MOTION IN A STRAIGHT LINE. time interval t and as t approaches zero, the ratio. has a finite limiting value.(where x is the distance

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1 KINEMATICS Rest and Motion, Distance and Displacement, Speed, Velocity, and Acceleration, Averae Velocity, Averae Acceleration, Instantaneous Velocity, Instantaneous Acceleration, Motion with Constant Acceleration, Motion Under Gravity, Equations of Motion Vectors and Scalars, Addin Vectors, Vector Subtraction, Resolution of Vectors, Unit Vectors, Addin Vectors by Components, Multiplication of Vectors, Dot Product, Cross Product, Projectile Motion, Time of Fliht, Rane, Maximum Heiht Reached, Relative Velocity Uniform Circular Motion. MOTION IN A STRAIGHT LINE We shall study here some properties of motion in a straiht line. This branch of physics which deals with the study of motion without referrin to the cause of the motion is called kinematics. time interval t and as t approaches zero, the ratio Δx has a finite limitin value.(where x is the distance Δt travelled) Vinst = Lt t 0 Rest and Motion The position of an object is specified relative to a coordinate system or frame of reference. A convenient way is to choose three mutually perpendicular axes and specify the coordinates of the object. If a body doesn t chane its position with time, then it is said to be at rest. If a body chanes its position with time, then it is said to be in motion. In actual motion of an extended body, different parts of the body move alon the same direction and at the same rate. We can therefore consider the motion, as that of a sinle particle. The simplest motion of a body is motion alon a straiht line. The line may be vertical (that of a fallin object), horizontal (that of a vehicle on road), or inclined. x (.) t Averae Velocity: The averae velocity of a particle in a time interval Δ t = (t t) is defined as its displacement divided by the time interval. Vav is a vector quantity. Distance and Displacement Suppose a particle is at A at time t and at B at time t with respect to a iven coordinate system. The actual lenth of the path AB durin the time interval t to t is called the distance travelled by the particle. The shortest path from A to B covered by the particle durin the time interval t to t is called the displacement of the particle i.e. the straiht line from the initial position to the final position. Vav = Δs Δt (.3) Y A B X Distance is a scalar quantity (has only manitude) where as displacement is a vector quantity (has both manitude and direction). Speed and Velocity Averae Speed: It is a way of describin how fast a particle moves. The averae speed is the total distance travelled divided by the time interval. If the particle travels a distance Δ x in time interval Δ t = (t t) then averae speed is defined as Δx Vav = Δt Vinst = (.) Instantaneous Speed: The averae speed is defined for a time interval and the instantaneous speed is defined at a particular instant. This is obtained by decreasin the KINEMATICS Instantaneous Velocity: It is the velocity of the particle at a iven instant. Let s be the displacement of the particle in the time interval t. As t approaches zero, the averae velocity approaches to a limitin value, which is the velocity at that instant Lt t 0 s t (.4) Averae Acceleration: Let v be the velocity of the particle at time t, and v be the velocity of the particle at time t. The averae acceleration can be defined as the ratio of chane in velocity to the time interval. 3

2 v -v aav t - t (.5) Averae velocity vav u v (.8) The displacement of the body is s = Instantaneous Acceleration: Let V be the chane in the velocity of the particle in the time interval t. As t approaches zero, the averae acceleration approaches a limitin value, which is the instantaneous acceleration. u+v t --(.9) From eqn (.7), we et u (u at ) s.t or s ut at (.0) By eliminatin t in eqn (.7) & (.0), we et v = u + as (.) The three equations (.7), (.0), (.) are called the equations of motion in a straiht line with a constant acceleration. Eqn (.7) is raphically represented in Fi. a inst = Lt t 0 v t (.6) When the acceleration is constant the averae acceleration is equal to the instantaneous acceleration. Motion with Constant Acceleration: Let us consider a particle havin a velocity u at time t = 0. Let v be the velocity at any later time t and let a be the constant acceleration. From the definition of acceleration, we can write it as a= v u or v = u + at (.7) t Under these conditions, the averae velocity over the time interval t is the averae of the velocity at the beinnin of the journey and at the end of the journey. KINEMATICS The area under the v t curve represents the displacement. (We need calculus to prove this statement). However, for motion with constant acceleration, we can show that the area under the v t curve is equal to the displacement x. Area between the instants t = 0 and t = t = OABD = Area of ABC + Area of rectanle OACD v u t ut at ut ( (v u) at) =x ** These equations of motion can be derived usin the principles of calculus as described below. However, this section can be taken up for learnin after learnin the basic principles of calculus Acceleration is defined as dv a= dv = a dt dt dv a dt or v = at + c.(i) t t = 0, let v = u c = u and v = at + u or v = u + at dx Further, v = dx = v dt dt interatin both the dx v dt dx (u at) dt sides, ( v u at) 4

3 at + c' at t= 0, if x = x0, we et c' = x0 x = x0 + ut + at dv dv dv dx Also, we can write, a = = = v dx dt dx dt v dv = a dx x = ut + v u a(x x 0 ) or v dv a dx u x0 v u a(x x 0 ) v x Points to Remember: (i) If a body starts from rest, its initial velocity, u = 0 (ii) If a body comes to rest, its final velocity, v = 0 (iii) If a body moves with a uniform velocity, its acceleration, a = 0 The displacement of a body in the nth second is iven by S (n) = Sn - Sn- = [un + an] [u (n-) + a (n - )] = u + a (n -) Freely fallin Body The most common example of motion with constant acceleration is that a body fallin freely towards the earth. In the absence of air resistance it is found that all the bodies, reardless of their size or weiht, fall with the same acceleration in a straiht line towards the earth s surface, if the distance covered is small as compared to the radius of the earth. This constant acceleration is called acceleration due to ravity and is denoted by. It always acts vertically downwards. Its value is approximately equal to 9.8 m/s or 3 ft/s.it is the same for all objects. The equations of motion derivedfor constant acceleration are also valid for free fall near earth s surface. P Q R Two vector quantities are said to be equal if they have the same manitude and direction, no matter where they are located in space. In the fi above P and Q are equal vectors (havin the same lenth and direction). The vector R is not equal to P, Q ( because its direction is opposite to that of P and Q). The neative of a vector quantity is a vector havin the same manitude but the opposite direction. For example, P = R or R = P. A parallel translation of a vector does not brin about any chane in it. The manitude of a vector quantity is a scalar and is always positive.. Manitude of P = P Addin Vectors Suppose that a particle moves from A to B and then later from B to C. We can represent its overall displacement with two successive displacement vectors, AB and BC. The effect of these two displacements is a net displacement AC (from A to C). It can be called as the resultant (or vector sum) of the vectors AB and BC, This rule of addition is called the trianle rule of addition. Points to Remember (i) When a body is dropped freely from a heiht, its initial velocity, u = o (ii) When a body is thrown vertically upwards, its final velocity, v = 0. (iii) The time taken by the body to rise to the hihest point is equal to the time taken by the body to fall from the same heiht to the initial point VECTORS We can represent it as s = a + b Vectors and Scalars Some physical quantities are completely described by a numerical (manitude) value alone. To perform any operations on these physical quantities, we can use ordinary rules of alebra. Such quantities are termed as scalars. Ex: mass, time, density, temperature etc.,quantities, which require manitude as well as direction to represent them, are termed as vector quantities. Ex: displacement, velocity, acceleration, etc., KINEMATICS 5

4 Vector S is the vector sum of vectors a and b. Dependin on the value of, the components of a vector may be positive, neative or zero. Unit Vectors A unit vector is a vector havin a manitude of unity. Its purpose is to describe a direction in space. The unit vectors in the positive directions of the x, y and z axes are labeled as i, j and k. Vector addition has two properties:. The order of addition does not matter i.e., a + b = b + a (commutative law). If there are more than two vectors, it does not matter how we roup them, as we add them. Thus if we want to add three vectors a, b and c, we can add a and b first and then add their vector sum to c or on the other hand, add b and c first and then add that sum to a. It can be represented by (a + b) +c = a + (b + c) ( Associative law) Vector Subtraction The difference of two vectors A and B is defined as the vector sum of A and B i.e. To subtract B from A, invert the direction of B and add to A. A B = A + ( B) Unit vectors are very useful for expressin other vectors, for example we can express a and b as a = ax i + ay j and b = bx i + by j The quantities ax i, ay j are the vector components of a, and in the same way bx i and by j are the components of b. Addin Vectors by Components We can add vectors by combinin their components, axis by axis. Let us consider r = a + b In terms of unit vectors r = rx i + ry j + rz k, a = ax i + ay j + az k, b = bx i + by j + bz k r = (ax +bx) i + (ay+ by) j + (az + bz) k rx = (ax + bx) etc. Multiplication of Vectors There are three ways in which vectors can be multiplied. They are (a) multiplyin a vector by a scalar (b) Vector by a vector Scalar product (c) Vector by a vector Vector product Resolution of Vectors Vector a in fi (a) is in the XY plane. If we drop perpendicular lines from the ends of a to the coordinate axes, the quantities ax and ay so formed are called the components of a in the x and y directions. The process of formin them is called resolvin the vector. In eneral, a vector will have three components. The components are written by usin trionometric functions as ax = a cos and ay = a sin ay a x +a y Thus, tan = and a = ax KINEMATICS Multiplication of a Vector by a number (Scalar) Suppose A is a vector of manitude A and k is a scalar. We define the vector B = ka as a vector of manitude ka If k is positive, the direction of vector B, is the same as that of A. If k is neative, the direction of B is opposite to A. In particular, multiplication (-) just inverts the direction of the vector. Then, vector A and B have equal manitudes but opposite directions. The Dot Product or Scalar Product : The scalar product or dot product of two vectors a and b is defined as a.b = ab cos (i) Where a is manitude of a, b is the manitude of b and is the anle between a and b. Equ (i) can be written as a.b = a (b cos ) = (a cos ) b (ii) The dot product between two mutually perpendicular vectors is zero. The dot product between two mutually parallel vectors is maximum. The dot product is commutative and distributive i.e.,a.b = b. a and a. (b + c) =a.b+a.c If the components of A and B are known, their scalar product can be calculated as A. B = (Ax i + Ay j + AZ k) (Bx i + By j+ Bz k) 6

5 A. B = AxBx +AyBy + AzBz Since i, j, k are mutually perpendicular. we have i. j = i. k = j. i =j. k = k. i = k. j = 0 Since i, j, k mutually orthoonal also i. i = j. j = k. k = The Cross Product or Vector Product: The vector product or cross product of two vectors a and b is denoted by a x b, which ives another vector. The manitude of this vector is c = a x b = ab sin Where a and b are the manitudes of the vectors a and b respectively and is the anle between the two vectors. If a and b are parallel or anti- parallel, a x b = 0. The manitude of a x b, is maximum when a and b are perpendicular to each other. The direction of a b is perpendicular to both a and b. Thus, it is perpendicular to the plane formed, by a and b. a, b, c form a riht handed system of co-ordinates. The commutative law does not apply to a vector product axb=-bxa The vector product follows the distributive law a x (b + c) = a b + a c The vector product does not follow the associative law a (b x c) ( a b) c In unit vector notation a x b = (ax i +ay j +az k ) x (bx i+ by j +bz k) = (ay bz az by) i + (az bx ax bz) j + (ax by ay bx) k. If i x j = k In such a system j x k = i and k x i = j, i x i = j x j = k x k = 0 Zero Vector A vector havin zero manitude is called a zero or null Vector. The direction of zero vector is indeterminate For any vector A + 0 = A A 0 = 0 For any number k, k 0 = 0 PROJECTILE MOTION Any object that is iven an initial velocity and follows a path determined by the ravitational force actin on it is called a projectile. The path followed by projectile is called its trajectory. KINEMATICS The above fi shows the path of a projectile, which is launched with some initial velocity u. It is makin an anle with the horizontal. The components of velocity can be resolved as ux = u cos : uy = u sin Durin its motion, its position and velocity will chane continuously, but its acceleration is constant and is always directed vertically downward. In projectile motion, the horizontal motion and the vertical motion are independent of each other i.e., neither motion effects the other. Now, we analyse the projectile motion horizontally and vertically ; because there is no acceleration in the horizontal direction, the horizontal component ux of the projectile s initial velocity remains unchaned throuhout the motion. Horizontal Motion vx = ux + ax t = u cos and x = ux t+ ax t = u (cos ) t (i) Vertical Motion The vertical motion is that of a particle in free fall. The y component of the initial velocity is uy y = uy t and y = uyt - t y = u (sin ) t t (ii) Substitutin the value of t from eqn (i) y = u sin x x u cos θ u cos θ x (iii) y = (tan ) x u cos The above equation is in the form y = ax + b x, which a and b are constants. This represents equation of a parabola, so the path is parabolic. in the Time of Fliht: The particle is projected from the point O, and reaches the same horizontal plane at the point B. The total time taken to reach B is the time of fliht. OB = R = ut cos The y coordinate at the point B is zero. Thus, from the equation ( ii) 0 = ut sin t u sin Thus, either t = 0 or, t = Now t = 0 corresponds to the point O of the particle. The time at which it reaches B is thus, 7

6 u sin. This is the time of fliht. Rane: It is the distance travelled by the particle time u sin t=. From the equation of horizontal motion, u sin R = u cos. t R = (u cos ) t = in u sin R has maximum value when sin = i.e at = 450 R = Maximum Heiht Reached: At the maximum heiht, the velocity of the particle is horizontal. The vertical component of velocity is thus, zero at the hihest point. y = uy - t v = u + as 0 = u sin - t 0 = (usin ) H u sin At the maximum heiht H = and t = r( ), V Averae acceleration of the particle alon direction ax x Vqx Vpx. t Averae acceleration of the particle alon y direction = w + v - wv cos and tan v sin v cos w V B v cos w KINEMATICS t u sin Relative Velocity: Consider two motor cars A and B movin in the same direction on a road. If A is movin with 30 mph and B with 40 mph a person in A would observe the car B to be drawin away from him at the rate of 0 mph. This represents the velocity of B relative to A. If however B is movin opposite to the direction of A with velocity 40 mph, for a person in A, B appears to be movin towards him at the rate of 70 mph. This therefore represents the velocity of B relative to A. Let us suppose that the two cars A and B are movin with velocities w and v in directions inclined at an anle as shown in Fiure below. Resolve the velocity of B into a component v cos parallel to the direction of A's motion and a component v sin at riht anles to the direction of A's motion. Now, the velocity of B relative to A consists of a component (v cos - w) in the direction of A's motion and a component (v sin - 0) or v sin at riht anles to the direction of A's motion. If V is the resultant of these components, then V = (v cos - w) + v sin v sin UNIFORM CIRCULAR MOTION A particle is in uniform circular motion ( possible only in horizontal plane ) if it travels around a circle or circular arc at constant speed. Althouh the speed does not vary, the particle has acceleration as v chanes in direction. Let a particle move in uniform circular motion with speed v in a circle of radius r. Consider two points P and Q in the path. Let us draw the velocity vectors at these two points. These vectors Vp and Vq have the same manitude V but they are pointin in different directions. S Vpx = +v cos, Vpy = + v sin and Vqx = + v cos, Vqy = - v sin. The time required for the particle to move from p to q at constant speed v is A ay Vqy Vpy t Here the neative sin indicates that the y acceleration points in the ve y -direction. By reducin t 0, we et the instantaneous acceleration. sin = As t 0, 0 and Lt and ay (inst) = Thus,when a particle moves at constant speed Vin a circle of radius r, the acceleration of the particle is always directed towards the center of the circle and has a constant manitude of the centripetal acceleration. Period of revolution T = V. This is called r r. V It is the time taken by the particle, to complete one revolution or to o round the circle once. In uniform circular motion, the acceleration is always directed radially towards the centre of the circle where as the velocity is always tanential to the circle in the direction of motion. Centripetal Force: The force which causes the centripetal acceleration is called the centripetal force and is directed towards the center of the circle. The manitude F of this force is constant and is iven by m v F = ma = ( centripetal force ) r 8

7 BOARD EXAM QUESTIONS I. VERY SHORT ANSWER QUESTIONS :-. Define displacement. Is it a scalar or a vector quantity? Give an example in support of your answer.. What does the slope of velocity time raph represent? 3. Define one dimensional motion of a particle. Give at least one example. 4. What are the characteristics of uniform motion? 5. What do you mean by relative velocity? State its units. 6. Which motion is exactly represented by =. 7. Give an example of a motion which even thouh is accelerated motion yet it is called uniform motion. 8. When is the averae velocity over an interval of time becomes equal to instantaneous velocity? 9. Under what condition the manitude of averae velocity of a particle is equal to the averae speed? 0. What will be the relative velocity between two objects movin parallel with equal velocities. Show their position-time raphs. Can we multiply two vectors?. Give an example of a physical quantity which has a direction but is not a vector. 3. Can we multiply a vector by a scalar? 4. What condition should be satisfied for vectors to be equal? 5. Distinuish between localized and non localized vectors. 6. What are co-initial and collinear vectors? 7. Write any two properties of zero vectors. 8. A vector is multiplied by a pure real number. Will its unit chane? 9. Is it true that vector addition is applicable to any two vectors? 0. How many directions of projection are possible for the same horizontal rane?. What is the rane of a projectile?. What is the anle of projection of maximum horizontal rane? 3. Is it true that instantaneous velocity of a projectile is tanential to its parabolic path? 4. Give two examples of projectile motion. 5. Can we use the equations of kinematics to find the heiht attained by a body projected upwards with any velocity? 6. Is it possible that the brakes of a car are so perfect that the car stops instantaneously. If no, why? 7. Why a projectile fixed alon the horizontal not follow a straiht line path? 8. Why does a tennis ball bounce hiher on hills than in plains? 9. The velocity of a particle is constant in manitude but not in direction. What is the nature of trajectory? II. SHORT ANSWER QUESTIONS :. Plot position-time raph for a uniform motion. What does the slope of the raph indicate?. Define instantaneous velocity. How is it expressed mathematically? 3. What are the characteristics of uniform motion? KINEMATICS BOARD EXAM PREPARATION 4. Draw the position-time raph for a stationary object. 5. Under what condition will the distance and displacement of a movin object will have the same manitude? 6. If the distance covered by a movin object varies directly as the time, what conclusions could you draw about the motion and the forces? 7. Under what condition is the averae velocity equal to instantaneous velocity? 8. Is it possible that displacement is zero but not the distance? 9. Sketch the velocity-time raph of uniform motion. What does it represent? 0. Can a body have zero velocity and finite acceleration?. Draw position-time raphs for two objects havin zero relative velocity.. Why do we not consider rate of chane of acceleration in the study of linear motion? 3. What do you understand by (a) resolution of a vector and (b) rectanular component of vector? 4. If,, are three consecutive vectors formin a trianle show that + + =. 5. Can you associate vectors with (a) lenth of a wire bent into a loop (b)a plane area (c) a sphere explain? 6. Prove that. =. 7. Can any of the components of a iven vector have reater manitude than that of the vector itself? 8. What is the final check to see that the iven physical quantity is a vector? 9. A vector is multiplied by a scalar. Is it necessary that its units will never chane? 0. Is it true that vector addition is applicable to any two vectors?. Is it possible to add vectors representin force of 5 N and 00 dyne?. Are the commutative and associative laws applicable to vector subtraction? 3. Is the acceleration of a projectile equal to zero when it reaches the top of its trajectory? If not, why not? 4. At what point on the projectile trajectory is the (a) potential enery maximum and (b) kinetic enery minimum? 5. Is it true that instantaneous velocity of a projectile is tanential to its parabolic path? 6. A bullet is dropped from a certain heiht and at the same time, another bullet is fired horizontally from the same heiht which one will hit the round earlier and why? 7. A projectile is fixed at an anle of to the horizontal with the speed v. If another projectile is projected with the same speed, then at what anle with the horizontal it must be projected so as to have the same rane? 8. What will be the effect on horizontal rane of a projectile when its initial velocity is doubled, keepin the anle at projector same? 9. Is the maximum heiht attained by projectile is larest when its horizontal rane is maximum? 30. A body is projected with speed u at an anle to the horizontal to have maximum rane. What is the velocity at the hihest point? 9

8 III. LONG ANSWER QUESTIONS:-. Acceleration is defined as the rate of chane of velocity. Suppose we call the rate of chane of acceleration as SLAP. What is the unit of SLAP?. What is the acceleration of a body when its velocity time raph is (a) perpendicular to time axis and (b) parallel to time axis? 3. Define the followin : (a) unit vector, (b) equal vectors, (c) null vector, and (d) displacement vector 4. Can two vectors of the same manitude have the resultant equal to either of them? Explain. 5. (a) Can two vectors of different manitudes be combined to ive zero resultant? (b) Can three vectors of different manitudes be combined to ive zero resultant? 6. Can any of the components of a iven vector have reater manitude than that of the vector itself? 7. Do + lie in the same plane? Explain. 8. State the paralleloram law of vector addition. 9. Prove that the horizontal rane is same when anle of projection is (i) reater than 45o by certain value, and (ii) less than 45o by the same value. 0. In lon jumpin, does it matter how hih you jump? What factors determine the span of the jump?. An aircraft executes a horizontal loop of radius.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to ravity.. A projectile is fired at an anle with the horizontal with velocity v. Derive the expression for maximum heiht attained by it. 3. Two bodies are thrown with the same initial velocity at anle and (9 to the horizontal. Determine the ratio of the maximum heihts reached by the bodies. IV. NUMERICALS. A car movin at a speed of 60 km/h is accelerated at a rate of 0 km/ ℎ. How much will be the distance covered by the car in 3 h?. Two cars A and B are runnin at velocities of 60 km/h and 45 km/h, respectively. Calculate the relative velocity of car A if (a) they are both travellin eastward and (b) car A is travellin eastward and car B is travellin westward. 3. A jet airplane travellin at the speed of 500 km/h ejects its products of combustion at the speed of 500 km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the round? 4. Two trains each of lenth 00 m, movin in opposite directions alon parallel lines, meet their speeds 60 km/h and 30 km/h. If their accelerations are 0.3 m/ and 0. m/, respectively, find the time taken by them to pass each other. 5. A car at rest accelerates uniformly to a speed of 44 km/h in 0 s. How much distance is covered by the car? 6. An electron is emitted with a velocity of 5 x 6 m/s. It is accelerated by an electric field in the direction of initial velocity at 3 x m/. If its final velocity is 7 x 0 m/s, calculate the time taken by the electron to attain the final velocity and the distance covered by it. 7. Find the resultant force of the followin forces which act on a particle: (a) (30) N due east; (b) 0 N due north; (c) 50 N due west; and (d) 40 N due south. 8. The three vectors,,and are represented in manitude and direction by Fi. If,,, + = A, show that S is the midpoint of PQ. 9. The position of a particle is iven by = [ + + ] m. Find (a) the velocity and acceleration of the particle, (b) the manitude of velocity and acceleration of the particle at =. 0. Find the anle between = 4 + = =.. Ball thrown vertically upward with a speed 9.6 m/s from the top of lower returns to Earth in 6 s. Find the heiht of the tower.. A man can swim with a speed of 4 km/h in still water. How lin does he take to cross a river km wide if the river flows steadily at 3 km/h and he makes his strokes normal to the river current? How far down the river does he o when he reaches the other bank? 3. A cricketer can throw a ball to a maximum horizontal distance of 00 m. How much hih above the round can the cricketer throw the same ball? 4. An aircraft is flyin at a heiht of 3400 m above the round. If the anle subtended at a round observation point by the aircraft positions 0.0 s apart is, what is the speed of the aircraft? 5. A missile is fired for maximum rane with an initial velocity of 0 m/s. if = 0 m/, the rane of the missile is. KINEMATICS BOARD EXAM PREPARATION 0

9 IIT- JEE/CET PREPARATION KINEMATICS a = 0.5 m/s CLASSWORK (Objectives) MOTION IN A STRAIGHT LINE. Vc = VA + a(t + t) A bus starts to move with an acceleration of m/s. A man who is 48 m behind the bus runs to catch the bus with a constant velocity of 0 m/s. In how much time, will he catch the bus? (a) 0 s (b) 8 s (c) 5 s (d) s Ans. (b) OA = S, t, v AB = S, t, v BC = S3, t3 Hence, Av speed S S S3 t t t3 900 km / hr m/s = A particle moves alon a straiht path ABC with a uniform acceleration of 0.5 m/s. While it crosses A its velocity is found to be 5 m/s. If it reaches C with a velocity 40 m/s, 30 seconds after it has crossed B in its path. The distance AB is (a) 800 m (b) 600 m (c) 400 m (d) 00 m Ans. (b) at = 600 m A particle movin with uniform acceleration is found to travel 35 m in the 8th second and 5 m in the th second. Its velocity in m/s at the beinnin of th second is (a) 49 (b) 45 (c) 47 (d) 5 Ans. (b) Displacement in the nth second a S n u n. () n = 8, Sn = 35 m & n =, Sn = 5m. a 35 u 8 () a 5 u (3) a = 4m/s and u = 5 m/s Velocity at the beinnin of the th second (at the end of the 0th second) (i.e. t = 0 s) v = u + at = 45 m/s A particle starts from rest. Its acceleration (a) versus time (t) is as shown in the fiure. The maximum speed of the particle will be a) 0 m/s b) 55 m/s c) 550 m/s d) 660 m/s Ans. (b) Area under accelerationtime raph ives the maximum velocity. Vm = u + aav.t AB ut Let We et t = 8s and s (8s is the answer) The speed of a car was 50 km/hr for the first 900 s, then 40 km/hr for the next 50 km and then the car decelerated uniformly at 0 km/hr till it came to rest. The averae speed of the car was (a) 50 km/hr (b) 7. m/s (c) 30 km/hr (d) 9 m/s Ans. (b) t + t = 70 s, t = 40 s as t = 30 s the man catch the bus after t sec. at point C. x ut at..t.. () 48 + x = V. t = 0. t. (). t t 6. A small block slides without friction down an inclined plane startin from rest. Let S n be the distance travelled from t = n to t = n. Then a) b) is c) d) Ans. (c) Distance travelled in nth second is,

10 7. A particle is movin eastwards with a velocity of 5 m/s. In 0 s the velocity chanes to 5 m/s northwards. The averae acceleration in this time is a) Zero m/s towards north-east b) b) Solvin this equation, we et vr = 3 km/h d) 0. A particle P is slidin down a frictionless hemispherical bowl. It passes the point A at t = 0. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at t = 0 alon the horizontal strin AB, with the speed v. Friction between the bead and the strin may be nelected. Let tp and tq be the respective times taken by P and Q to reach the point B. Then a) tp < tq b) tp = tq m/s towards north-west c) m/s towards north Ans. (c) c) tp > tq d) Ans. (b) in north-west direction. m/s ( in north-west direction) A river is flowin from west to east at a speed of 5m/min. A man on the south bank of the river, capable of swimmin at 0 m/min in still water, wants to swim across the river in the shortest time. He should swim in a direction. a) Due nort b) 30 east of north c) 30 west of north d) 60 east of north Ans. (c) To cross the river in shortest time resultant should be to swim perpendicular to the river current A boat which has a speed of 5 km/h in still water crosses a river of width km alon the shortest possible path in 5 min. The velocity of the river water n km/h is a) b) 3 Ans. (b) Shortest possible path comes when resultant velocity of boatman becomes perpendicular to river current as shown in fiure. c) 4 d) The horizontal component of the velocity is v for both P and Q. Horizontal displacements for both the particles are equal. Therefore tp = tq. MOTION UNDER GRAVITY. A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three seconds of its motion. The stone remains in the air for (a) 6 s (b) 5 s (c) 7 s (d) 4 s Ans. (b) Let the last second be the nth sec. Then (n ) 9 or n = 5 s. Ball A is dropped from the top of a buildin. At the same instant ball B is thrown vertically upwards from the round. When the balls collide, they are movin in opposite directions and the speed of A is twice the speed of B. At what fraction of the heiht of the buildin does the collision occur? (a)/3 (b)/3 (c)/4 (d)/5 Ans. (b)

11 h fraction of the heiht = (h is measured from the round) 3. A parachutist drops freely from an aeroplane for 0 s before the parachute opens out. Then he descends with a net retardation of.5 ms -. If he bails out of the plane at a heiht of 495 m and = 0 ms -, his velocity on reachin the round will be (a).5 ms - (b) 7.5 ms - (c) 5 ms - (d) 0 ms - Ans. (c) The velocity of the parachutist at the end of 0 seconds = 0 = 00 ms - and the distance fallen in 0 seconds =500 m. The distance travelled after he bails out is s = = 995 m. For this distance u = 00 ms - and a = -.5 ms -. Therefore, the final velocity v is iven by v = u + as v = (00) which ives v = 5 ms A particle is dropped vertically from rest, from a heiht. The time taken by it to fall throuh successive distances of km each will then be ( / ) second (a) all equal, bein equal to x 0 (b) in the ratio of the square roots of the inteers,, 3,. in the ratio of the difference in the square roots of (c) the inteers, i.e., (d),( ), ( 3 Ans. (a) For upward motion, v = u - h. For downward motion, v = h (u = 0). Since v h, the raph of v vs h is parabolic. Hence raphs (b), (c) and (d) are wron. 6. A bullet is fired vertically upwards with an initial velocity of 50 ms -. It covers a distance h durin the first second and a distance h durin the last 3 seconds of its upward motion. If = 0 ms, h and h will be related as h (a) h = 3h (b) h = h (c) h = h (d) h 3 Ans. (c) h ut t 50 0 () m No w, the velocity of the bullet at the end of first seconds is v = u - t = 50-0 = 30 ms -. This is the initial velocity for the last 3 seconds of upward motion. The distance covered in the last 3 seconds is h (3) = 45 m Thus h = h 5s )( 4 3 ) in the ratio of the reciprocals of the square roots of h the inteers i.e., /. /, / 3, / 4,.. Ans. (c) The time taken to cover n meters is iven by tn = n n= (n ) now tn + tn = [ (n ) n] so, the desired ratio is 3) 5. A ball is dropped vertically from a heiht h above the round. It hits the round and bounces up vertically to a heiht h/. Nelectin subsequent motion and air resistance, its velocity v varies with the heiht h as,( ), ( 3 )( 4 s s tn similarly, the time taken to cover (n + ) meters is iven by tn + = v 7. h u = 50 m/s A stone dropped from a buildin of heiht h reaches the round after t seconds. From the same buildin if two stones are thrown (one upwards and the other downwards) with the same velocity u and they reach the round after t and t seconds respectively, then the time interval t is t t (a) t = t - t (b) t (c) t tt (d) t t t Ans. (c) The displacement of the stone in the three cases is the same, equal to h. h t. (i) 3

12 h ut t.. (ii) VECTORS. (iii) t tt [ Hint : (ii) t + (iii) t ] 8. From a buildin two balls A and B are thrown such that A is thrown vertically upward and B vertically downwards, both with the same speed. If va and vb are their respective velocities on reachin the round, then a) vb > va b) vb = va c) va > vb d) their velocities depend on their masses Ans. (b) Displacement is same for both A and B ( use v = u + as) 9. A parachutist after bailin outfalls 50 m without friction. When he opens the parachute, he decelerates at ms-. He reaches the round with a speed of 3 ms-. At what heiht did he bail out? c) 9 m b) 8 m c)93 m d) m Ans. (c) Let H be the heiht from the round when the parachutist bails out. The velocity after fallin throuh h = 50 m is. a) b) c) is d) AB sinθ = 0. A particle has an initial velocity of acceleration of a) 7 units c) and an. Its speed after 0 s is: b) 8.5 units 0 units d) 7 units Ans (d) v = u +at = and A vector A is alon the positive z-axis and its vector product with another vector B is zero, then vector B could be b) a) c) d) Ans (d) A ball is released from rest from the top of a tower of heiht h metre. It takes T seconds to reach the round. What was the heiht of the ball from the round in T/3 seconds? a) π and (Given) directed downwards. After openin the parachute, he falls a distance (H 50). For this distance, initial, final velocity v = 3, then the anle between Ans (a) 3. velocity u = acceleration a = -m. If b) c) d) Hence vector B also must be alon ± z-axis. The anle between vectors A = 3 6 is c) Zero b) 45 Ans (d) We know that Ans. (c) d) 80.(i)...(ii) The manitudes of A and B are -h = 0 - (iii) If the ball travels a distance h in time t = T/3, and -h = 0 - (iv) Substitutin for A, B and A. B in (i) we have h = Distance of the ball from the round = h h =h- c) 90 and cos θ = or θ = zero, which is choice (a). 5. If A B = C + D, select the correct alternative (a) B is parallel to C + D. (b) A is perpendicular to C. 4

13 (c) Component of C alon A = Component of D alon A (d) Component of C alon A = - Component of D alon A Ans: (d) (A + B) = (C + D) C + D is perpendicular to both A and B i.e., A. (C + D) = 0 or A (Component of C alon A) + A (Component of D alon A) = 0 3 B by usin this in (Bcosθ = -A), we et from (i) and (ii) A = 30. (B C) = (i+j-3k) (3i+pj+5k) = i (0+ 3p)- j(5 + 9) +k(p 6) A.(B C) = (i-j+k). (i (0+ 3p)- j(5 + 9) +k(p 6)) Solvin this we et, p = If A =, B = 5 and A B = 8 then A B is (a) 6 (b) 3 (c) 4 (d) 7 Ans. (a) cos sin or A B AB cos = = 6 The resultant of two vectors A and B is perpendicular to the vector A and its manitude is equal to half of the manitude of vector B. The anle between A and B is (a) 900 (b) 00 (c) 500 (d) 450 Ans. (c) Given (A + B). A = 0 i.e A + AB cosθ = 0 (i) B cosθ = - A Also A+B B (a) (b) 70 (c) 0 (d) none A B ( i + j- k ) ( i j k ) = i j k Manitude of ( A 3 B) ( 4) (5) ( 7) = PROJECTILES 3. Two tall buildins are 40 m apart. With what speed must a ball be thrown horizontally from a window of 45 m above the round in one buildin so that it will enter a window.5 m from the round in the other? (a) 8 m/s (b) 6 m/s (c) 9.8 m/s (d) 9.6 m/s Ans. (a) For motion in the vertical direction h t 9.8 t and t = 5 s For motion in the horizontal direction, 40 = ut and u = 8 m/s 3. The speed of projection of a projectile is increased by 5%, without chanin the anle of projection. The percentae increase in the rane will be (a).5% (b) 5% (c) 7.5% (d) 0% Ans. (d).5 v R u =0% v R u Four projectiles are projected with the same speed at anles 00, 450, 600 and 750 with the horizontal. The rane will be the lonest for the projectile whose anle of projection is (a) 00 (b) 450 (c) 600 (d) 750 Ans. (b) Rane R u sin /. For the B (ii) ( 4 90 Ans. (a) or (A + B). (A + B) = or AB cosθ = A i j k and B i j k. 6. If A B = 0 also A C = 0, the anle between B and C is (a) zero (b) 450 (c) 900 (d) 800 Ans. (c) A B 0 ; A B A C 0 ; A C C is also perpendicular to B 7. The value of p so that the vectors i j + k, i + j 3 k and 3 i + p j + 5 k are coplanar should be (a) 6 (b) - 4 (c) 4 (d) - 8 Ans. (b) Vectors A, B and C are coplanar when their scalar triple product is zero. i.e. A. B C = 0 3 => θ = 500 Obtain the manitude of A 3 B if cosθ = A + AB cosθ = 0) Same u, R sin. sin is maximum for = 450. A ball rolls off the top of a stair way with a horizontal velocity u m / s. If the steps are h meters hih and b meters wide, the ball with hit the ede of the nth step. If (a) n = h u / b (b) n = h u / b (c) n = h u / b (d) n = h u / b 5

14 Ans. (d) The situation is shown in fiure If the ball hits the nth step, then horizontal distance traversed = nb, vertical distance traversed = n h. Here velocity alon horizontal direction = h, velocity alon vertical direction = 0 n b = u t and n h = 0 + t nb nb or t = and n h = u u n= 35. hu c) Yes, 60 b) Yes, 30 c) No d) Yes, 45 Ans (a) The person will catch the ball if the horizontal rane = distance covered by the person in the time of fliht, i.e., if 38. A projectile can have the same rane for two anles of projection. If T and T are the times of fliht in the two cases, then the product TT is proportional to b d) R and 39. The initial velocity = 0 sin 300 = 0 ms - (upwards). The time to reach the round is iven by 40 0 t 0 t or t - t - 8 = 0 t = 4s, -s t : tf = : 36. A particle is projected at 60 to the horizontal with a kinetic enery K. The kinetic enery at the hihest point is [AIEEE] c) K b) zero c) K/4 d) K/ Ans (c) Let u be the velocity of projection. The component of u alon the horizontal is ux = ucosθ, where θ = 60. Given Hence A boy playin on the roof of a 0 m hih buildin throws a ball with a speed of 0 m/s at an anle of 30 with the horizontal. How far from the throwin point will the ball be at a heiht of 0 m from the round? [Take = 0 m/s] [AIEEE ] c) 5.0 m b) 4.33 m c).60 m d) 8.66 m Ans (a) The ball will be at point P at a heiht of 0 m from the round OP = R = = At the hihest point the vertical component uy = usinθ = 0, but the horizontal component remains the same. Hence, kinetic enery at the hihest point is c) R Ans (c) A projectile has the same rane for anles of projection θ and (90 - θ). Therefore, From the top of a tower of heiht 40 m, a ball is projected upwards with a speed of 0 m/s at an anle of elevation of 300. The ratio of the total time taken by the ball to hit the round to its time of fliht (time taken to come back to the same elevation) is (Take = 0 m/s ) (a) : (b) 3 : (c) 3 : (d).5 : Ans. (a) : u sin The time of fliht t f s. 0 b) a) 40. A ball is projected with kinectic enery E at an anle of 45 to the horizontal. The kinectic enery of the ball at the hihest point of its trajectory will be a) E b) c) d) zero Ans (c) At the hihest point, the velocity has only horizontal 37. A ball is thrown from a point with a speed u at an anle θ to the horizontal. From the same point and at the same instant, a person starts runnin with a constant speed u/ to catch the ball. Will he able to catch the ball? If yes, what should be the value of θ? component vx = ux = u cosθ = u cos 45 = u/. K.E. = 6

15 CLASSWORK - Subjectives 3. MOTION IN A STRAIGHT LINE. A bus starts from rest with a constant acceleration of 5 m/sec. At the same time a car travellin with a constant velocity of 50 m/sec overtakes and passes the bus. Find (i) at what distance will the bus overtake the car? (ii) How fast will the bus be travellin then? : Bus u = 0 a = 5 m/s A motor cycle and a car start from rest at the same place at the same time and travel in the same direction. The cycle accelerates uniformly at m/s upto a speed of 36 km/h and the car at 0.5 m/s upto a speed of 54 km/h. Calculate the time and distance at which the car overtakes the cycle. : u=0 a = m/s MC S (M) 0 m/s S (M) 54 km/hr u=0 a = 0.5 m/s S(C) 5 m/s S(C) When the car overtakes the motor cycle, the two have travelled the same distance in the same time. Let the time taken be t, while the total distance travelled be s. For motor cycle : 0 = 0 + t => t = 0 s S ( M ) 0..t..(0) = 50 m () S (M) = 0 (t - 0) = (0 t 00) m () Total distance s = s(m) + s(m) = (0t - 50) m (a) For car : 5 = t or t = 30 sec. (3) s (C) (30) 5 meter s (C) 5(t 30) 5t 450 (4) Car 50 m/s. t=0 B t and s are the same for both - car and bus. for bus s 0 5t and for car s = 50t t = 0 sec. and s = 000 m velocity of the bus at t = 0 s = 5 0 = 00 m/s A car accelerates from rest at a constant rate for sometime after which it decelerates at a constant rate to come to rest. If the total time elapse is t seconds, evaluate (i) the maximum velocity reached and (ii) the total distance travelled. : vm = 0 + t () 0 = vm - t (v = 0, u = vm, a = - and t = t) 0 = t - t or t = t () t = (t t) t or t (3) vm = t s = s + s = t t t 4. Total distance s = s(c) + s (C) = 5 t - 5 m (b) Equatin eqs. (a) and (b) 0t 50 = 5 t 5 or t = 35 s s = 300 m A particle movin with uniform acceleration in a straiht line covers a distance of 3 m in the 8th second and 5 m in the 6th second of its motion. What is the displacement of the particle from the beinnin of the 6th second to the end of 5th second? S n u a (n ) 5a ( n = 8 ) () 3 u a (6 ) u 3a (n = 6) () 5 u a (3 ) u 7

16 9 m / s, u m / s 4 8 Now, the velocity at the end of 5s (velocity at the beinnin of 6th second) = v = u + 5a Velocity at the end of 5th s, v = u + 5 a Averae velocity durin this interval of 0 v v seconds = = u + 0 a Distance travelled durin the 0s interval = S a 7. Two trains A and B, each of lenth 00 m, movin in opposite directions on parallel tracks, meet with speeds 0 ms - and 0 ms - respectively. If their respective accelerations are 0.6 ms - and 0.4 ms-, find the time taken by them to cross each other. : S = averae velocity time (u 0a) t = 36.5 m 5. Assume that a car is able to stop with a retardation of 8 m/s and that a driver can react to an emerency in 0.5 s. Calculate the overall stoppin distance of the car for a speed of 60 km/hr of the car. (8 km/hr = 5m/s) The distance travelled durin 0.5 s (before brakes are applied). Initial relative speed (u) = = 30 ms -. Relative displacement (s) = = 00 m. Relative acceleration (a) = =.0 ms -. The time t for trains to cross each other is obtained from or 00 30t.0 t or t + 60t = 0 or / 60 t Time cannot be neative. 5 5 m Velocity time = 60 (0.5) 8 3 Distance travelled after the application of brakes 5 u = 7.36 m 60 s 8 8 a Total distance travelled 6. 5 = 7.36 m 5.69m 3 A train travellin at 7 km/hr is checked by track repairs. It retards uniformly for 00 m coverin the next 400 m at constant speed and accelerates uniformly to 7 km/hr in a further 600 m. If the time at the constant speed is equal to the sum of the times taken in retardin and acceleratin, find the total time taken. Total displacement 0 V 0 V = t Vt t Hence t = A car accelerates from rest with α = m/sec in a straiht track. Then it comes to rest applyin its brakes after time t = 0 seconds, coverin a distance of 00 metres. Usin raphical method find the maximum speed of the car. The area of V/t raph ives the displacement, (straiht line) 0 V t t 3 Vt 00 but V = 400/t and (t + t3) = t t = 60 s = (t + t3) S = 00 (iven) or Vmax = 0 m/s as t = 0 s Total time = t + t + t3 = s MOTION UNDER GRAVITY 8

17 . A steel ball is dropped from the roof of a buildin. A man standin in front of a m hih window in the buildin noticed that the ball takes 0. s to fall from the top to bottom of the window. The ball continues to fall and strikes the round. On strikin the round, the ball bounces up with the same speed with which it hits the round. If the ball reappears at the bottom of the window s after passin the bottom of the window on the way down, find the heiht of the buildin. : The situation is shown in fiure. ( u = 0). (3) t 3 multiplyin equation () by t and () by t, and then addin, we et h t t and from eq (3) h t 3 tt 3. From a point A 80 m above the round, a particle is projected vertically upwards with a velocity of 9.4 m/s. Five seconds later another particle is dropped from a point B, 34.3 m vertically below A. Determine when and where one overtakes the other. : Let the two particles cross at a point C at time t. Consider the journey from the top to the bottom of the window u (0.) u = 9.5 m/sec velocity at round = = 0.9 m/s (0.9) 9.8 = m (v = u + as) A ball projected vertically upwards from A, the top of a tower, reaches the round in t sec. If it is projected vertically downwards from A with the same velocity, it reaches the round in t seconds. If it falls freely from A, show that it heiht of buildin =. would reach the round in t t seconds. : Let h be the heiht of the tower and u be the initial velocity of the ball. (i) When the ball is projected vertically upwards, h ut t.. () (ii) When the ball is projected downwards,.. () h ut t (iii) When the ball falls freely. 4. AB = AC BC 34.3 = (-9.4t + 4.9t) 4.9 (t 5) = 9.6t.5 or t = 8 s Three second after the second particle is released from B, the first one overtakes it at C at a position above the round. A body falls from O passin the points A, B and C the distance AB and BC bein equal. The time to fall from A to B is twice that to fall from B to C. If the velocity of the particle at B is 49 m/s, find the distance AB and the time taken to fall from A to B. 9

18 t h t (t - n) The two stones will reach the round simultaneously, if the above condition is satisfied. Limitin value of n t i.e., 7. AB = BC = h and tab = tbc = t 49 = va + (t) = v t h = v(t) + 4.9( t) h = v(3t) + 4.9(3 t) () () (3) Solvin vt = 4.9 t or v = 4.9 t From () 49 = 4.9 t t; t = s t = 4 s and h = 7.6 m A stone is dropped at a heiht of 6.5 m from a balloon which rises up from the round. Find the velocity of the balloon at that moment if the stone reaches the round 5 seconds after it was dropped. Initial velocity of the stone = velocity of the balloon = u m/s (upward). Displacement of the stone when it reaches the round = 6.5 m (downwards) a = - = m/s; t = 5 s ; S = = u ( S ut at ) u =.5 m/s A stone is dropped from the top of a cliff of heiht h, n seconds later, a second stone is projected downwards from the same cliff with a vertically downward velocity u. Show that the two stones will reach the bottom of the cliff toether, if h(u n) = n(u n). What can you say about the limitin value of n? : u = 0 u = u h i.e., n (h/) A balloon is risin vertically upwards at a velocity of 0 ms. When it is at a heiht of 45 m from the round, a parachutist bails out from it. After 3 s he opens the parachute and decelerates at a constant rate of 5 ms. (i) What was the heiht of the parachutist above the round when he opened his parachute? (ii) How far was he from the balloon at this instant of time? (iii) With what velocity would he hit the round? (iv) After how lon would he strike the round after his exit from the balloon? Take = 0 ms. : (i) Initial velocity = 0 ms - upwards Displacement in 3 s = (3) = - 5 m (downward) heiht from the round (at t = 3s) = 45-5 = 30 m. (ii) In 3 s, the balloon has risen throuh 30 m ( as the velocity of the balloon is 0 ms upwards). Hence, the parachutist is = 45 m away from the balloon at time t = 3 s. (iii) Parachute velocity at (t = 3 s) is iven by v = ut + at = 0 + (-0) 3 = - 0 ms To calculate the time taken to hit the round, we use S ut at 30 0t 5 t or (t 6) (t ) = 0 or t = s or 6 s. t = s, V = = - 0 ms - It will hit the round with a downward velocity which is acceptable. For t = 6 s, V will be positive velocity (upwards) which is not possible. So t = s. (iv) The total time he taken (after his exit) to strike the round is 3 s + s = 5 s. 8. A particle is projected vertically upwards from a point X on the round. It takes a time t to reach a point A at a heiht h above the round. It continues to move and takes a time t to reach the round. Find (a) the heiht h, (b) the 0