Mathematics-28.Unit-25 Statics & Dynamics

SECTION - A Statics SYLLABUS force, resultant of forces acting at a point, parallelogram law of forces, resolved parts of a force, Equilibrium of a particle under three concurrent forces, triangle law of forces and its converse, Lami's theorem and its converse, Two parallel forces, like and unlike parallel forces, couple and its moment. Resultant Force: If two or more forces P, Q, S,… act upon a rigid body and if a single force, R, can be found whose effect upon the body is the same as that of the forces P, Q, S, … this single force R is called the resultant of the other forces. Resultant of forces acting in the same direction (same straight line) is equal to their sum. LAW OF PARALLELOGRAM OF FORCES If two forces, acting at a point, are represented in magnitude and direction by the two sides of a parallelogram drawn from one of its angular points, their resultant is represented both in magnitude and direction by the diagonal of the parallelogram passing through that angular point. Magnitude and Direction of the Resultant of Two Forces: Let OA and OB represent the forces P and Q acting at a point O and inclined to each other at an angle  then the resultant R and direction ‘’ (shown in figure) will be given by and Case (i): If P = Q, then tan = tan (/2) ⇒  = /2 Case (ii): If the forces act at right angles, so that  = 90°, we have and tan = Example 1: The resultant of two forces P and Q is R. If Q is doubled, R is doubled and when Q is reversed, R is again doubled, show that P : Q :R :: . Solution: Let α be the angle between the forces P and Q. Now from the given conditions, we have …(1) and …(2) and ⇒ 4R2 = P2+Q2 2PQ cos …(3) Adding (1) and (3), we get …(4) Eliminating  from (2) and (3), we get P2 + 2Q2 4R2 = 0 …(5) From (4) and (5), we have ⇒ ⇒ P:Q:R:: . Problem 2: If the line of action of the resultant of two forces P and Q divides the angle between them in the ratio 1 : 2 then the magnitude of resultant is (A) (B) (C) (D) Solution: Let 3 be the angle between the forces P and Q. This means that the resultant make an angle  with the direction of P and angle 2 with the direction of Q Therefore, ⇒ ….(1) Also, ⇒ ….(2) From (1) and (2), we get R = . RESOLUTION OF A FORCE A force may be resolved into two components in an infinite number of ways. The most important case of the resolution of forces occurs when we resolve a force into two components at right angles to one another. Components of a Force in Two Directions: Let F be the given force represented in magnitude and direction by OC and let the directions of the two components be along OL and OM. Also ∠COL =  and ∠COM = . Then and Remark: ● When a given force is resolved into two directions which are at right angles to each other, then its component in these directions are called resolved parts of the given force. Clearly, the resolved part of a given force in a given direction is obtained by multiplying the given force by the cosine of the angle between the given force and the given direction. Now in the above situation if + = 90°, then F1 = F cos  and F2 = F sin . Example 3: The resultant of two forces P and Q is equal to and makes an angle of 30° with the direction of P; show that P is either equal to Q or double of Q. Solution: Let R = Q be the resultant of two forces and α = angle between the forces P and Q. Clearly R cos 30° = P+Q cos and R sin30° = Q sin ⇒ Q cos = R cos30° - P and R sin30° = Q sin ⇒ (R cos30°-P )2+ (R sin30°)2 = Q2 ⇒ R2 + P2 – 2PR cos 30° = Q2 ⇒ 3Q2+P2 – 2P.Q. = Q2 ⇒ P2-3PQ+2Q2 = 0 ⇒ (P-Q) (P-2Q) = 0 ⇒ P = Q or 2Q. Triangle of Forces: If three forces, acting at a point, are represented in magnitude and direction by the sides of a triangle, taken in order, they will be in equilibrium. Remark: ● In the triangle of forces it must be carefully noted that the forces must be parallel to the sides of a triangle taken in order, i.e. taken the same way round. Lami’s Theorem: If three forces acting on a particle keep it in equilibrium, each is proportional to the sine of the angle between the other two. Thus if the forces are P, Q and R and , ,  be the angles between Q, R; R, P; and P, Q respectively then if the forces are in equilibrium, we have Example-4: Forces P, Q and R acting along OA, OB and OC, where O is the circumcentre of the triangle ABC are in equilibrium. Show that . Solution: ∠ BOC = 2A, ∠COA = 2B, ∠AOB = 2C Applying Lami's Theorem , we get ⇒ Apply sine rule and cosine rule to get ⇒ . Polygon of Forces: If any number of forces, acting on a particle, be represented, in magnitude and direction, by the sides of a polygon, taken in order, the forces will be in equilibrium. IMPORTANT DEDUCTIONS ● The resultant of two forces, acting at a point O in directions OA and OB and represented in magnitude by . OA and .OB is represented by (+). OC, where C is a point in AB such that CA = .CB. ● The sum of the resolved parts of two forces in a given direction is equal to the resolved part of their resultant in the same direction. ● If the forces acting on a particle are in equilibrium, the algebraic sum of their resolved parts in two directions at right angles are separately zero. ● Conversely, if the sum of their resolved parts in two directions at right angles separately vanish, the forces are in equilibrium. PARALLEL FORCES Forces whose lines of action are parallel are called parallel forces. Parallel forces are of two type: (i). Like parallel forces: If two forces acting at different points of a rigid body be such that their lines of action are in one direction, we say that the forces are like parallel forces. In the figure, two forces P and Q are called like parallel forces, where P > Q. (ii). Unlike parallel forces: If two forces acting at different points of a rigid body be such that their lines of action are in opposite direction, we say that the forces are unlike parallel forces. In the figure, two forces P and Q are called unlike parallel forces, where P > Q. Notes: ● The resultant of two like parallel forces P and Q acting at A and B equivalent to a force P + Q acting in the same direction at C, a point in AB such that P . AC = Q . BC. ● If P ≥ Q, then CA < BC, i.e. the resultant R = P + Q will act at a point nearer to the larger force and farther from the smaller force. ● i.e. each force P, Q and R respectively is proportional to the distance between the other two. ● The resultant of two unlike parallel forces of magnitude P and Q (P > Q), acting at points A and B of a rigid body is a force of magnitude P –Q, parallel to them in the direction of the greater force P and acts at point C, which divides AB externally in the ratio Q : P i.e. Example-5: P and Q are two like parallel forces acting at A and B respectively. If P is moved parallel to itself through a distance d on AB, show that the resultant of P and Q moves through a distance . Solution: Let the resultant of two parallel forces P and Q act at a point C, then ⇒ ……(1) Now let P be moved parallel to itself from the point A to A′ and the resultant in this case is also moved to a point C′. Then ⇒ ……(2) From (1), BC = ……(3) From (2), BC′ = ……(4) Now, CC′ = BC′ – BC = (A′B – AB) = , where AA′ = d (given). Example 6: The resultant of two unlike parallel forces of 10 N and 18 N act along a line at a distance of 12 cm from the line of action of the smaller force. Find the distance between the lines of action of two forces. Solution: Let the unlike forces 18 N and 10 N act at A and B respectively and let C be the point through which their resultant R passes. Clearly C lies out side AB and nearer to the greater force 18 N. Distance of resultant R from the line of action of smaller force i.e. BC = 12 cm Hence, 18  AC = 10  BC ⇒ AC = = 6 cm ⇒ AB = BC –AC = 5 cm. MOMENTS: The moment of a force about a point is the product of the force and perpendicular distance of its line of action from the point. If F be a force and p the perpendicular distance of its line of action from the fixed point O, then moment of F about O = F  p. ● Positive and Negative Moments: The moment of a force about a point measures the tendency of the force to cause rotation about that point. If the tendency of the force is to turn the lamina in anti-clockwise direction then usual convention is to regard the moment as positive and that in clockwise direction as negative. ● The algebraic sum of the moments of a set of forces about a given point is the sum of the moments of the forces, each moment having its proper sign prefixed to it. ● The algebraic sum of the moments of any two forces about any point in their plane is equal to the moment of their resultant about the same point. ● If any number of forces in one plane acting on a rigid body have a resultant, the algebraic sum of their moment about any point in their plane is equal to the moment of their resultant. ● The moment of a force about an axis is the product of the resolved part of the force in a direction perpendicular to the axis (the other component being the product of force parallel to the axis and the shortest distance between the axis and the line of the force. Example 7: A uniform beam AB is 18 feet long and weighs 30 lb. Masses of 20 and 45 lb are suspended from A and B respectively. At what point must the beam be supported so that it may rest horizontally? Solution: Let the weight of the beam AB acts at its middle point C. when masses of 20 and 45 lb are suspended from A and B respectively, then let resultant of these three passes through O. Since, AB = 18 ft ⇒ AC = BC = 9ft Let CO = x ft, then AO = (9 + x) ft and BO = (9 – x) ft Since, the resultant passes through O, therefore moments of the acting force at point A, B and C about O will be zero. ⇒ 20 . AO + 30 . CO – 45 . BO = 0 ⇒ 20 (9 + x) + 30 x –45(9 –x) = 0 ⇒ 19x –45 = 0 ⇒ x = ⇒ Beam should have support at a distance of ft i.e. 11 ft from A. COUPLES A system of two equal and unlike parallel forces, whose lines of action are not the same, is called a couple or a torque. ● Moment of Couple: The moment of a couple is the product of one of the forces forming the couple and the arm of the couple. The perpendicular distance between the forces is called the arm of the couple. ● The moment of the couple is regarded as positive or negative according as it has a tendency to turn the body in the anti-clockwise of clockwise direction. ● The algebraic sum of the moments of the two forces forming a couple about any point in their plane is constant, and equal to the moment of the couple. ● Two couples, acting in one plane upon a rigid body, whose moments are equal and opposite balance one another. ● Any number of couples in the same plane acting on a rigid body are equivalent to a single couple, whose moment is equal to the algebraic sum of the moments of the couples. ● If three forces, acting upon a rigid body, are represented in magnitude, direction, and line of action by the sides of a triangle taken in order, they are equivalent to a couple whose moment is represented by twice the area of the triangle. Example 8: Two unlike parallel forces each of magnitude 20 units acting on a rigid body form an anticlock couple. If these forces lie in the xy-plane and act at the points A (–1, 0), B (3, 0) and are inclined at 600 to the x-axis, find the moment of the couple. Solution: From B, draw BM perpendicular to the line of action of the force passing through A. Since, AB = 4 units and ∠MAB = 600 From right angled triangle AMB MB = AB sin 600 = 2 units Therefore, the moment of the couple = 20  MB = 20  2 units = 120 units SECTION - B Dynamics Speed and velocity, average speed, instantaneous speed, acceleration and retardation, resultant of two velocities, Motion of a particle along a line, moving with constant acceleration, Motion under gravity, Laws of motion, Projectile motion. RECTILINEAR MOTION When a particle or a body moves in a straight line, the motion is said to be rectilinear motion. Here we will study the rectilinear motion ofa a particle when it moves with uniform or constant acceleration. Equations of Motion: A particle moves along a straight line with initial velocity u and constant acceleration ‘a’ from a fixed point. If v be the velocity acquired in time ‘t’ and s be the distance travelled by the particle in time ‘t’. Then (i). v = u + at (ii). s = ut + at2 (iii). v2 = u2 + 2as Remark: If the particle moves in a straight line with a retardation g then we put –a in place of a. ⮚ v = u – at ⮚ s = ut – at2 ⮚ v2 = u2 – 2as Distance Travelled in a Particular Time If a particle starts with initial velocity u m/sec and uniform acceleration a m/sec2, the distance travelled in nth second is given by Sn = u + a(2n –1) Average Velocity Suppose a particle moves in a straight line with initial velocity u m/sec and constant or uniform acceleration a m/sec2. If v is the velocity of the particle at time t, then its average velocity during the interval of time t is given by and distance travelled in t time is  t. MOTION OF A TRAIN BETWEEN TWO STATIONS When a train starts from station A to station B on a non-stop journey, its speed can have the patterns. When at station A, it is at rest. Then it moves and gains velocity. As the distance increases the velocity also increases and at one point of time it reaches the maximum, which is maintained for same distance i.e. the velocity remains constant and then the velocity starts decreasing and finally becomes zero, when the train reaches to station B. Example 1: A body has velocity of 15 m/sec at a certain instant and 10 seconds later it has a velocity 45 m/sec. If the velocity changes uniformly, find the space described. Solution: We have, u = initial velocity of the particle = 15 m/sec a= acceleration of the particle v = final velocity of the particle = 45 m/sec the particle takes 10 seconds to acquire the velocity of 45 m/sec so v = u + at ⇒ 45 = 15 + 10 a ⇒ a = 3 m/sec2 Let the space described be s meter, then v2 = u2 + 2as so, (45)2 = (15)2 + 2  3  s ⇒ s = 300 meters. Example 2: A particle starting from rest moves with an acceleration of 4 cm/sec2. Find in what time it acquire a velocity of 60 cm/s. Also, find the distance described in this time. Solution: It is given that, u = initial velocity of the particle = 0 a = acceleration of the particle = 4 cm/s2 v = final velocity of the particle = 60 cm/sec Suppose the particle takes t seconds to acquire the velocity of 60 cm/s, then v = u + at ⇒ 60 = 0 + 4t ⇒ t = 15 seconds Let s be the distance described by the particle in 15 seconds. Then s = ut + at2 =  4  (15)2 = 450 cm. Thus, the particle takes 15 seconds to acquire a velocity of 60 cm/s and distance described in this time is 450 cm. Example 3: A body starts from rest and at the end of 10 seconds, it is moving at the rate of 22 m/sec. 5 seconds later, its velocity is 105.6 km/hr. Is the acceleration constant? Solution: Let the acceleration of the body be constant and equal to f m/sec2. Since the body starts from rest and acquired a velocity of 22 m/sec at the end of 10 seconds. Therefore, v = u + at ⇒ 22 = 0 + a  10 ⇒ a = 2.2 m/sec2 Let v be the velocity acquired by the body in the next 5 seconds, then v = 22 + 2.2  5 = 33 m/sec Clearly, this velocity is not equal to 105.6 km/hr or m/sec. Hence, the acceleration is not constant. Example 4: A point travels 7 meters in the first second, 11 meters in the third second and 17 meters in the sixth second. Is the acceleration uniform? If yes, find it. Solution: Let u be the initial velocity and a m/sec2 be the acceleration of the particle. Since the particle traverses 7 meters in first second, 11 meters in the third second and 17 meters in the sixth second, therefore 7 = u + a(2  1 –1) ⇒ 14 = 2u + a ……(1) 11 = u + a(2  3 –1) ⇒ 22 = 2u + 5a ……(2) and 17 = u + a(6  2 –1) ⇒ 34 = 2u + 11a ……(3) from equation (1), (2) and (3) u = 6 m/sec, a = 2 m/sec2 Values of u and f satisfy equation (3). Hence, the particle moves with uniform acceleration of 2 m/sec2. Example 5: A body covers half of its journey with a speed of a m/sec and the other half with a speed of b m/sec. Calculate the average speed of the body during the whole of journey. Solution: Let t1 and t2 seconds be the time taken by the body to cover the first half and the other half respectively. Let s meters be the total distance, then t1 = and t2 = average speed = = . Example 6: A car travelling with a uniform acceleration has a velocity 18 km/hr at the instant of observation and 54 km/hr after covering a distance of 400 meters. How much further must it travel to attain a velocity of 72 km/hr. Solution: Let the uniform acceleration of the car be a km/hr, we have u = initial velocity of the car = 18 km/hr v = final velocity of the car = 54 km/hr s = distance covered = 400 meters = 0.4 km v2 = u2 + 2as ⇒ (54)2 = (18)2 + 2a  0.4 a = 3240 km/hr2 Thus the acceleration of the car is 3240 km/hr2 Suppose the car travels s km to attain a speed of 72 km/hr. Then (72)2 = (54)2 + 2  3240  s ⇒ s = 350 meters. MOTION IN A STRAIGHT LINE UNDER GRAVITY All bodies which are allowed to fall from a height at a given place in vacuum have the same constant acceleration. This acceleration is due to the gravitational attraction of the body by the Earth towards its centre. It is denoted by ‘g’. In ‘SI' system the value of g is 9.8 m/sec2. Equations for Vertical Motion: Case I: When a body is projected vertically downwards with initial velocity u then the equation describing motions are v = u + gt ……(1) h = ut + gt2 ……(2) v2 = u2 + 2gh ……(3) hn = distance covered in nth second = u + g(2n –1) Case II: When body is falling down freely, then v = gt ……(1) h = gt2 ……(2) v2 = 2gh ……(3) hn = g(2n –1) ……(4) Case III: When body is projected vertically upward with initial velocity u, then v = u –gt ……(1) h = ut – gt2 ……(2) v2 = u2 –2gh ……(3) hn = u – g(2n –1) ……(4) Greatest height attained = ……(5) Time to the greatest height = ……(6) Time to a given height h =  . Notes: ● Time from any point on the path to the highest point is same as the time from the highest point to the given point when the body is returning. MOTION OF A BODY RELEASED FROM A BALLOON OR A LIFT we will be studying the vertical motion of a body when it is released from an ascending balloon or ascending or descending lift, which is moving with uniform velocity or acceleration. In such cases the initial velocity of the body is same as the velocity of the moving object from which the body is released. The subsequent motion of the body will be the motion under gravity. Some Important Results: Case I: When a lift is ascending with uniform acceleration of a m/sec2 and after t seconds a body is dropped from it. Then at the time when the body is dropped. (i) Initial velocity of the body is same as that of the lift and is in the same direction, so, the velocity of the body is at m/sec. (ii) Initial velocity of the body relative to the lift = velocity of the body – velocity of the lift = at – at = 0. (iii) Acceleration of the body = g m/sec2 in downward direction (iv) Acceleration of the lift = a m/sec2 in upward direction (v) Acceleration of the body relative to the lift = Acceleration of the body – Acceleration of the lift = g – (–a) = a + g in downward direction. Case II: When a lift is ascending with uniform acceleration of a m/sec2 and t seconds a body is thrown vertically upwards with velocity v m/sec. Then at that time (i) Initial velocity of the body = v + velocity of the lift = v + at in upward direction (ii) Initial velocity of the body relative to the lift = velocity of the body – velocity of lift = (v + at) –at = v m/sec. (iii) Acceleration of the body relative to the lift in vertically upward direction is (a + g) m/sec2 Case III: When a lift is descending with uniform acceleration a m/sec2 and after time t a body is dropped from it. Then at that time (i) Velocity of the body = velocity of the lift = a t m/sec in downward direction (ii) Initial velocity of the body relative to the lift = 0 (iii) Acceleration of the body relative to the lift in downward direction = acceleration of the body – acceleration of lift = (g – a) m/sec2. Example 7: From what height did a body fall if it dropped 75 m during the last second of the fall (g = 10 m/sec2). Solution: Let the body fall for t seconds, then Distance covered in tth second = g(2t –1) =  10  (2t –1) = 75 ⇒ t = 8 sec. Thus, body falls for 8 seconds. Let h be the height from which the body falls. Then h = gt2 =  10  (8)2 = 320 meters. Hence the body falls from a height of 320 meters. Example 8: A body is thrown vertically upward and rises to a height of 10 meters. Calculate (i) the velocity with which the body was thrown upwards and (ii) the time taken by the body to reach the highest point. Solution: Let the body be thrown upwards with a velocity of 4 m/sec. Thus greatest height attained = meters. (i) We have, greatest height = 10 meters ⇒ = 10 ⇒ u2 = 196 ⇒ u = 14 m/sec Hence, the required velocity of projection = 14 m/sec (ii) Time taken by the body to reach the greatest height = sec = = sec. Example 9: The greatest height attained by a particle projected vertically upwards is 19.6 meters. Find how soon after projection the particle will be at a height of 14.7 meter. Solution: Let u be the velocity of projection since at the greatest height, the velocity of the particle is zero. 02 = u2 –2g  19.6 ⇒ u2 = 19.6  19.6 ⇒ u = 19.6 m/sec Suppose after time t the particle is at a height of 14.7 meter, then 14.7 = 19.6t –  9.8  t2 ⇒ t2 –4t+ 3 = 0 ⇒ (t –1) (t –3) = 0 ⇒ t = 1, 3 Thus, the particle will be at a height of 14.7 meter after 1 sec of its projection. It will also attain the same height after 3-seconds while coming downward after attaining the maximum height. PROJECTILES BASIC TERMS When a particle is projected into the air, ● then the particle is called a projectile. ● the point from which the particle is projected is called the point of projection. ● the angle that the direction in which it is projected makes with the horizontal plane through the point of projection is called the angle of projection; ● the path which the particle describes is called its trajectory; ● the distance between the point of projection and the point where the path meets any plane drawn through the point of projection is its range on the plane; ● the time that elapses before it again meets the horizontal plane through the point of projection is called the time of flight. ● the initial velocity with which the projectile is projected is called the velocity of projection. ● the maximum height reached by a projectile during its motion is called greatest height. LIST OF FORMULAE ● Velocity and direction of motion after time ‘t’ v2 = u2 – 2ugt sin + g2t2 and tan = ● Velocity and direction of motion at height ‘h’ v2 = u2 – 2gh and tan  = ● Greatest height attained by a projectile h = ● Time for the greatest height = ● Time of flight t = ● Range = ● Maximum Range = Here u : initial velocity of the body  : Angle of projection  : Angle which the velocity of the projectile at a certain point makes with the horizontal (It is the tangent to the path of the projectile at the point) g : acceleration due to gravity Note: ● The path of a projectile is a parabola, given by y = x tan  i.e. with vertex ≡ latus rectum = and height of directrix = ● The velocity at any point is equal in magnitude, to that which would be acquired by a particle in falling freely from the directrix to that point. ● There are two direction for a given range and velocity and also two times for a given height. Range and Time of Flight on an Inclined Plane: If from a point on a plane which is inclined at an angle  to the horizon, a particle is projected with a velocity u, at an angle  with the horizontal, in a plane passing through the normal to the inclined plane and the line of greatest slope, then ● Time of flight, T= ● Range = ● Maximum range up the plane = Note: ● Range, time of flight and maximum range down the plane is obtained by putting – for . ● For a given range and given velocity there are two different directions of projection which are equally inclined to the direction of projection for maximum range. Example 10: If the maximum range of a particle is R, show that the greatest height attained is . A body can throw a ball 60 meters. How long is the ball in the air, and what height does it attain. Solution: Let the particle be projected with a velocity v at an angle  with the horizontal. Horizontal range = Greatest height = The range is maximum where sin2 = 1 ⇒  = 45 ⇒ R = ⇒ H = Now R = 60 metres ⇒ h = 15 metres Now time t of flight = = ● t2 = ⇒ t = 3.5 sec aprox. RELATIVE VELOCITY When two points P and Q move in the same plane, the rate of change of position of P with respect to Q (i.e. as seen from Q) is defined as the relative velocity of P with respect to Q and it is denoted by . DETERMINATION OF RELATIVE VELOCITY If two points P and Q are moving with velocities u and v respectively along two straight lines intersecting at an angle  then vPQ = ….(1) and tan = ….(2) The relations (1) and (2) give the magnitude and direction of vPQ. Notes: ● If the points P and Q be moving in the same direction with velocity u and v ( < u) then vPQ = u – v ( in the direction of P). ● If the points P and Q be moving in the opposite directions, then vPQ = u + v ( in the direction of P) ● In case the two points P, Q are moving with velocities u, v at angles ,  with x–axis, and v be the velocity of P and relative to Q inclined at an angle  to the x–axis then v cos = u cos – v cos v sin = u sin – v sin Example 11: A boat is sailing due east with uniform velocity of 15 km/m and another boat is sailing due south with uniform velocity of 20 km/m. Find the magnitude and direction of the velocity of the second relative to the first. Solution: v1 = 15 km/m; v2 = 20 km/m ;  = v21 = = = 25 km/m tan = i.e  = tan–1 . Example 12: A watchman on a light house observes two ships A and B. A sailing north at 20 km/hr and B sailing north–east at 20 km/hr. In what direction and with what speed does the ship B appear to be sailing to a person on the ship A. Solution: u = 20, v = 20 ,  = vBA = = 20 km/hr. tan = – = – 1  =135 i.e at 20 km/hr sailing south-east. Determination of true velocity when the apparent velocity is given When the relative velocity of a point Q with respect to an observer P is given, as also the velocity of the observer P, the true velocity of Q is obtained by compounding these two given velocities. ⇒ i.e vQ = where  is the angle between and tan = . Example 13: To a man walking due east at the rate of 4 km/hr, the wind appears to blow from the North East at 4 km/hr. Find the true velocity of the wind. Solution: u = 4, v = 4 ,  = 135 u = velocity of observer v = apparent velocity of wind w = true velocity of wind w = = = 4 km/hr. tan = i.e.  = Hence the wind is actually blowing from north at 4 km/hr. SHORTEST DISTANCE BETWEEN TWO POINTS IN MOTION The shortest distance between them can be obtained by taking the perpendicular distance of one of the points from the relative path of the second with respect to the first. Example 14: At a particular instant two aeroplanes are at a distance of 250 kilometre, one due east of the other. The first one is moving towards west at the rate of 100 km/hr and the second with velocity of 75 km/hr towards the south. Find the time when they are nearest to each other and the least distance between them. Solution: AB = 250 km v1 = 100 km/hr, v2 = 75 km/hr is respect by v21= =125 km/hr From ABM, AM = (AM) sin = 250  (From BP′R, sin = ) i.e least distance = 150 km also BM = (AB) cos = 250  = 200 km ⇒ time = hr i.e. 1 hour 36 minutes. OBJECTIVE ASSIGNMENT 1. Resultant of two forces each equal to P and inclined at an angle of 120° is (A) 2P (B) (C) P (D) None of these Solution : Using Sesultant Hence (C) is the correct answer. 2. The angle between two equal forces acting on a particle when the square of their resultant is three times their product is : (A) 90° (B) 60° (C) 120° (D) None of these Solution : Let each force be P and  is the angle between them. Then, Also, R2 = 3P × P , 150° ⇒  = 60° or 300° But 300° also gives 60°. Hence (B) is the correct answer. 3. Two forces P and Q are such that P : Q = 3 : 4. If their resultant is inclined at an angle 30° to P, then angle between R and Q is : (A) (B) (C) 60° (D) None of these Solution : If  is the angle between R and Q, then resolving R along P and Q, we get and But (given) Hence (B) is the correct answer. 4. Two forces P + Q, P – Q inclined at 120° with each other are such that their resultant makes an angle 30° with the bisector then P + Q : P – Q is : (A) 3 : 1 (B) 1 : 1 (C) 2 : 1 (D) 4 : 1 Solution : From OAC, by sine formula, Hence (C) is the correct answer. 5. Forces 7, 5 and 3 acting on a particle are in equilibrium. The angle between the pair of forces 5 and 3 is : (A) 120° (B) 90° (C) 60° (D) 30° Solution : Since R2 = P2 + Q2 + 2PQ cos ∴ (7)2 = (5)2 + (3)2 + 2.5.3 cos ⇒  = 60° Hence (C) is the correct answer. 6. If forces of 12, 5 and 13 units weight balance at a point, two of them are inclined at : (A) 30° (B) 45° (C) 60° (D) 90° Solution : Since (13)2 = (12)2 + (5)2, forces of 12 units and 5 units are perpendicular. Hence (D) is the correct answer. 7. The sum of the two forces is 18 and their resultant perpendicular to the lesser of the forces is 12, then the lesser force is (A) 5 (B) 3 (C) 7 (D) 15 Solution : Let the two forces be P, Q (P < Q).  P + Q = 18. Let  be the angle between the resultant and Q. Since the resultant is perpendicular to P. . . Also 12 = Q cos  ( Resultant is 12) . Hence (A) is the correct answer. 8. Forces of magnitudes 5, 10, 15 and 20 act on a particle in the directions of North, East and West respectively. The magnitude of their resultant is (A) (B) 10 (C) (D) Solution : R cos  = 15 cos 0 + 5 cos 90 + 20 cos 180 + 10 cos 270 = – 5 . . . (1) R sin  = 15 sin 0 + 5 sin 90 + 20 sin 180 + 10 sin 270 = – 5 . . . (2) . Hence (C) is the correct answer. 9. The resultant R of two forces P and Q act at right angles to P. Then the angle between the forces is (A) (B) (C) (D) Solution : Hence (B) is the correct answer. 10. The moment of a force about a point is (A) (B) (C) (D) Solution : Vector moment  Moment = . Hence (B) is the correct answer. 11. When a particle be kept at rest under the action of the following forces (A) (B) (C) (D) Solution : Since 8 N + 5 N = 13 N  13 N ↓ is equal and opposite to the resultant of 8 N ↑ and 5 N ↑. Hence (A) is the correct answer. 12. Three forces P, Q, R act along the sides BC, CA, AB of a ABC, then (A) (B) P cos A + Q cos B + R cos C = 0 (C) P + Q + R = 0 (D) aP + bQ + cR = 0 Solution : Since distances of the incentre from the sides are equal  P + Q + R = 0. Hence (C) is the correct answer. 13. A body moving with a speed of 36 km/hr is brought to rest in 10 seconds. What is the negative acceleration? (A) 2 m/sec2 (B) 3 m/sec2 (C) 1 m/sec2 (D) None of these Solution: u = 36 km/hr = 10 m/sec, (v = 0 final velocity) v = u –ft ⇒ u = ft ⇒ f = = 1 m/sec2 Hence (C) is correct answer. 14. O is the circumcentre of ABC. If forces P, Q and R acting along OA, OB, OC are in equilibrium, then P : Q : R is (A) (B) (C) (D) Solution :  P : Q : R = 2 sin A cos A : 2 sin B cos B : 2 sin C cos C = a cos A : b cos B : c cos C (By sine formula). Hence (C) is the correct answer. 15. Two like parallel forces P and 3P act on a rigid body at point A and B respectively. If the forces are interchanged in position, the resultant will be displaced through the distance of (A) (B) (C) (D) Solution: Case I: If the resultant act at C, then AC = Case II: If the resultant acts at D, then AD = ⇒ CD = AC – AD = . 16. The least velocity with which a cricket ball can be thrown 20 m horizontally is (A) 28 m/sec (B) 14 m/sec (C) 21 m/sec (D) 42 m/sec Solution: Let the required velocity be u m/sec. Then, Maximum horizontal range = 20 m ⇒ ⇒ u2 = 20  9.8 ⇒ u = 14 m/sec 17. If T be the time of flight, R be the horizontal range and  the angle of projection of a particle, then tan  = (A) (B) (C) (D) Solution: we have T = and R = ⇒ ⇒ tan  = 18. The maximum horizontal range of a projectile when the velocity of projection is 28 m/sec is (A) 40 m (B) 80 m (C) 100 m (D) none of these Solution: Maximum horizontal range = = 80 m Hence (B) is the correct answer. 19. The horizontal range of a projectile is 4 times its maximum height. The angle of projection is (A) 300 (B) 450 (C) 600 (D) none of these Solution: Given that ⇒ tan  = ⇒  = 300. Hence (A) is the correct answer. 20. If r and r′ be the maximum ranges up and down the inclined plane respectively and R be maximum range on horizontal plane, then r, R, r′ are in (A) A.P (B) G.P (C) H.P (D) none of these Solution: Consider (1 + sin  + 1 – sin ) ⇒ ⇒ are in A.P. ⇒ r, R, r′ are in H.P. Hence (C) is the correct answer. 21. A shot fired from a gun on top of a tower, 272 feet high hits the ground at a distance of 4352 feet in 17 seconds. The velocity and direction of projection are (A) 256, 300 (B) 256 , 450 (C) 256 , 600 (D) none of these Solution: Given that time t = 17 second, the shot goes down 272 ft and describes horizontal distance 4352 ft. ⇒ 272 = –u sin  . t + gt2 ⇒ 17 u sin  = 16  17(17 –1) ⇒ u sin  = 256 and 4352 = u cos  t = 17 u cos  ⇒ u cos  = 256 ⇒ u = 256 and tan  = 1 ⇒  = 450. 22. A particle, starting with initial velocity of 50 m/sec, moves in a straight line with a uniform acceleration of 40 m/sec2. Then velocity of the particle after 4 seconds (A) 210 m/sec (B) 110 m/sec (C) 150 m/sec (D) 160 m/sec Solution: v = u + ft ⇒ v = (50 + 40  4) = 210 m/sec Hence (A) is correct answer. 23. A body moving with a speed of 36 km/hr is brought to rest in 10 seconds. What is the negative acceleration and distance travelled by the body before coming to rest (A) 60 m (B) 70 m (C) 50 m (D) 40 m Solution: u = 36 km/hr = 10 m/sec, (v = 0 final velocity) v = u –ft ⇒ u = ft ⇒ f = = 1 m/sec2 so v2 = u2 –2fs ⇒ 0 = (10)2 –2  1  s ⇒ s = 50 meter. Hence (C) is correct answer. 24. A body whose velocity is being constantly accelerated has at certain instant a velocity of 22 m/sec and in the following minute, it travels a distance of 10320 m. The acceleration is: (A) 4 m/sec2 (B) 5 m/sec2 (C) 6 m/sec2 (D) 3 m/sec2 Solution: s = ut + ft2 ⇒ f = = 5 m/sec2 Hence (B) is correct answer. 25. A body has an initial velocity of 200 cm/sec and is subject to a retardation of 2 cm/sec2. At what time will its velocity be zero and how far will it then travel. (A) 10 sec, 100 meter (B) 100 sec, 100 meter (C) 20 sec, 200 meter (D) 200 sec, 2000 meter Solution: v = u –ft = 0 ⇒ 0 = 200 –2t ⇒ t = 100 sec And v2 = u2 –2fs ⇒ (200× 10–2)2 = 2  2 10–2 × s s = 100 meters. Hence (B) is correct answer. 26. A particle describes 650 meters in 10 seconds and its velocity at the end of that time is 80 m/sec. Find the initial velocity and acceleration (A) 50 m/sec, 3 m/sec2 (B) 10 m/sec, 30 m/sec2 (C) 10 m/sec, 5 m/sec2 (D) 5 m/sec, 10 m/sec2 Solution: 650 = 10u +  f  102 ⇒ u + 5f = 65 ……(1) and 80 = u + 10f ……(2) so from equation (1) and (2) f = 3 m/sec2 and u = 50 m/sec Hence (A) is correct answer. 27. The velocity of a train increases from 15 km/hr to 60 km/hr while it moves through a distance of km. If the acceleration is uniform, find its magnitude (A) m/sec2 (B) m/sec2 (C) m/sec2 (D) m/sec2 Solution: u = 15 km/hr = m/sec, v = 60 km/hr = m/sec s = km = 125 m so v2 = u2 + 2fs ⇒ + 2  f  125 ⇒ f = m/sec2. Hence (C) is correct answer. 28. Find the distance described during the tenth second by a particle having initial velocity of 60 m/sec and moving with a retardation of 2 m/sec2. Find the time in which it comes to rest. (A) 31 m, 20 sec (B) 41 m, 30 sec (C) 21 m, 20 sec (D) 51 m, 20 sec Solution: Sn = u – f(2t –1) = 60 –  2  (2  10 –1) = 60 –19 = 41 meters And v = u –ft ⇒ 0 = 60 –2  t ⇒ t = 30 seconds Hence (B) is correct answer. 29. A body moving with a uniform acceleration describes 55 m in the sixth second from rest. How much distance will it move in the 8th second (A) 55 meters (B) 65 meters (C) 75 meters (D) 85 meters Solution: S = u + f(2t –1) ⇒ 55 = 0 +  f (12 –1) ⇒ f = 10 m/sec2 Distance = u + f (2t –1) = 0 +  10  (2  8 –1) = 75 meters. Hence (C) is correct answer. 30. Resultant velocity of two velocities 30 km/hr and 60 km/hr making an angle 60° with each other is : (A) 90 km/hr (B) 30 km/hr (C) (D) None of these Solution : units. Hence (C) is the correct answer. 31. A body of mass 20 kg is pulled along a smooth horizontal table by a constant force. It describes 18 m from rest in 3 sec. The magnitude of force is : (A) 20 N (B) 40 N (C) 80 N (D) 120 N Solution : u = 0, s = 18 m, t = 3 sec . Also, m = 20 kg.  . Hence (C) is the correct answer. 32. A body is projected vertically upwards from a tower of height 192 ft. If it strikes the ground in 6 seconds, then the velocity with which the body is projected is : (A) 64'/sec (B) 32'/sec (C) 16'/sec (D) None of these Solution : Using – s , we get . Hence (A) is the correct answer. 33. A stone is thrown vertically upward with a velocity of 96 m/sec. The time interval between the two instances when it is 80 m from the ground is : (A) 18.33 sec (B) 17.46 sec (C) 0.87 sec (D) None of these Solution : u = 96 m/sec, s = 80 m, g = 10 m/sec2 Using , we get 5t2 – 96t + 80 = 0 ⇒ t = 18.33, 0.87.  Required time interval is 18.33 – 0.87 = 17.46 sec. Hence (B) is the correct answer. 34. After a ball has been falling under gravity for 5 seconds, it passes through a pane of glass and looses half of its velocity, and now reaches the ground in 1 second. The height of the glass above the ground is : (A) 2900 cm (B) 2543 cm (C) 2943 cm (D) None of these Solution : v = gt = 5g After passing through the glass pane, its velocity becomes .  = 3 (981) = 2943 cm. Hence (C) is the correct answer. 35. If you want to kick a football to the maximum distance, the angle at which it should be kicked is (assuming no resistance) (A) 45° (B) 90° (C) 30° (D) 60° Solution : For maximum range, angle of projection is 45. Hence (A) is the correct answer. 36. If h1 and h2 are the greatest heights for the two paths of a projectile with a given horizontal range R, then (A) (B) (C) (D) Solution : For a given horizontal range, there are always two angles of projection, namely  and .  and . Hence (A) is the correct answer. 37. A point moves with uniform acceleration and v1, v2, v3 denote the average velocities in three successive intervals at time t1, t2, t3, then (A) (B) (C) (D) Solution : Since average velocity = velocity at the middle of the time interval.  . and v2 – v3 = . Hence (B) is the correct answer. 38. A particle is dropped under gravity from rest from a height h(g = 9.8 m/sec2) and then it travels a distance 9h/25 in the last second. The height h is : (A) 100 m (B) 122.5 m (C) 145 m (D) 167.5 m Solution : and Reject , we get t = 5. . Hence (B) is the correct answer. 39. A dyne is the force which can produce an acceleration of 1 cm per sec when acting on a mass of : (A) 1 mg (B) 1 dg (C) 1 g (D) 1 kg Solution : 1 g. Hence (C) is the correct answer. 40. A bullet of mass 0.006 kg travelling at 120 metres/sec penetrates deeply into a fixed target and is brought to rest in 0.01 sec. The distance through which it penetrates the target is : (A) 3 cm (B) 6 cm (C) 30 cm (D) 60 cm Solution : u = 120 m/sec, v = 0, t = 0.01 Using v = u + ft ⇒ Using v2 – u2 = 2fs, we get 0 – (120)2 = – 2 (120) (100 s) . Hence (D) is the correct answer. 41. A cricket ball of mass 200 grams moving with a velocity of 20 metres/sec is brought to rest by a player in 0.1 sec. The average force applied by the player is : (A) 4 × 103 dynes (B) 4 × 104 dynes (C) 4 × 105 dynes (D) 4 × 106 dynes Solution : u = 20 m/sec, v = 0, t = 0.1 sec. Using v = u + ft ⇒ f = – 200 m/sec2.  retardation = 200 m/sec2 = 200  100 = 4  106 dynes.  P = mf = 200  200  100 = 4  106 dynes. Hence (D) is the correct answer. 42. The position at time of a particle moving along x-axis given by the relation x = t3 – 9t2 + 24t + 6, where x denotes the distance in metre from the origin. The velocity v of the particle at the instant when the acceleration becomes zero, is given by (A) v = 3 (B) v = –3 (C) v = 0 (D) v = 6 Solution : x = t3 – 9t2 + 24t + 6  f = 0 ⇒ t = 3  v = 3 (3)2 – 18 (3) + 24 = – 3. Hence (B) is the correct answer. 43. If the greatest height attained by a projectile is one quarter of its range on the horizontal plane, then the angle of projection is : (A) (B) (C) (D) Solution : . Hence (A) is the correct answer. 44. A man can throw a stone to a maximum horizontal distance of 36 metres. Then the maximum height (in metres) to which it may rise is : (A) 9 (B) 12 (C) 15 (D) 18 Solution : Maximum horizontal range  Max height . Hence (A) is the correct answer. 45. The courses of two cars A and B are due north and they have same speed of 2o km/hr. At noon, B is due north of A and 10 km away. The time when they are nearest is (A) 15 min (B) 20 min (C) 25 min (D) None of these. Solution : Let A, B be the positions of the cars at noon. After t, let their positions be C and D respectively.  AC = BD = 20 t  CD2 = CB2 + BD2 = (10 – 20t)2 + (20t)2 ( AB = 10) Differentiating w.r.t. t,  CD is least when min. Hence (A) is the correct answer. 46. The angle of projection of a particle when its range on a horizontal plane is times the greatest height attained by it is (A) 15 (B) 30 (C) 45 (D) 60. Solution : Range = and greatest height  By given condition, . Hence (B) is the correct answer. 47. If the position of the resultant of two like parallel forces P and Q is unaltered, when the positions of P and Q are interchanged, then (A) P = Q (B) P = 2Q (C) 2P = Q (D) none of these Solution: We have AC = Since the position of the resultant does not alter ⇒ AC = AD ⇒ P = Q. 48. Two unlike parallel forces P and Q act at points 5 metre apart. If the resultant force is 9N and acts at a distance of 10 metre from the greater force P, then (A) P = 16N, Q = 7N (B) P = 15N, Q = 7N (C) P = 27N, Q = 18N (D) P = 18N, Q = 9N Solution: The resultant of P and Q acts at C. Therefore, AC = ⇒ 10 = ⇒ Q = 18 Putting Q = 18 in P – Q = 9, we get P = 27. 49. If R and R′ are the resultant of two forces ( P > Q) according as they are like or unlike such that R : R′ = 25 : 7, then P : Q = (A) 2 : 1 (B) 3 : 4 (C) 4 : 3 (D) 1: 2 Solution: We have, and ⇒ and ⇒ ⇒ = ⇒ ⇒ . 50. Three like parallel forces P, Q, R act at the corner points of a triangle ABC. Their resultant passes through the circumcentre if (A) (B) P = Q = R (C) P + Q + R = 0 (D) none of these Solution: Since the resultant passes through the circumcentre of ABC, therefore, the algebraic sum of the moments about it is zero. Hence, P + Q + R = 0 51. Forces forming a couple are of magnitude 12N each and the arm of the couple is 8 m. The force of an equivalent couple whose arm is 6m is of magnitude (A) 8N (B) 16 N (C) 12N (D) 4N Solution: Since the two couples are equivalent. Therefore, their moments are equal. Hence, 128 = F6 ⇒ F = 16N. 52. The two like parallel forces, acting at a distance of 3 m apart which is equivalent to a force 12 N acting at a distance of 0.5 m from one of the forces is (A) 10 N, 2 N (B) 8 N, 3 N (C) 4 N, 8 N (D) 12 N, 3 N Solution: Let the like parallel forces be P and Q act at A and B respectively. Let the resultant 12 N act at C where AC = 0.5 m. ⇒ P =  2.5 = 10 N Q =  0.5 = 2 N. 53. Three forces of magnitude 30, 60 and P acting at a point are in equilibrium if the angle between the first two is 60, the value of P is (A) 30 (B) 30 (C) 20 (D) 25 Solution: Since the forces are in equilibrium i.e. one of the forces must be equal in magnitude but opposite in direction to the resultant of the other two forces . The angle between the first two forces is 60 and their resultant is of magnitude = . 54. If the resultant of two forces of magnitude P and 2P is perpendicular to P, then the angle between the forces is (A) (B) (C) (D) Solution: Let the angle between the forces P and 2P be . Since the resultant of P and 2P is perpendicular to P. Therefore tan ⇒ P + 2P cos = 0 ⇒  = . 55. Forces M and N acting at a point O make an angle 150. Their resultant acts at O, has magnitude 2 units and is perpendicular at to M. Then, in the same unit, the magnitudes of M and N are (A) 2 , 4 (B) (C) 3, 4 (D) 4, 5 Solution: We have 22 = M2 + N2 + 2MN cos150 …..(1) and tan ⇒ M + N cos150 = 0 ….(2) Solving (1) and (2), we get M = and N = 4. 56. The resultant R of two forces P and Q act at right angles to P. Then the angle between the forces is (A) cos–1 (B) cos–1 (C) sin–1 (D) sin–1 Solution: Let  be the angle between the forces P and Q. ⇒ P + Q cos = 0 ⇒  = cos–1 . 57. The maximum resultant of two forces is P and the minimum resultant is Q, the two forces are at right angles, the resultant is (A) P + Q (B) P – Q (C) (D) Solution: Let the magnitude of the given forces be X and Y. Then X + Y = P and X – Y = Q Let R be the resultant of P and Q , then R = = . 58. Three forces P, Q, R are acting at a point in a plane. The angle between P and Q, Q and R are 1500 and1200 respectively, then for equilibrium, forces P, Q, R are in the ratio (A) 1 : 2 : 3 (B) 1 : 2 : (C) 3 : 2 : 1 (D) : 2 : 1 Solution: Forces are as marked in the figure. By Lami’s Theorem, we have or, or, ∴ P : Q : R = : 2 : 1. 59. If the resultant of two forces acting on a particle be at right angles to one of them and its magnitude be one third of the other. The ratio of the larger forces to the smaller is (A) 5 : 2 (B) 2 : 2 (C) 3 : 2 (D) none of these Solution: Let  be the angle between two forces P and Q and their resultant be , perpendicular to P. Then = P2 + Q2 + 2PQ cos  ……(1) and tan 900 = ⇒ P = –Q cos  ……(2) From (1) and (2) = P2 + Q2 + 2P(–P) ⇒ ⇒ P : Q = 2 : 3 Hence the ratio of longer to smaller is 3 : 2 . 60. If , where R is the resultant of two forces P and Q acting at a point, then the angle between R and P is (A) (B) (C) (D) none of these Solution: Let = k Let the angle between P and Q is  P2 + Q2 + 2PQ cos  ⇒ (4k)2 = (2k)2 + (3k)2 + 2.2k.3k cos  ⇒ 3k2 = 12k2 cos  ⇒ cos  = Let R makes angle  with P tan  = ⇒  = .