4. WORK, ENERGY AND POWER

Transcription

1 4. WORK, ENERGY AND POWER Work is said to be done by a force when the force produces a displacement in the body on which it acts in any direction except perpendicular to the direction of the force. Work is the product of the applied force and the displacement of the body in the direction of force. W = F s cos= F.s * If F F ˆ ˆ ˆ x i Fy j Fzk and s xi ˆ yj ˆ zkˆ Then W = F.s = Fxx Fy y Fz z Dimension M L T UNIT S I : joule C.G.S. : erg joule = 0 7 erg NATURE OF WORK DONE Positie work : W = Fs cos If the angle is acute ( < 90 ) then the work is said to be posit ie. T he positie work signifies that the external force faours the mot ion of the body. * When a body falls freely under the action of graity ( = 0 ), the work done by graity is positie. * When a spring is stretched, stretching force and the displacement both are in the same direction. So work done by stretching force is positie. < 90 positie work F s motion Negatie work : * If the angle is obtuse ( > 90 ). Then the work is said to be negatie. It signifies that the direction of force is such that it opposes the motion of the body. * Work done by frictional force is negatie when it opposes the motion. * Work done by breaking force on the car is negatie. > 90 F negatie work motion > 90 s Zero work W = Fs cos Work done will be zero if F = 0 or s = 0 = 90 = 90 work done is zero F motion s * Work done by coolie is zero (against graity). * In circular motion work done by centripetal force is zero. * When = 0 : A force does maximum positie work. * When = 80 force does maximum negatie work. F O W=0 =90 WORK DONE BY A VARIABLE FORCE When magnitude and direction of the force aries with position, the work done by force for infinitesimal displacement ds is dw = F.ds The total work done for displacement from A to B is W B F.ds B (F cos )ds AB A A

2 In terms of rectangular components F F ˆi F ˆj F kˆ, ds dxi ˆ dyj ˆ dzkˆ x y z ds F B B W (F ˆi F ˆj F k).(dxi ˆ ˆ dyj ˆ dzk) ˆ AB x y z A WORK DONE BY SEVERAL FORCES xb yb zb F dx F dy F dz xa x y z y A za When seeral forces acts on a body then the net work done on the body is the algebric sum of work done by indiidual force. W net = F.s F.s...F n.sn here s s,,... s n are the displacement of points of application of forces F, F,... F respectiely. n If a graph is constructed of the component F cos of a ariable force, then the work done by the force can be determined by measuring the area between the cure and the displacement axis. GOLDEN KEY POINTS * Work is defined for an interal or displacement there is no term like instantaneous work similar to instantaneous elocit y. * For a particular displacement work is independent of time, work will be same for same displacement whether the time taken is small or large. * When seeral forces acts, work by a force for a particular displacement is independent of other forces. * Displacement depends on referance frame so work done by a force is referance frame dependent so work done by a force can be different in different reference frame. * Effect of work is change in kinetic energy (K.E.) * Work is done by the source or agent that applies the force. Ex. A force F = ( x) N acts on a particle in x direction, where x is in metre. Find the work done by this force during a displacement from x = 0 to x =. Sol. Work done for small displacement dx is dw F. dx = ( x) dx A Total work done W = x (0 0.5x)dx = x0 0.5x 0x = 0.5 0( 0) ( 0) = 0 Ex. Calculate the amount of workdone in raising a glass of water weighing 0.5 kg through a height of 0 cm. (g = 0 m/s ) Sol. Workdone W = Force distance = mgh = = J Ex. A body of mass 0 kg is at rest. A force of 5N is applied on it. Calculate the work done in first second. Sol. Displacement in the direction of force is d = at = F t m (s ut at ) so, work done is W = F t m = 5 5 J 0 8 Ex. Sol. A spring of force constant 00 N/m is stretched upto 5 cm. Find out work done. Spring is streched by length = 5 cm = 0.05 m Work done for small displacment dx is dw = Fdx = kx dx Total work done for length is W Fdx kx dx k 0 0 = 00(.05) 0.5 J

3 POWER Power of a body is defined as the rate at which the body can do the work. work Power time or W P t W is the amount of the work done in total time t. This power P is also called aerage power. INSTANTANEOUS POWER (P) Instantaneous power is the power at any gien instant. Suppose an agent does an infinitesimally amount of work dw in an infinitesimally time dt then, dw P dt or F.ds P ( dw F.ds) dt or ds P F. dt or Pt F. Dimensions : [M L T 3 ] W (from P ) t SI UNIT : joule watt second Bigger UNITS of power : kilo watt (kw) = 000 watt = 0 3 W maga watt = 000,000 watt or MW = 0 3 kw horse power (h.p.) = 746 W = 550 ft. lb/s FPS UNIT : (foot pound)/second, W = ft. lb/s A conenient rule of thumb approximation is hp ~ 4 3 kw Note : * When an agent deliers power at a uniform rate, the aerage power is equal to the instantaneous power. * Power is the ratio of two scalars (Also' it is the scalar product of two ectors) so power is scalar quantity. * When the time taken to complete a gien amount of work is important we measure the power of agent doing work. * The slope of work time graph gies the instantaneous power slope = tan= dw dt = P (instantaneous power) * Area under power time graph gies the work done. Area under P t graph = Area ABCD = (PR PQ) =P t dt Area under power time graph = Pdt W Efficiency : Machines are designed to conert energy into some of the useful work howeer, because of frictional effects, the work performed by the machine is always less than the energy put into mechine. The efficiency of a machine work done energy input P A D R S dt P Q C B t Ex. Sol. The force required to tow a boat at constant elocity is proportional to the speed. If a speed of 4.0 km/h requires 7.5 kw, how much power does a speed of km/h require? Let the force be F =, where is speed and is a constant of proportionality. the power required is P = F = Let P be the power required for speed and P be the power required for speed. 3

4 P = 7.5 kw and = 3, P P P = (3) 7.5 kw = 67.5 kw. Ex. Sol. A kg airplane takes off at a speed of 50 m/s, and 5 min later it is at an eleation of 3 km and has a speed of 00 m/s. What aerage power is required during this 5 min, if 40 percent of the power is used in oercoming dissipatie forces? Energy supplied to plane in 300 s (5 min) is (PE + KE) at altitude KE ground = ( ) + (30000) (00) (30000) (50) = = MJ For Aerage Power 0.6 P = MJ 300 s so = 5.55 MW Ex. In unloading grain from the hold of a ship, an eleator lifts the grain through a distance of m. Grain is discharged at the top of the eleator at a rate of.0 kg each second and the discharge speed of each grain particle is 3.0 m/s. Find the minimum-horsepower motor that can eleate grain in this way. Sol. The work done by the motor each second, i.e. power = mgh + m here the mass of grain discharged (and lifted ) is one second. m = =.0 kg, = 3.0 m/s, and h = m power = 49 W = 0.33 hp The motor must hae an output of at least 0.33 hp CONSERVATIVE AND NON CONSERVATIVE FORCE CONSERVATIVE FORCE A force is said to be conseratie if work done by the force on a particle moing between two points does not depends on the path taken by the particle. Example s * Graitational force, not only due to earth but in its general form as gien by the uniersal law of graitation is a conseratie force. * Elastic force in a stretched or compressed spring is a conseratie force. * Electrostatic force bet ween two stationary electric charges is a conseratie force. * Magnetic force between two magnetic poles is a conseratie force. NONCONSERVATIVE FORCE A force is said to be non conseratie, if work done or against the force in moing a body from one position to another, depends on the path. Example * The elocity dependent forces such as air resistance iscous force magnetic force *Retarding forces such as friction force and drag force. 4

5 Difference bet ween Conserat ie and non-conseratie forces : Conseratie force Non-conseratie force * Work done does not depend upon path * Work done depends upon path * Work done in a round trip is zero * Work done in a round trip is not zero * When only a conseratie force acts within * Work done against a non-conseratie a system, the kinetic energy and potential energy force may be dissipated as heat energy. can change. Howeer their sum, the mechanical energy of the system does not change. * Work done is completely recoerable. * Work done is not completely recoerable.g GOLDEN KEY POINTS * Forces acting along the line joining the centres of two bodies are called central forces. Graitational and Electrostatic forces are central forces. * All central forces are conseratie but all conseratie forces are not central force s. * The concept of potential energy exists only in the case of conseratie force. Ex. An eleator weighing 500 kg is to be lifted up at a constant speed of 0.4m/s what should be the minimum horse power of the motor to be used. Sol. P = F = (mg) = = 960 W or 960 P =.6 H.P. 746 Ex. A car of mass 000 kg is lifted up a distance of 30 m by a crane in minute. A second crane does the same job in minutes. Do crane consume same or different amounts of fuels? What is the power supply by each crane? Sol. M = 000 kg and h = 30m Work done in lifting car upto height h is Mgh = = J From the principle of conserat ion of energy, both the crane will consume fuel equialent to J. Two crane consume same amount of fuel. For first crane power P W t 60 5 = 9800W For second crane power P W t 0 5 = 4900W Ex. A tube-well pump out 400 kg of water per minute. If water is coming out with a speed of 3 m/s. What is power of the pump? How much work is done if pump runs for 0 hour. Sol. Mass of water pumped out per sec is m = = 40 kg/s, elocity of water = 3m/s Ex. kinetic energy of water coming out per second = Power of pump = 80 W m = Work done = Power time = = J 40 (3) = 80 J A pump can take out 700 kg of water per hour from a well 00 m deep. Calculate the power of the pump, assuming that its efficiency is 50%. (g = 0 m/s ) Sol. Output power = mgh t = = 000 W Efficiency output power input power 5

6 Input power = output power = = 4 kw Ex. A truck of mass 0, 000 kg moes up an inclined plane rising in 50 with a speed of 36 km/hr. Find the power of the engine (g = 0 m/s ) Sol. Force against which work is done F = mg sin = 0, = 000 N speed = = 0 m/s so P = = 0 kw ENERGY Energy is defined as the internal capacity for doing work. When we say that a body has energy it mean that it can do work. * Different forms of energy : Mechanical energy, electrical energy, optical (light) energy, acoustical (sound), molecular, atomic and necular energy. These forms of energy can change from one form to the other. * Mass energy relation : According to Einstein mass energy equialence principle mass and energy are inter conertible i.e. they can be changed into each other. Equialent energy corresponding to mass m is E = mc where, m : mass of the particle c : speed of light In mechanics we only concerned with mechanical energy which is of two type. (a) kinetic energy (b) Potential energy * Energy is a scalar quantity * Dimensions : [M L T ] * S I UNIT : joule Other units erg = 0 7 joule kwh = joule ev = joule cal = 4. joule Kinetic energy : Definitions : () Kinetic energy is the internal capacity of doing work of the object by irtue of its motion. () The K.E. of a body is the energy possesed by the body by irtue of its motion. * K.E. of a body can be calculated by the amount of work done in stopping the moing body. or from the amount of work done in giing the present elocity to the body from the state of rest. * Kinetic energy is a scalar quantity that is associated with state of motion of an object. * If a particle of mass m is moing with elocity '' much less than the elocity of the light then the kinetic energy K.E. is gien by K.E. = m * As mass m and or. are always positie so K.E. can neer be negatie. * The kinetic energy depends on the frame of reference. * Work done on the body is the measure of K.E. of the body. * The expression K.E. = m holds een when the force applied aries in magnitude or in direction or in both. 6

7 * Relation between K.E. (K) and linear momentum (p) : p = m and K = m = m (m ) = p m K = * Graphs p m or p mk K = constant p = constant K p K m = constant m m p K K K p p p Work Energy theorem According to this theorem work done by net force on a body is equal to change in its kinetic energy. W = KE or W m m If 'dw' be the small amount of work done in giing an infinitesimally displacement dw = F.ds If force and displacement hae same direction, then dw = Fds cos0 = Fds d s to the body, then d dw = (ma)ds = m ds = md dt If W is the work done in increasing the elocity from to then W md m d m m m m ( ) W = K K W = KE W = change in K.E. of body. GOLDEN KEY POINTS * If there is no change in the speed of a particle, there is no change in kinetic energy. So work done by the resultant force is zero. * If K.E. of the body decreases then work done is negatie i.e. the force opposes the motion of the body. * If K.E. of the body increases then work done is positie * In aboe discussion we hae assumed that the work done by the force is effectie only in changing the kinetic energy of the body. It should howeer be remembered that work done on a body may also be stored in the body in the form of potential energy. 7

8 Ex. Kinetic energy of a par ticle is increased by 300%. Find the percentage increase in momentum. Sol. Kinetic energye = m momentum p = m When E is increased by300% E ' = E + 3E = 4E = ( m )4 = m If ' is elocity of body, then m ' = m ' = So p' = m' = m hence, percentage change in momentum = m m 00 = 00% m Ex. Sol. A bullet weighing 0 g is fired with a elocity 800 m/s. After passing through a mud wall m thick, its elocity decreases to 00 m/s. Find the aerage resistance offered by the mud wall. Work done by the aerage resistance offered by the wall = change in K.E. of the bullet. F s = m mu F Resistance offered = 350 N m( u ) 0.0( ) s = 350 N Ex.3 The displacement x of a particle moing in one dimension, under the action of constant force is related to time t by equation t x 3 where x is in metre and t in sec. Calculate : (a) The displacement of the particle when its elocity is zero; (b) The work done by the force in the first 6 sec. dx Sol. As t x 3 i.e., x = (t 3) (t 3)...(i) dt (a) will be zero when (t 3) = 0 i.e., t = 3 x = (3 3) = 0 i.e., when elocity is zero, displacement is also zero. (b) From eqn. (i), () t = 0 = (0 3) = 6 and () t = 6 = (6 3) = 6 So by Work - Energy Theorem W K m[ F I ] m[6 ( 6) ] 0 Potential Energy (P.E.) Definition :- Potential energy is the internal capacity of doing work of a system by irtue of its configuration. Definition :- The energy which a body has by irtue of its position or configuration in a conseratie force field. In case of conseratie forces. dw = du = F.dr U r du U r Wheneer and whereer possible, we take the reference point at and assume potential energy to be zero at. r If r = and U = 0then U F.dr F.dr * Potential energy may be +e or -e depending upon nature of field. For attraction = -e For repulsion = +e U for repulsion * Relation between F and U :- F U = gradient U r for attraction where ˆ i ˆ j kˆ x y z 8

9 U ˆ U ˆ U F i j k ˆ x y z U x U y U z = Par tial differentiation of U w.r.t. x (keeping y, z constant) = Partial differentiation of U w.r.t. y (x, z constant) = Partial differentiation of U w.r.t. z (x, y constant) POTENTIAL ENERGY IN DIFFERENT CONSERVATIVE FIELDS (a) Graitational field :- for large distance i.e. P.E. at infinity = 0 potential energy at a point whose distance with center of earth is r for at point P r > R (R = Radius of earth) P.E. = GMm r (b) Electrostatic force field :- Potential energy at infinity = 0 P.E. = 4 0 qq r r separation between charge particle Note : Take q q with sign q q (c) Spring force field :- Reference point is taken at its natural length. If x is expa ns io n/ compre ssion i n spr in g t hen P.E. of spri ng ( Energy stored i n spri ng) is r U = Kx and spring force F = - Kx [directed towards natural Length] Kx Natural length x Kx x x expansion/compression from natural length (d) Intermolecular force :- (i) Reference is taken as infinit y where potential energy is zero. (ii) If r = seperation between molecules. a b U = 6 r r a and b are constant. (iii)in shown here graph at r = r 0, P.E. = min. so particle is in stable equilibrium for r < r 0 Net Intermolecular force repulsie r > r 0 Net Intermolecular force attractie r = r 0 Net Intermolecular force 0 U r = r 0 r 9

10 EQUILIBRIUM Body is said to be in equilibrium, when seeral forces act on a body simultaneously in such a way that the resultant force on the body is zero, i.e., F 0 with F F As if a ector is zero all its components must anish, i.e., in equilibrium as F F 0 Fx 0 Fy 0 and Fz 0 i i i.e., if a body is in translatory equilibrium it will be either at rest or in uniform motion. If it is rest, the equilibrium is called static, otherwise dynamic. Static equilibrium can be diided into following three type s. Stable equilibrium : If on slight displacement from equilibrium position a body has tendency to regain its original position, it is said to be in stable equilibrium. In case of stable equilibrium potential energy is centre of graity is lowest. Fig. shows some example of stable equilibrium. M D M D M D M D Unstable Equilibrium : If on slight displacement from equilibrium position body moes in the direction of displacement, the equilibrium is said to be unstable. In this situation potential energy of the body is maximum (d U/dr = negatie) and so centre of grait y is highe st, Fig. depicts some examples of ustable equilibrium. Neutral equilibrium : If on slight displacement from equilibrium position a body has no tendency to come back to original position or to moe in the direction of displacement, it is said to be in neutral equilibrium. In this situation potential energy of the body is constant (d U/dr = 0) and so centre of graity remains at constant height. Fig. depicts some examples of neutral equilibrium. In case of stable equilibrium lesser the potential energy or lower the centre of graity, i.e., greater the base area and more stable is the equilibrium. In following Fig. situation (A) has highest stability. 0

11 (A) (B) If we plot graphs between F ersus r and U ersus r, at equilibrium F will be zero while U will be optimum (max. or min. or constant). If U = min. i.e., d U dr = positie, equilibrium is stable (C) U = max. i.e., d U dr = negatie, equilibrium is unstable U = const. i.e., d U dr = zero, equilibrium is neutral GOLDEN KEY POINTS * Potential energy is a relatie quantit y. * Potential energy is defined only for conseratie force field. * Potential energy of a body at any position in a conseratie force field is defined as a work done by an external agent against the action of conseratie force in order to shift it form reference point (PE = 0) to the present position. * Potential energy of a body in a conseratie force field is equal to the work done by the body in moing from its pre sent position to reference position. * At reference position, the potential energy of the body is zero. * Potential energy can be positie or negatie or zero. * Potential energy depends on frame of reference but change in potential energy is independent of reference frame. * Potential energy should be considered to be a property of the entire system, rather than assigning it to any specific particle. * It's a function of position does not depends on the path. Ex. Sol. A uniform rod of length 4m and mass 0 kg is lying horizontal on the ground. The work done in keeping it ertical will be? As the rod is kept ertical position the shift in the centre of graity is equal to the half length = so Work done W = mgh = mg = = 39 J Ex. Sol. Ex. Sol. A particle is moing in a potential region gien by U = K(x + y + z ). Find out the force acting on the particle. ˆU ˆU ˆ U F i j k x y z U x kx, U ky y F k(xi ˆ yj ˆ zk) ˆ, U kz z F k(xi ˆ yj ˆ zk)n ˆ A uniform chain of length L and mass M is held on a smooth table with one fourth of its length hanging oer the edge. Find the work required to pull the hanging part of the chain? Weight of the hanging part of the chain = Mg 4

12 This weight acts at the centre of graity of the hanging part which is at a distance L 8 below the edge of the table work done W = Mg 4 L 8 = MgL 3 MECHANICAL ENERGY Sum of kinetic and potential energy is called mechanical energy. E = K + U * Mechanical energy is scalar quantity and measured in Joule. * Mechanical energy of a body depends upon frame of reference. * Negatie alue of Mechanical energy indicates a bound state e.g. electron in an atom, and a satellite reoling around a planet both are in bound state haing negatie mechanical energy. LAW OF CONSERVATION OF ENERGY According to law of conseration of energy, energy may be transformed from one form to another but it can not be created or destroyed i.e. the total energy of a system always remains constant. LAW OF CONSERVATION OF MECHANICAL ENERGY For conseratie forces the sum of kinetic and potential energie s at any point remains constant through out the motion. It does not depend upon time, and position this is known as the law of conseration of mechanical energy. (K.E. f + P.E. f ) = (K.E. i + P.E. i ) LAW OF CONSERVATION OF LINEAR MOMENTUM According to Newton's Second law of motion the rate of change of momentum is equal to the applied force. if F = dp dt F = 0 then we hae dp = 0 i.e. p = constantnt dt This leads to the law of conseration of momentum which is" In the absence of external forces, the total momentum of the system is consered." * For an isolated system, the initial momentum of the system is equal to the final momentum of the system. If the system consists of n bodies haing momentum p,p,p 3,...p n, then p p p... p = constant 3 n * As linear momentum depends on frame of reference, obserers in different frames would find different alues of linear momentum of a gien system but each would agree that his own alue of linear momentum doe s not change with time proided the system is isolated and closed, i.e., law of conseration of linear momentum is independent of frame of reference though linear momentum depends on frame of reference. * Conseration of linear momentum is equialent to Newton's III law of motion for a system of two particles in absence of external force by law of conseration of linear momentum. p p = constant i.e., m m = constant Differentiating aboe with respect to time d d m m 0 dt dt [as m is constant]

13 m a m a 0 or F F 0 or d [as a ] dt [as F ma or F F i.e., for eery action there is equal and opposite reaction which is Newton's III law of motion. * This law is uniersal, i.e., it applies to body macroscopic as well as as microscopic systems. COLLISION OF BODIES A collision is said to take place when either two bodies physically collide against each other or when the path of one body is changed by the influence of the other body. As a result of collision, the momentum and kinetic energy of the interacting bodies change. Forces inoled in a collision are action-reaction forces, i.e., the internal force s of the system. The total momentum remains consered in any type of collision. Type s of col lision according to t he direct ion o f col lision : (a) Head on collision : direction of elocities of bodies is similar to the direction of collision. u u A B B A m Before collision m m After collision m (b) Oblique collision : direction of elocities of bodies is not similar to the direction of collision. * Types of collision according to the conseration law of kinetic energy : (a) Elastic collision : kinetic energy is consered. KE before collision = KE after collision (b) Inelastic collision : kinetic energy is not consered. Some energy is lost in collision KE before collision > KE after collision (c) Perfect inelastic collision : Two bodies stick together after the collision. momentum remains consered in all types of collisions. Coefficient of rest itut ion (e) The coefficient of re stitution is defined as the ratio of the impulses of recoery and deformation of either body. impulse of recoery e = impulse of deformation elocity of separation along line of impact e elocity of approach along line of impact Value of e is for elastic collision, 0 for perfectly inelastic collision, 0 < e < for inelastic collision. HEA D ON ELASTIC COLLISION The elastic collision in which the colliding bodies moe along the same straight line path before and after the collision. 3

14 u u A B A B A B m m Before collision Collision m After collision m Assuming initial direction of motion to be positie and u > u (so that collision may take place) and applying law of conseration of linear momentum, we get m u + m u = m + m no force : they collide due difference of their elocities only i.e., m (u ) = m ( u )... () For collision is elastic, kinetic enegy before collision must be equal to KE after collision, i.e., mu m u m m... () or m (u ) m ( u ) Diiding equation () by () u + = + u or (u u ) = ( )...(3) In -D elastic collision 'elocit y of approach' before collision is equal to the 'elocit y of recession' after collision, no matter what the masses of the colliding particles be. This law is called Newton's law for elastic collision Now if we multiply equation (3) by m and substracting it from () (m m ) u + m u = (m + m ) or m m u m u m m m m...(4) Similarly multiplying equation (3) by m and adding it to equation () m u + (m m )u = (m + m ) m m u m u m m m m... (5) SPECIAL CASES (a) If the two bodies are of equal masses : m = m = m = u and = u Thus, if two bodies of equal masses undergo elastic collision in one dimension, then after the collision, the bodies will exchange their elocitie s. (b) If two bodies are of equal masses and second body is at rest. m = m and initial elocity of second body u = 0 = 0, = u When body A collides against body B of equal mass at rest, the body A comes to rest and the body B moes on with the elocity of the body A. In this case transfer of energy is hundred percent eg. Billiard's Ball, Nuclear moderation. (c) If the mass of a body is negligible as compared to other. m >> m and u = 0 = u = u When a heay body A collides against a light body B at rest, the body A should keep on moing with same elocity and the body B will moe with elocity double that of A. if m >> m and u = 0 then = 0 4

15 When light body A collides against a heay body B at rest, the body A should start moing with same elocity just in opposite direction while the body B should practically remains at rest. Ex. Two ball bearing mass 5kg each is moing in opposite directions with equal speed 5m/s. collides head on with each other. Find out the final elocities of the balls if collision is elastic Sol. Here m = m = 5kg u = 5 m/s u = 5 m/s In such type of condition elocity get interchange so = u = 5 m/s = u = 5 m/s Ex. A ball of 0. kg makes an elastic head on collision with a ball of unknown mass that is initially at rest. If the 0.kg ball rebounds at one third of its original speed. What is the mass of other ball Sol. Here m = 0. kg m =? u = 0 u = u = u/3 As m m u m u m m m m u 0. m 3 0. m ( )u m = 0. kg HEA D ON INELASTIC COLLISION OF TWO PARTICLES ` Let the coefficient of restitution for collision is e () Momentum is consered m u + m u = m + m...() () Kinetic energy is not consered. (3) According to Newton's law By soling eq. () and () u u = e...() = (m em )u m m + m ( e)u m m, = (m em )u m m + m ( e)u m m PERFECT INELASTIC COLLISION In case of inelastic collision, after collision two bodies moe with same elocity (or stick together) If two particles of masses m and m, moing with elocities u and u (u < u ) respectiely along the same line collide 'head on ' and after collision they hae same common elocity, then by conseration of linear momentum, m u + m u = m + m or m u m u (m m )...() Now as the KE of the system before collision is K m u m u i And after collision is K f (m m ) Loss in KE during collision K K i K f [ mu m u ] (m m )...() Substituting the alue of from eq. (), (m u m u ) K [(m u m u ) ] (m m ) or mm (u u uu ) K [ ] (m m ) m m K (u u ) (m m ) if the target is initially at restu = 0 and u = u 5

16 mm K u (m m ) or K m K (m m ) [K = m u ] Now if target is massie, i.e.,, m >> m K in percentage loss in KE = 00% K i.e., if a light moing body strikes a heay target at rest and sticks to it, practically all its KE is lost. EL ASTIC COLLISION IN TWO DIMENSION : * Two perfectly elastic bodies A and B of masses m and m mo i ng along t he same straight li ne (x-a xis) with elocities u and u. After the collision, the two bo die s A and B tr ael w it h elocit ie s a nd along direction making angles and with the incident direction * Collision is perfectly elastic, so according to conseration of K.E. x' m m A B u y' y u m A B m x mu m u m m. * Momentum of the two bodies separately consered along x-axis and y-axis The component of momentum of body A. After collision along x-axis = m cos The component of moment of body B. After collision along x-axis = m cos * Applying the law of conseration of momentum along x-axis m u + m u = m cos + m cos * The component of momentum of body A after collision along y-axis = m sin * The component of momentum of body B after collision along y-axis = m sin * Applying the conseration of linear momentum along y-axis = m sin ( m sin or m sin = m sin GOLDEN KEY POINTS * Momentum remains consered in all t ype s of collisions. * Total energy remains consered in all types of collisions. * Only conseratie force s works in elastic collisions. * In inelastic collisions all the forces are not conseratie. Ex. Sol. A simple pendulum of length m has a wooden bob of mass kg. It is struck by a bullet of mass 0 kg moing with a speed of X 0 m/s. The bullet gets embeded into the bob. Obtain the height to which the bob rises before swinging back. Applying principle of conseration of linear momentum mu = (M + m) 0 ( 0 ) = ( +.0).0 K.E. of the block with bullet in it, is conerted into P.E. as it rises through a height h (M m) (M m)gh = gh m u M h (M + m) Ex. h = g h = = 0. m A body falling on the ground from a height of 0m, rebounds to a height.5m calculate (i) The percentage loss in K.E. (ii) Ratio of the elocities of the body just before and just after the collision. 6

17 Sol. let and be the elocity of the body just before and after the collision K = m = mgh...() and K = m mgh...() diiding we get h 0 = 4 h.5 Ex. Sol. percentage loss in K.E. = e h e t e g e = mg(h h ) 00 = = 75% mgh 0 A body strikes obliquely with another identical stationary rest body elastically. Proe that they will moe perpendicular to each other after collision. Conseration of linear momentum in X-direction gies mu = m cos + m cos m m u = cos + cos... () Conseration of linear momentum in Y-direction gies m u x 0 = m sin m sin 0 = sin sin...() Conseration of kinetic energy gies mu m m...(3) () + () gies Before collision After collision u + 0 = cos cos cos cos + sin sin sin sin u = (cos sin ) (cos + sin ) + (cos cos sin sin u + cos ( ) u = + } cos ( ) = 0 + = 90 Ex. A ball is dropped freely from height 'h'. This ball is continuously rebounding, then find out : (i) elocity of ball after 'n' rebound (ii) height attained by the ball after 'n' rebound (iii) time taken by the ball in n th rebound (i) total distance coered by the ball before it stops rebounding. Sol. (i) Velocity after 'n' rebound : h h h up to No. of collision 4 5 By Newton formula ( ) = e (u u ) = gh here = 0, u = 0 (surface at rest) = e (opposite direction) = u = e... () = e... () = e(e)... (3) [() & ()] 7

18 = e similarly 3 = e 3, 4 = e 4... n = e n n = e n gh (ii) Height attained by the ball after 'n' rebound : = e gh = e gh h = e h = e gh = e gh h = e 4 h Similarly h n = e n h (iii)time taken in n th rebound : h = e h gt = e gt t = e t t = et t = e h g... () h = e 4 h gt = e 4 ( gt ) t = e 4 t, t = e t t = e h g... () Similarly t n = e n h g t n = e n t Special Case :- Total time in 'n' rebound (T) T = t + t + t +... T = t + et + e t + e 3 t +... T = t + t (e + e + e ) T = t + t e e = e h e t e g e (i) Distance coered by the ball before it stops : S = h + h + h = h + e h + e 4 h + e 6 h +... = h + e h( + e + e 4 + e ) S = h + e h e = h e e, S = h 8 e e

19 STD. XII TIME : Prof. SAMEER UNIA S PHYSICS TUTORIALS TOPIC : WORK, ENERGY AND POWER - TUTORIAL SHEET - I STUDENT NAME : PHYSICS DATE :. A bomb of mass m = kg thrown ertically upwards with a speed u = 00 m/s explodes into two parts after t = 5s. A fragment of mass m = 400 g moes downwards with a speed = 5 m/s, then speed and direction of another mass m will be () 40 m/s downwards () 40 m/s upwards (3) 60 m/s upwards (4) 00 m/s upwards. A suspended simple pendulum of length is making an angle with the ertical. On releasing, its elocity at lowest point will be () g( cos ) () g sin (3) g( cos ) (4) g 3. A kg stationary bomb is exploded in three parts haing mass ratio : : 3. Parts haing same mass moe in perpendicular directions with elocity 30 m/s, then the elocity of bigger part will be () 0 m/s () 0 m/s (3) 5 m/s (4) 5 m/s 4. A body of mass 5 kg has momentum of 0 kg m/s. When a force of 0. N is applied on it for 0 seconds, what is the change in its kinetic energy ().J ().J (3) 3.3J (4) 4.4J 5. Which of the following is true () Momentum is consered in all collisions but kinetic energy is consered only in inelastic collision () Neither momentum nor kinetic energy is consered in inelastic collisions. (3) Momentum is consered in all collisions but not kinetic energy (4) Both momentum and kinetic energy are consered in all collisions. 6. A.0 H.P. motor pumps out water from a well of depth 0m and fills a water tank of olume 38 liters at a height of 0 m from the ground. The running time of the motor to fill the empty water tank is (g = 0ms ) () 5 minutes () 0 minutes (3) 5 minutes (4) 0 minutes

20 7. A mass of 0 kg moing with a speed of 0 m/s collides with another stationary mass of 5 kg. As a result of the collision, the two masses stick together. The kinetic energy of the composite mass will be () 600 Joule () 800 Joule (3) 000 Joule (4) 00 Joule 8. The work done against graity in taking 0 kg. mass at m height in s will be () 49 J () 98 J (3) 96 J (4) None of these 9. A force of 0N displaces an object by 0m. If work done is 50J then direction of force make an angle with direction of displacement () 0 () 90 (3) 60 (4) None of these 0. A heay body moing with a elocity 0 ms and another small object at rest undergo an elastic collision. The latter will moe with a elocity of () 0 m/s. () 40 m/s. (3) 60 m/s. (4) Zero. A body of mass 5 kg moing with a elocity 0 m/s collides with another body of the mass 0 kg at rest and comes to rest. Velocity of the second body due to the collision is ().5m/s () 5m/s (3) 7.5m/s (4) 0m/s. The kinetic energy of a body becomes four times its initial alue. The new linear momentum will be:- () Four times the initial alue () Thrice the initial alue (3) Twice the initial alue (4) Same as the initial alue 3. If a spring extends by x on loading then energy stored by the spring is :- (T is tension in the spring) () T x () T k (3) k T (4) T k 4. A bullet of mass m and elocity is fired into a large block of wood of mass M. The final elocity of the system is () + m m + M () F m + MI HG M K J F m + I (3) HG M K J F (4) HG m m + M I K J 5. A nucleus of mass number A originally at rest, emits alpha particle with speed. The recoil speed of the daughter nucleus is () 4 A 4 () 4 A 4 (3) A 4 (4) A

21 6. Water is falling on the blades of a turbine at a rate of 00 kg/s from a certain spring. If the height of the spring be 00 metres, the power transferred to the turbine will be () 00 kw () 0 kw (30 kw (4) 000 kw 7. A body A experiences perfectly elastic collision with a stationary body B. If after collision the bodies fly apart in the opposite direction with equal elocities, the mass ratio of A and B is () () 3 (3) 4 (4) 5 8. A big ball of mass M, moing with elocity u strikes a small ball of mass m, which is at rest. Finally small ball attains elocity u and big ball. What is the alue of () M m M u () m M m u (3) m M m (4) M M m 9. A body of mass kg falls from a height of 0 m. What is the loss in potential energy :- () 400 J () 300 J (3) 00 J (4) 00 J 0. A collision is said to be perfectly inelastic when :- () Coefficient of restitution = 0 () Coefficient of restitution = (3) Coefficient of restitution = (4) Coefficient of restitution <. A heay nucleus at rest breaks into two fragments which fly off with elocities 8 :. The ratio of radii of the fragments is () : () : 4 (3) 4 : (4) :. A particle falls from a height h upon a fixed horizontal plane and rebounds. If e is the coefficient of restitution the total distance traelled before rebounding has stopped is F HG () h e e I KJ F HG () h e e I KJ F HG (3) h e e I F (4) KJ h e HG e I KJ 3. A body of mass 6 kg under a force which causes displacement in it gien S = t 4 metres where t is time. The work done by the force in seconds is () J () 9J (3) 6J (4) 3J...

22 A B 4. If the potential energy of two molecules is gie by, U r r 6 then at equilibrium position, its potential energy is equal to () A 4B () B 4A (3) B A (4) 3A 5. A man m = 80 kg is standing on a trolley of mass 30 kg on a smooth surface. If man starts walking on trolley along rails at a speed of ms, then after 4 sec, his displacement relatie to ground is () 4 m () 4.8 m (3) 3. m (4) 6 m 6. A stationary particle explodes into two particles of masses m and m which moe in opposite directions with elocities and. The ratio of their kinetic energies E /E is () m /m () m /m (3) (4) m /m 7. A neutron makes a head on elastic collision with a stationary deuteron. The fractional energy loss of the neutron in the collision is () 6/8 () 8/9 (3) 8/7 (4) /3 8. If the force applied is F and the elocity gained is, then the power deeloped is () F () F (3) F (4) F 9. Two particles of mass M A and M B and there elocities are V A and V B respectiely collides. After collision they inter changes their elocities then ratio of M A M B is () V V A B () V V B A (3) (VA V B ) (V V ) B A (4) 30. A body of mass 4m at rest explodes into three pieces. Two of the pieces each of mass m moe with a speed each in mutually perpendicular directions. The total kinetic energy released is () m () m (3) 3 m 5 m (4) 3. If the kinetic energy of a body is double of its initial kinetic energy, then the momentum of the body will be () times () times (3) times (4) none of these...

23 F 5i 4 j N 3. A force acts ˆ ˆ on a body and produced a displacement ˆ ˆ ˆ S 6i 5 j 3k m. The work done will be () 30 J () 40 J (3) 0 J (4) 0 J 33. A mass of 0.5 kg moing with a speed of.5 m/s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k=50n/m. The maximum compression of the spring would be () 0. m ().5 m (3) 0.5 m (4) 0.5 m 34. A ball of mass kg and another of mass 4 kg are dropped together from a 60 feet tall building. After a fall of 30 feet each towards earth, their respectie kinetic energies will be in the ratio of () : 4 () : (3) : (4) : 35. A bomb of mass 3.0 kg explodes in air into two pieces of masses.0 kg and.0 kg. The smaller mass goe s at a speed of 80 m/s. The total energy impar ted to the t wo fragments is ().07 kj ().4 kj (3).4 kj (4) 4.8 kj 36. A bomb of mass 30Kg at rest explodes into two pieces of masses 8 kg and kg. The elocity of 8kg mass is 6ms. The kinetic energy of the other mass is () 54 J () 56 J (3) 486J (4) 34J 37. A force F acting on an object aries with distance x as shown here. The force is in N and x in m. The work done by the force in m oing the object from x = 0 to x = 6m is () 8.0 J () 3.5 J (3) 4.5 J (4) 9.0 J 38. A block of mass 0 kg. is moing in x direction with a constant speed of 0m/sec. It is subjected to a force F = 0 x joules/meter during its trael from x=0 meters to x=30 meters. Its final kinetic energy will be () 475 joules () 450 joules (3) 75 joules (4) 50 joule s 39. Two bodies of mass kg and 4kg hae equal K.E. then the ratio of their momentum is () : () : (3) 4 : (4) :

24 40. A stone of mass m is tied to a string of length at one end and by holding second end it is wh irled i nto a hor izontal circle, t hen work done w il l be () 0 () F HG I m (3) (mg) (4) KJ F HG m I KJ 4. A body of mass 3 kg is under a constant force which causes a displacement s in metres in it, gien by the relation s = 3 t, where t is in seconds. Work done by the force in seconds is () 5 9 J () 3 8 J (3) 8 3 J (4) 9 5 J J o f work is do ne in s lid in g a kg blo ck up an i ncli n ed pla n e o f height 0 m. Taki ng g = 0 m/s, work done against friction is () 00 J () 00 J (3) Zero (4) 000 J 43. For inelast ic collision bet ween t wo spherical rigid bodie s () the total kinetic energy is consered () the total potential energy is consered (3) the linear momentum is not consered (4) the linear momentum is consered 44. Two identical balls, one moes with m/s and second is at rest, collides elastically. After collision elocity of second and first ball will be () 6m/s, 6m/s () m/s, m/s (3) m/s, 0m/s (4) 0m/s, m/s 45. A body of mass m haing an initial elocity makes head on collision with a stationary body of mass M. After the collision, the body of mass m comes to rest and only the body haing mass M moes. This will happen only when () m >> M () m << M (3) m = M (4) m = M 46. In stretching a spring by cm energy stored is gien by U, then stretching by 0 cm energy stored will be () U () 5U (3) U 5 (4) 5U 47. A ertical spring with force constant k is fixed on a table. A ball of mass m at a height h aboe the free upper end of the spring falls ertically on the spring, so that the spring is compressed by a distance d. The net work done in the process is () mg(h d) kd () mg(h d) kd (3) mg(h d) kd (4) mg(h d) kd..4..

25 48. If the kinetic energy is increased by 300%, the momentum will increase by () 00% () 00% (3) 50% (4) 300% 49. If the kinetic energy of a body increases by 4% the momentum () increases by % () increases by 4% (3) increases by 8% (4) increases by 6% 50. An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, kg first part moing with a elocity of ms and kg second part moing with a elocity of 8 ms. If the third part flies off with a elocity of 4ms, its mass would be () 3kg () 5 kg (3) 7 kg (4) 7 kg 5. A block of mass M is attached to the lower end of a ertical spring. The spring is hung from a ceiling and has force constant alue k. The mass is released from rest with the spring initially unstretched. the maximum extension produced in the length of the spring will be () Mg/k () Mg/k (3) Mg/k (4) 4 Mg/k 5. An engine pumps water continuously through a hose. Water leaes the hose with a elocity and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water () m () m3 (3) m 3 (4) m 53. A body of mass kg is thrown upwards with a elocity 0 m/s. It momentarily comes to rest after attaining a height of 8 m. How much energy is lost due to air friction? (g = 0 m/s ) () 0 J () 0 J (3) 30 J (4) 40 J 54. An engine pumps water through a hose pipe. Water passes through the pipe and leaes it with a elocity of m/s. The mass per unit length of water in the pipe is 00 kg/m. What is the power of the engine? () 800 W () 400 W (3) 00 W (4) 00 W 55. A ball moing with elocity m/s collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their elocities (in m/s) after collision will be () 0, () 0, (3), (4), The potential energy of a system increase s if work is done () Upon the system by a nonconseratie force () By the system against a conseratie force (3) By the system against a nonconseratie force (4) Upon the system by a conseratie force..5..

26 57. A body projected er tically from the earth reaches a height equal to ear th's radius before returning to the ear th. The power exer ted by the graitational force is greate st () At the highest position of the body () At the instant just before the body hits the earth (3) It remains constant all through (4) At the instant just after the body is projected 58. Force F on a particle moing in a straight line aries with distance d as shown in the figure. The work done on the particle during its displacement of m is F(N) () 8 J () J (3) 6 J (4) 3 J d(m) A 59. The potential energy of a particle in a force field is : U = r B r where A and B are positie constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the par ticle is () A/B () B/A (3) B/A (4) A/B 60. Two spheres A and B of masses m and m respectiely collide. A is at rest initially and B is moing with elocity along x-axis. After collision B has a elocity in a direction perpendicular to the original direction. The mass A moe s after collision in the direction. () = tan (/) to the x-axis () = tan ( /) to the x-axis (3) same as that of B (4) opposite to that of B 6. A car of mass m starts from rest and accelerates so that the instantaneous power deliered to the car has a constant magnitude P 0. The instantaneous elocity of this car is proportional to () t / () t/ m (3) t P 0 (4) t / 6. A stone is dropped from a height h. It hits the ground with a certain momentum P. If the same stone is dropped from a height 00% more than the preious height, the momentum when it hits the ground will change by () 00 % () 00 % (3) 68% (4) 4%..6..

27 STD. XII TIME : Prof. SAMEER UNIA S PHYSICS TUTORIALS TOPIC : WORK, ENERGY AND POWER - TUTORIAL SHEET - II STUDENT NAME : PHYSICS DATE :. Which of the following statements is true for work done by conseratie forces () It does not depend on path () It is equal to the difference of final and initial energy function (3) It can be recoered completely (4) All of the aboe. Which of the following statement is incorrect for a conseratie field? () Work done in going from initial to final position is equal to change in kinetic energy of the particle. () Work done depends on path but not on initial and final positions. (3) Work done does not depend on path but depends only on initial and final positions (4) Work done on a particle in the field for a round trip is zero. 3. As shown in the diagram a particle is to be carried from point A to C ia paths (I), (II) and (III) in graitational field, then which of the following statements is correct () Work done is same for all the paths () Work done is minimum for path (II) (3) Work done is maximum for path (I) (4) None of the aboe 4. The relation bet ween conseratie force and potential energy U is gien by () F = du () z dx F = U dx (3) F = du (4) F = du dx dx 5. Which of the following is a non conseratie force () Electric force () Graitational force (3) Spring force (4) Viscous force 6. A constant force F is applied to a body of mass m moing with initial elocity u. If after the body undergoes a displacement S its elocity becomes, then the total work done is () m[ + u ] () m [u + ] (3) m [ u ] (4) m[ u ] 7. The law of conseration of momentum is based on Newton s () First law of motion () Second law of motion (3) Third law of motion (4) Law of graitation..7..

28 8. A ball is dropped from a height of 0m. If 40% of its energy is lost on collision with the earth then after collision the ball will rebound to a height of () 0 m () 8 m (3) 4 m (4) 6 m 9. A ball strikes the floor and after collision rebounds back. In this state () Momentum of the ball is consered () Mechanical energy of the ball is consered (3) Momentum of ball earth system is consered (4) The kinetic energy of ball ear th system is not consered 0. If the alues of force and length are increased four times then the unit of energy will increase by () 4 times () times (3) 8 times (4) 6 times. A bullet of mass P is fired with elocity Q in a large body of mass R. The final elocity of the system will be () R P R () PQ P R (3) ( P Q) R (4) ( P R) Q P. A sphere of mass m moing with a constant elocity collides with another stationary sphere of same mass. The ratio of elocities of two spheres after collision will be, if the co-efficient of restitution is e- () e e () e e. (3) e e (4) e e 3. An electric motor produces a tension of 4500N in a load lifting cable and rolls it at the rate of m/s. The power of the motor is () 9KW () 5KW (3) 5KW (4) HP 4. A ball falls from a height of 5m and strikes the roof of a lift. If at the time of collision, lift is moing in the upward direction with a elocity of m/s, then the elocity with which the ball rebounds after collision will be (e = ) () m/s downwards () m/s upwards (3) 3 m/s upwards (4) m/s downwards 5. A motor of 00 H.P. is moing with a constant elocity of 7 km/hour. The forward force exerted by the engine on the car is () N () N (3) N (4) None of the aboe 6. Two elastic bodies P and Q haing equal masses are moing along the same line with elocities of 6 m/s and 0 m/s respectiely. Their elocities after the elastic collision will be in m/s..8..

29 () 0 and 5 () 5 and 0 (3) 0 and 6 (4) 0 and 5 7. If the momentum of a body is incresed n times, its kinetic energy increases. () n times () n times (3) n times (4) n times 8. A metal ball does not rebound when struck on a wall, whereas a rubber ball of same mass when thrown with the same elocity on the wall rebounds. From this it is inferred that () Change in momentum is same in both () Change in momentum in rubber ball is more (3) Change in momentum in metal ball is more (4) Initial momentum of metal ball is more than that of rubber ball 9. The unit of the co efficient of restitution is () m/s () s/m (3) m s (4) None of the aboe 0. A bomb of mass 9 kg explode s into t wo piece s of 3kg and 6 kg. The elocity of 3 kg piece is 6 m/s. The kinetic energy of 6 kg piece is () 768 Joule () 786 Joule (3) 9 Joule (4) 687 Joule. Two solid balls of rubber A and B whose masses are 00 gm and 400 gm respectiely, are moing in mutually opposite directions. If the elocity A is 0 3 m/s and both the balls come to rest after collision, then the elocity of ball B is () 0.5 m/s () 0 5 m/s (3).5 m/s (4) None of the aboe. The graph between E k and p is (E = kinetic energy and p = momentum) K () () (3) (4) 3. A Kg ball falls from a height of 5 cm and rebounds upto a height of 9 cm. The co efficient of restitution is () 0.6 () 0.3 (3) 0.40 (4) The graph between potential energy U and displacement X in the state of stable equilibrium will be () () (3) (4)..9..