find the acceleration of body of mass 2 kg from the ﬁgures shown. [Refer to TB page 36 Q2, Try These] A. From ﬁgure on left side The weight of the 2 kg block acts downwards. W = mg = 2 × 10 = 20N A 30 N force acts on this block upwards. So net force on the 2 kg block is, F = 30 – 20 = 10 N upwards Acceleration, a = F = 10 = 5m/s2 m 2 From ﬁgure on right side The weights of the 2 kg and 3 kg blocks act downwards. W for 2 kg block = 20 N, and W for 3 kg block = 30 N So, net force on 2 kg block = 30 N upwards – 20 N downwards = 10 N upwards Acceleration, a = F/ m = 10/ (2+3) = 2 m/s2 SESSION 2. SECOND LAW OF MOTION 43

Q5. Two rubber bands stretched to the standard length cause an object to accelerate at 2m/s2 . Suppose another object with twice the mass is pulled by four rubber bands stretched to the standard length, what is the acceleration of the second object? [Refer to TB page 36 Q3, Try These] A. Let m1kg be the mass of the ﬁrst object. The force with which the ﬁrst object is accelerated, F = ma m = m1, and acceleration a = 2m/s2 Force, F1 = m1 × 2 = 2m1N Since four rubber bands are used (i.e., twice the number of rubber bands over the ones used on the ﬁrst object) on the second object, force acting on the second object F2 = 2F1 Newton T he mass o f the second object, m2 = twice the mass o f the f irst object m2 = 2 × m1 = 2m1 Acceleration of the second object, a2 = F2 = 2F1 = F1 = a = 2m/s2 m2 2m1 m1 Q6. A force of 5 N produces an acceleration of 8ms−2 on a mass m1 and an acceleration of 24ms−2 on a mass m2 . What acceleration would the same force provide if both the masses are tied together? [Refer to TB page 36 Q5, Try These] A. Force, F = 5 N Acceleration produced on the ﬁrst mass, a1 = 8ms−2 In case of ﬁrst object, mass m1 = F = 5 kg a1 8 In case of second object, mass m2 = F = 5 kg a2 24 If both are tied together, total mass, M = 5 + 5 = 15 + 5 = 20 = 5 kg 8 24 24 24 24 6 Acceleration produced by same force, A = F = 5 =6 M 5 6 SESSION 2. SECOND LAW OF MOTION 44

Thus, acceleration produced on the combined masses = 6 m/s2 Q7. 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 36 Q6, Try These] 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 Q8. System is shown in ﬁgure. Assume there is no friction.Find the acceleration of the blocks and tension in the string. Take g = 10m/s2. [Refer to TB page 36 Q7, Try These]] A. Force on each block= mg = 3 × 10N Acceleration, a = F = 3 × 10 = 30 = 5m/s2 m 3+3 6 ∴ Acceleration of system = 5m/s2 Force, F = ma = 3 × 5N = 15N Q9. Three identical blocks, each of mass 10 kg, are pulled as shown on the horizontal fric- tionless 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 36 Q8, Try These] SESSION 2. SECOND LAW OF MOTION 45

A. Recall: F = ma ⇒ a = F m Force acting on the system, F = 30 N Total mass on which the force acts, m = 10kg × 3 = 30kg ∴ a = 30N = 1m/s2 30kg ∴ Acceleration of each block = 1m/s2 Again,T1 = Force acting on ﬁrst block T1 = 30N = 10N 3 T2 = Tension acting on second block T2 = 30N×2 = 20N 3 Q10. Illustrate the second law of motion with an example.( [Refer to TB page 35 Q7] A. Second law: The harder you hit the ball, the more quickly it moves away from you, because you impart a greater acceleration to it. Q11. 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 35 Q4] A. i. Impulse = Force × time = 200N × 3s = 600N s. ii. Acceleration = Force ÷ mass = 200N = 1/6ms−2 1200kg Change in velocity = acceleration × time = 0.5ms−1 SESSION 2. SECOND LAW OF MOTION 46

2.6 Application to Daily Life, Concern to Bio Diversity Q1. 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 35 Q12] 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 Q2. 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 1.7 ms-2 ? [Refer to TB page 37 Q11, Try These] A. Force applied, F = ma m, mass of automobile = 1500 kg negative acceleration, a = –1.7 ms-2 Force between vehicle and road F = ? F, force = 1500 x –1.7 N = –2550 N (Note: The negative sign indicates the direction of force is opposite to the direction of motion.) Q3. A truck is moving under a hopper with a constant speed of 20 m/s. Sand falls on the truck at the rate 20 kg/s. What is the force acting on the truck due to falling of sand? [Refer to TB page 37 Q12, Try These] A. Constant speed = 20 m/s Time, t = 1 second SESSION 2. SECOND LAW OF MOTION 47

Acceleration of truck; a = ? We have v = u+at (and u=0) 20 = 0 + a × 1 a = 20 m/s2 Force, F = ma m = mass of sand falling per second = 20 kg and a = 20 m/s2 Force acting on truck = 20 × 20 N = 400 N. Q4. A ball of mass ‘m’ moves perpendicular to a wall with speed v, strikes it and rebounds with the same speed in the opposite direction. What is the direction and magnitude of the average force acting on the ball due to the wall? [Refer to TB page 37 Q14, Try These] A. According to Newton’s third law of motion, Force exerted by the ball on the wall = –( Force exerts by the wall on the ball) ∴ FBW = –FWB Force exerted by the ball : Mass of ball = m, speed = v, FBW = ma = m.v/t As the wall is at rest and exerts some force on the ball of mass m, then the ball moves in the other direction with the same speed. SESSION 2. SECOND LAW OF MOTION 48