These physical quantities can be classified into two categories, depending on their dependent or independent existence.

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1 Chapter Motion The branch of Science which deals with the study of nature and natural phenomena is called Physics. In Physics, we deal with inanimate matter, energy and radiation. The quantities in terms of which laws of Physics are described are called physical quantities.in physics, we will be forming several mathematical relationships of two or more physical quantities. These relationships help us to understand how nature works. To use these physical quantities we need to measure them. Some Examples of Physical Quantities Length Mass Time Electric current Temperature Luminous intensity Amount of substance. These physical quantities can be classified into two categories, depending on their dependent or independent existence. a) Fundamental Physical Quantities b) Derived Physical Quantities. Fundamental Physics Quantities: The physical quantities which are independent of other physical quantities are called Fundamental Physical Quantities. Mass, length and time are examples of fundamental physical quantities. Fundamental Physical Quantities are also called base quantities. Derived Physical Quantities: The physical quantities which can be obtained from fundamental physical quantities are known as Derived Physical Quantities. Work, energy, velocity are few examples of derived physical quantities. The basic S.I. Units Units: The physical quantities are quantitatively expressed in terms of units. Units are also used to differentiate between two or more different physical quantities. The most widely used system of unit is SI, which is known as International System of Units. Quantities Unit Symbol Length metre m Mass kilogram kg Time second s Temperature kelvin K Amount of substance mole mol Electric current ampere A Luminous Intensity candela cd

2 Motion Units obey basic algebraic rules i.e., addition, subtraction, multiplication and division. For example: Volume = length breadth height. We measure all the three in metre hence the unit of volume is m 3. Similarly, m 3 = (00 cm) 3. = (00 cm) (00 cm) (00 cm) = ( ) (cm cm cm) = 0 6 cm 3. mass Density = volume unit of mass = kg unit of volume = m 3. kg unit of density = 3 m Again, speed = distance time metre unit of speed = = ms second Vectors and Scalars The physical quantities which have magnitude only and no direction are scalar quantities. Speed, mass, time, current, work, etc., are the examples of scalar quantities. The physical quantities which have both magnitude and direction are known as vectors. Velocity, which has magnitude as well direction is an example of a vector. The other examples are displacement, force, momentum, impulse, magnetic field, etc. Vector quantities are those quantities which obey laws of vector addition, subtraction etc. Motion In our daily life, we have been observing the motion of various objects around us, e.g., moving vehicles, flying aeroplanes, running animals, human beings, etc. An object is said to be at rest (relatively) if it does not change its position with time and / or surroundings. An object is said to be in motion if it changes its position with time and / or surroundings. Rest and motion are with respect to surrounding, not absolute, they are relative. It means an object in one situation can be at rest but in another situation the same object can be in motion.

3 Motion Mainly, the motions of a body are of three types: i) Linear Motion ii) Circular Motion iii) Oscillatory Motion. Motion along a Straight Line During such type of motion, the object occupies a definite position on the straight path at a certain interval of time. For example, a person leaving his home O at 0 am by car through a straight road (Here O is reference point). He crosses the city A at a distance of 50 km from his home at am, reaches another city B at a distance of 00 km from his home pm, finally he reaches his destination C at distance of 40 km from his home at pm. t = 0 0 am O t = hr t = hr A am B pm 50 km 50 km 40 km 00 km 40 km t = hr pm The total path length covered by the object is OA + AB + BC, i.e., 50 km + 50 km + 40 km = 40 km. This is the distance covered by the object. Distance: The total path length covered by an object during motion in a given interval of time is called distance. It can never be zero or negative for a body in motion. Displacement: The shortest distance measured between the initial and the final position of an object is known as of displacement. It may be zero as well as negative. C Speed: If an object covers s metre in t second, then speed is defined as Speed = distance travelled time taken or v = t s There is a problem with defining speed this way i.e., if an object covers s metres in t second. It is not necessary that the object has maintained a speed of s/t every second. Actually the s/t is an average speed of the trip, and average speed during a very short time interval is known as instantaneous speed, the time period is so short that the speed is practically unchanging during the interval The speedometer of a car reads this speed and this speed is known as instantaneous speed. Average Speed: The average speed of an object is defined as the ratio of the total distance travelled by the object to the total time taken, i.e., Average speed = Total distance Total time travelled taken. 3

4 Motion Velocity: Velocity means speed with direction. To define velocity of an object we must specify the direction towards which the object is moving along with the speed. It is defined as Velocity = displacement time taken Average Velocity: If the velocity of an object is changing at uniform rate, then average velocity is defined as the arithmetic mean of initial velocity and final velocity for a given period of time i.e., Average Velocity = Initial velocity Final velocity Total displacement Total time taken v av u v Speed and velocity have the same units, i.e. m/s or km/hr. Speed is a scalar quantity whereas velocity is a vector quantity. Accelerated Motion Any object whose velocity is changing is said to be accelerated. In an accelerated motion if the change in velocity of an object in each unit of time is constant, the object is said to be moving with uniformly accelerated motion. On the other hand, if the change in velocity is not constant, then such a motion is called non-uniformly accelerated motion. Acceleration is defined as the change in velocity of an object per unit time. i.e., Acceleration = change in velocity time taken a = Final velocity Initial velocity time taken or a = v u t ( zero acceleration, uniform acceleration with non zero initial velocity; 3 non-uniform retardation, 4 uniform acceleration, 5 non-uniform acceleration, 6 infinite acceleration and 7 uniform retardation) The acceleration is taken to be positive if it is in the direction of velocity and negative when it is opposite to the direction of velocity. The SI unit of acceleration is m/s. 4

5 Motion Graphs Through graph we can show two varying quantities which are dependent on each other. For example, in cricket match, runs scored by a team and number of overs bowled by the opponent team are varying quantities and they are dependent, also we can show the variation by taking scores along one axis and overs along the other axis. (from the graph we can find information about run scored in every over) Runs Overs Similarly, we can plot graph between velocity- time, displacement-time and acceleration-time graph as they are varying quantities for a moving body. Distance-Time Graphs Suppose a body covers equal distances in equal intervals of time, this means, if we take any two intervals of same duration, we will find the distance covered by an object is the same. Distance (m) s B C F We will plot graph of such a body taking y-axis as the distance axis and x-axis as the time axis. From the graph, we see that as the object moves from A to B, it covers a distance s s in time t t. The slope of the distancetime graph will give us speed. s A E D t t t 3 Time (s) BE Slope of the graph = tan = = speed. AE Velocity-Time Graphs This graph is plotted taking the velocity of an object moving along a straight line on the vertical axis and time along the horizontal axis. Velocity in m/sec Uniform acceleration Velocity in m/sec Non-uniform acceleration (Time in seconds) (Time in seconds) 5

6 Motion Note: Area under the velocity-time graph is displacement and area under speed-time graph is distance. Velocity (m/s) uniform acceleration A velocity (constant) B uniform deceleration Graphical Understanding O C D E Time (s) = 0, a = 0, v = constant i.e., the line parallel to time axis represents that the particle is moving with constant velocity = 90, a =, v = increasing i.e. line perpendicular to time axis represents that the particle is increasing its velocity, but time does not change. It means the particle possesses infinite acceleration. Practically it is not possible. = constant, so a = constant and v is increasing uniformly with time i.e., line with constant slope represents uniform acceleration of the particle. increasing so acceleration increasing i.e., line bending towards velocity axis represent the increasing acceleration in the body. increasing so acceleration decreasing i.e. line bending towards time axis represents the decreasing acceleration in the body. 6

7 Motion Positive constant acceleration because is constant and < 90 but initial velocity of the particle is negative. Positive constant acceleration because is constant and < 90 but initial velocity of particle is positive. Negative constant acceleration because is constant and > 90 but initial velocity of the particle is positive. Negative constant acceleration because is constant and > 90 but initial velocity of the particle is zero. Negative constant acceleration because is constant and >90 but initial velocity of the particle is negative. Three Equations of Motions Motion in a straight line is known as linear motion, for a linear motion with constant acceleration, we can form equations relating velocity, displacement, time and acceleration. Equation : This establishes the relationship between initial velocity, final velocity, acceleration and time. Suppose the initial velocity u, the final velocity v, time t and acceleration a, then according to the definition of acceleration change in velocity a = time taken v u a = t v u at 7

8 Motion Equation : This equation establishes the relation between displacement, time and acceleration. This equation gives displacement S a body will cover initially moving with velocity u after accelerating at the rate a for t second. u = initial velocity v = final velocity Average velocity = So, displacement = S = (v u) (v u) t S = ( u at u) t S ut at ( v = u + at) Equation 3: u v S t u v v u as t v u a a v u a or v u as Equation of Motion by Graphical Method i) Equation for Velocity-Time Relation: Consider an object moving along a straight line with uniform acceleration a. v (m/s) C Let u be the initial velocity of the object at t = 0 and v be the final velocity after time t, S is the displacement travelled by the object in time t. From graph, we can see that velocity changes uniformly with time t and the motion is a straight line AB as shown in the figure. Where, u A B D OA = ED = u OC = EB = v, and OE = AD = t O E ii) 8 Now we know that the slope of the velocity-time graph for uniformly accelerated motion represents acceleration of the object. Acceleration = Slope of the velocity-time graph AB tan = a = or a = v u t BD BE ED OC OA AD OE OE or v u at Equation for Position-Time Relation: As we know that the area bounded by the velocity-time graph and time axis for a given time, interval represents the displacement covered by the object for uniformly accelerated motion. From graph Acceleration, a = BD BD AD t

9 Motion or BD = at Now, the displacement S, travelled by the object is S = Area of trapezium (OABE) = Area of rectangle OADE + Area of triangle ADB = OA OE + BD AD = ut + at t = ut + at S ut at iii) Equation for Position-Velocity Relation: From v-t graph, displacement, S travelled by the object in time t moving under uniform acceleration, a is given by S = Area of trapezium (OABE) = (EB + OA) OE u v = (OC + OA) OE = t = u v v u. a [from the equation v = u + at, v u = at or v u t ] a = (v u a ) or v u as. Uniform Circular Motion When an object moves along a circular path with constant speed (i.e., it covers equal distances on the circumference of the circle in equal intervals of time), the motion of the object is called uniform circular motion. For example, we take a piece of thread and a small piece of stone at one of its ends. Now, move the stone with hand to describe a circular path with constant speed by holding the thread at the other end. The motion of the athlete moving along a circular path is also an example of uniform circular motion. In circular motion, V Speed = v = Distance covered time taken r = r t Where r is the radius of the circle, r is the circumference of the circle. For uniform circular motion, speed is constant but not velocity. 9

10 Motion Time period: In circular motion, the time period is defined as the time taken by the object to complete one revolution on its circular path. Frequency: In circular motion, the frequency is defined as the number of revolutions completed by the object on its circular path in unit time. It is generally denoted by v, its unit is s or Hertz (Hz) T = or v = v T Also angular velocity, = =v. T T Projectile Motion When a body is thrown obliquely under the influence of gravity, the body is called projectile and the motion is called projectile motion. Consider a body is thrown with velocity u at an angle with the horizontal as shown in figure: Point O is called point of projection. The lenght CA is called maximum height. The length OB is called range. Time of flight : The time during which body is in motion is called time of flight. It is denoted by T. usin T g Maximum height, H : The maximum height attained by a body is called maximum height. It is given by H. sin H u g Range: The maximum horizontal distance covered by a body is called range. It is given by R. x ( u cos ) T u sin ( u cos ) g u (sin cos ) g u sin g Equation for the displacement in n th second a Sn u (n ) (where, n = n th second) 0

11 Motion SOLVED PROBLEMS Example : Solution: A person goes out for a bike ride to a nearby town. A record of the trip is as follows: 30 minutes at 30 km/h, 5 minutes at 40 km/h, 5 minutes at 0 km/h for a break for rest, and 0 minutes at 5 km/h. a) What is the distance the person travelled? b) What is the average speed? Remember to make the units consistent, you should convert the times from minutes to hours At, first you need to figure out how far the rider went in each segment of the ride: distance = s = vt S S = (30 km/h) (0.5 h) = 5 km = (40 km/h) (0.5 h) = 0 km S 3 = (0 km/h) ( h) = 0 km S 4 = (5 km/h) ( 3 h) = 5 km Total distance, S = S + S + S 3 + S 4 = 30 km Total distance travelled Average speed = Total time taken = km/h = 30 4 = 5.7 km/h Example : Solution: Train A starts at 4 miles South of a bridge and heads North at a constant speed of 30 miles per hour. Train B starts 6 miles North of the bridge. What velocity must Train B have so that the two trains cross the bridge at the same time? Train B covers a distance of 6 miles in same time in which train A covers the distance of 4 miles at a speed of 30 miles/hr. t A = t B VB = 45 miles/hour Example 3: A car increases its velocity from 5 m/s to 5 m/s and covers the distance of 0 m. a) Find the magnitude of this acceleration b) Find the time it takes for the car to travel this distance

12 Motion Solution: a) The car had an initial velocity (u) of 5 m/s, the final velocity (v) of 5 m/s, and accelerated across the distance (s) of 0 metres. We need to use this to find the acceleration. v = u + as a = v u S a = (5m / s) (5m / s) (0m) a = 0 m/s b) Now, we need to find the time v = u + at t = v u a 5 m/s 5 m/s t = 0 m/s t =.0 s Example 4: A car starts from rest and travels for 0 seconds with a constant acceleration of 3.0 m/s. The driver then applies the brakes causing a constant negative acceleration of 4.0 m/s. Assuming the brakes are applied for.0 seconds: a) How fast is the car going at the end of braking? b) How far has the car gone at the end of braking? Solution: a) At first, you calculate what s going on after the 0 seconds of positive acceleration. We have a = 3.0 m/s t = 0 sec. u = 0 m/s Using the equation v = u + at v = 0 + (3.0 m/s ) (0 s) v = 30 m/s Using the above results we get v = u + at v = 30 m/s + ( 4.0 m/s ) (.0 s) v = m/s

13 Motion b) Now, first while accelerating: S = ut + at S = (3.0 m/s ) (0 s) S = 50 m Then, while decelerating: S = ut + at = 30 4 = 5 m S = S + S = 50 m + 5 m = 0 m Example 5: Solution: A train starts from the rest and attains a velocity of 7 km h in 5 minutes. Assuming that the acceleration is uniform, find the a) acceleration b) distance travelled by the train for attaining this velocity. We have, u = 0; v = 7 km h = 0 ms and t = 5 minutes = 300 s. a) From equation of motion we have a = (v u) = t 0 ms 0 ms 300s b) From equation of motion we have as = v u = v 0 = 5 ms Thus, s = v (0 ms ) = a (/5) ms = 3000 m = 3 km The acceleration of the train is 5 ms and the distance travelled is 3 km. Example 6: The brakes applied to a car produce an acceleration of 6 ms in the opposite direction to the motion. If the car takes s to stop after the application of brakes, calculate the distance it travels during this time. Solution: We have, a = 6 ms ; t = s and v = 0 ms. From equation of motion we have v = u + at 0 = u + ( 6 ms ) s or u = ms. Again, S = ut + at = ( ms ) ( S) + ( 6 ms ) (S) = 4 m m = m. Thus, the car will move m before it stops after the application of brakes. Can you now appreciate why drivers are cautioned to maintain some distance between vehicles while travelling on the road? 3

14 Motion TEST YOURSELF. A car is going due north at a speed of 3 km/h. It makes a 90 left turn without changing the speed. Its change is velocity is about. (a) 30 km/h (b) 4.4 km/h north-west (c) zero (d) 4.4 km/h south-west. A body travels on a straight line with uniform velocity u for some time and velocity v for next equal time. The average velocity v av is given by (a) vav uv (b) vav u v (c) vav u v (d) v av u v 3. A body travels on a straight line with the uniform velocity u for some distance and with uniform velocity v for same distance. The average velocity v av is given by (a) vav uv (b) vav u v (c) vav u v (d) v av u v 4. A body travels half of a certain distance with uniform velocity 4 m/s and other half of the distance in two equal time intervals with velocity 5.5 m/s and 6.5 m/s. Average velocity is (a) 6 3 m/s (b).4 m/s (c) 4.8 m/s (d) m/s 5. A ball is released from an elevator going up with an acceleration a. The acceleration of the ball after the release is (a) a upward (b) (g a) downward (c) (g a) upward (d) g downward IMPORTANT DEFINITIONS. Distance is the actual length of the path covered by a moving object irrespective of its direction.. Displacement is the shortest distance measured between initial and final position in a particular direction. 3. Speed is distance travelled by an object per unit time. 4. Average speed is total distance travelled by the object divided by total time taken to cover this distance. 5. Uniform speed (or constant speed) if the object covers equal distances in equal intervals of time, no matter how small these time intervals may be, then its speed is said to be uniform. e.g. A car running at constant speed of 30 m/s covers equal distances of 30 m in every second. 6. Non-uniform speed if the object covers unequal distances in equal intervals of time, no matter how small these time intervals may be, then its speed is said to be non-uniform speed. e.g. motion of a car on a crowded road. 4

15 Motion 7. Average Velocity: If the velocity of an object is changing at uniform rate then average velocity is defined as the arithmetic mean of initial velocity and final velocity for a given period of time. 8. Uniform Circular Motion:When an object moves along a circular path with constant speed (i.e., it covers equal distances on the circumference of the circle in equal intervals of time), the motion of the object is called uniform circular motion. 9. Time period: In circular motion, the time period is defined as the time taken by the object to complete one revolution on its circular path. 0. Frequency: In circular motion, the frequency is defined as the number of revolutions completed by the object on its circular path in unit time. IMPORTANT FORMULAE. Speed = distance travelled time taken or v = S t. Velocity = v 3. av u displacement time taken v 4. Acceleration = 5. v u at change in velocity time taken or a = v u t 6. S = ut + at 7. v u = as 8. Projectile Motion along x-axis Initial velocity: v ox = v 0 cos Acceleration: a x = 0 The motion is uniform 9. Projectile Motion along y-axis Initial velocity: v oy = v 0 sin Acceleration: a x = g The motion is uniformly accelerated 0. Time of Flight: It is the total time taken by the projectile to go up and came down to the same level from where it was projected. It is given by v0 sin T g 5

16 Motion. Maximum of Height of the Projectile: When a projectile attains its maximum height its vertical velocity becomes zero. i.e. v y = 0 The maximum height of the projectile is given by. Range: v sin 0 H g u sin R g 3. The maximum height H and the range R are related to each other as, R = 4 H cot. a 4. Sn u (n ) (where, n = n th second) S n is displacement in n th second REMEMBER! Equations of motion are valid only for constant acceleration. In equations of motion, S is displacement and not distance. Be careful about the direction of velocity and acceleration. When acceleration and initial velocity are in opposite directions then, particle reverses its direction of motion at some instant. All the physical quantities in equations of motion are measured w.r.t. the same frame of reference. IMPORTANT TIPS FOR COMPETITIVE EXAMS For solving the questions in which the particle is moving with constant acceleration, keep in mind the following points. Make a diagramatical representation of the given situation. Decide which direction you will consider as the positive and which one as the negative according to your convenience with respect to the chosen origin. In this way write down the values of given variables with appropriate signs (+ or ) and units. Then look for the things you have to find and/or prove. First of all see whether the initial velocity is zero or non-zero. If it is zero then the particle will move along the direction of acceleration without any change in direction of its motion. In this situation distance and magnitude of displacement would be always equal. If the initial velocity is not zero and is in the same direction as that of acceleration, then in this case also the particle will move along its initial direction of motion without any change in direction and hence distance and magnitude of displacement would always be the same. If initial velocity is not zero and its direction is oppsoite to that of acceleration, then it is advisable first to find the time when its velocity becomes zero. Upto this instant, distnace and displacement would be same, afterwards distance and displacement would be different. Remember equations of motion give you the displacement and not distance. 6

17 Motion There may be possible answers for a kinematics question you have, so look for the interpretation of these two answers it may be that one answer is irrelevant. For one dimensional motion the angle between acceleration and velocity is either 0 or 80 and it does not change with time. For two dimensional motion, the angle between acceleration and velocity is other than 0 or 80 and also it may change with time. If the angle between a and v is 90, the path of the particle is a circle. The particle speed up, that is the speed of the particle increases when the angle between a and v lies between 0 and +90. The particle speed down, that is the speed of the particle decreases, when the angle between a and v lies between +90 and 80. The speed of the particle remains constant in uniform circular motion when the angle between a and v is equal to 90. The distance covered by a particle never decreases with time, it always increases. Since distance Displacement average speed of a body is equal or greater than the magnitude of the average velocity of the body. The average speed of a body is equal to its instantaneous speed if the body moves with a constant speed. Velocity of the body is positive if it moves to the right side of the origin. Velocity is negative if the body moves to the left side of the origin. When a body reverses its direction of motion while moving along a straight line, then the distance travelled by the body is greater than the magnitude of the displacement of the body. In this case, average speed of the body is treater than its average velocity. Speedometer measures the instantaneous speed of a vehicle. When particle moves with speed v upto half time of its total motion and in rest time it is moving v v with speed v then vav. (equal times) When particle moves the first half of a distance at a speed of v and second half of the distance vv at speed v then vav v v.(equal distances) When particle covers one-third distance at speed v, next one third at speed v and last one third 3vv v3 at speed v 3, then vav v v v v v v 3 3 For two particles having displacement time graph with slopes and possesses velocities v v tan and v respectively then v tan. Velocity of a particle having uniform motion = slope of displacement time graph. Greater the slope of displacement-time graph, greater is the velocity and vice-versa. Slope of velocity time graph = acceleration. 7

18 Motion The slope of the line v t graph is a measure of acceleration. Larger the slope; higher the acceleration, and vice versa. a a a a tan tan Area under v t graph = displacement. If a body starts from rest and moves with uniform acceleration then distance covered by the body in t sec is proportional to t (i.e. (i.e. S t ) If a body starts from rest and moves with uniform acceleration then distance covered by the body in nth sec is proportional to (n ) (i.e. S (n ) So, we can say that the ratio of distance covered in st, nd and 3 rd second is : 3 : 5. n A particle moving with uniform acceleration from A to B along a straight line has velocities v and v at A and B respectively. If C is the mid-point between A and B then velocity of the particle v v at C is equal to v. The body returns to its point of projection with the same magnitude of the velocity with which it was thrown vertically upward, provided air resistance is neglected. All bodies fall freely with the same acceleration. The acceleration of the falling bodies does not depend on the mass of the body. If two bodies are dropped from the same height, they reach the ground in the same time and with the same velocity. If a body is thrown upwards with velocity u from the top of a tower and another body is thrown downwards from the same point and with the same velocity, then both reach the ground with the same speed. When a particle returns to the starting point, its average velocity is zero but the average speed is not zero. If both the objects A and B move along parallel lines in the same direction, then the relative velocity of A w.r.t. B is given by v AB = v A v B and the relative velocity of B w.r.t. A is given by v BA = v B v A. If both the objects A and B move along parallel lines in the opposite direction, then the relative velocity of A w.r.t. B is given by v AB = v A ( v B ) = v A + v B and the relatively velocity of B w.r.t. A is given by v BA == v B v A. 8

19 Motion Suppose a body is projected upwards from the ground with the velocity u. It is assumed that the friction of the air is negligible. The characteristics of motion of such a body are as follows: i) The maximum height attained = H = u /g. ii) Time taken to go up (ascent) = Time taken to come down (descent) = t = u/g iii) Time of flight T = t = u/g iv) The speed of the body on returning to the ground = speed with which it was thrown upwards. If the friction of air is taken into account, then the motion of the object thrown upwards will have the following properties: a) Time taken to go up (ascent) < time taken to come down (descent) b) The speed of the object on returning to the ground is less than the initial speed. Same is true for velocity (magnitude), momentum (magnitude) and kinetic energy. c) Maximum height attained is less than u /g. A ball is dropped from a building of height h and a reaches after t seconds on earth from the same building if two balls are thrown (one upwards and other downwards) with the same velocity, u and they reach the earth surface after t and t seconds respectively then t t t A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distance of m each will then be in the ratio of the difference in the square roots of the integers i.e., ( ),( 3 )...( 4 3),... The students commonly believe that the fundamental equation of kinematics such as v v at 0 x v 0t at v v ax 0 are applicable for any kind of motion along a straight line. It is important to remember that these equations are not the general equations to be applicable for any kind of motion along the straight line. These are specifically derived for uniformly accelerated motion in which acceleration of the particle is constant, that is, acceleration is not a function of time, displacement or velocity. If water drops are allowed to fall from a certain height at regular intervals, then the distances between two successive drops are always in the ratio : 3 : 5 : 7 : 9... If three balls are thrown with same speed such that one is thrown vertically downward (ball A) the second is thrown vertically upward (ball B), and the third is thrown horizontally (ball C) then all the three balls reach the ground with same speed. If t A, t B and t C are the respective time taken by the balls A, B and C, then t C t t A B 9

20 Motion The students usually have the following misconceptions. A heavy body falls at a faster rate as compared to a lighter body. (Misconception) All the bodies fall at the same rate irrespective of their masses, if they are identical in shape and size. The rate of fall is measured in terms of acceleration due to gravity. The apparent difference in the rate of fall of the light and heavy bodies appears due to air resistance. Strictly speaking, in vacuum all the bodies irrespective of their masses, shapes and sizes will fall at the same rate. In a free fall, the initial velocity of a body is always zero. (Misconception) In a free fall, the initial velocity of a body may or may not be zero. Free fall means the vertical acceleration of the body is equal to g. It is a common misconception that whenever a ball is dropped, its initial velocity is always equal to zero. Whenever a ball is dropped its initial velocity is equal to the velocity of the body from where it is being dropped. For example, when a ball is dropped from a rising balloon, its initial velocity is equal to the velocity of the balloon at the instant. 0 PART - I: MISCELLANEOUS DOMAIN. Are the following speeds increasing, decreasing, or unchanging? a) A car covers longer and longer distances in equal times. b) A car takes longer and longer times to cover equal distances. c) A car covers equal distances in equal times. d) A car covers equal distances in shorter and shorter times. e) A car takes equal times to cover equal distances. f) In equal times, a car covers shorter and shorter distances.. Two bicyclers pass each other on a straight road, one moving north and the other south. Could they possibly have the same velocities? The same speeds? 3. Which of the following is true for displacement? a) It cannot be zero b) Its magnitude is greater than the distance travelled by the object. 4. Give examples of an object that has a) a high velocity and low acceleration b) a low velocity and high acceleration c) an unchanging speed but high acceleration. 5. How much farther does a freely falling object fall in three times as much initial time? In four times as much initial time? Would your answers be the same if the object were falling freely on the surface of Mars?

21 Motion 6. You drop a rock off a cliff and note that it hits the ground below in 6 s. How high is the cliff, assuming that the air is so thin that air resistance can be neglected during the entire fall? How fast is the rock moving when it hits the ground? Unless the air is extremely thin, air resistance will not be negligible. What does this tell you about your calculation of the cliff s height: Is the cliff actually higher, or is it lower, than your calculated answer? (Take g = 0 m/s ) 7. A stone is thrown in a vertically upward direction with a velocity of 5 ms. If the acceleration of the stone during its motion is 0 ms in the downward direction, what will be the height attained by the stone and how much time will it take to reach there? 8. A train is travelling at a speed of 90 km h. Brakes are applied so as to produce a uniform acceleration of 0.5 ms. Find how far the train will go before it is brought to rest. 9. An athlete completes one round of a circular track of diameter 00 m in 40 s. What will be the distance covered and the displacement at the end of minutes 0 s? 0. A ball is gently dropped from a height of 0 m. If its velocity increases uniformly at the rate of 0 ms, with what velocity will it strike the ground? After what time will it strike the ground? HIGHER ORDER THINKING SKILLS (HOTS). A particle is moving in a straight line with initial velocity 0 m/s and constant retardation of 4 m/s. Calculate the distance travelled by it in 3rd second.. A stone is thrown upward with an initial velocity of 45 m/s. Calculate the distance travelled by the stone in 5th second. 3. A stone is thrown vertically upward with a velocity u from the top of a building of height h. Find the total distance travelled by the stone, before coming to ground and the total time taken to reach the ground. 4. A ball is dropped from a building. It took time t to cover the first half height of the building and took time t for the next half. Find t /t. 5. A ball is thrown upward with a velocity u such that it travelled a height more than the height of a building h. If t and t are the two times in which the displacement of the ball was h. Find. a) t + t b) t t 6. A ball is projected with a velocity of 0 m/s at an angle of 30 with the horizontal. Find i) Horizontal range ii) Maximum height iii) time of flight 7. The maximum horizontal range attained by a ball is 00 m. What could be the maximum height, attained by the same ball?

22 Motion VALUE BASED QUESTIONS 8. An old woman crossing the road was holding a money purse. A pickpocket snatches away her purse. A school student of Class - IX having seen this incident tries to help her. He informs the police inspector who stands nearby the inspector collects the purse from the pickpocket and hands it over to the old lady. a) What values do you find in the school student? b) The inspect in jeep is chasing the pick pocket on a straight road. The jeep is going at its maximum speed v. The picketpocket rides on the motorcycle of a friend waiting for him when the jeep is at a distance d. The motorcycle starts with constant acceleration a. Show that pickpocket will be caught if v ad. 9. Ram, a student of Class - IX, was suffering from malaria. The area is full of mosquitoes. He was not having a mosquito net. His friend, Shyam has an extra net. Also he took Ram to a doctor, got him medicines for him. After a week, Ram became normal. a) Comment upon qualities of Shyam. b) The mosquito net over a 7 m 4 m bed is 3 m high. It has a hole at one corner of the bed through which a mosquito enters the net. It flies and sits at the diagonally opposite corner of the net. (i) find the magnitude of the displacement of mosquito (ii) Taking the hole as the origin, length of bed as x-axis, width as y-axis and vertically up as z-axis. Write the components of the displacement vector. PART - II: MULTIPLE CHOICE QUESTIONS. A body moves km north and then 5 km east in 3 h. Its average speed and average velocity in m/s are (a) 4.3 km/h at 5 tan east of north,.57 m/s (b).57 m/s, 4.3 km/h at 5 tan east of north (c) 8.5 km/h, 4.3 km/h at 5 tan north of east (d).57 m/s, 4.3 km/h at 5 tan west of north. Velocity of a body moving in a straight line varies with displacement as v displacement at t = 0 is s = 0. Then its displacement at t = s is (a) 4 m (b) m (c) 0 m (d) 8 m 3. Acceleration time graph of a body moving in a straight line is shown in figure. Velocity of the body at t = 0 is m/s. Velocity at end of second is (a) m/s (b) m/s (c) 3 m/s (d) 4 m/s 4 4s m/s. Its

23 Motion 4. A policeman chases a thief 30 m ahead of him and gains 3 m in 5s after the chase began. After 0s, distance between them is (a) 4 m (b) 8 m (c) 4 m (d) 6 m 5. A ball is thrown upwards from ground with certain speed. It is at a height of 80 m at two times. The time interval being 6s. (g = 0 ms ). Its speed is (a) 5 m/s (b) 50 m/s (c) 75 m/s (d) 00 m/s 6. Speed time graph of a body moving along a fixed direction is shown in figure. Average speed of the body between t = to 6s is (a) 5 m/s (b) 0 m/s (c) 9 m/s (d) 4 m/s 7. A body moving along a straight line covers 35 m in 4th second and 40 m in 5th second, then (a) Its initial velocity and acceleration are 7.5 ms and 5 ms (b) Its initial velocity and acceleration are 0 ms and.5 ms (c) Its velocity after 0.5s is 0 ms (d) Both (a) and (c) are correct 8. A body goes from P to Q with a velocity of m/s and comes back to P with a velocity of 3 m/s. Its average velocity during entire journey is (a).5 m/s (b) m/s (c).4 m/s (d) zero 9. A body moves along a quadrant of a circle of radius 4m from P to Q with constant speed of m/s as shown in the figure. Its average velocity is (a) m/s (b) zero (c) 4 m/s (d) 4 m/s 0. A hovercraft flies for 8 seconds with a velocity of t 4 ms in a straight line, where t is time in seconds. Its covers a distance of (a) m (b) 4 m (c) 8 m (d) 6 m. The driver of a train travelling at 0 km/h sees on the same track 00 m infront of him a slow train travelling in the same direction at 30 km/h. The least retardation that must be applied to the faster train to avoid a collision is (a) 6.5 m/s (b) 3.5 m/s (c).5 m/s (d) 5 m/s. A body starts from rest with a constant acceleration. At time t second, its speed is 50 m/s and one second later, its speed is 00 m/s. Distance travelled during (t + )th second is (a) 37.5 m (b) 75 m (c) 5 m (d) 87.5 m 3

24 Motion 3. A body is thrown from ground with a speed of 5 m/s at an angle of 45 with the horizontal. At what distance does it hit the ground again? (a) 5 m (b) 0 m (c).5 m (d) 5 m 4. A ball is kicked with a velocity of 0 m/s at an angle of 45 with the horizontal. Time taken by it to reach the ground is (a) 0.7s (b).4s (c).8s (d) 3.5s 5. For Q.4, maximum height reached is (a).5 m (b) 5m (c) 7.5 m (d) 0 m 6. A projectile is fired horizontally with a velocity of 98 m/s from top of a hill 490 m high. The speed with which it hits the ground and the angle its velocity makes with the horizontal are (a) 49 m/s, 45 (b) 98 m/s, 45 (c) 98 m/s, 30 (d) 49 m/s, The horizontal range of a projectile is 4 3 times its maximum height. Its angle of projection is (a) 30 (b) 45 (c) 60 (d) At the top of trajectory of a projectile, the acceleration is (a) zero (b) g 9. A body is thrown upwards and it returns to ground describing a parabolic path. Which of the following remains constant? (a) speed of body (b) kinetic energy of body (c) vertical component of velocity + body (d) horizontal component of body (c) g (d) g 0. A projectile fired with initial velocity is at same angle has a range R. If initial velocity is doubled at same angle of projection, its range will be (a) R (b) R (c) R (d) 4R PREVIOUS YEARS QUESTIONS. A ball is projected at an angle of 45 with horizontal. In the absence of air resistance the ball follows: (a) elliptical orbit (b) sinusoidal path (c) parabolic path (d) linear path. A ball is thrown up vertically in still air with a velocity of 0 ms. It comes back to ground. The velocity - time graph is (a) (b) (c) (d) 4

25 Motion 3. The distance between two spots A and B on the opposite banks of river is 75 km. Speed of the boat in still water is twice as much as that of speed of the water current of river. The boat travels in the river from A to B and returns back to the spot in 6 hours. What is speed of boat in still water? (a).5 km h (b) 5 km h (c) 6 km h (d) 8 km h 4. An object starting from rest moves on a straight road for time t and comes to rest. Finally, the distance is covered in two parts. In first part, it is accelerated at constant acceleration and than decelerates at rate. Then, maximum velocity is (a) t (b) t 5. Velocity time graph of an object is (c) t (d) t (a) (b) (c) (d) 6. A particle starting from rest is moving with uniform acceleration in a straight line. The % increase of displacement of particle in 9th second compared to that in immediate previous second is about. (a) 8.3% (b) 0.6% (c) 3.3% (d) 4.5% 7. A particle is moving along a straight line. Its v-t graph is as shown Match the following: Physical Quantity (i) Acc. at 4 second (ii) Velocity at 4 second Remarks (p) +ve (q) ve (iii) Direction of motion at s (r) 0 (a) (i) (p), (ii) (q), (iii) (r) (c) (i) (q), (ii) (r), (iii) (p) (b) (i) (r), (ii) (r), (iii) (p) (d) (i) (q), (ii) (p), (iii) (r) 5

26 Motion 8. Identify the type of motion of an object that has a constant speed but var ying velocity (a) Circular motion (b) Linear motion (c) Vertical motion (d) Slow motion 9. Which velocity time graph represents the non-uniform acceleration of the truck? (a) (b) (c) (d) 0. A man walks 8 m to the east in s followed by another 6 m to the south in 3s. What is the average speed and velocity of the man? Average speed (ms ) Average velocity ms (a).8.0 (b).8.8 (c) (d) A driver takes 0.0s to apply brakes soon after he sees a need for it. If he is driving a car at a speed of 54 kmh and brakes cause a deceleration of 6.0 ms, then the distance travelled by the car, after he sees the need to apply the brakes is: (a).75 m (b) 4.50 m (c) 0.55 m (d) 6.5 m. A block accelerates down a slope, as shown in the figure. On the lower portion, i) Speed of block may increase, decrease or remain same ii) Acceleration of block reduces iii) Mass of block redcues Which of the following is/are correct? (a) (i) only (b) (i) and (ii) only (c) (ii) and (iii) only (d) (i), (ii) and (iii) 3. A ball dropped on ground from a height of.00 m rises to a height of 75 cm on rebounce. When thrown down from the same height with a velocity of.0 m/s, it would rise to (g = 0 ms ). (a) 80 cm (b) 90 cm (c) 85 cm (d) 95 cm 6

27 Motion 4. A particle is moving in positive x-direction with its velocity varying as v x. Assume that at t = 0, particle was located at x = 0. Determine (i) time dependence of velocity (ii) acceleration (iii) mean velocity of particle averaged over same time that particle takes to cover first S metres of path. 5. A 00 m sprinter increases her speed from rest uniformly at the rate of m/s upto 40 m and covers remaining distance with uniform speed. The sprinter covers first half of the run in t s and second half in t s. Then (a) t > t (b) t < t (c) t = t (d) information given is incomplete 6. Rain is falling vertically with a speed of.7 m/s. A girl is walking with a speed of.0 m/s in N E (North-East) direction. To shield herself, she holds her umbrella making an approximate angle with vertical in a certain direction then (a) = 60 in N - E direction (b) = 30 in N - E direction (c) = 60 in S - W direction (d) = 30 in S - W direction 7. A ball in released from rest above a horizontal surface. The graph shows the variation with time of its velocity (not to scale) the scale in this graph is changed at every impact A, B, C, D and E represent areas. Which of the following are correct. (a) A = B and B = C (b) A = C and C = E (c) B = C and D = E (d) All of the above 8. An arrow shot vertically upward loses its initial speed by 60% in 3 seconds. The maximum height reached by the arrow is (g = 9.8 m/s ) (a).5 m (b) 44. m (c) 00 m (d) 45 m. 9. A ball is dropped from a height of 7. m. It bounces back to 3. m after striking the floor. It remains in contact with floor for 0 ms. Average acceleration of ball during contact is (g = 0 m/s ) [ (a) 00 m/s (b) 00 m/s (c) 600 m/s (d) 000 m/s ANSWERS TEST YOURSELF. (b). (d) 3. (c) 4. (c) 5. (d) PART - I. (a) Increasing (b) Decreasing (c) Unchanging (d) Increasing (e) Unchanging (f) Decreasing. No, Yes 3. (a) False; (b) False 7

28 Motion 4. (a) Any fast-moving object whose speed or direction of motion is changing slowly, such as a high velocity bullet that is gradually slowing down (b) Any slow moving object that is quickly speeding up, such as a car starting up rapidly; (c) Any object turning a sharp corner at a high but unchanging speed. 5. Eight times as far. Fifteen times as far. Yes, both answers would be the same on mars m, 60 m/s, lower 7..5 m, 0.5s m 9. Distance = 00 m, displacement = 00 m 0. Velocity = 0 ms, time = s. m..5 m (a) u g, (b) h g u u hg h, g g u 6. (i) 0 3m, (ii) 5m, (iii) sec 7. 5 m 8. (a) The student is sympathetic towards others, helping and applies his presence of mind is solving the problems, knows to use public services. 9. (a) He has caring attitude, and concern for others. (b) 74 m i.e (c) 7m, 4m, 3m PART - II. (b). (d) 3. (c) 4. (a) 5. (b) 6. (c) 7. (d) 8. (d) 9. (c) 0. (d). (b). (c) 3. (c) 4. (c) 5. (d) 6. (b) 7. (a) 8. (d) 9. (d) 0. (d) PREVIOUS YEARS QUESTIONS. (c). (c) 3. (a) 4. (d) 5. (b) 6. (c) 7. (c) 8. (a) 9. (d) 0. (a). (a). (b) 3. (b) 4. (i) v t (ii) a (iii) v S S Average velocity = 5. (a) 6. (b) 7. (c) 8. (a) 9. (d) 8

29 Chapter Force & Laws of Motion In the previous chapter, we learnt about motion. In motion, we also learnt about the concepts of displacement, velocity and acceleration and formed equations with which we can determine displacement, velocity and acceleration. In this section, we are going to learn about the cause of motion and laws governing the motion of bodies. The Meaning of Force A force is a push or pull upon an object resulting from the object s interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. When the interaction ceases, the two objects no longer experience the force. Forces only exist as a result of an interaction. For simplicity, all forces (interactions) between objects can be placed into two broad categories:. Contact forces and. Forces resulting from action-at-a-distance (Non-contact forces). Contact forces are types of forces in which the two interacting objects are physically in contact with each other. Examples of contact forces include frictional forces, tension forces, normal forces, air resistance forces, and applied forces.. Action-at-a-distance forces are types of forces in which the two interacting objects are not in physical contact with each other, yet are able to exert a push or pull despite a physical separation. Examples of action-at-a-distance forces include gravitational forces (e.g., the sun and planets exert a gravitational pull on each other despite their large spatial separation; even when your feet leave the earth and you are no longer in contact with the earth, there is a gravitational pull between you and the Earth), electric forces (e.g., the protons in the nucleus of an atom and the electrons outside the nucleus experience an electrical pull towards each other despite their small spatial separation), and magnetic forces (e.g., two magnets can exert a magnetic pull on each other even when separated by a distance of a few centimeters). Contact Forces Frictional Force Tension Force Normal Force Air Resistance Force Action-at-a-Distance Forces (Non-contact forces) Gravitational Force Electrical Force Magnetic Force Nuclear Force Applied Force Spring Force 9

30 Force & Laws of Motion Balanced and Unbalanced Forces If two or more forces act on a body such that their resultant is zero, then the forces are said to be balanced forces. In this case, the body will be either at rest or will move with a uniform velocity. i.e. the acceleration of the body will be zero. If two or more forces act on a body such that their resultant is not zero, then the forces are said to be unbalanced forces. In this case, the body will have accelerated motion. Consider an example of a balanced force, a person standing upon the ground. There are two forces acting upon the person. The force of gravity exerts a downward force. The floor exerts an upward force. Since these two forces are of equal magnitude and in opposite directions, they balance each other. The person is in equilibrium. There is no unbalanced force acting upon the person and thus the person maintains his state of rest. Now consider a book sliding from left to right across a tabletop. In this case, the force of gravity - pulling downward and the force of the table pushing upwards on the book are of equal magnitude but opposite in directions. These two forces balance each other. Yet there is no force present to balance the force of friction. As the book moves to the right, friction acts to the left to slow the book down. There is an unbalanced force and as such, the book changes its state of motion. The book is not in equilibrium and subsequently retards to come to rest. Balanced force Unbalanced force Newton s First Law of Motion According to this law, an object remains in a state of rest or of uniform motion in a straight line unless an unbalanced external force compels to change the state of the object. This statement means that the external unbalanced force applied on a body alone can change its state of rest or state of uniform motion along a straight line, or we can say that in the absence of external unbalanced force, the velocity of the body cannot change i.e., the body cannot accelerate. Inertia: It is the property of a body due to which it resists a change in its state of rest or of uniform motion. That is why the first law of motion is also known as the law of inertia. Quantitatively inertia of a body is 30

31 Force & Laws of Motion measured by the mass of a body. Heavier the body, greater is the force required to change its state and hence greater is its inertia. Example: When a bus or train starts suddenly, the passenger sitting inside tends to fall backward. This is so because the lower part of the body starts moving with the bus or train but the upper part tries to remain at rest due to inertia of rest. When a bus or train stops suddenly, the passenger sitting inside tends to fall forward. This is so because the lower part of the body comes to rest, but the upper part tends to continue its motion due to inertia of motion. Linear Momentum Linear Momentum of a body is the quantity of motion contained in the body. We define momentum of a body as the product of the mass and velocity The linear momentum of a body depends on its mass and velocity. The larger the mass of a body, greater is its momentum. Also larger the velocity of a body, greater is its momentum. Thus, i.e., Momentum = mass velocity or p = mv Momentum is a vector quantity. Its direction is same as that of velocity. The S.I. unit of momentum is kilogram metre per second (kgms ) Further, when two bodies of unequal masses m, m have the same linear momentum i.e., p = p or m v = m v or v v m m i.e., velocities of the bodies having equal linear momenta vary inversely as their masses i.e., the heavier body has smaller velocity and the lighter body has the larger velocity. Newton s Second Law of Motion Newton s second law of motion gives us a measure of force (i.e., quantitative definition of force). According to Newton, an object will only accelerate if there is net or unbalanced force acting upon it. Second law of motion states that the rate of change of linear momentum of a body is directly proportional to the applied unbalanced external force and the change takes place along the direction of applied unbalanced force. Mathematical Formulation of Second Law of Motion Suppose, an object of mass, m is moving along a straight line with an initial velocity, u. It is uniformly accelerated to velocity, v in time t. Initial momentum, p = mu Final momentum, p = mv The change in linear momentum = p = p p = m(v u) m( v u) The rate of change in linear momentum = p ma...(i) t 3

32 Force & Laws of Motion 3 Now, according to Newton s second law of motion, where k is constant of proportionality From (i) and (ii) kf net = ma or F net ma k p F t net or p kf t Where k is a constant of proportionality and its value depends on the units adopted for measuring force. The unit of force is so chosen that the value of k becomes. F net ma Thus, the Newton s second law of motion gives us a method of measuring the force in terms of mass and acceleration. The absolute unit of force in SI unit is newton (N) As F = ma N = kg ms Thus, one newton force is that force which produces an acceleration of ms in a body of mass kg in the direction of force. In c.g.s. system, the unit of force is dyne ( dyne = g cms ). As, N = kg ms = 000 g 00 cms = gcms N = 0 5 dyne The gravitational unit of force in SI is kilogram weight (kg wt) or kilogram force (kgf). It is that force which produce an acceleration of 9.8 ms in a body of mass kg. kgwt = 9.8 N. Note: i) No force is required to move a body uniformly along a straight line. ii) Accelerated motion is always due to an external unbalanced force. Impulse: The force which acts on bodies for short time is called impulsive force. Impulse is a measure of total effect of the force and it is given by the product of force and the time for which the force acts on the body, i.e., Impulse = Force time or I = F t [if F remains constant for time t] Impulse is a vector quantity and its direction is same as that of the force. According to Newton s second law of motion m (v u) F = ma = or F t = mv mu t Impulse = Force time or I = F t net...(ii)

33 Force & Laws of Motion Thus, impulse of a force during an impact is equal to the change in linear momentum of the body. Applications (i) (ii) A cricket player lowers his hands while catching a cricket ball to avoid injury. In this case, the player increases time to reduce the momentum of the ball to zero and therefore, has to apply smaller force against the ball in order to stop it. Thus ball in turn exerts a smaller force on his hands. In a high jump, the athelete is made to fall on a cushioned bed or on heap of sand. This is to increase the time of the athelete s fall to stop after making the jump. This decreases the rate of change of momentum and hence force. Therefore, the athelete is not hurt. Newton s Third Law of Motion "For every action there is an equal and opposite reaction" This law is exemplified by what happens if we step off a boat onto the bank of a lake: as we move in the direction of the shore, the boat tends to move in the opposite direction (leaving us facedown in the water, if we aren t careful). A force is a push or a pull on an object, which results from its interaction with another object. Forces result from interactions. As discussed earlier some forces result from contact interactions (normal, frictional, tension, and applied forces are examples of contact forces) and other forces are the result of action-at-adistance interactions (gravitational, electrical, and magnetic forces). According to Newton, whenever objects A and B interact with each other, they exert forces upon each other. When we sit in a chair, our body exerts a downward force on the chair and the chair exerts an upward force on our body. There are two forces resulting from this interaction a force on the chair and a force on our body. These two forces are called action and reaction forces and are the subject of Newton s third law of motion. Newton s third law states that: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The magnitude of the force on the first object equals the magnitude of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs that is equal and opposite, action-reaction force pairs. Consider the flying motion of birds. A bird flies by use of its wings. The wings of a bird push air downwards. In turn, the air reacts by pushing the bird upwards. The magnitude of the force on the air equals the magnitude of the force on the bird, the direction of the force on the air (downwards) is opposite to the direction of the force on the bird (upwards). Thus action-reaction force pairs make it possible for birds to fly. When a gun is fired, it exerts a forward force on the bullet. The bullet exerts an equal and opposite reaction force on the gun. This results in the recoil of the gun. Principle of Conservation of Linear Momentum Consider two balls A and B of masses m A and m B travelling in the same direction along a straight line with velocities u A and u B respectively. 33

34 Force & Laws of Motion A B F AB m A m B ua u B m A m B va v B Before collision F BA After collision Let u A > u B and the two balls collide with each other and collision lasts for time t. During collision, the ball A exerts a force F AB on ball B and the ball B exerts a force F BA on ball A. Suppose v A and v B are the velocities of the two balls after collision. Now, momentum of the ball A before and after collision are m A u A and m A v A respectively. (va u ) The rate of change of momentum (or F AB, action) of ball A during collision = m A A. t (vb u ) Similarly, the rate of change of momentum of (or F BA, reaction) ball B = m B B. t The force F AB exerted by ball A on ball B (action) and the force F BA by the ball B on ball A (reaction) must be equal and opposite to each other. or or or F AB = F BA ma (v AuA) mb (v BuB) 0 t t m A u A + m B u B = m A v A + m B v B Total momentum before collision = Total momentum after collision This means that the total momentum of two balls remains constant (or conserved). If the external force is zero on a body or system of bodies then the linear momentum of the body or system of bodies remains conserved. This is called principle of conservation of linear momentum. Additional Information Type of Force (and Symbol) Applied Force (F app ) Description of Force An applied force is a force which is applied to an object by a person or another object. If a person is pushing a desk across the room, then there is an applied force acting upon the object. The applied force is the force exerted on the desk by the person. Gravity Force (also known as Weight) (F grav ) The force of gravity is the force with which the earth, moon, or other massively large objects attracts another object towards itself. By definition, this is the weight of the object. All objects upon earth 34

35 Force & Laws of Motion experience a force of gravity, which is directed downward towards the center of the earth. The force of gravity on earth is always equal to the weight of the object as given by the equation: F = m g grav g is called acceleration due to gravity where g = 9.8 m/s (on Earth) and m = mass (in kg) (Caution: do not confuse weight with mass.) Normal Force (F norm ) Friction Force (F frict ) The normal force is the support force exerted upon an object, which is in contact with another object. For example, if a book is resting upon a surface, then the surface is exerting an upward force upon the book in order to support the weight of the book. On occasions, a normal force is exerted horizontally between two objects, which are in contact with each other. The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. The friction force opposes the relative motion of the object. For example, if a book moves across the surface of a desk, then the desk exerts a friction force in the opposite direction of its motion. Friction results from the two surfaces being pressed together closely, causing intermolecular attractive forces between molecules of different surfaces. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. The friction force can be calculated using the equation: F = µ N frict norm where µ = coefficient of friction Air Resistance Force (F air ) Tensional Force (F tens ) The air resistance is a special type of frictional force, which acts upon objects as they travel through the air. Like all frictional forces, the force of air resistance always opposes the motion of the object. This force will frequently be neglected due to its negligible magnitude. It is most noticeable for objects, which travel at high speeds (e.g., a skydiver or a downhill skier), or for objects with large surface areas. The tension is the force, which is transmitted through a string, rope, or wire when it is pulled by forces acting from each end. The tension force is directed along the wire and pulls equally on the objects on either end of the wire. 35

36 Force & Laws of Motion Spring Force (F spring ) The spring force is the force exerted by a compressed or stretched spring upon any object, which is attached to it. An object, which compresses or stretches a spring is always acted upon by a force which restores the object to its rest or equilibrium position. For most springs (specifically, for those which are said to obey Hooke s Law ), the magnitude of the force is directly proportional to the amount of stretch or compression. So far we have seen the situation for force is applied on a single body. Let us learn when some bodies are connected by some string and then the system is in motion. Here we use Free Body Diagram (FBD) to study the motion of induvidual object in the system. In free body diagram, a body is isolated from the system and we draw all the forces acting on the body when it was in the system. Following are the systems. :- Atwood s Machine (A single fixed ideal pulley i.e. a frictionless pulley with negligible mass). i) When masses m and m (m > m ), attached to a string of negligible mass, pass over a massless pulley and then released from position of rest : For mass A : m g T = m a For mass B : T m g = m a...(i)...(ii) m m a = g m m and T = mm m m g ii) In the above case, when pulley is moving up with acceleration f, then m m a = ( g f ) m m mm and T = ( g f ) m m iii) In the above case, when pulley is moving down with acceleration f, then m m a = ( g f ) m m mm and T = ( g f ) m m. When m lies on inclined smooth surface and m hangs vertically downwards For A : m g T = m a For B : T m g sin = m a m m sin a = g m m and T = mm ( sin g m m m g sin m B T A m g T m 3. When mass m lies on horizontal smooth surface and m hangs vertically downward For A : m g T = m a For B : T 0 = m a 36

37 Force & Laws of Motion m a = g m m m B T mm and T = g m m A T m 4. Some Special Cases m g i) Let m move down and m move up with acceleration a; and both the inclined surfaces are smooth: For A : m g sin T = m a For B : T m g sin = m a m sin m sin a = g m m and T = mm (sin sin g m m m T B T A m ii) Let F = force acting on interface of masses A and B F = force acting on interface of masses B and C When three bodies A, B and C are placed on a smooth surface and these are pushed by force F, then F = m m3 m m m 3 F and F = m3 F m m m 3 m 3 m T m T iii) C B A F Let three bodies A, B and C be connected through strings of negligible masses, as shown. Let the surface be smooth. Then tensions T and T are given by the relations: T = m m3 m m m 3 F and T = m3 F m m m 3 SOLVED EXAMPLES Example : Let a force act on a mass m. A mass m is added and you observe that the acceleration dropped to /5 of its former level. Assuming the force remains same, find the ratio of m to m Solution: The force is observed to be the same in both cases: F = m a = (m + m ) a The acceleration in each case is different, with a = a /5. Substituting for a 37

38 Force & Laws of Motion m a = (m + m ) (a /5) 5m = m + m 4m = m m /m = /4 Example : A train of mass kg is moving at a speed of 80 km/h. The brakes, which produce a net backward force of N, are applied for 5.0 seconds. a) What is the new speed of the train? b) How far has the train travelled in this time? Solution: F = ma a = F/m We know that F = N m = kg So a = Given, N kg = 0.8 m/s u = 80 km/h = Now, v = u + at 5 80 m/s =. m/s 8 v =. m/s + ( 0.8 m/s²) (5.0 s) v = 5.5 m/s b) We know the acceleration and initial velocity. Applying the formula s = ut + at s = (. m/s) (5.0 s) + ( 0.8 m/s ) (5.0 s) s = 467 m Example 3: Solution: 38 A constant force acts on an object of mass 5 kg for a duration of s. It increases the object s velocity from 3 ms to 7 ms. Find the magnitude of the applied force. Now, if the force was applied for a duration of 5 s, what would be the final velocity of the object? We have been given that u = 3 ms and v = 7 ms, t = s and m = 5 kg m (v u) F = t Substitution of values in this reaction gives

39 Force & Laws of Motion F = 5 kg (7 ms 3ms )/ s = 0 N. Now, if this force is applied for a duration of 5 s (t = 5s), then the final velocity can be calculated by rewriting v = u + at Ft F v = u + a m m On substituting the values of u, F, m and t, we get the final velocity, v = 3 ms. Example 4: Solution: A bullet of mass 0 g is fired horizontally with a velocity 50 ms from a pistol of mass kg. what is the recoil velocity of the pistol? We have the mass of bullet m =0 g (= 0.0 kg) and the mass of the pistol, m = kg; initial velocities of the bullet (u ) and pistol (u ) = 0, respectively. The final velocity of the bullet, v = + 50 ms. The direction of bullet is taken from left to right (positive, by convention). Let v be the recoil velocity of the pistol. Total momentum of the pistol and bullet before the firing, when the gun is at rest = ( + 0.0) kg 0 ms = 0 kg ms Total momentum of the pistol and bullet after it is fired = 0.0 kg (+50 ms ) + kg v ms = (3 + v) kg ms According to the law of conservation of momentum:- Total momentum after firing = Total momenta before the firing 3 + v = 0 v =.5 ms. Negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet, that is, right to left. Example 5: Solution: A girl of mass 40 kg jumps with a horizontal velocity of 5 ms onto a stationary cart with frictionless wheels. The mass of the cart is 3 kg. What is the velocity of the girl as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction. Let v be the velocity of the girl on the cart as the cart starts moving. The total momenta of the girl and cart before the interaction. = 40 kg 5 ms + 3 kg 0 ms = 00 kg ms. 39

40 Force & Laws of Motion Total momenta after the interaction = (40 + 3) kg v ms = 43 v kg ms. According to the law of conservation of momentum, the total momentum is conserved during the interaction. That is, 43 v = 00 v = 00/43 = ms. The girl on cart would move with a velocity of 4.65 ms in the direction in which the girl jumped. Example 6: Two blocks of masses m and m lie on a smooth horizontal surface in contact with each other. Force F is applied to the mas m in figure and then to mass m as in figure. Calculate the contact force between the blocks in the two cases. Solution: F m m a m F m Now, the contact force between the blocks is making mass m move. So contact force = mass acceleration mf m m a m F m (same as in previous case) Now the contact force between the blocks is making mass m move. mf So, contact force = mass acceleration = m m TEST YOURSELF. A train accelerates on a horizontal rail due to the force exerted by (a) the earth (b) driver of the car (c) the rail (d) the engine of the train. A 50 N force acts on a body of mass 0 kg and produces a constant velocity on a rough surface. The force of friction exerted on the body is (a) 50 N (b) 00 N (c) 50 N (d) 00 N 40

41 Force & Laws of Motion 3. A machine gun mounted on a 000 kg car, on a horizontal frictionless surface fires 0 bullets/ second. Mass of each bullet is 0 g and velocity of each bullet is 50 m/s. Acceleration of the car is (a) m/s (b) 0 m/s (c) 0 m/s (d) 40 m/s 4. A grenade having mass of 0 kg flying horizontally with a velocity of 0 m/s explodes into two fragments. The larger fragment has velocity of 0 m/s in the direction of initial velocity of grenade while smaller fragment has a velocity of 0 m/s in the opposite direction. The masses of fragments are (a) 5 kg, 5 kg (b) 40 3 kg, 0 3 kg (c) 0 kg, 0 kg (d) 6 kg, 4 kg 5. A uniform rod PQ of mass M and length L is pulled with a constant force F on a smooth horizontal surface. The tension in the rod at the distance a from the other end P is (a) (c) F LF a (b) af L P Q (d) F L IMPORTANT DEFINITIONS. A force is a push or pull upon an object resulting from the object s interaction with another object.. If two or more forces act on a body such that their resultant is zero, then the forces are said to be balanced forces. 3. If two or more forces act on a body such that their resultant is not zero, then the forces are said to be unbalanced forces. 4. Linear Momentum of a body is the quantity of motion contained in the body. We define momentum of a body as the product of the mass and velocity 5. Impulse is a measure of total effect of the force and it is given by the product of force and the time for which the force acts on the body,. Momentum = mass velocity or p = mv Fnet ma 3. Impulse = Force time or I = F t 4. I = F t = mv mu = p p IMPORTANT FORMULAE a L F 4

42 Force & Laws of Motion IMPORTANT TIPS FOR COMPETITIVE EXAMS. Four types of forces exist in nature. They are - gravitation g force Fw and nuclear force n F. F, electromagnetic F. A system or a body is said to be in equilibrium, whe the net force acting on it is zero. em, weak 3. If a number of forces F,F,F 3... act on the body, then it is in equilibrium when F F F A body in equilibrium cannot change the direction of motion. 5. In the absence of the force, a body moves along a straight line path with uniform velocity. 6. Inertia is proportional to mass of the body. 7. If a body moves along a curved path, then it is certainly acted upon by a force. 8. Force cause acceleration. 9. A single isolated force cannot exist. 0. Newton s first law of the motion defines the force.. Force is a vector quantity.. Absolute units of force are dyne in CGS system and newton (N) in SI N 0 dyne. 4. Gravitational units of force are gf (or gwt) in CGS system and kgf (or kgwt) in SI. 5. gf = 980 dyne and kgf = 9.8 N 6. Absolute units of force remains the same throughout the universe while the gravitational units of force varies from place to place as they depend upon the value of g. 7. Newton s second law of motion gives the measure of force i.e. F = ma. 8. Acceleration of a horse-cart system is H F a M m where H = horizontal component of reaction F = force of friction M = mass of horse m = mass of cart 9. The weight of the body measured by the spring balance in a lift is equal to the apparant weight. 0. Apparent weight of a freely falling body = ZERO (State of weightlessness). If the person climbs up along the rope with acceleraton a, then tension in the rope will be m (g + a). If the person climbs down along the rope with acceleration, then tension in the rope will be m (g - a) 3. When the person climbs up or down with uniform speed tention in the string will be mg. 4. A body starting from rest moves along a smooth inclined plane of length l, height h and having angle of inclinationθ. 4

43 Force & Laws of Motion (i) Its acceleration down the plane is g sin θ. (ii) Its velocity at the bottom of the inclined plane will be gh gl sin θ. (iii) Time taken to reach the bottom will be t l l h g sinθ sinθ g (iv) If the angle of inclination is changed keeping the height constant then t t sin θ sin θ 5. For an isolated system (on which no external force acts), the total momentum remains conserved (Law of conservaton of momentum). 6. The change in momentum of a body depends on the magnitude and direction of the applied force and the peiod of time over which it is applied i.e. it depends on its impulse. 7. Guns recoil when fired, because of the law of conservation of linear momentum. The positive momentum gained by the bullet is equal to negative recoil momentum of the gun and so the total momentum before and after the firing of the gun is zero. m 8. Recoil velocity of the gun is V υ M where m = mass of bullet, M = mass of gun and = muzzle velocity of bullet. 9. Impulse, I Ft change in momentum 30. Unit of impulse is N-s. 3. Action and reaction forces never act on the same body. They act on different bodies. If they act on the same body, the resultant force on the body will be zero i.e., the body will be in equilibrium. 3. Action and reaction forces are equal in magnitude but opposite in direction. 33. Action and reaction forces out along the line joining the centres of two bodies. 34. Newton s third law is applicable whether the bodies are at rest or in motion. 35. The non-inertial character of the earth is evident from the fact that a falling object does not fall straight down but slightly deflects to the east due to coryolis (pseudo) force. 36. Force of friction is non-conservative force. 37. Force of friction always acts in a direction opposite to that of the relative motion between the surfaces. 38. Rolling friction is much less than the sliding friction. This knowledge ws used by man to invent the wheels. 39. The friction between two surfaces increase (rather than to decrease) when the surfaces are made highly smooth. 40. The atomic and molecular forces of attraction between the two surfaces at the point of contact give rise to friction between the surfaces. 4. Whereever the observer is situated in space that is called the frame of reference. The reference frame is associated with a co-ordinate system and a clock to measure the position and time of the events happening in space. 43

44 Force & Laws of Motion Inertial Reference Frame A reference frame in which Newton s first law is valid is called an inertial reference frame. An inertial frame is either at rest or moving with uniform velocity. Any frame moving at constant velocity relative to a known inertial frame is also called an inertial frame. In an inertial frame, an object subject to no net force will stay at rest or move at contant velocity. If the acceleration of a particle is zero in one inertial frame, it is zero in all inertial frames. Ideally, no inertial frame exists in the universe. For practical purpose, a frame of reference may be considered as inertial if its acceleration is negligible with respect to the accelerations of the object to be observed. For example, to measure the acceleration of a falling apple, earth can be considered as an inertial frame. On the contrary, to observe the motion of planets, earth cannot be considered as an inertial frame but for this purpose the sun may be assumed as an inertial frame. 4. Non-Inertial Frame: An accelerated frame of reference is called a non-inertial frame. Objects in non-inertial frames do not obey Newton s first law. 43. Pseudo Force: It is an imaginary force which is recognized only by a non-inertial observer to explain the physical situation according to Newton s laws. The magnitude of this force F p is equal to the product of the mass m of the object and acceleration a of the frame of reference. The direction of the force is opposite to the direction of acceleration F ma. 44. A weighing machine measures the weight of a body - misconception The weighing machine measures the reaction between the machine and the body. The body exerts a force N on the machine and the machine also exert an equal and opposite force on the body. 45. Weightlessness means the absence of weight - misconception Weightlessness usually means an absence of reaction between the body and the surface with which it is in contact. The weight W of a body of mass m is the gravitational force exerted upon it by the earth, W = mg N may be same as W. It may also be larger or smaller than W depending upon whether the elevator is accelerating up or down. 46. (i) A body moving with constant speed in a circular path is continuously accelerated towards the centre of rotation. The magnitude of acceleration is given by v a r ω r Where v is the constant speed v ωr and r is the radius of the circular path. (ii) According to Newton s Second Law, a body moving in a circular path with constant speed must be acted upon by an unbalanced force which is always directed towards the centre. This necessary unbalanced force is called the centripetal force. mv F r mω r 44

45 Force & Laws of Motion (iii) A body moving with constant speed in a circle is not in equilibrium. (iv) It should be remembered that in the absence of the centripetal force the body will move in a straight line with constant speed. (v) It is not a new kind of force which acts on bodies. In fact, any force which is directed towards the centre may provide the necessary centripetal force. For example : (i) When a stone tied with a string is rotated in a circular path, the tention in the string provides the centripetal force (See fig. ). (ii) When a car negotiates a curve the force of friction between the tyres and the road provides the necessary centripetal force (See fig. ) The gravitation force of attraction between the sun and a planet provides the centripetal force (See fig. 3). The electric force of attraction between the nucleus and the revolving electron provides the centripetal force in an atom (See fig. 4). 47. The most common misconception about the friction force is that it opposes the motion of a body. Friction force opposes the relative motion between the two surfaces in contact. It opposes motion and yet makes the motion possible. For example, We are able to walk only because of the force of friction. With our feet we push the ground backward, which in turn, pushes us forward. We are able to push the ground only because of friction. On a smooth surface one cannot imagine to walk. A car acceleratates on the ground because of friction. The engine of the car rotates the wheels and the friction force pushes the car forward. In the absence of friction the wheel will keep on rotating and the car will not move. 45

46 Force & Laws of Motion If a small block is gently placed on a plank moving on a smooth horizontal surface. Then, the frictional force between the block and the plank accelerates the block and decelerates the plank so that fianlly, they start moving together. In this case, frciton opposes the motion of the plank but produces the motion of block, but in general, opposes the relative motion between the block and the plank. 48. When we know that the momentum conservation is applicable in all types of collisions (elastic or inelastic), then why do we requrie an equation of kinetic energy conservation or the equation of coefficient of restitution to completely solve a collision problem. Collision is a phenomenon in which exchange of motion occurs between the colliding particles. Both momentum and kinetic energy are the quantitative measures of motion and they togetehr describes the exchange of motion between colliding particles. Momentum conservation, being a vector equation, describes the direction of motion of the separating particles after collision. Kinetic energy conservation, being a scalar equation, describes the magnitude of velocities of the separating particles after collision. REMEMBER!. Every body of the system is independent.. When two forces act on a body in opposite directions, their resultant is difference of the two and body moves in the direction of greater force. 3. For two surfaces in contact, there are action and reaction forces which are equal in magnitude but opposite in direction. 4. i) When there is string or spring between two bodies, there is tension (tension is the force setup in a string or spring when pulled at both ends) in it. ii) If mass of string or spring is negligible, then tension in it is same throughout. iii) Tension always acts away from the point or body under consideration. 5. i) Friction always acts in a direction opposite to the direction in which body moves or tends to move. ii) Static friction is self adjusting force and increases from zero to maximum value called limiting force of friction (N = mg). Where N is normal reaction and is coefficient of friction). iii) Kinetic friction remains constant provided velocity of body is not very large. Comparable to velocity of light). 46

47 Force & Laws of Motion iv) Frictional forces act in two different direction for two bodies in contact. Frictional force is a mutual force. Mutual forces are equal and opposite. Consider pulling body. (with force F) kept on body. Direction of frictional force on body is in a direction opposite to that of F. Direction of frictional force on body is in same direction as that of F. These two forces are equal. 6. a) i) If a body is accelerating upwards, its weight increases, given by the relation R = m(g + a). Where R is normal reaction on the body of mass m by the surface accelerating up with acceleration a ii) If body is moving up with retardation, a may be negative. b) i) If a body is moving down with acceleration a its weight decreases, given by relation R = m(g a). Where R is normal reaction on the body of mass m by the surface accelerating down with acceleration a. ii) If a body is moving down with retardation, sign of a will change. c) If a body is moving down freely, (a = g) then its effective weight is zero or it experiences weightlessness. 7. Choosing the system on which we have to apply laws of motion. Your system may be a single particle, a body, a combination of bodies, bodies connected by a string or spring. Thumb rule is that all parts of the system should have same acceleration. Example: A, (C + D), (A + C + D) can be taken as a system but (A + B) or (B + C + D) cannot be taken as a system as A, C, D more horizontally but B moves vertically. 8. Identifying the forces: Identify forces only after deciding the system. Do not include any force applied by the system in the list of forces. Make a list of the forces acting on the system due to all other objects other than the system. 9. Making a free body diagram (F.B.D): Represent the system by a point and draw vectors representing the forces acting on the system with this point as the common origin. Forces may be along a line, in a plane or in space. 47

48 Force & Laws of Motion Example: For the system shown in the figure, we have to calculate (i) acceleration of the system (ii) tension(a) between A and B and (b) between B and C. 48 Consider (A + B + C) as a system unbalanced external force 5 5 Acceleration = ms Total mass 5 3 To calculate tension between A and B Let the tension between A and B be T, Draw F.B.D. of A remember tension acts away from point under consideration. Again applying II law 5 T = 5a = 5 T = 0 N To calculate tension between B and C Let tension between B and C be T. Draw F.B.D. of C Again apply II law T 5 = a = = T = 7 N You will get same result by drawing F.B.D. of B on applying II law (T T = 3a, 0 T = 3, T = 7 N). PART - I: MISCELLANEOUS DOMAIN. a) Would it be easier to lift this book on Earth or on the moon? b) Neglecting friction, would it be easier to set this book into horizontal motion at 5 m/s on the Earth or on the moon? c) Neglecting friction, would it be easier to set this book into horizontal motion at 5 m/s on Earth or in distant space?. A large truck collides head-on with a small car. Which force is bigger: the force by the truck on the car or the force by the car on the truck? 3. A -N book rests on the palm of your hand. Describe the two forces that act on the book. How strong is each force and in what direction does each act? Are these two forces equal but opposite to each other? Are these two forces part of one force pair?

49 Force & Laws of Motion 4. From each of the following forces, describe the other member of the force pair. a) The force a player exerts on a baseball while throwing it. b) A bat hitting a baseball. c) A baseball hitting a catcher s mitt. d) An apple s weight. e) Your weight f) Your arms pulling on a box as you slide it across the floor. g) A rope pulling forward on a water skier. 5. A car weighing 0,000 N moves along a straight level road at an unchanging speed of 80 km/hr. Air resistance is 300 N, and rolling resistance is 400 N. Find the net force on the car and the strength and direction of the perpendicular force and the drive force. 6. Is an object s acceleration always in the same direction as its velocity? If not, give an example in which it is not. Is an object s acceleration always in the same direction as the net force on the object? If not, give an example in which it is not. 7. Draw a force diagram showing the forces on an apple at rest on a table. Find the net force on the apple. 8. Is any force exerted on you when you speed up along a straight line? When you slow down along a straight line? How do you know? 9. A ball weighing 8 N is thrown straight upward. Disregarding air resistance, find the direction and strength of the net force on the ball as it moves upward. What is the direction of the ball s acceleration? Are the net force and the acceleration in the same direction in this case? Can they ever be in different directions? 0. An apple is accelerated upward by your hand. Which is larger, the apple s weight or the upward force by your hand? What if you accelerate the apple downward while it is in the palm of your hand? What if you lift the apple at an unchanging velocity? What if you lower the apple at an unchanging velocity?. A stone of mass kg is thrown with a velocity of 0 ms 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?. A 8000 kg engine pulls a train of 5 wagons, each of 000 kg, along a horizontal track. If the engine exerts a force of N and the track offers a friction force of 5000 N, then calculate: a) the net accelerating force b) the acceleration of the train 3. Two objects, each of mass.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is.5 ms before the collision. What will be the velocity of the combined object after collision if they stick to each other. 49

50 Force & Laws of Motion 4. A bullet of mass 0 g travelling horizontally with a velocity of 50 ms 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. 5. An object of mass 00 kg is accelerated uniformly from a velocity of 5 ms to 8 ms in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object. 6. M = M = 5 kg HIGHER ORDER THINKING SKILLS (HOTS) M 3 = 0 kg. Find T 7. A block A of mass 5 kg is pulled by a force of 00 N as shown in the figure. Find the acceleration of the block. 8. Find the acceleration of the 500g block 9. A small block B is placed on another block A of mass 5 kg and 50 length 0 cm. Initially B is near the right end of Block A. A constant horizontal force of 0 N is applied to the block A. All the surfaces are assumed frictionless. Find the time elapsed before the block B separates from block A.

51 Force & Laws of Motion 0. Find the force exerted by the 0 cm part of the block on the 0 cm part, if the mass of entire block is 3 kg. VALUE BASED QUESTIONS. Vivek and Mohan conduct an experiment in one physics lab. Body A and B are connected by a massless string. Body A is pulled by a force which is measured with the help of a spring balance. Mohan says that if action is the force with which body A pulls body B then, according to Newton s II law of motion, body B also pulls body A with equal force in opposite direction, called reaction. Mohan is baffled by the fact that the system of bodies moves as according to him, the sum of the forces defined as action and reaction is zero but still the system of bodies moves. (a) Vivek gives the correct explanation to Mohan. What is the explanation? (b) What is the value shown by Vivek?. Sohan and Mohan are walking together towards market. They take a shortcut and find lot of sand on the road. Sohan is puzzled over the fact that it is difficult to walk on sand. He asks Mohan. Mohan explains the correct reason. (a) What are the values shown by them? (b) What is the explaination n given by Mohan? 3. Sohan gets hurt while catching a fast moving cricket ball. Mohan advices him to move his hands backwards while catching the ball and explains the reason. (a) What is the value shown by Mohan? (b) What is the reason behind it? 4. Sohan wanted to win the interschool sports. So he practiced high jump every day for about a month. He participated and won I position in the interschool sports. Sohan didn t understand why he had to run some steps before taking a jump. He asked his sports teacher. His teacher explained the reason (a) Comment upon their values (b) What was the explanation. PART - II: MULTIPLE CHOICE QUESTIONS. Choose the incorrect statement. (a) A body can have constant speed yet have a varying velocity (b) A body s acceleration is zero if its velocity is constant (c) A body s acceleration is constant and non-zero if net force on the body is zero (d) A body s velocity is constant or zero if net force on the body is zero. 5

52 Force & Laws of Motion. A ball of mass m is connected to a ball of mass M by a massless spring. The balls are pressed so that the spring is compressed. When released ball of mass m moves with acceleration a. Acceleration of mass M is (a) Ma m (b) ma M (c) Ma M m (d) ma M m 3. A boy standing in a stationary lift drops a ball from height h. Time taken by ball to reach the floor of the lift is t. Now lift moves up with uniform acceleration a. Time taken by the ball, when dropped from same height to reach the floor is t (a) t a g a (b) t (c) g 4. Two forces each of magnitude F act on a body placed at point O as shown. The force necessary to keep it at rest is a g (d) t (a) F along X axis (b) F along + X axis (c) F along X axis (d) F along + X axis 5. A car moving at speed v is stopped by retarding force F in distance x. If speed of car becomes 3 times, the force needed to stop it within same distance x is (a) F (b) 9F (c) 6F (d) 3F 6. A car moving at speed v is stopped by a retarding force F in distance x. If retarding force were 3F. The stopping distance is (a) x (b) 9 x (c) x 6 (d) x 3 7. A body of mass kg is suspended through two light spring balances, then (a) The reading of upper balance will be kg and the reading of lower balance will be zero (b) The reading of both balances will be kg (c) The reading of both balances will be kg (d) Sum of readings both balances will be kg and individual readings may be anything. 8. A force x acts on a body to accelerate it from rest to velocity v. The force y acts on it to decelerate it to rest, then (a) x must be unequal to y (b) x must be equal to y (c) x may be equal to y (d) none of the above 5

53 Force & Laws of Motion 9. The magnitude of horizontal force F needed to keep the body of mass m stationary on the smooth inclined plane is (a) F (b) mg sin (c) mg cos (d) mg tan 0. If the force on a rocket remains the same, its acceleration (a) still decreases (b) still increases (c) first increases then decreases (d) remains the same. A body of mass x strikes a wall normally with a speed y m/s and rebounds will same speed but in opposite direction. The magnitude of change in linear momentum is (a) 4xy (b) xy (c) xy (d) zero. A bullet of mass m and velocity u is fired into a block of mass M. Bullet gets embedded in the block, then velocity of the system just after the collision is (a) M m u m (b) mu M m (c) mu m M (d) m M u m 3. When a body undergoes acceleration (a) its linear momentum increases (b) its speed always increases (c) an unbalanced force always acts on it (d) all of the above 4. Choose the incorrect statement (a) Newtons III law talks about nature of the force (b) Newton s I law is the law of inertia (c) Mass is a measure of inertia (d) Unbalanced force produces constant velocity 5. Two balls of same mass and size fall from same height. The first ball rebounds to same height and second ball does not rebound, then (a) first ball experiences greater change of linear momentum (b) second ball experiences greater change of linear momentum (c) both balls experience same change of linear momentum (d) both balls experience no change of linear momentum. PREVIOUS YEARS QUESTIONS 6. A.00 kg ball and a.00 kg ball collide with each other. The data from their collision is shown on the table given below :.0 0 k g ball.0 0 kg ball T ime Px Py Px Py Befo re co llisio n A fter collis ion Here px and py are x and y components respectively of linear momentum. The angle between the balls after collision is (a) 50º (b) 0º (c) 90º (d) 60º 53

54 Force & Laws of Motion 7. A monkey is holding onto one end of a light rope which passes over a frictionless pulley and at the other end there is a plane mirror which has a mass equal to the mass of the monkey. At equilibrium the monkey is able to see her image in the mirror. Consider three situations : I. The monkey climbs up the rope. II. The money tries to push the rope down. III. The monkey lets go of the rope. Under which of the above conditions does the monkey continue to see her image? (a) I only. (b) II only (c) III only. (d) I, II and III. 8. Two identical balls are released simultaneously from on equal heights h. Ball A is thrown horizontally with velocity v and the ball B is just released. Choose the alternative that best represents the motion of A and B with respect to an observer who moves with velocity v/ with respect to the ground as shown in the figure. A B h v/ h ground A B A B A B A B (a) A (b) B (c) C (d) D 9. The maximum tension in the string of a simple pendulum is. times the minimum tension. If θ0 is the angular amplitude, then θ 0 is : (a) cos 5 4 (b) cos 4 3 (c) 5 cos 6 (d) cos A force F, acts on a particle to accelerate it from rest to a velocity v. The force F is then replaced by force F which decelerates it to rest. Which of the following statements is true? (a) F must be equal to F (b) F may be equal to F (c) F must be unequal to F (d) None of these. A Parachutist of weight W strikes the ground with his legs fixed and come to rest with an upward 54 acceleration of magnitude 3g. Force exerted on him by the ground during landing is : (a) W (b) W (c) 3 W (d) None of the above