Physics Class IX Reference Study Material PDF

PHYSICS REFERENCE STUDY MATERIAL for CLASS – IX CHAPTER WISE CONCEPTS, FORMULAS AND NUMERICALS INLCUDING HOTS PROBLEMS Prepared by M. S. KUMARSWAMY, TGT(MATHS) M....

PHYSICS REFERENCE STUDY MATERIAL for CLASS – IX CHAPTER WISE CONCEPTS, FORMULAS AND NUMERICALS INLCUDING HOTS PROBLEMS Prepared by M. S. KUMARSWAMY, TGT(MATHS) M. Sc. Gold Medallist (Elect.), B. Ed. Kendriya Vidyalaya GaCHIBOWLI Prepared by: M. S. KumarSwamy, TGT(Maths) Page - A - Prepared by: M. S. KumarSwamy, TGT(Maths) Page - B -  DEDICATED TO MY FATHER LATE SHRI. M. S. MALLAYYA Prepared by: M. S. KumarSwamy, TGT(Maths) Page - C -  CHAPTER – 8 MOTION MOTION Motion means movement. The motion of an object is perceived when its position changes continuously with respect to some stationary object. DISTANCE The distance travelled by an object is the length of actual path travelled by the object during the motion. DISPLACEMENT The displacement of an object is the shortest distance travelled between the initial and final position of the object. When final position coincides with the initial position, displacement is 0 but distance is not equal to 0. Both, the distance and displacement are measure in meter or cm or km. Distance is a scalar quantity having only. Displacement is a vector quantity having both magnitude and direction. The distance travelled by an object in motion can never be zero or negative. The displacement can be positive, zero or negative. INTEXT QUESTIONS PAGE NO. 100 1. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds? Ans. The farmer takes 40 s to cover 4 × 10 = 40 m. 40 In 2 min and 20 s (140 s), he will cover a distance =  140  140m 40 140 Therefore, the farmer completes  3.5 rounds (3 complete rounds and a half round) of 40 the field in 2 min and 20 s. That means, after 2 min 20 s, the farmer will be at the opposite end of the starting point. Now, there can be two extreme cases. Case I: Starting point is a corner point of the field. In this case, the farmer will be at the diagonally opposite corner of the field after 2 min 20 s. Therefore, the displacement will be equal to the diagonal of the field. Hence, the displacement will be 102  10 2  14.1m Case II: Starting point is the middle point of any side of the field. In this case the farmer will be at the middle point of the opposite side of the field after 2 min 20 s. Therefore, the displacement will be equal to the side of the field, i.e., 10 m. For any other starting point, the displacement will be between 14.1 m and 10 m. 2. 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. Ans. (a) Not true. Displacement can become zero when the initial and final position of the object is the same. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 1 -  (b) Not true. Displacement is the shortest measurable distance between the initial and final positions of an object. It cannot be greater than the magnitude of the distance travelled by an object. However, sometimes, it may be equal to the distance travelled by the object. UNIFORM MOTION A body is said to have a uniform motion if it travels equal distances in equal intervals of time, no matter how small these intervals may be. Eg. A vehicle running at a constant speed of 10m/sec ,will cover equal distances of 10metres every second, so its motion will be uniform. NON-UNIFORM MOTION A body is said to have a non- uniform motion if it travels unequal distances in equal intervals of time, no matter how small these intervals may be. Eg. A freely ball from a certain height covers unequal distances in equal intervals of time, so its motion is non uniform. Non uniform motion is also called accelerated motion. SPEED Speed of a body is defined as the distance travelled by the body in unit time. distance travelled(s ) Speed (v)  time taken(t ) Speed is a scalar quantity. Uniform Speed: When a body travels equal distances in equal intervals of time, the speed of the body is said to be uniform. Non-uniform Speed: When a body travels unequal distances in equal intervals of time, the speed of the body is said to be non-uniform. VELOCITY Velocity of a body is the distance travelled by the body in unit time in a given direction. distance travelled in a given direction displacement Velocity   time taken time Velocity is a vector quantity. Uniform Velocity: When a body travels equal distances in equal intervals of time in a particular direction, the velocity of the body is said to be uniform. Non-uniform Velocity: When a body travels unequal distances in equal intervals of time in a particular direction, the velocity of the body is said to be non-uniform. Both speed and velocity are measured in m/s or cm/s or km/hr AVERAGE VELOCITY Average Velocity: When velocity of a body is changing at a uniform rate, average velocity is Initial velocity(u ) + final velocity(v ) given by vav  2 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 2 - INTEXT QUESTIONS PAGE NO. 102 1. Distinguish between speed and velocity. Ans. Speed Velocity Speed is the distance travelled by an object Velocity is the displacement of an object in in a given interval of time. It does not have a given interval of time. It has a unique any direction. direction. The speed of an object can never be The velocity of an object can be negative, negative. At the most, it can become zero. positive, or equal to zero. This is because This is because distance travelled can never displacement can take any of these three be negative. values. 2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed? Total distance covered Ans. Average speed  and Total time taken Displacement Average velocity  Total time taken If the total distance covered by an object is the same as its displacement, then its average speed would be equal to its average velocity. 3. What does the odometer of an automobile measure? Ans. The odometer of an automobile measures the distance covered by an automobile. 4. What does the path of an object look like when it is in uniform motion? Ans. An object having uniform motion has a straight line path. 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 × 108 m s−1. Ans. Time taken by the signal to reach the ground station from the spaceship = 5 min = 5 × 60 = 300 s Speed of the signal = 3 × 108 m/s Distance travelled Speed  Time taken ∴Distance travelled = Speed × Time taken = 3 × 108 × 300 = 9 × 1010 m Hence, the distance of the spaceship from the ground station is 9 × 1010 m. ACCELERATION Acceleration of a body is defined as the rate of change of velocity of the body with time. It is v u given by formula: a t It is measured in m/s2 or cm/s2 or km/hr2. Acceleration is a vector quantity. If the velocity of the body increases with time, the acceleration is positive, and the kind of motion is called accelerated motion. If the velocity of the body decreases with time, the acceleration is negative (retardation), and the motion is called decelerated motion. Uniform Acceleration: When velocity of the body changes by equal amounts in equal intervals of time in a particular direction, the acceleration of the body is said to be uniform. Non-uniform Acceleration: When velocity of the body changes by unequal amounts in equal intervals of time in a particular direction, the acceleration of the body is said to be non- uniform. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 - INTEXT QUESTIONS PAGE NO. 103 1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration? Ans. (i) A body is said to have uniform acceleration if it travels in a straight path in such a way that its velocity changes at a uniform rate, i.e., the velocity of a body increases or decreases by equal amounts in an equal interval of time. (ii) A body is said to have non-uniform acceleration if it travels in a straight path in such a way that its velocity changes at a non-uniform rate, i.e., the velocity of a body increases or decreases in unequal amounts in an equal interval of time. 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. 5 Ans. Initial speed of the bus, u = 80 km/h = 80   22.22m / s 18 5 Final speed of the bus, v = 60 km/h = 60   16.66m / s 18 Time take to decrease the speed, t = 5 s v  u 16.66  22.22 Acceleration, a    1.12m / s 2 t 5 Here, the negative sign of acceleration indicates that the velocity of the car 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. Ans. Initial velocity of the train, u = 0 (since the train is initially at rest) 5 Final velocity of the train, v = 40 km/h = 40   11.11m / s 18 v  u 11.11  0 Acceleration, a    0.0185m / s 2 t 5 Time taken, t = 10 min = 10 × 60 = 600 s Hence, the acceleration of the train is 0.0185 m/s2. NUMERICALS 1. A particle is moving in a circle of diameter 5m. Calculate the distance covered and the displacement when it competes 3 revolutions. 2. A body thrown vertically upwards reaches a maximum height ‘h’. It then returns to ground. Calculate the distance travelled and the displacement. 3. A body travels a distance of 15m from A to B and then moves a distance of 20m at right angles to AB. Calculate the total distance travelled and the displacement. 4. An object is moving in a circle of radius ‘r’. Calculate the distance and displacement (i) when it completes half the circle (ii) when it completes one full circle. 5. An object travels 16m in 4s and then another 16m in 2s. What is the average speed of the object? Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 - 6. Vishnu swims in a 90m long pool. He covers 180m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Vishnu. 7. In along distance race, the athletics were expected to take four rounds of the track such that the line of finish was same as the line of start. Suppose the length of the track was 200m. (a) What is the total distance to be covered by the athletics? (b) What is the displacement of the athletics when they touch the finish line? (c) Is the motion of the athletics uniform or non-uniform? (d) Is the displacement of an athletic and the distance covered by him at the end of the race equal? 8. Starting from a stationary position, Bhuvan paddles his bicycle to attain a velocity of 6m/s in 30s. Then he applies brakes such that the velocity of bicycle comes down to 4m/s in the next 5s. Calculate the acceleration of the bicycle in both the cases. 9. Amit is moving in his car with a velocity of 45km/hr. How much distance will he cover (a) in one minute and (b) in one second. 10. The odometer of a car reads 2000 km at the start of a trip and 2400km at the end of the trip. If the trip took 8 hr, calculate the average speed of the car in km/hr and m/s. 11. An electric train is moving with a velocity of 120km/hr. How much distance will it move in 30s? 12. A body is moving with a velocity of 15m/s. If the motion is uniform, what will be the velocity after 10s? 13. A train travels some distance with a speed of 30km/hr and returns with a speed of 45km/hr. Calculate the average speed of the train. 14. A train 100m long moving on a straight level track passes a pole in 5s. Find (a) the speed of the train (b) the time it will take to cross a bridge 500m long. 15. A car travels along a straight line for first half time with speed 40km/hr and the second half time with speed 60km/hr. Find the average speed of the car. 16. A body starts rolling over a horizontal surface with an initial velocity of 0.5m/s. Due to friction, its velocity decreases at the rate of 0.05m/s2. How much time will it take for the body to stop? 17. A car traveling at 36km/hr speeds upto 70km/hr in 5 seconds. What is its acceleration? If the same car stops in 20s, what is the retardation? 18. A scooter acquires a velocity of 36km/hr in 10seconds just after the start. It takes 20 seconds to stop. Calculate the acceleration in the two cases. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 - 19. On a 120km track, a train travels the first 30 km at a uniform speed of 30 km/hr. How fast must the train travel the next 90 km so as to average 60 km/hr for the entire trip? 20. A train travels at 60 km/hr for 0.52 hr; at 30 km/hr for the next 0.24 hr and at 70 km/hr for the next 0.71 hr. What is the average speed of the train? GRAPHICAL REPRESENTATION OF MOTION A graph represents the relation between two variable quantities in pictorial form. It is plotted between two variable quantities. The quantity that is varied our choice is called independent variable. The other quantity, which varies as a result of this change, is called dependent variable. For example, in distance-time graph, time is independent variable and distance is dependent variable. Similarly, in velocity-time graph, time is independent variable and velocity is dependent variable. DISTANCE-TIME GRAPH The distance-time graph represents the change in the position of a body with time. In this graph, we take time along the x-axis and the distance along the y-axis. The distance-time graph under different conditions are shown below. 1. When the body is at rest The position of the body does not change with time. Its distance from the origin continues to be same at all instants of time. Therefore, we obtain a straight line parallel to x-axis(time axis) (see the below figure) 2. When the body is in uniform motion When an object travels equal distances in equal intervals of time, it moves with uniform speed. This shows that the distance travelled by the object is directly proportional to time taken. Thus, for uniform speed, a graph of distance travelled against time is a straight line, as shown in below figure. The portion OB of the graph shows that the distance is increasing at a uniform rate. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 - On the graph, AC denotes the time interval (t2 – t1) while BC corresponds to the distance (s2 – s1). We can see from the graph that as the object moves from the point A to B, it covers a distance (s2 – s1) in time (t2 – t1). The speed, v of the object, therefore can be represented as distance travelled speed  time taken s2  s1  v t2  t1 3. When the body is in non-uniform motion When a body moves with non-uniform speed it covers unequal distance in equal intervals of time. The distance-time graph for a body which is moving with non-uniform speed is not a straight line. It is a curve. The speed of the object will change with respect to time. VELOCITY-TIME GRAPH The geometrical relationship between the velocity of an object and the time taken by the object is called the velocity-time graph. The velocity-time graph of an object can be drown by taking the time taken along the X-axis and the velocity along the Y-axis. The ratio of the velocity and the time taken will give the acceleration of the object. Therefore, the slope of the velocity-time graph gives the acceleration Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 - of the given object. That is, by using this graph one can find the acceleration of an object. The velocity-time graph under different conditions are shown below. 1. When the body is moving with uniform velocity If the object moves at uniform velocity, the height of its velocity-time graph will not change with time. It will be a straight line parallel to the x-axis. We know that the product of velocity and time give displacement of an object moving with uniform velocity. The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of the displacement. To know the distance moved by the car between time t1 and t2 using the above figure, draw perpendiculars from the points corresponding to the time t1 and t2 on the graph. The velocity of v km/h is represented by the height AC or BD and the time (t2 – t1) is represented by the length AB. So, the distance s moved by the car in time (t2 – t1) can be expressed as s = AC × CD = v(t2 – t1) = Area of the rectangle ABDC (shaded portion) 2. When the body is moving with uniform acceleration In this case, the velocity-time graph for the motion of the car is shown in below figure. The nature of the graph shows that velocity changes by equal amounts in equal intervals of time. Thus, for all uniformly accelerated motion, the velocity-time graph is a straight line. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -  The area under the velocity-time graph gives the distance (magnitude of displacement) moved by the car in a given interval of time. If the car would have been moving with uniform velocity, the distance travelled by it would be represented by the area ABCD under the graph. Since the magnitude of the velocity of the car is changing due to acceleration, the distance s travelled by the car will be given by the area ABCDE under the velocity-time graph. That is, s = area ABCDE = area of the rectangle ABCD + area of the triangle ADE 1  AB  BC  ( AD  DE ) 2 3. When the body is moving with a variable acceleration In the case of non-uniformly accelerated motion, velocity-time graphs can have any shape. In below Fig. (a) shows a velocity-time graph that represents the motion of an object whose velocity is decreasing with time while Figure (b) shows the velocity-time graph representing the non-uniform variation of velocity of the object with time. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 - INTEXT QUESTIONS PAGE NO. 107 1. What is the nature of the distance−time graphs for uniform and non-uniform motion of an object? Ans. The distance−time graph for uniform motion of an object is a straight line (as shown in the following figure). The distance−time graph for non-uniform motion of an object is a curved line (as shown in the given figure). Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 10 - 2. What can you say about the motion of an object whose distance−time graph is a straight line parallel to the time axis? Ans. When an object is at rest, its distance−time graph is a straight line parallel to the time axis. A straight line parallel to the x-axis in a distance−-time graph indicates that with a change in time, there is no change in the position of the object. Thus, 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? Ans. Object is moving uniformly. A straight line parallel to the time axis in a speed−time graph indicates that with a change in time, there is no change in the speed of the object. This indicates the uniform motion of the object. 4. What is the quantity which is measured by the area occupied below the velocity−time graph? Ans. Distance Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 11 -  The graph shows the velocity−time graph of a uniformly moving body. Let the velocity of the body at time (t) be v. Area of the shaded region = length × breath Where, Length = t, Breath = v Area = vt = velocity × time …(i) Distance We know, Velocity  Time ∴ Distance = Velocity × Time…(ii) From equations (i) and (ii), Area = Distance Hence, the area occupied below the velocity−time graph measures the distance covered by the body. NUMERICALS 1. The right-sided figure is the distance-time graph of an object. Do you think it represents a real situation? If so, why? If not, why not? 2. The graph in below figure shows the positions of a body at different times. Calculate the speed of the body as it moves from (i) A to B (ii) B to C and (iii) C to D. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 12 - 3. The velocity time graph of an ascending passenger lift is given below. What is the acceleration of the lift: (i) during the first two seconds (ii) between 2nd and 10th second (iii) during the last two seconds. 4. A body is moving uniformly with a velocity of 5m/s. Find graphically the distance travelled by it in 5 seconds. 5. Study the speed-time graph of a body shown in below figure and answer the following questions: (a) What type of motion is represented by OA? (b) What type of motion is represented by AB? (c) What type of motion is represented by BC? (d) Calculate the acceleration of the body. (e) Calculate the retardation of the body. (f) Calculate the distance travelled by the body from A to B. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 13 - 6. In the above question, calculate (i) distance travelled from O to A (ii) distance travelled from B to C. (iii) total distance travelled by the body in 16 sec. 7. A car is moving on a straight road with uniform acceleration. The following table gives the speed of the car at various instants of time: Time(t) 0 10 20 30 40 50 Speed(m/s) 5 10 15 20 25 30 Draw the speed time graph choosing a convenient scale. Determine from it (i) the acceleration of the car (ii) the distance travelled by the car in 50 sec. 8. The graph in below figure shows the positions of a body at different times. Calculate the speed of the body as it moves from (i) A to B (ii) B to C and (iii) C to D. 9. A car is moving on a straight road with uniform acceleration. The speed of the car varies with time as follows: Time(t) 0 2 4 6 8 10 Speed(m/s) 4 8 12 16 20 24 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 14 -  Draw the speed time graph choosing a convenient scale. Determine from it (i) the acceleration of the car (ii) the distance travelled by the car in 10 sec. 10. The graph given below is the velocity-time graph for a moving body. Find (i) velocity of the body at point C (ii) acceleration acting on the body between A and B (iii) acceleration acting on the body between B and C. EQUATIONS OF MOTION BY GRAPHICAL METHOD When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. There are three such equations. These are: v = u + at --------------- (1) 1 s  ut  at 2 ------------- (2) 2 2a s = v2 – u2 ----------- (3) where u is the initial velocity of the object which moves with uniform acceleration a for time t, v is the final velocity, and s is the distance travelled by the object in time t. Eq. (1) describes the velocity-time relation and Eq. (2) represents the position-time relation. Eq. (3), which represents the relation between the position and the velocity, can be obtained from Eqs. (1) and (2) by eliminating t. These three equations can be derived by graphical method. EQUATION FOR VELOCITY-TIME RELATION In the graph, AC gives the initial velocity (u). BE gives the final velocity (v). CE represents the time taken t. DF gives the change in velocity. Change in velocity Acceleration= Time DF OF  OD a  CE OE  OC But OE – OC = t v u a t Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 15 -  v – u = at …… (i)  v = u + at ……(I) EQUATION FOR POSITION-VELOCITY RELATION Let ‘s’ be the displacement of the body in a time t. In the graph, Displacement = Area CABE s = Area of the rectangle CAGE + Area of the triangle ABG. 1 s = AC x CE + (AG x GB) 2 Here AC = u CE = t AG = t GB = v – u = at [from(i)] 1 s = ut + x t x at 2 1 2 s = ut + at …………… (II) 2 EQUATION FOR POSITION-TIME RELATION In the graph, Displacement = Area of the trapezium CABE 1 s = (AC+EB) x CE 2 Here AC = u, EB = v, CE = t uv s  t.............(ii ) 2 v u From (i), we have t  a Substituting the value of t, Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 16 -  u v vu s  2 a v  u2 2 s 2a 2 2  v  u  2as v2 = u2 + 2as …………….(III) (I), (II) and (III) are the equations of motion. Acceleration due to gravity The velocity of the body gradually decreases and becomes zero at a maximum height.The body is decelarated or retarded. When the body is allowed to fall down, the velocity gradually increases. Now the body is accelerated. The decelaration or acceleration due to the gravitational force of earth is known as accelaration due to gravity, denoted as ‘g’. The average value of ‘g’ is 9.8 m/s2. The velocity of the body thrown vertically upwards will decrease by 9.8m for every second and the velocity of a body falling down increases by 9.8m for every second. The equations of motion for this body can be obtained from the equations of motion. v =u + at 1 2 s= ut + at 2 v2 = u2 + 2as For the body thrown upwards, equations can be obtained by substituting a = –g and s = h we get, v = u – gt 1 2 h = ut – gt 2 v2 = u2 – 2gh When a body allowed to fall freely, u = 0. a = g and s = h. Now, the equations will be v = gt 1 2 h= gt 2 v2 = 2gh INTEXT QUESTIONS PAGE NO. 109 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. Ans. (a) Initial speed of the bus, u = 0 (since the bus is initially at rest) Acceleration, a = 0.1 m/s2 Time taken, t = 2 minutes = 120 s Let v be the final speed acquired by the bus. v u v0 a   0.1  t 120 ∴v = 12 m/s Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 17 -  (b) According to the third equation of motion: v2 − u2 = 2as Where, s is the distance covered by the bus (12)2 − (0)2 = 2(0.1) s s = 720 m Speed acquired by the bus is 12 m/s. Distance travelled by the bus 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. Ans. Initial speed of the train, u = 90 km/h = 25 m/s Final speed of the train, v = 0 (finally the train comes to rest) Acceleration = −0.5 m s−2 According to third equation of motion: v2 = u2 + 2 as (0)2 = (25)2 + 2 (−0.5) s Where, s is the distance covered by the train 252 s  625m 2  0.5 The train will cover a distance of 625 m before it comes to rest. 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? Ans. Initial velocity of the trolley, u = 0 (since the trolley was initially at rest) Acceleration, a = 2 cm s−2 = 0.02 m/s2 Time, t = 3 s According to the first equation of motion: v = u + at Where, v is the velocity of the trolley after 3 s from start v = 0 + 0.02 × 3 = 0.06 m/s Hence, the velocity of the trolley after 3 s from start is 0.06 m/s. 4. A racing car has a uniform acceleration of 4 m s−2. What distance will it cover in 10 s after start? Ans. Initial velocity of the racing car, u = 0 (since the racing car is initially at rest) Acceleration, a = 4 m/s2 Time taken, t = 10 s 1 According to the second equation of motion: s  ut  at 2 2 Where, s is the distance covered by the racing car 1 400 s  0   4  102   200m 2 2 Hence, the distance covered by the racing car after 10 s from start is 200 m. 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? Ans. Initially, velocity of the stone,u = 5 m/s Final velocity, v = 0 (since the stone comes to rest when it reaches its maximum height) Acceleration of the stone, a = acceleration due to gravity, g = 10 m/s2 (in downward direction) There will be a change in the sign of acceleration because the stone is being thrown upwards. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 18 -  Acceleration, a = −10 m/s2 Let s be the maximum height attained by the stone in time t. According to the first equation of motion: v = u + at 0 = 5 + (−10) t 5 t   0.5s 10 According to the third equation of motion: v2 = u2 + 2 as (0)2 = (5)2 + 2(−10) s 52 s  1.25m 20 Hence, the stone attains a height of 1.25 m in 0.5 s. UNIFORM CIRCULAR MOTION An athlete runs along the circumference of a circular path. This type of motion is known as circular motion. The movement of an object in a circular path is called circular motion. When an object moves in a circular path with a constant velocity, its motion is called uniform circular motion. In uniform circular motion, the magnitude of the velocity is constant at all points and the direction of the velocity changes continuously. We know that the circumference of a circle of radius r is given by 2  r. If the athlete takes t seconds to go once around the circular path of radius r, the velocity v is given by 2 r v t NUMERICALS 1. A car increases its speed from 20 km/h to 50 km/h in 10 seconds. What is its acceleration? 2. A ship is moving at a speed of 56km/h. One second later, it is moving at 58km/h. What is its acceleration? 3. A scooter acquires a velocity of 36km/h in 10 seconds just after the start. Calculate the acceleration of the scoter. 4. A racing car has uniform acceleration of 4m/s2. What distance will it cover in 10 seconds after start? 5. A car acquires a velocity of 72km/h in 10 seconds starting from rest. Find (a) the acceleration (b) the average velocity (c) the distance travelled in this time. 6. A body is accelerating at a constant rate of 10m/s2. If the body starts from rest, how much distance will it cover in 2 seconds? 7. An object undergoes an acceleration of 8m/s2 starting from rest. Find the distance travelled in 1 second. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 19 - 8. A moving train is brought to rest within 20 seconds by applying brakes. Find the initial velocity, if the retardation due to brakes is 2m/s2. 9. A car accelerates uniformly from 18km/h to 36 km/h in 5 seconds. Calculate (i) acceleration and (ii) the distance covered by the car in that time. 10. A body starts to slide over a horizontal surface with an initial velocity of 0.5 m/s. Due to friction, its velocity decreases at the rate of 0.05 m/s2. How much time will it take for the body to stop? 11. A train starting from the rest moves with a uniform acceleration of 0.2 m/s2 for 5 minutes. Calculate the speed acquired and the distance travelled in this time. 12. A bus was moving with a speed of 54 km/h. On applying brakes, it stopped in 8 seconds. Calculate the acceleration and the distance travelled before stopping. 13. A motor cycle moving with a speed of 5 m/s is subjected to an acceleration of 0.2 m/s2. Calculate the speed of the motor cycle after 10 seconds and the distance travelled in this time. 14. The brakes applied to a car produce an acceleration of 6 m/s2 in the opposite direction to the motion. If the car takes 2 seconds to stop after the application of brakes, calculate the distance it travels during this time. 15. A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity. 16. Calculate the speed of the tip of second’s hand of a watch of length 1.5 cm. 17. A cyclist goes once round a circular track of diameter 105m in 5 minutes. Calculate his speed. 18. A cyclist moving on a circular track of radius 50m complete revolution in 4 minutes. What is his (i) average speed (ii) average velocity in one full revolution? 19. The length of minutes hand of a clock in 5 cm. Calculate its speed. 20. A car starts from rest and moves along the x-axis with constant acceleration 5m/s2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest? 21. An object is dropped from rest at a height of 150m and simultaneously another object is dropped from rest at a height 100m. What is the difference in their heights after 2 seconds if both the objects drop with same acceleration? How does the difference in heights vary with time? 22. Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds. 23. Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of u12:u22 (Assume upward acceleration is – g and downward acceleration to be +g). Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 20 - 24. An object starting from rest travels 20m in first 2 seconds and 160m in next 4 seconds. What will be the velocity after 7 seconds from the start? 25. An electron moving with a velocity of 5 x 104 m/s enters into a uniform electric field and acquires a uniform acceleration of 104m/s2 in the direction of its initial motion. (i) Calculate the time in which the electron would acquire a velocity double of its initial velocity. (ii) How much distance the electron would cover in this time? EXERCISE QUESTIONS PAGE NO. 112 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? Ans. Diameter of a circular track, d = 200 m d Radius of the track, r =  100m 2 Circumference = 2πr = 2π (100) = 200π m In 40 s, the given athlete covers a distance of 200π m. 200 In 1 s, the given athlete covers a distance = m 40 The athlete runs for 2 minutes 20 s = 140 s 200  22 ∴Total distance covered in 140 s  140  2200m 40  7 The athlete covers one round of the circular track in 40 s. This means that after every 40 s, the athlete comes back to his original position. Hence, in 140 s he had completed 3 rounds of the circular track and is taking the fourth round. He takes 3 rounds in 40 × 3 = 120 s. Thus, after 120 s his displacement is zero. Then, the net displacement of the athlete is in 20 s only. In this interval of time, he moves at the opposite end of the initial position. Since displacement is equal to the shortest distance between the initial and final position of the athlete, displacement of the athlete will be equal to the diameter of the circular track. ∴ Displacement of the athlete = 200 m Distance covered by the athlete in 2 min 20 s is 2200 m and his displacement is 200 m. 2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 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? Ans. (a) From end A to end B Distance covered by Joseph while jogging from A to B = 300 m Time taken to cover that distance = 2 min 50 seconds = 170 s Total distance covered Average speed  Total time taken Total distance covered = 300 m Total time taken = 170 s Displacement = shortest distance between A and B = 300 m Time interval = 170 s 300 Average speed   1.765m / s 170 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 21 -  Displacement Average velocity  Total time taken Displacement = shortest distance between A and B = 300 m Time interval = 170 s 300 Average velocity   1.765m / s 170 The average speed and average velocity of Joseph from A to B are the same and equal to 1.765 m/s. (b) From end A to end C Total distance covered Average speed  Total time taken Total distance covered = Distance from A to B + Distance from B to C = 300 + 100 = 400 m Total time taken = Time taken to travel from A to B + Time taken to travel from B to C = 170 + 60 = 230 s 400 Average speed   1.739m / s 230 Displacement Average velocity  Total time taken Displacement from A to C = AC = AB − BC = 300 − 100 = 200 m Time interval = time taken to travel from A to B + time taken to travel from B to C = 170 + 60 = 230 s 200 Average velocity   0.87m / s 230 The average speed of Joseph from A to C is 1.739 m/s and his average velocity is 0.87 m/s. 3. Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 40 km/h. What is the average speed for Abdul’s trip? Ans. Case I: While driving to school Average speed of Abdul’s trip = 20 km/h Total distance covered Average speed  Total time taken Total distance = Distance travelled to reach school = d Let total time taken = t1 d d  20   t1 ..........(i ) t1 20 Case II: While returning from school Total distance = Distance travelled while returning from school = d Now,total time taken = t2 d d  40   t1 ..........(ii) t2 40 Total distance covered in the trip Average Speed for Abdul's trip = Total time taken Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 22 -  Where, Total distance covered in the trip = d + d = 2d Total time taken, t = Time taken to go to school + Time taken to return to school = t1 + t2 2d  Average speed = t1 +t 2 From equations (i) and (ii), 2d 2 80 Average speed =    26.67m / s d d 2 1 3 + 20 40 40 Hence, the average speed for Abdul’s trip is 26.67 m/s. 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s2 for 8.0 s. How far does the boat travel during this time? Ans. Initial velocity, u = 0 (since the motor boat is initially at rest) Acceleration of the motorboat, a = 3 m/s2 Time taken, t = 8 s According to the second equation of motion: 1 s  ut  at 2 2 Distance covered by the motorboat, s 1 s  0   3  82  96m 2 Hence, the boat travels a distance of 96 m. 5. A driver of a car travelling at 52 km/h applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km/h 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? Ans. Case A: Initial speed of the car, u1 = 52 km/h = 14.4 m/s Time taken to stop the car, t1 = 5 s Final speed of the car becomes zero after 5 s of application of brakes. Case B: Initial speed of the car, u2 = 3 km/h = 0.833 m/s 0.83 m/s Time taken to stop the car, t2 = 10 s Final speed of the car becomes zero after 10 s of application of brakes. Plot of the two cars on a speed−time graph is shown in the following figure: Distance covered by each car is equal to the area under the speed−time graph. Distance covered in case A, 1 1 S1   OP  OR   14.4  5  36m 2 2 Distance covered in case B, 1 1 S 2   OS  OQ   0.83  10  4.15m 2 2 Area of ΔOPR > Area of ΔOSQ Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 23 -  Thus, the distance covered in case A is greater than the distance covered in case B. Hence, the car travelling with a speed of 52 km/h travels farther after brakes were applied. 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? Distance Ans. Speed  Time y-axis Distance Slope of the graph=  x-axis Time ∴Speed = slope of the graph Since slope of object B is greater than objects A and C, it is travelling the fastest. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 24 -  (b) All three objects A, B and C never meet at a single point. Thus, they were never at the same point on road. (c) On the distance axis: 7 small boxes = 4 km 4 ∴1 small box  km 7 Initially, object C is 4 blocks away from the origin. 16 ∴Initial distance of object C from origin  km 7 Distance of object C from origin when B passes A = 8 km 16 56  16 40 Distance covered by C  8     5.714km 7 7 7 Hence, C has travelled a distance of 5.714 km when B passes A. (d) 4 36 Distance covered by B at the time it passes C for 9 boxes   9   5.143km 7 7 Hence, B has travelled a distance of 5.143 km when B passes A. 7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s2, with what velocity will it strike the ground? After what time will it strike the ground? Ans. Distance covered by the ball, s = 20 m Acceleration, a = 10 m/s2 Initially, velocity, u = 0 (since the ball was initially at rest) Final velocity of the ball with which it strikes the ground, v According to the third equation of motion: v2 = u2 + 2 as v2 = 0 + 2 (10) (20) v = 20 m/s According to the first equation of motion: v = u + at Where, Time, t taken by the ball to strike the ground is, 20 = 0 + 10 (t) Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 25 -  t=2s Hence, the ball strikes the ground after 2 s with a velocity of 20 m/s. 8. The speed-time graph for a car is shown is 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? (a) 1 The shaded area which is equal to  4  6  12m represents the distance travelled by the car in 2 the first 4 s. (b) The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car. 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 in a certain direction with an acceleration in the perpendicular direction. Ans. (a) Possible When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s2. (b) Possible When a car is moving in a circular track, its acceleration is perpendicular to its direction. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 26 - 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? Ans. Radius of the circular orbit, r = 42250 km Time taken to revolve around the earth, t = 24 h Speed of a circular moving object, 2 r v t 2  3.14  42250 v 24  v  1.105  104 km / h  3.069km / s Hence, the speed of the artificial satellite is 3.069 km/s. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 27 -  ASSIGNMENT QUESTIONS MOTION Multiple Choice Questions 1. If the displacement of an object is proportional to square of time, then the object moves with (a) uniform velocity (b) uniform acceleration (c) increasing acceleration (d) decreasing acceleration 2. The distance time graph of a body coincides with its time axis. The body must be (a) in uniform motion (b) at rest (c) in uniformly accelerated motion (d) in zig-zag motion 3. From the given v – t graph (see below Fig.), it can be inferred that the object is (a) in uniform motion (b) at rest (c) in non-uniform motion (d) moving with uniform acceleration 4. The velocity time graph of a body is parallel to the time axis. The body is (a) at rest (b) having uniform acceleration (c) having zero acceleration (d) having non-uniform acceleration 5. A particle is moving in a circular path of radius r. The displacement after half a circle would be: (a) Zero (b) π r (c) 2 r (d) 2π r 6. A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is, (a) u/g (b) u2/2g (c) u2/g (d) u/2g Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 28 - 7. The numerical ratio of displacement to distance for a moving object is (a) always less than 1 (b) always equal to 1 (c) always more than 1 (d) equal or less than 1 8. Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 m/s. It implies that the boy is (a) at rest (b) moving with no acceleration (c) in accelerated motion (d) moving with uniform velocity 9. Area under a v – t graph represents a physical quantity which has the unit (a) m2 (b) m (c) m3 (d) m/s 10. Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in below Fig.. Choose the correct statement (a) Car A is faster than car D. (b) Car B is the slowest. (c) Car D is faster than car C. (d) Car C is the slowest. 11. Slope of a velocity – time graph gives (a) the distance (b) the displacement (c) the acceleration (d) the speed 12. In which of the following cases of motions, the distance moved and the magnitude of displacement are equal? (a) If the car is moving on straight road (b) If the car is moving in circular path (c) The pendulum is moving to and fro (d) The earth is revolving around the Sun Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 29 - 13. Which of the following figures (see below Figure) represents uniform motion of a moving object correctly? SHORT ANSWER QUESTIONS 14. The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Justify you answer. 15. How will the equations of motion for an object moving with a uniform velocity change? 16. A car starts from rest and moves along the x-axis with constant acceleration 5 m/s2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest? 17. A motorcyclist drives from A to B with a uniform speed of 30 km/h and returns back with a speed of 20 km/h. Find its average speed. 18. Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height. 19. The velocity-time graph (see below Figure) shows the motion of a cyclist. Find (i) its acceleration (ii) its velocity and (iii) the distance covered by the cyclist in 15 seconds. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 30 - 20. A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement–time graph is shown in below figure. Plot a velocity– time graph for the same. LONG ANSWER QUESTIONS 21. An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start. 22. An electron moving with a velocity of 5 × 104 m/s enters into a uniform electric field and acquires a uniform acceleration of 104 m/s2 in the direction of its initial motion. (i) Calculate the time in which the electron would acquire a velocity double of its initial velocity. (ii) How much distance the electron would cover in this time? 23. Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds. 24. Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of u12 : u22 ( Assume upward acceleration is –g and downward acceleration to be +g ). 25. An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time? Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 31 -  CHAPTER – 9 FORCE AND LAWS OF MOTION FORCE A force is anything that can cause a change to objects. Forces can: change the shape of an object move or stop an object change the direction of a moving object. A force can be classified as either a contact force or a non-contact force. A contact force must touch or be in contact with an object to cause a change. Examples of contact forces are: the force that is used to push or pull things, like on a door to open or close it the force that a sculptor uses to turn clay into a pot the force of the wind to turn a windmill A non-contact force does not have to touch an object to cause a change. Examples of non- contact forces are: the force due to gravity, like the Earth pulling the Moon towards itself the force due to electricity, like a proton and an electron attracting each other the force due to magnetism, like a magnet pulling a paper clip towards itself The unit of force is the newton (symbol N). This unit is named after Sir Isaac Newton who first defined force. Force is a vector quantity and has a magnitude and a direction. EFFECT OF FORCE: 1. Force can make a stationary body in motion. For example a football can be set to move by kicking it, i.e. by applying a force. 2. Force can stop a moving body – For example by applying brakes, a running cycle or a running vehicle can be stopped. 3. Force can change the direction of a moving object. For example; By applying force, i.e. by moving handle the direction of a running bicycle can be changed. Similarly by moving steering the direction of a running vehicle is changed. 4. Force can change the speed of a moving body – By accelerating, the speed of a running vehicle can be increased or by applying brakes the speed of a running vehicle can be decreased. 5. Force can change the shape and size of an object. For example -– By hammering, a block of metal can be turned into a thin sheet. By hammering a stone can be broken into pieces. Forces are can also divided into two types: 1. Balanced Forces 2. Unbalanced Forces Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 32 - BALANCED FORCES If the resultant of applied forces is equal to zero, it is called balanced forces. Example : - In the tug of war if both the teams apply similar magnitude of forces in opposite directions, rope does not move in either side. This happens because of balanced forces in which resultant of applied forces become zero. Balanced forces do not cause any change of state of an object. Balanced forces are equal in magnitude and opposite in direction. Balanced forces can change the shape and size of an object. For example - When forces are applied from both sides over a balloon, the size and shape of balloon is changed. UNBALANCED FORCES If the resultant of applied forces are greater than zero the forces are called unbalanced forces. An object in rest can be moved because of applying balanced forces. Unbalanced forces can do the following:  Move a stationary object.  Increase the speed of a moving object.  Decrease the speed of a moving object.  Stop a moving object.  Change the shape and size of an object. LAWS OF MOTION: NEWTON’S LAWS OF MOTION:  Newton's First Law of Motion - Any object remains in the state of rest or in uniform motion along a straight line, until it is compelled to change the state by applying external force.  Newton's Second Law of Motion - The rate of change of momentum is directly proportional to the force applied in the direction of force.  Newton's Third Law of Motion - There is an equal and opposite reaction for evrey action NEWTON’S FIRST LAW OF MOTION: Any object remains in the state of rest or in uniform motion along a straight line, until it is compelled to change the state by applying external force. Explanation: If any object is in the state of rest, then it will remain in rest untill a exernal force is applied to change its state. Similarly an object will remain in motion untill any external force is applied over it to change its state. This means all objects resist to in changing their state. The state of any object can be changed by applying external forces only. NEWTON’S FIRST LAW OF MOTION IN EVERYDAY LIFE: a. A person standing in a bus falls backward when bus is start moving suddenly. This happens because the person and bus both are in rest while bus is not moving, but as the bus starts moving the legs of the person start moving along with bus but rest portion of his body has tendency to remain in rest. Because of this person falls backward; if he is not alert. b. A person standing in a moving bus falls forward if driver applies brakes suddenly. This happens because when bus is moving, the person standing in it is also in motion along with bus. But when driver applies brakes the speed of bus decreases suddenly or bus comes in the state of rest suddenly, in this condition the legs of the person which are in the contact with bus come in rest while the rest parts of his body have tendency to remain in motion. Because of this person falls forward if he is not alert. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 33 - MASS AND INERTIA: The property of an object because of which it resists to get disturbed its state is called Inertia. Inertia of an object is measured by its mass. Inertia is directly proportional to the mass. This means inertia increases with increase in mass and decreases with decrease in mass. A heavy object will have more inertia than lighter one. In other words, the natural tendency of an object that resists the change in state of motion or rest of the object is called inertia. Since a heavy object has more inertia, thus it is difficult to push or pul a heavy box over the ground than lighter one. INTEXT QUESTIONS PAGE NO. 118 1. Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five-rupees coin and a one-rupee coin? Ans. Inertia is the measure of the mass of the body. The greater is the mass of the body; the greater is its inertia and vice-versa. (a) Mass of a stone is more than the mass of a rubber ball for the same size. Hence, inertia of the stone is greater than that of a rubber ball. (b) Mass of a train is more than the mass of a bicycle. Hence, inertia of the train is greater than that of the bicycle. (c) Mass of a five rupee coin is more than that of a one-rupee coin. Hence, inertia of the five rupee coin is greater than that of the 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. The velocity of the ball changes four times. As a football player kicks the football, its speed changes from zero to a certain value. As a result, the velocity of the ball gets changed. In this case, the player applied a force to change the velocity of the ball. Another player kicks the ball towards the goal post. As a result, the direction of the ball gets changed. Therefore, its velocity also changes. In this case, the player applied a force to change the velocity of the ball. The goalkeeper collects the ball. In other words, the ball comes to rest. Thus, its speed reduces to zero from a certain value. The velocity of the ball has changed. In this case, the goalkeeper applied an opposite force to stop/change the velocity of the ball. The goalkeeper kicks the ball towards his team players. Hence, the speed of the ball increases from zero to a certain value. Hence, its velocity changes once again. In this case, the goalkeeper applied a force to change the velocity of the ball. 3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch. Ans. Some leaves of a tree get detached when we shake its branches vigorously. This is because when the branches of a tree are shaken, it moves to and fro, but its leaves tend to remain at rest. This is because the inertia of the leaves tend to resist the to and fro motion. Due to this reason, the leaves fall down from the tree when shaken vigorously. 4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest? Ans. Due to the inertia of the passenger. Every body tries to maintain its state of motion or state of rest. If a body is at rest, then it tries to remain at rest. If a body is moving, then it Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 34 -  tries to remain in motion. In a moving bus, a passenger moves with the bus. As the driver applies brakes, the bus comes to rest. But, the passenger tries to maintain his state of motion. As a result, a forward force is exerted on him. Similarly, the passenger tends to fall backwards when the bus accelerates from rest. This is because when the bus accelerates, the inertia of the passenger tends to oppose the forward motion of the bus. Hence, the passenger tends to fall backwards when the bus accelerates forward. MOMENTUM Momentum is the power of motion of an object. The product of velocity and mass is called the momentum. Momentum is denoted by ‘p’. Therefore, momentum of the object = Mass x Velocity. Or, p = m x v Where, p = momentum, m = mass of the object and v = velocity of the object. NEWTON'S SECOND LAW OF MOTION Newton's second Law of Motion states that The rate of change of momentum is directly proportional to the force applied in the direction of force. For example; when acceleration is applied on a moving vehicle, the momentum of the vehicle increases and the increase is in the direction of motion because the force is being applied in the direction of motion. On the other hand, when brake is applied on the moving vehicle, the momentum of the vehicle decreases and the decrease is in the opposite direction of motion because the force is being applied in the opposite direction of motion. Mathematical formulation of Newton’s Second Law of Motion: Let mass of an moving object = m. Let the velocity of the object changes from ‘u’ to ‘v’ in the interval of time ‘t’. This means, Initial velocity of the object = u. Final velocity of the object = v. We know that momentum (p) = Mass x velocity Therefore, Momentum (p) of the object at its initial velocity u = m x u = mu Momentum (p) of the object at its final velocity v = m x v = mv The change in momentum = mv – mu mv  mu Rate of change of momentum  ………………… (i) t According to the Newton’s Second Law of motion force is directly proportional to the rate of change of momentum. This means, Force ∝ Rate of change of moentum After substituting the value of rate of change of momentum from equation (i) we get. mv  mu Force  t m (v  u ) F t  v u   F  ma  a    t   F  k.ma ……………. (ii) where, k is proportionality constant. Since, 1 unit force is defined as the mass of 1kg object produces the acceleration of 1m/s2 Therefore, 1 unit of Force = k x 1 kg x 1m/s2 Thus k = 1. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 35 - By substituting the value of ‘k = 1’ in equation (ii) we get F = m.a ……………… (iii) ⇒ Force = mass × acceleration Thus Newton’s Second Law of Motion gives the relation between force, mass and acceleration of an object. According to the relation obtained above, Newton’s Second Law can be modified as follows: The product of mass and acceleration is the force acting on the object. The SI unit of Force: Newton (N) Since Force = Mass x Acceleration The unit of mass = kg and The unit of acceleration = m/s2 If force, mass and acceleration is taken as 1 unit. Therefore, 1 Newton (N) = 1kg x 1m/s2 Thus, Newton (N) = kg m/s2 Equation (iii) can be also written as F a ………………… (iv) m This equation is the form of Newton’s Second Law of Motion. According to this equation, Newton’s Second Law of Motion can also be stated as follow: The acceleration produced by a moving body is directly proportional to the force applied over it and inversely proportional to the mass of the object. From the above relation it is clear that Acceleration increases with increase in force and vice versa. Acceleration decreases with increase in mass and vice versa. That’s why a small vehicle requires less force to attain more acceleration while a heavy vehicle requires more force to get the same acceleration. NEWTON’S SECOND LAW OF MOTION IN EVERYDAY LIFE: a. A fielder pulls his hand backward; while catching a cricket ball coming with a great speed, to reduce the momentum of the ball with a little delay. According to Newton’s Second Law of Motion; rate of change of momentum is directly proportional to the force applied in the direction. b. While catching a cricket ball the momentum of ball is reduced to zero when it is stopped after coming in the hands of fielder. If the ball is stopped suddenly, its momentum will be reduced to zero instantly. The rate of change in momentum is very quick and as a result, the player’s hand may get injured. Therefore, by pulling the hand backward a fielder gives more time to the change of momentum to become zero. This prevents the hands of fielder from getting hurt. NUMERICAL 1. What is acceleration produced by a force of 12 Newton exerted on an object of mass 3 kg? 2. What force would be needed to produce an acceleration of 4m/s2 on a ball of mass 6 kg? 3. A force of 5 N gives a mass m1, an acceleration of 8 m/s2, and a mass m2, an acceleration of 24 m/s2. What acceleration would give if both the masses are tied together? Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 36 - 4. Calculate the force required to impart a car a velocity of 30m/s in 10 seconds. The mass of the car is 1500 kg. 5. A motorcycle is moving with a velocity of 90km/h and it takes 5 seconds to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcycle if its mass along with the rider is 200kg. 6. What is the momentum of a man of mass 75kg when he walks with a velocity of 2m/s? 7. What would be the force required to produce an acceleration of 2m/s2 in a body of mass 12 kg? What would be the acceleration it the force were doubled? 8. A man pushes a box of mass 50 kg with a force of 80N.What will be the acceleration of the box? What would be the acceleration if the mass were doubled? 9. A certain force exerted for 1.2 second raises the speed of an object from 1.8m/s to 4.2 m/s. Later, the same force is applied for 2 second. How much does the speed change in 2 second? 10. A constant force acts on an object of mass 5 kg for duration of 2 second. It increases the object’s velocity from 3cm/s to 7m/s. Find the magnitude of the applied force. Now if the force were applied for a duration of 5 seconds, what would be the final velocity of the object? 11. A motorcar is moving with a velocity of 108km/h and it takes 4 seconds to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar it its mass along with the passengers is 1000 kg. 12. A force of 5 N gives a mass m1, an acceleration of 10 m/s2, and a mass m2, an acceleration of 20 m/s2. What acceleration would it give if both the masses were tied together? 13. For how long should a force of 100 N act on a body of mass 20 kg so that it acquires a velocity of 100 m/s? 14. A 150 g ball traveling at 30m/s strikes the palm of a players hand and is stopped in 0.06 sec. Calculate the force exerted by the ball on the hand. 15. A body of mass 1 kg is kept at rest. A constant force of 6.0 N starts acting on it. Find the time taken by the body to move through a distance of 12m. 16. A force of 4 N acts on a body of mass 2 kg for 4 s. Assuming that the body to be initially at rest, find (i) its velocity when the force stops acting (ii) the distance covered in 10 s after the force starts acting. 17. A feather of mass 10 g is dropped from a height. It is found to fall down with a constant velocity. What is the net force acting on it? Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 37 - 18. A hockey ball of mass 200g traveling from west to east at 10m/s is struck by a hockey stick. As a result, then ball gets turned back and now has a speed of 5m/s. If the ball and hockey stick were in contact for 0.2 s, calculate (i) initial and final momentum of the ball (ii) rate of change of momentum of the ball (iii) the force exerted by hockey stick on the ball. 19. A stone of mass 500 g is thrown with a velocity of 20m/s across the frozen surface of a lake. It comes to rest after traveling a distance of 0.1 km. Calculate force of friction between the stone and frozen surface of lake. 20. A body starts from rest and rolls down a hill with a constant acceleration. If its travels 400 m in 20 seconds, calculate the force acting on the body if its mass is 10kg. 21. The velocity time graph of a ball of mass 20g moving along a straight line on a long table is given in below figure. How much force does the table exert on the ball to bring it to rest? 22. The speed time graph of a ball of mass 30g moving along a straight line is shown in below figure. Calculate the opposing force that brings the ball to rest. What will be the percentage change in momentum of a body when both its mass and velocity are doubled? Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 38 - 23. A force of 2 N gives a mass m1 an acceleration of 5m/s2 and a mass m2, an accelerated of 7m/s2. What acceleration would be produced if both the masses are tied together? 24. A body of mass 2 kg moving with a velocity of 10m/s is brought to rest in 5 sec. Calculate the stopping force applied. NEWTON'S THIRD LAW OF MOTION Newton’s Third Law of Motion states that there is always reaction for every action in opposite direction and of equal magnitude. Explanation: Whenever a force is applied over a body, that body also applies same force of equal magnitude and in opposite direction. Example – (a) Walking of a person - A person is able to walk because of the Newton’s Third Law of Motion. During walking, a person pushes the ground in backward direction and in the reaction the ground also pushes the person with equal magnitude of force but in opposite direction. This enables him to move in forward direction against the push. (b) Recoil of gun - When bullet is fired from a gun, the bullet also pushes the gun in opposite direction, with equal magnitude of force. This results in gunman feeling a backward push from the butt of gun. (c) Propulsion of a boat in forward direction – Sailor pushes water with oar in backward direction; resulting water pushing the oar in forward direction. Consequently, the boat is pushed in forward direction. Force applied by oar and water are of equal magnitude but in opposite directions. CONSERVATION OF MOMENTUM – Law of Conservation of Momentum – The sum of momenta of two objects remains same even after collision. In other words, the sum of momenta of two objects before collision and sum of momenta of two objects after collision are equal. Mathematical Formulation of Conservation of Momentum: Suppose that, two objects A and B are moving along a straight line in same direction and the velocity of A is greater than the velocity of B. Let the initial velocity of A=uA Let the initial velocity of B= uB Let the mass of A= mA Let the mass of B=mB Let both the objects collide after some time and collision lasts for ' t' second. Let the velocity of A after collision= vA Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 39 - Let the velocity of B after collision= vB We know that, Momentum = Mass x Velocity Therefore, Momentum of A (FA) before collision = mA x uA Momentum of B (FB) before collision = mB x uB Momentum of A after collision = mA x vA Momentum of B after collision = mB x vB Now, we know that Rate of change of momentum = mass x rate of change in velocity Change in velocity  Rate of change of momentum =mass  time  v  uA  Therefore, rate of change of momentum of A during collision, FAB  mA  A   t   v  uB  Similarly, the rate of change of momentum of B during collision, FBA  mB  B   t  Since, according to the Newton’s Third Law of Motion, action of the object A (force exerted by A) will be equal to reaction of the object B (force exerted by B). But the force exerted in the course of action and reaction is in opposite direction. Therefore, FAB   FBA  v u   v u  or , m A  A A   mB  B B   t   t   mA (v A  u A )   mB (vB  uB )  mAv A  mAu A   mB vB  mB uB  mAv A  mB vB  mAu A  mB uB ……………. (i) Above equation says that total momentum of object A and B before collision is equal to the total momentum of object A and B after collision. We observe that the total momentum of the two balls remains unchanged or conserved provided no other external force acts. As a result of this ideal collision experiment, we say that the sum of momenta of the two objects before collision is equal to the sum of momenta 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. CONSERVATION OF MOMENTUM – PRACTICAL APPLICATION  Bullet and Gun – When bullet is fired from a gun, gun recoils in the opposite direction of bullet. The momentum of bullet is equal to momentum of gun. Since, the bullet is has very small mass compared to the gun, hence velocity of bullet is very high compared to the recoil of gun. In the case of firing of bullet, law of conservation of momentum is applied as usual.  When a cricket ball is hit by bat, the Law of Conservation of Momentum is applied. NUMERICAL 1. The velocity of a body of mass 10kg increases from 4m/s to 8m/s when a force acts on it for 2s. (a) What is the momentum before and after the force acts? (b) What is the gain in momentum per second? (c) What is the value of the force? Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 40 - 2. A boy pushes a wall with a force of 20N. What is the magnitude and direction of the force experienced by the body? 3. A 20 g bullet is shot from a 5 kg gun with a velocity of 500m/s. What is the speed of the recoil of the gun? 4. A 10 g bullet is shot from a 5 kg gun with a velocity of 400m/s. What is the speed of the recoil of the gun? 5. When two bodies A and B interact with each other, A exerts a force of 10N on B, towards east. What is the force exerted by B on A? 6. A man weighting 60kg runs along the rails with a velocity of 18km/h and jumps into a car of mass 1 quintal standing on the rails. Calculate the velocity with which car will start traveling along the rails. 7. The car A of mass 1500kg, traveling at 25m/s collides with another car B of amss 1000kg traveling at 15m/s in the same direction. After collision, the velocity of car A becomes 20m/s. Calculate the velocity of car B after collision. 8. A bullet of mass 10g is fired from a gun of mass 6 kg with a velocity of 300m.s. Calculate the recoil velocity of the gun. 9. A bullet of mass 50g is fired from a gun of mass 6 kg with a velocity of 400m.s. Calculate the recoil velocity of the gun. 10. A bullet of mass 10g is moving with a velocity of 400m/s gets embedded in a freely suspended wooden block of mass 900g. What is the velocity acquired by the block? 11. A gun of mass 3 kg fires a bullet of mass 30g. The bullet takes 0.003s to move through the barrel of the gun and acquires a velocity of 100m/s. Calculate (i) the velocity with which the gun recoils (ii) the force exerted on gunman due to recoil of the gun. 12. A heavy car of mass 200kg traveling at 10m/s has a head on collision with a sports car B of mass 500kg. If both cars stop dead on colliding, what was the velocity of car B? 13. A machine gun fires 25h bullet at the rate of 600 bullets per minute with a speed of 200m/s. Calculate the force required to keep the gun in position. 14. A bullet of mass 20g is moving with a velocity of 300m/s gets embedded in a freely suspended wooden block of mass 880g. What is the velocity acquired by the block? 15. A girl of mass 50kg jumps out of a rowing boat of mass 300kg on to the bank with a horizontally velocity of 3m/s. With what velocity does the boat begin to move backwards? Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 41 - 16. A truck of mass 2500kg moving at 15m/s collides with a car of mass 1000kg moving with at 5m/s in the opposite direction. What is their common velocity? 17. A bullet of mass 20 g is fired horizontally with a velocity of 150m/s from a pistol of mass 2kg. What is the recoil velocity of the pistol? 18. A body of mass 60kg running at 3m/s jumps on to a trolley of mass 140kg moving with a velocity of 1.5m/s in the same direction. What is their common velocity? 19. A girl of mass 40kg jumps with a horizontal velocity of 5m/s onto a stationary cart with frictionless wheels. The mass of the cart is 3kg. What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction? 20. Two hockey players of opposite teams, while trying to hit a hockey ball on the ground collide and immediately become entangled. One has a mass of 60kg, and was moving with a velocity 5m/s, while the other has a mass 55kg and was moving faster with a velocity of 6m/s towards the first player. In which direction and with what velocity will they move after they become entangled? Assume that the frictional force acting between the feet of the two players and ground is negligible. INTEXT QUESTIONS PAGE NO. 126 1. If action is always equal to the reaction, explain how a horse can pull a cart. Ans. A horse pushes the ground in the backward direction. According to Newton’s third law of motion, a reaction force is exerted by the Earth on the horse in the forward direction. As a result, the cart moves forward. 2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity. Ans. Due to the backward reaction of the water being ejected. When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, then a reaction force is exerted on him by the ejecting water in the backward direction. This is because of Newton’s third law of motion. As a result of the backward force, the stability of the fireman decreases. Hence, it is difficult for him to remain stable while holding the hose. 3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s−1. Calculate the initial recoil velocity of the rifle. Ans. Mass of the rifle, m1 = 4kg Mass of the bullet, m2 = 50g = 0.05 kg Recoil velocity of the rifle = v1 Bullet is fired with an initial velocity, v2 = 35m/s Initially, the rifle is at rest. Thus, its initial velocity, v = 0 Total initial momentum of the rifle and bullet system =(m1 + m2)v = 0 Total momentum of the rifle and bullet system after firing = m1v1 + m2v2 = 4(v1) + 0.05 x 35 = 4v1 + 1.75 According to the law of conservation of momentum: Total momentum after the firing = Total momentum before the firing 1.75 4v1  1.75  0  v1    0.4375m / s 4 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 42 -  The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s. 4. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s−1 and 1 m s−1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m s−1. Determine the velocity of the second object. Ans. Mass of one of the objects, m1 = 100 g = 0.1 kg Mass of the other object, m2 = 200 g = 0.2 kg Velocity of m1 before collision, v1 = 2 m/s Velocity of m2 before collision, v2 = 1 m/s Velocity of m1 after collision, v3 = 1.67 m/s Velocity of m2 after collision = v4 According to the law of conservation of momentum: Total momentum before collision = Total momentum after collision  m1v1 + m2v2 = m3v3 + m4v4  (0.1) x 2 + (0.2) x 1 = (0.1) x 1.67 + (0.2) x v4  0.4 = 0.167 + 0.2v4  v4 = 1.165 m/s Hence, the velocity of the second object becomes 1.165 m/s after the collision. EXERCISE QUESTIONS PAGE NO. 128 1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason. Ans. Yes. Even when an object experiences a net zero external unbalanced force, it is possible that the object is travelling with a non-zero velocity. This is possible only when the object has been moving with a constant velocity in a particular direction. Then, there is no net unbalanced force applied on the body. The object will keep moving with a non-zero velocity. To change the state of motion, a net non-zero external unbalanced force must be applied on the object. 2. When a carpet is beaten with a stick, dust comes out of it. Explain. Ans. Inertia of an object tends to resist any change in its state of rest or state of motion. When a carpet is beaten with a stick, then the carpet comes to motion. But, the dust particles try to resist their state of rest. According to Newton’s first law of motion, the dust particles stay in a state of rest, while the carpet moves. Hence, the dust particles come out of the carpet. 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope? Ans. When the bus accelerates and moves forward, it acquires a state of motion. However, the luggage kept on the roof, owing to its inertia, tends to remain in its state of rest. Hence, with the forward movement of the bus, the luggage tends to remain at its original position and ultimately falls from the roof of the bus. To avoid this, it is advised to tie any luggage kept on the roof of a bus with a rope. 4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because (a) the batsman did not hit the ball hard enough. (b) velocity is proportional to the force exerted on the ball. (c) there is a force on the ball opposing the motion. (d) there is no unbalanced force on the ball, so the ball would want to come to rest. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 43 -  Ans. (c) A batsman hits a cricket ball, which then rolls on a level ground. After covering a short distance, the ball comes to rest because there is frictional force on the ball opposing its motion. Frictional force always acts in the direction opposite to the direction of motion. Hence, this force is responsible for stopping the cricket ball. 5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg.) Ans. Initial velocity, u = 0 (since the truck is initially at rest) Distance travelled, s = 400 m Time taken, t = 20 s 1 According to the second equation of motion: s  ut  at 2 2 Where, Acceleration = a 1 1 400  0  a(20) 2  400  a(400) 2 2 2  a = 2m/s 1 metric tonne = 1000 kg (Given) ∴ 7 metric tonnes = 7000 kg Mass of truck, m = 7000 kg From Newton’s second law of motion: Force, F = Mass × Acceleration F = ma = 7000 × 2 = 14000 N Hence, the acceleration of the truck is 2 m/s2 and the force acting on the truck is 14000 N. 6. A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice? Ans. Initial velocity of the stone, u = 20 m/s Final velocity of the stone, v = 0 (finally the stone comes to rest) Distance covered by the stone, s = 50 m According to the third equation of motion: v2 = u2 + 2as Where, Acceleration, a (0)2 = (20)2 + 2 × a × 50 a = −4 m/s2 The negative sign indicates that acceleration is acting against the motion of the stone. Mass of the stone, m = 1 kg From Newton’s second law of motion: Force, F = Mass × Acceleration F = ma F = 1 × (− 4) = −4 N Hence, the force of friction between the stone and the ice is −4 N. 7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force; (b) the acceleration of the train; and (c) the force of wagon 1 on wagon 2. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 44 -  Ans. (a)Force exerted by the engine, F = 40000 N Frictional force offered by the track, Ff = 5000 N Net accelerating force, Fa = F − Ff = 40000 − 5000 = 35000 N Hence, the net accelerating force is 35000 N. (b)Acceleration of the train = a The engine exerts a force of 40000 N on all the five wagons. Net accelerating force on the wagons, Fa = 35000 N Mass of the wagons, m = Mass of a wagon × Number of wagons Mass of a wagon = 2000 kg Number of wagons = 5 ∴ m = 2000 × 5 = 10000 kg Mass of the engine, m′ = 8000 kg Total mass, M = m + m′ = 18000 kg From Newton’s second law of motion: Fa = Ma F 35000 a a   1.944m / s 2 M 18000 Hence, the acceleration of the wagons and the train is 1.944 m/s2. (c)Mass of all the wagons except wagon 1 is 4 × 2000 = 8000 kg Acceleration of the wagons = 3.5 m/s2 Thus, force exerted on all the wagons except wagon 1 = 8000 × 3.5 = 28000 N Therefore, the force exerted by wagon 1 on the remaining four wagons is 28000 N. Hence, the force exerted by wagon 1 on wagon 2 is 28000 N. 8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m/s2? Ans. Mass of the automobile vehicle, m = 1500 kg Final velocity, v = 0 (finally the automobile stops) Acceleration of the automobile, a = −1.7 ms−2 From Newton’s second law of motion: Force = Mass × Acceleration = 1500 × (−1.7) = −2550 N Hence, the force between the automobile and the road is −2550 N, in the direction opposite to the motion of the automobile. 9. What is the momentum of an object of mass m, moving with a velocity v? (a) (mv)2 (b) mv2 (c) ½ mv2 (d) mv Ans. (d) mv Mass of the object = m Velocity = v Momentum = Mass × Velocity Momentum = mv 10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet? Ans. A force of 200 N is applied in the forward direction. Thus, from Newton’s third law of motion, an equal amount of force will act in the opposite direction. This opposite force is the fictional force exerted on the cabinet. Hence, a frictional force of 200 N is exerted on the cabinet. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 45 - 11. Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m s-1 before the collision during which they stick together. What will be the velocity of the combined object after collision? Ans. Mass of one of the objects, m1 = 1.5 kg Mass of the other object, m2 = 1.5 kg Velocity of m1 before collision, v1 = 2.5 m/s Velocity of m2, moving in opposite direction before collision, v2 = −2.5 m/s (Negative sign arises because mass m2 is moving in an opposite direction) After collision, the two objects stick together. Total mass of the combined object = m1 + m2 Velocity of the combined object = v According to the law of conservation of momentum: Total momentum before collision = Total momentum after collision m1v1 + m2 v1 = (m1 + m2) v 1.5(2.5) + 1.5 (−2.5) = (1.5 + 1.5) v 3.75 − 3.75 = 3 v v=0 Hence, the velocity of the combined object after collision is 0 m/s. 12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move. Ans. The truck has a large mass. Therefore, the static friction between the truck and the road is also very high. To move the car, one has to apply a force more than the static friction. Therefore, when someone pushes the truck and the truck does not move, then it can be said that the applied force in one direction is cancelled out by the frictioal force of equal amount acting in the opposite direction. Therefore, the student is right in justifying that the two opposite and equal cancel each other. 13. A hockey ball of mass 200 g travelling at 10 m/s is struck by a hockey stick so as to return it along its original path with a velocity at 5 m/s. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick. Ans. Mass of the hockey ball, m = 200 g = 0.2 kg Hockey ball travels with velocity, v1 = 10 m/s Initial momentum = mv1 Hockey ball travels in the opposite direction with velocity, v2 = −5 m/s Final momentum = mv2 Change in momentum = mv1 − mv2 = 0.2 [10 − (−5)] = 0.2 (15) = 3 kg m s−1 Hence, the change in momentum of the hockey ball is 3 kg m s−1. 14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet. Ans. Now, it is given that the bullet is travelling with a velocity of 150 m/s. Thus, when the bullet enters the block, its velocity = Initial velocity, u = 150 m/s Final velocity, v = 0 (since the bullet finally comes to rest) Time taken to come to rest, t = 0.03 s According to the first equation of motion, v = u + at Acceleration of the bullet, a 0 = 150 + (a ×0.03 s) Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 46 -  150 a  5000m / s 2 0.03 (Negative sign indicates that the velocity of the bullet is decreasing.) According to the third equation of motion: v2 = u2 + 2as 0 = (150)2 + 2 (−5000) s (150) 2 22500 s   2.25m 2(5000) 10000 Hence, the distance of penetration of the bullet into the block is 2.25 m. From Newton’s second law of motion: Force, F = Mass × Acceleration Mass of the bullet, m = 10 g = 0.01 kg Acceleration of the bullet, a = 5000 m/s2 F = ma = 0.01 × 5000 = 50 N Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N. 15. An object of mass 1 kg travelling in a straight line with a velocity of 10 m/s collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object. Ans. Mass of the object, m1 = 1 kg Velocity of the object before collision, v1 = 10 m/s Mass of the stationary wooden block, m2 = 5 kg Velocity of the wooden block before collision, v2 = 0 m/s ∴ Total momentum before collision = m1 v1 + m2 v2 = 1 (10) + 5 (0) = 10 kg m s−1 It is given that after collision, the object and the wooden block stick together. Total mass of the combined system = m1 + m2 Velocity of the combined object = v According to the law of conservation of momentum: Total momentum before collision = Total momentum after collision m1 v1 + m2 v2 = (m1 + m2) v 1 (10) + 5 (0) = (1 + 5) v 10 5 v  m/ s 6 3 The total momentum after collision is also 10 kg m/s. Total momentum just before the impact = 10 kg m s−1