,... m n. , m 2. , m 3. 2, r. is called the moment of mass of the particle w.r.t O. and m 2

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2 CENTRE OF MASS CENTRE OF MASS Every physical syste has associated with it a certain point whose otion characterises the otion of the whole syste. When the syste oves under soe external forces, then this point oves as if the entire ass of the syste is concentrated at this point and also the external force is applied at this point for translational otion. This point is called the centre of ass of the syste. CENTRE OF MASS OF A SYSTEM OF 'N' DISCRETE PARTICLES Consider a syste of N point asses,,,... n whose position vectors fro origin O are given by r, r, r,... respectively. Then the position vector of the centre of ass C of the r n syste is given by. r r... r r c... r c n M i r i i n n n ; rc n i n i r i i i where, i r i is called the oent of ass of the particle w.r.t O. n M i i is the total ass of the syste. Note: n If the origin is taken at the centre of ass then i r i 0. hence, the COM is the point about which i the su of ass oents of the syste is zero. POSITION OF COM OF T WO PARTICLES Centre of ass of two particles of asses and separated by a distance r lies in between the two particles. The distance of centre of ass fro any of the particle (r) is inversely proportional to the ass of the particle () i.e. r / or r r or r r or r r and r r Here, r distance of COM fro and r distance of COM fro Fro the above discussion, we see that r r / if, i.e., COM lies idway between the two particles of equal asses. Siilarly, r > r if < and r < r if <, i.e., COM is nearer to the particle having larger ass. AIEEE_CENTRE OF MASS -

3 Exaple. Two particles of ass kg and kg are located at x 0 and x. Find the position of their centre of ass. Since, both the particles lies on x-axis, the COM will also lie on x-axis. Let the COM is located at x x, then r distance of COM fro the particle of ass kg x and r distance of COM fro the particle of ass kg ( x) Exaple. Using r r x or or x x Thus, the COM of the two particles is located at x. Ans. The position vector of three particles of asses kg, kg and kg are (î 4 ĵ kˆ), (î ĵ kˆ) and (î ĵ kˆ) respectively. Find the position vector r r of their centre of ass. The position vector of COM of the three particles will be given by r r r rcom Substituting the values, we get r r COM ()(î 4ĵ kˆ) ()(î ĵ kˆ) ()(î ĵ kˆ) ( î ĵ kˆ ) Ans. Exaple. Four particles of ass kg, kg, kg and 4 kg are placed at the four vertices A, B, C and D of a square of side. Find the position of centre of ass of the particles. Assuing D as the origin, DC as x -axis and DA as y-axis, we have kg, (x, y ) (0, ) kg, (x, y ) (, ) kg, (x, y ) (, 0) and 4 4 kg, (x 4, y 4 ) (0, 0) Co-ordinates of their COM are x COM x x 4x 4 4 ()(0) () () 4(0) AIEEE_CENTRE OF MASS -

4 Siilarly, y COM y y y 4y 4 4 ()() () (0) 4(0) (x COM, y COM ) (0.5, 0. ) Ans. Thus, position of COM of the four particles is as shown in figure. Exaple 4. Consider a two-particle syste with the particles having asses and. If the first particle is pushed towards the centre of ass through a distance d, by what distance should the second particle be oved so as to keep the centre of ass at the sae position? Consider figure. Suppose the distance of fro the centre of ass C is x and that of fro C is x. Suppose the ass is oved through a distance d towards C so as to keep the centre of ass at C. Then, x x...(i) and (x d) (x d)....(ii) Subtracting (ii) fro (i) d d or, d d, CENTRE OF MASS OF A CONTINUOUS MASS DISTRIBUTION For continuous ass distribution the centre of ass can be located by replacing suation sign with an integral sign. Proper liits for the integral are chosen according to the situation x c x d, y c d y d, z c d z d d d M (ass of the body) r c M rd. x x coponent of centre of ass of eleent Note: If an object has syetric ass distribution about x axis then y coordinate of COM is zero and vice-versa CENTRE OF MASS OF A UNIFORM ROD Suppose a rod of ass M and length L is lying along the x-axis with its one end at x 0 and the other at x L. Mass per unit length of the rod L M Hence, d, (the ass of the eleent dx situated at x x is) L M dx The coordinates of the eleent dx are (x, 0, 0). Therefore, x-coordinate of COM of the rod will be AIEEE_CENTRE OF MASS -

5 x COM L 0 x d d x0 dx xx xl L M (x) dx 0 L M L L x dx L 0 The y-coordinate of COM is y COM y d d 0 Siilarly, z COM 0 i.e., the coordinates of COM of the rod are L,0, 0, i.e. it lies at the centre of the rod. Exaple 5. A rod of length L is placed along the x-axis between x 0 and x L. The linear density (ass/length) of the rod varies with the distance x fro the origin as Rx. Here, R is a positive constant. Find the position of centre of ass of this rod. Mass of eleent dx situated at x x is d dx Rx dx The COM of the eleent has coordinates (x, 0, 0). Therefore, x-coordinate of COM of the rod will be x COM L xd 0 d y x0 dx xx xl x L 0 L (x)(rx)dx 0 (Rx)dx R R L 0 L 0 x dx x dx x x L 0 L 0 L The y-coordinate of COM of the rod is y COM Siilarly, z COM 0 y d d 0 (as y 0) L Hence, the centre of ass of the rod lies at,0, 0 Ans. AIEEE_CENTRE OF MASS - 4

6 CENTRE OF MASS OF A SEMICIRCULAR RING Figure shows the object (sei circular ring). By observation we can say that the x-coordinate of the centre of ass of the ring is zero as the half ring is syetrical abnout y-axis on both sides of the origin. Only we are required to find the y-coordinate of the centre of ass. Y Rd y c d yrsin X To find y c we use y c M d y...(i) Here for d we consider an eleental arc of the ring at an angle fro the x-direction of angular width d. If radius of the ring is R then its y coordinate will be R sin, here d is given as d M R R d So fro equation ---(i), we have y c M 0 M R Rd (R sin) R 0 sin d y c R...(ii) CENTRE OF MASS OF SEMICIRCULAR DISC Figure shows the half disc of ass M and radius R. Here, we are only required to find the y- coordinate of the centre of ass of this disc as centre of ass will be located on its half vertical diaeter. Here to find y c, we consider a sall eleental ring of ass d of radius x on the disc (disc can be considered to be ade up such thin rings of increasing radii) which will be integrated fro 0 to R. Here d is given as M d R ( x)dx Y y c dx x R X x Now the y-coordinate of the eleent is taken as, as in previous section, we have derived that the centre of ass of a sei circular ring is concentrated at R Here y c is given as y c R M d 0 x R 4M M x R 0 dx y c 4R AIEEE_CENTRE OF MASS - 5

7 . Centre of ass of a unifor rectangular, square or circular plate lies at its centre. Axis of syetry plane of syetry.. For a lainar type (-diensional) body with unifor negligible thickness the forulae for finding the position of centre of ass are as follows : r r... Atr A tr... rcom... A t A t... ( At) or r COM Ar A r... A A... Here, A stands for the area,. If soe ass of area is reoved fro a rigid body, then the position of centre of ass of the reaining portion is obtained fro the following forulae: (i) r COM r r or r COM Ar A r A A (ii) x COM x x or x COM Ax A x A A y COM y y or y COM Ay A y A A and z COM z z or z COM Az A z A A Here,, A, r, x, y and z are the values for the whole ass while, A, r, x, y and z are the values for the ass which has been reoved. Let us see two exaples in support of the above theory. Exaple 6. Find the position of centre of ass of the unifor laina shown in figure. Y O a X Here, A area of coplete circle a a A area of sall circle a 4 (x, y ) coordinates of centre of ass of large circle (0, 0) AIEEE_CENTRE OF MASS - 6

8 a and (x, y ) coordinates of centre of ass of sall circle,0 Using x COM Ax A x A A we get x COM a a 4 a a a 6 a and y COM 0 as y and y both are zero. Therefore, coordinates of COM of the laina shown in figure are CENTRE OF MASS OF SOME COMMON SYSTEMS a, 0 6 Ans. A syste of two point asses r r The centre of ass lies closer to the heavier ass. Rectangular plate (By syetry) x c b y c L A triangular plate (By qualitative arguent) at the centroid : y c h A sei-circular ring y c R x c 0 A sei-circular disc y c 4R x c 0 AIEEE_CENTRE OF MASS - 7

9 A heispherical shell y c R x c 0 A solid heisphere y c R 8 x c 0 A circular cone (solid) y c 4 h A circular cone (hollow) y c h Exaple 7. A unifor thin rod is bent in the for of closed loop ABCDEFA as shown in the figure. The y- coordinate of the centre of ass of the syste is Ans. Solution. r 6r (A) (B) (C) r (D) Zero (B) r The centre of ass of seicircular ring is at a distance fro its centre. (Let ass/ length) Y c r 4r r r r r r r 6r MOTION OF CENTRE OF MASS AND CONSERVATION OF MOMENTUM : Velocity of centre of ass of syste v c dr dt dr dr... dt dt M n drn dt v v v M... Here nuerator of the right hand side ter is the total oentu of the syste i.e., suation of oentu of the individual coponent (particle) of the syste n v AIEEE_CENTRE OF MASS - 8 n

10 Hence velocity of centre of ass of the syste is the ratio of oentu of the syste to the ass of the syste. P Syste M v c Acceleration of centre of ass of syste a c dv dt dv dv... dt dt M n dv dt n a a a M... a n n Net force on syste Net External Force Net internal Force Net External Force M M M (action and reaction both of an internal force ust be within the syste. Vector suation will cancel all internal forces and hence net internal force on syste is zero) F ext M a c where F ext is the su of the 'external' forces acting on the syste. The internal forces which the particles exert on one another play absolutely no role in the otion of the centre of ass. If no external force is acting on a syste of particles, the acceleration of centre of ass of the syste will be zero. If a c 0, it iplies that v c ust be a constant and if v c is a constant, it iplies that the total oentu of the syste ust reain constant. It leads to the principal of conservation of oentu in absence of external forces. If F ext 0 then v c constant If resultant external force is zero on the syste, then the net oentu of the syste ust reain constant. Motion of COM in a oving syste of particles: () COM at rest : If F ext 0 and V c 0, then COM reains at rest. Individual coponents of the syste ay ove and have non-zero oentu due to utual forces (internal), but the net oentu of the syste reains zero. (i) All the particles of the syste are at rest. (ii) Particles are oving such that their net oentu is zero. exaple: (iii) (iv) (v) (vi) (vii) (viii) A bob at rest suddenly explodes into various saller fragents, all oving in different directions then, since the explosive forces are internal & there is no external force on the syste for explosion therefore, the COM of the bob will reain at the original position and the fragent fly such that their net oentu reains zero. Two en standing on a frictionless platfor, push each other, then also their net oentu reains zero because the push forces are internal for the two en syste. A boat floating in a lake, also has net oentu zero if the people on it changes their position, because the friction force required to ove the people is internal of the boat syste. Objects initially at rest, if oving under utual forces (electrostatic or gravitation)also have net oentu zero. A light spring of spring constant k kept copressed between two blocks of asses and on a sooth horizontal surface. When released, the blocks acquire velocities in opposite directions, such that the net oentu is zero. In a fan, all particles are oving but COM is at rest AIEEE_CENTRE OF MASS - 9

11 () COM oving with unifor velocity : If F ext 0, then V c reains constant therefore, net oentu of the syste also reains conserved. Individual coponents of the syste ay have variable velocity and oentu due to utual forces (internal), but the net oentu of the syste reains constant and COM continues to ove with the initial velocity. (i) All the particles of the syste are oving with sae velocity. e.g.: A car oving with unifor speed on a straight road, has its COM oving with a constant velocity. (ii) (iii) (iv) Internal explosions / breaking does not change the otion of COM and net oentu reains conserved. A bob oving in a straight line suddenly explodes into various saller fragents, all oving in different directions then, since the explosive forces are internal & there is no external force on the syste for explosion therefore, the COM of the bob will continue the original otion and the fragent fly such that their net oentu reains conserved. Man juping fro cart or buggy also exert internal forces therefore net oentu of the syste and hence, Motion of COM reains conserved. Two oving blocks connected by a light spring on a sooth horizontal surface. If the acting forces is only due to spring then COM will reain in its otion and oentu will reain conserved. (v) Particles colliding in absence of external ipulsive forces also have their oentu conserved. () COM oving with acceleration : If an external force is present then COM continues its original otion as if the external force is acting on it, irrespective of internal forces. Exaple: Projectile otion : An axe thrown in air at an angle with the horizontal will perfor a c o plicated otion of rotation as well as parabolic otion under the effect of gravitation H co u sin g R co u sin g T usin g R co Exaple: Circular Motion : A rod hinged at an end, rotates, than its COM perfors circular otion. The centripetal force (F c ) required in the circular otion is assued to be acting on the COM. c F ù R COM F c g F c F c F c g g g AIEEE_CENTRE OF MASS - 0

12 Exaple 8. A projectile is fired at a speed of 00 /s at an angle of 7º above the horizontal. At the highest point, the projectile breaks into two parts of ass ratio :, the lighter piece coing to rest. Find the distance fro the launching point to the point where the heavier piece lands. Internal force do not effect the otion of the centre of ass, the centre of ass hits the ground at the position where the original projectile would have landed. The range of the original projectile is, x COM u sincos g The centre of ass will hit the ground at this position. As the saller block coes to rest after breaking, it falls down vertically and hits the ground at half of the range, i.e., at x 480. If the heavier block hits the ground at x, then x COM x x 960 ( )(480) ()(x ) ( ) x 0 Ans. Moentu Conservation : The total linear oentu of a syste of particles is equal to the product of the total ass of the syste and the velocity of its centre of ass. P M F ext dp dt If F ext v c 0 dp 0 ; P constant dt When the vector su of the external forces acting on a syste is zero, the total linear oentu of the syste reains constant. P + P + P P constant. n Exaple 9. A shell is fired fro a cannon with a speed of 00 /s at an angle 60º with the horizontal (positive x- direction). At the highest point of its trajectory, the shell explodes into two equal fragents. One of the fragents oves along the negative x-direction with a speed of 50 /s. What is the speed of the other fragent at the tie of explosion. As we know in absence of external force the otion of centre of ass of a body reains uneffected. Thus, here the centre of ass of the two fragents will continue to follow the original projectile path. The velocity of the shell at the highest point of trajectory is v M ucos 00 cos60º 50 /s. Let v be the speed of the fragent which oves along the negative x-direction and the other fragent has speed v,. which ust be along positive x-direction. Now fro oentu conservation, we have or v v + v or v v v v v + v ( 50) /s AIEEE_CENTRE OF MASS -

13 Exaple 0. Exaple. A an of ass is standing on a platfor of ass M kept on sooth ice. If the an starts oving on the platfor with a speed v relative to the platfor, with what velocity relative to the ice does the platfor recoil? Consider the situation shown in figure. Suppose the an oves at a speed w towards right and the platfor recoils at a speed V towards left, both relative to the ice. Hence, the speed of the an relative to the platfor is V + w. By the question, V + w v, or w v V...(i) w Taking the platfor and the an to be the syste, there is no V external horizontal force on the syste. The linear oentu ice of the syste reains constant. Initially both the an and the ///////////////////////////////////////////////// platfor were at rest. Thus, 0 MV - w or, MV (v V) [Using (i)] or, V v. M A flat car of ass M is at rest on a frictionless floor with a child of ass standing at its edge. If child jups off fro the car towards right with an initial velocity u, with respect to the car, find the velocity of the car after its jup. Let car attains a velocity v, and the net velocity of the child with respect to earth will be u v, as u is its velocity with respect to car. M v M u Exaple. Ans. Solution. /////////////////////////////////////////////////////////////////// /////////////////////////////////////////////////////////// Initially, the syste was at rest, thus according to oentu conservation, oentu after jup ust be zero, as (u v) M v u v M In a free space a rifle of ass M shoots a bullet of ass at a stationary block of ass M distance D away fro it. When the bullet has oved through a distance d towards the block the centre of ass of the bullet-block syste is at a distance of : (A) ( D d) fro the block (B) d M D fro the rifle M M (C) d DM M fro the rifle (D) (D d) M M (A,B,D) Rifle d D Bullet of ass() COM x D-d-x As; Mx (D d x) and Block M (D d) x fro the block M x' D d x (D d)m fro the bullet. M fro the bullet AIEEE_CENTRE OF MASS -

14 Exaple. The centre of ass of two asses & oves by distance 5 x when ass is oved by distance x and is kept fixed. The ratio is Ans. Solution (A) (B) 4 (C) /4 (D) None of these (B) ( + ) 5 x x + O Exaple ; 4 ; A unifor disc of ass and radius R is placed on a sooth horizontal floor such that the plane surface of the disc is in contact with the floor. A an of ass / stands on the disc at its periphery. The an starts walking along the periphery of the disc. The size of the an is negligible as copared to the size of the disc. Then the centre of disc. 4 (A) oves along a circle of radius R R (C) oves along a circle of radius Ans. Solution. Exaple 5. (C) oves along a circle of radius R (D) does not ove along a circle (A) The centre of ass of an + disc shall always reain at rest. Since the an is always at periphery of disc, the centre of disc shall always be at distance R/ fro centre of ass of two body syste. Hence centre of disc oves in circle of radius R/. A person P of ass 50 kg stands at the iddle of a boat of ass 00 kg oving at a constant velocity 0 /s with no friction between water and boat and also the engine of the boat is shut off. With what velocity (relative to the boat surface) should the person ove so that the boat coes to rest. Neglect friction between water and boat. Ans. Solution. Exaple 6. (A) 0 /s towards right (B) 0 /s towards right (C) 0 /s towards left (D) 0 /s towards left (A) Moentu of the syste reains conserved as no external force is acting on the syste in horizontal direction. ( ) 0 50 V V 0 /s towards right, as boat is at rest. V Pboat 0 /s Two en of asses 80 kg and 60 kg are standing on a wood plank of ass 00 kg, that has been placed over a sooth surface. If both the en start oving toward each other with speeds /s and /s respectively then find the velocity of the plank by which it starts oving. Solution. Applying oentu conservation ; (80) + 60 ( ) ( ) v 40 v 40 /sec. 6 AIEEE_CENTRE OF MASS -

15 Exaple 7. Each of the blocks shown in figure has ass kg. The rear block oves with a speed of /s towards the front block kept at rest. The spring attached to the front block is light and has a spring constant 50 N/. Find the axiu copression of the spring. Assue, on a friction less surface kg k50n/ kg ///////////////////////////////////////////////////////////////// Maxiu copression will take place when the blocks ove with equal velocity. As no net external horizontal force acts on the syste of the two blocks, the total linear oentu will reain constant. If V is the coon speed at axiu copression, we have, ( kg) ( /s) ( kg)v + ( kg)v or, V /s. Initial kinetic energy ( kg) ( /s) J. Final kinetic energy ( kg) (/s) + ( kg) ( /s) J The kinetic energy lost is stored as the elastic energy in the spring. Hence, (50 N/) x J J J or, x 0.. Exaple 8. Figure shows two blocks of asses 5 kg and kg placed on a frictionless surface and connected with a spring. An external kick gives a velocity 4 /s to the heavier block towards the lighter one. Deduce (a) velocity gained by the centre of ass and (b) the separate velocities of the two blocks with respect to centre of ass just after the kick. 5kg kg (a) Velocity of centre of ass is ////////////////////////////////////////////////////////////// (b) v c /s Due to kick on 5 kg block, it starts oving with a velocity 4 /s iediately, but due to inertia kg block reains at rest, at that oent. Thus, velocity of 5 kg block with respect to the centre of ass is v /s and the velocity of kg block w.r.t. to centre of ass is v /s Exaple 9. The two blocks A and B of sae ass connected to a spring and placed on a sooth surface. They are given velocities (as shown in the figure) when the spring is in its natural length : Ans. (A) (B) (C) (D) (A) the axiu velocity of B will be 0 /s the axiu velocity of B will be greater than 0 /s the spring will have axiu extension when A and B both stop the spring will have axiu extension when both ove towards left. AIEEE_CENTRE OF MASS - 4

16 Solution. Suppose B oves with a velocity ore than 0 /s a should ove at a velocity greater than 5 /s and increases the overall energy which is not possible since there is no external force acting on the syste. Hence B should ove with a axiu velocity 0 /s. Also both A and B can never stop so as to keep the oentu constant. Also both A and B can never ove towards left siultaneously for oentu reaining conserved. Hence only (A) is correct. IMPULSE Ipulse of a force F acting on a body for the tie interval t t to t t is defined as :- t F dt t F dv dt dt dt dv ( v - v ) Ä P change in oentu due to force F t Also, Re s F Ä P (ipulse - oentu theore) t Res dt Note: Ipulse applied to an object in a given tie interval can also be calculated fro the area under force tie (F-t) graph in the sae tie interval. Instantaneous Ipulse : There are any cases when a force acts for such a short tie that the effect is instantaneous, e.g., a bat striking a ball. In such cases, although the agnitude of the force and the tie for which it acts ay each be unknown but the value of their product (i.e., ipulse) can be known by easuring the initial and final oenta. Thus, we can write. Fdt P P f P i Iportant Points : () It is a vector quantity. () Diensions [MLT ] () S unit kg /s (4) Direction is along change in oentu. (5) Magnitude is equal to area under the F-t. graph. (6) F dt F av dt F av t (7) It is not a property of a particle, but it is a easure of the degree to which an external force changes the oentu of the particle. Exaple 0. The hero of a stunt fil fires 50 g bullets fro a achine gun, each at a speed of.0 k/s. If he fires 0 bullets in 4 seconds, what average force does he exert against the achine gun during this period. The oentu of each bullet (0.050 kg) (000 /s) 50 kg-/s. The gun has been iparted this uch aount of oentu by each bullet fired. Thus, the rate of change of oentu of the gun ( 50 kg / s) 0 4s 50 N. In order to hold the gun, the hero ust exert a force of 50 N against the gun. AIEEE_CENTRE OF MASS - 5

17 Ipulsive force : A force, of relatively higher agnitude and acting for relatively shorter tie, is called ipulsive force. An ipulsive force can change the oentu of a body in a finite agnitude in a very short tie interval. Ipulsive force is a relative ter. There is no clear boundary between an ipulsive and Non- Note: Ipulsive force. Usually colliding forces are ipulsive in nature. Since, the application tie is very sall, hence, very little otion of the particle takes place. Iportant points :. Gravitational force and spring force are always non-ipulsive.. Noral, tension and friction are case dependent.. An ipulsive force can only be balanced by another ipulsive force.. Ipulsive Noral : In case of collision, noral forces at the surface of collision are always ipulsive eg. N i Ipulsive; N g Non-ipulsive N N ///////////////////////////// Both norals are Ipulsive N N ////////////////////////////////////////////////////////////////// N, N Ipulsive; N non-ipulsive N N Both norals are Ipulsive AIEEE_CENTRE OF MASS - 6

18 . Ipulsive Friction : If the noral between the two objects is ipulsive, then the friction between the two will also be ipulsive. Friction at both surfaces is ipulsive Friction due to N is non-ipulsive and due to N and N are ipulsive. Ipulsive Tensions : W hen a string jerks, equal and opposite tension act suddenly at each end. Consequently equal and opposite ipulses act on the bodies attached with the string in the direction of the string. There are two cases to be considered. (a) One end of the string is fixed : The ipulse which acts at the fixed end of the string cannot change the oentu of the fixed object there. The object attached to the free end however will undergo a change in oentu in the direction of the string. The oentu reains unchanged in a direction perpendicular to the string where no ipulsive forces act. (b) Both ends of the string attached to ovable objects : In this case equal and opposite ipulses act on the two objects, producing equal and opposite changes in oentu. The total oentu of the syste therefore reains constant, although the oentu of each individual object is changed in the direction of the string. Perpendicular to the string however, no ipulse acts and the oentu of each particle in this direction is unchanged. T is Ipulsive T is non-ipulsive T is non-ipulsive /////////////////// All noral are ipulsive but tension T is ipulsive only for the ball A For this exaple: In case of rod, Tension is always ipulsive and in case of spring, Tension is always non-ipulsive. B A C T Exaple. Two identical block A and B, connected by a assless string are placed on a frictionless horizontal plane. A bullet having sae ass, oving with speed u strikes block B fro behind as shown. If the bullet gets ebedded into the block B then find : A C u B ///////////////////////////////////////////////////// (a) The velocity of A,B,C after collision. (b) Ipulse on A due to tension in the string (c) Ipulse on C due to noral force of collision. (d) Ipulse on B due to noral force of collision. AIEEE_CENTRE OF MASS - 7

19 (a) By Conservation of linear oentu v u (b) (c) u T dt u N dt u u (d) ( N T) dt Ndt T dt u COLLISION OR IMPACT u N dt Collision is an event in which an ipulsive force acts between two or ore bodies for a short tie, which results in change of their velocities. Note : (a) (b) (c) In a collision, particles ay or ay not coe in physical contact. The duration of collision, t is negligible as copared to the usual tie intervals of observation of otion. In a collision the effect of external non ipulsive forces such as gravity are not taken into a account as due to sall duration of collision (t) average ipulsive force responsible for collision is uch larger than external forces acting on the syste. The collision is infact a redistribution of total oentu of the particles. Thus, law of conservation of linear oentu is indispensable in dealing with the phenoenon of collision between particles. Line of Ipact The line passing through the coon noral to the surfaces in contact during ipact is called line of ipact. The force during collision acts along this line on both the bodies. Direction of Line of ipact can be deterined by: (a) Geoetry of colliding objects like spheres, discs, wedge etc. (b) Direction of change of oentu. If one particle is stationary before the collision then the line of ipact will be along its otion after collision. Classification of collisions (a) On the basis of line of ipact (i) Head-on collision : If the velocities of the colliding particles are along the sae line before and after the collision. (ii) Oblique collision : If the velocities of the colliding particles are along different lines before and after the collision. (b) On the basis of energy : (i) Elastic collision : In an elastic collision, the colliding particles regain their shape and size copletely after collision. i.e., no fraction of echanical energy reains stored as deforation potential energy in the bodies. Thus, kinetic energy of syste after collision is equal to kinetic energy of syste before collision. Thus in addition to the linear oentu, kinetic energy also reains conserved before and after collision. AIEEE_CENTRE OF MASS - 8

20 (ii) Inelastic collision : In an inelastic collision, the colliding particles do not regain their shape and size copletely after collision. Soe fraction of echanical energy is retained by the colliding particles in the for of deforation potential energy. Thus, the kinetic energy of the particles after collision is not equal to that of before collision. However, in the absence of external forces, law of conservation of linear oentu still holds good. (iii) Perfectly inelastic : If velocity of separation along the line of ipact just after collision becoes zero then the collision is perfectly inelastic. Collision is said to be perfectly inelastic if both the particles stick together after collision and ove with sae velocity, Note : Actually collision between all real objects are neither perfectly elastic nor perfectly inelastic, its inelastic in nature. Exaples of line of ipact and collisions based on line of ipact (i) Two balls A and B are approaching each other such that their centres are oving along line CD. Head on Collision (ii) Two balls A and B are approaching each other such that their centre are oving along dotted lines as shown in figure. Oblique Collision (iii) Ball is falling on a stationary wedge. Oblique Collision COEFFICIENT OF RESTITUTION (e) The coefficient of restitution is defined as the ratio of the ipulses of reforation and deforation of either body. AIEEE_CENTRE OF MASS - 9

21 e Ipulse of Ipulse of reforation deforation F dt F r d dt Velocity of seperation along line of ipact Velocity of approach along line of ipact The ost general expression for coefficient of restitution is e velocity of separation of points of contact along line of ipact velocity of approach of point of contact along line of ipact Exaple for calculation of e Two sooth balls A and B approaching each other such that their centres are oving along line CD in absence of external ipulsive force. The velocities of A and B just before collision be u and u respectively. The velocities of A and B just after collision be v and v respectively. Just Before collision u u Just After collision v v C A B Line of ipact D C A B Line of ipact D u > u v < v u v v u v v N D N D N R N R Deforation Reforation F ext 0 oentu is conserved for the syste. u + u ( + )v v + v v u u v v...() Ipulse of Deforation : J D change in oentu of any one body during deforation. (v u ) for ( v + u ) for Ipulse of Reforation : J R change in oentu of any one body during Reforation. (v v) for e (v v ) for Ipulse of Reforation ( JR ) v v Velocity of separation along line of ipact Ipulse of Deforation ( J ) u u Velocity of approach along line of ipact D AIEEE_CENTRE OF MASS - 0

22 Note : e is independent of shape and ass of object but depends on the aterial. The coefficient of restitution is constant for a pair of aterials. (a) e Ipulse of Reforation Ipulse of Deforation Velocity of separation along the LOI Velocity of approach along the LOI Kinetic energy of particles after collision ay be equal to that of before collision. Collision is elastic. (b) e 0 Ipulse of Reforation 0 Velocity of separation along the LOI 0 Kinetic energy of particles after collision is not equal to that of before collision. Collision is perfectly inelastic. (c) 0 < e < Ipulse of Reforation < Ipulse of Deforation Velocity of separation along the LOI < Velocity of approach along the LOI Kinetic energy of particles after collision is not equal to that of before collision. Collision is Inelastic. Note : In case of contact collisions e is always less than unity. 0 e Iportant Point : In case of elastic collision, if rough surface is present then k f < k i (because friction is ipulsive) Where, k is Kinetic Energy. Rough ///////////////////////////////////// A particle B oving along the dotted line collides with a rod also in state of otion as shown in the figure. The particle B coes in contact with point C on the rod. To write down the expression for coefficient of restitution e, we first draw the line of ipact. Then we resolve the coponents of velocities of points of contact of both the bodies along line of ipact just before and just after collision. Then e v u x x v u x x Collision in one diension (Head on) u u (a) Before Collision v v (b) After Collision e v v u u u > u v > v (u u )e (v v ) AIEEE_CENTRE OF MASS -

23 By oentu conservation, u + u v + v v v + e(u u ) and v v u u e(u u u u e(u u ) ) Special Case : () e 0 v v for perfectly inelastic collision, both the bodies, ove with sae vel. after collision. () e and, we get v u and v u i.e., when two particles of equal ass collide elastically and the collision is head on, they exchange their velocities., e.g. /s Before Collision v 0 () >> + and 0 v u No change and v u + e(u u ) Now If e v u u Exaple. Two identical balls are approaching towards each other on a straight line with velocity /s and 4 /s respectively. Find the final velocities, after elastic collision between the. /s 4/s The two velocities will be exchanged and the final otion is reverse of initial otion for both. 4/s /s AIEEE_CENTRE OF MASS -

24 Exaple. Exaple 4. Exaple 5. Three balls A, B and C of sae ass are placed on a frictionless horizontal plane in a straight line as shown. Ball A is oved with velocity u towards the iddle ball B. If all the collisions are elastic then, find the final velocities of all the balls. u A B C ////////////////////////////////////////// A collides elastically with B and coes to rest but B starts oving with velocity u u A B C ////////////////////////////////////////// After a while B collides elastically with C and coes to rest but C starts oving with velocity u Final velocities V A 0; V B 0 and V C u A B C u ////////////////////////////////////////// Ans. Four identical balls A, B, C and D are placed in a line on a frictionless horizontal surface. A and D are oved with sae speed u towards the iddle as shown. Assuing elastic collisions, find the final velocities. u u A B C D ///////////////////////////////////////////////////// A and D collides elastically with B and C respectively and coe to rest but B and C starts oving with velocity u towards each other as shown u u A B C D ///////////////////////////////////////////////////// B and C collides elastically and exchange their velocities to ove in opposite directions u A B C u D ///////////////////////////////////////////////////// Now, B and C collides elastically with A and D respectively and coe to rest but A and D starts oving with velocity u away fro each other as shown u u A B C D ///////////////////////////////////////////////////// Final velocities V A u ( ); V B 0; V C 0 and V D u ( ) Two particles of ass and oving in opposite directions on a frictionless surface collide elastically with velocity v and v respectively. Find their velocities after collision, also find the fraction of kinetic energy lost by the colliding particles. Ans. v v Let the final velocities of and be v and v respectively as shown in the figure: v By conservation of oentu: (v) + ( v) (v ) + (v ) or 0 v + v or v + v 0...() AIEEE_CENTRE OF MASS -

25 and since the collision is elastic: v v v ( v) or v v v...() Solving the above two equations, we get, v v and v v Ans. i.e., the ass returns with velocity v while the ass returns with velocity v in the direction shown in figure: v The collision was elastic therefore, no kinetic energy is lost, KE loss KE i - KE f or, (v) ()( v) ( v) ()v 0 Exaple 6. On a frictionless surface, a ball of ass oving at a speed v akes a head on collision with an identical ball at rest. The kinetic energy of the balls after the collision is /4th of the original. Find the coefficient of restitution. As we have seen in the above discussion, that under the given conditions : By using conservation of linear oentu and equation of e, we get, e e v ' v and v ' v Given that K f Ki or 4 Substituting the value, we get v + v 4 v e + e 4 or e Ans. Exaple 7. A block of ass kg is pushed towards a very heavy object oving with /s closer to the block (as shown). Assuing elastic collision and frictionless surfaces, find the final velocities of the blocks. kg 0/s /s very heavy object ///////////////////////////////////////////////////////////////////// Let v and v be the final velocities of kg block and heavy object respectively then, v u + (u u ) u u 4 /s v /s AIEEE_CENTRE OF MASS - 4

26 Exaple 8. A ball is oving with velocity /s towards a heavy wall oving towards the ball with speed /s as shown in fig. Assuing collision to be elastic, find the velocity of the ball iediately after the collision. The speed of wall will not change after the collision. So, let v be the velocity of the ball after collision in the direction shown in figure. Since collision is elastic (e ), separation speed approach speed or v + or v 4 /s Ans. Exaple 9. Two balls of asses kg and 4 kg are oved towards each other with velocities 4 /s and /s respectively on a frictionless surface. After colliding the kg ball returns back with velocity /s. Just before collision Just after collision kg 4/s /s 4kg /s kg 4kg v //////////////////////////////////////////// //////////////////////////////////////// Then find: (a) velocity of 4 kg ball after collision (b) coefficient of restitution e. (c) Ipulse of deforation J D. (d) Maxiu potential energy of deforation. (e) Ipulse of reforation J R. (a) By oentu conservation, (4) 4() ( ) + 4(v ) v /s velocity of separation (b) e velocity of approach ( ) 4 ( ) (c) At axiu defored state, by conservation of oentu, coon velocity is v 0. J D (v u ) (v u ) (0 4) 8 N -s 4(0 ) 8 N - s or 4(0 ) 8 N - s (d) Potential energy at axiu defored state U loss in kinetic energy during deforation. or U u u ( + )v or (4) 4() U 4 Joule ( + 4) (0) AIEEE_CENTRE OF MASS - 5

27 (e) or J R (v v) (v v ) ( 0) 4 N-s 4(0 ) 4 N-s or e J J R D J R ej D (0.5) ( 8) 4 N-s C ollisio n in tw o diension (o blique). A pair of equal and opposite ipulses act along coon noral direction. Hence, linear oentu of individual particles do change along coon noral direction. If ass of the colliding particles reain constant during collision, then we can say that linear velocity of the individual particles change during collision in this direction.. No coponent of ipulse act along coon tangent direction. Hence, linear oentu or linear velocity of individual particles (if ass is constant) reain unchanged along this direction.. Net ipulse on both the particles is zero during collision. Hence, net oentu of both the particles reain conserved before and after collision in any direction. 4. Definition of coefficient of restitution can be applied along coon noral direction, i.e., along coon noral direction we can apply Relative speed of separation e (relative speed of approach) Exaple 0. A ball of ass hits a floor with a speed v 0 aking an angle of incidence a with the noral. The coefficient of restitution is e. Find the speed of the reflected ball and the angle of reflection of the ball. The coponent of velocity v 0 along coon tangential direction v 0 sin will reain unchanged. Let v be the coponent along coon noral direction after collision. Applying, Relative speed of separation e (Relative speed of approach) along coon noral direction, we get v ev 0 cos v ( ev0cos ) v' v0sin Thus, after collision coponents of velocity v are v 0 sin and ev 0 cos v' (v0 sin) (ev 0 cos) and tan or tan v0 sin ev cos 0 tan e Ans. Note : For elastic collision, e v v 0 and Ans. AIEEE_CENTRE OF MASS - 6

28 Exaple. A ball of ass akes an elastic collision with another identical ball at rest. Show that if the collision is oblique, the bodies go at right angles to each other after collision. In head on elastic collision between two particles, they exchange their velocities. In this case, the coponent of ball along coon noral direction, v cos becoes zero after collision, while that of becoes v cos. While the coponents along coon tangent direction of both the particles reain unchanged. Thus, the coponents along coon tangent and coon noral direction of both the balls in tabular for are given below. Ball Coponent along coon tangent direction Coponent along coon noral direction Before collision After collision Before collision After collision v sin v sin v cos v cos Fro the above table and figure, we see that both the balls ove at right angle after collision with velocities v sin and v cos. Note : When two identical bodies have an oblique elastic collision, with one body at rest before collision, then the two bodies will go in directions. VARIABLE MASS SYSTEM : If a ass is added or ejected fro a syste, at rate kg/s and relative velocity v rel then the force exerted by this ass on the syste has agnitude v rel. Thrust Force ( F t ) F v t rel d dt (w.r.t. the syste), Suppose at soe oent t t ass of a body is and its velocity is v. After soe tie at t t + dt its ass becoes ( d) and velocity becoes v dv. The ass d is ejected with relative velocity v r. Absolute velocity of ass d is therefore ( v + v r ). If no external forces are acting on the syste, the linear oentu of the syste will reain conserved, or P i P f or v ( d) ( v + d v ) + d ( v + v r ) or v v + d v (d) v (d) (d v ) + (d) v + v r d AIEEE_CENTRE OF MASS - 7

29 The ter (d) (d v ) is too sall and can be neglected. d v v dv d r d or v r dt dt Here, dv dt thrust force F t and d d rate at which ass is ejecting or Ft vr dt dt Probles related to variable ass can be solved in following four steps. Make a list of all the forces acting on the ain ass and apply the on it.. Apply an additional thrust force F d t on the ass, the agnitude of which is v r dt and direction is given by the direction of v r in case the ass is increasing and otherwise the direction of v r if it is decreasing.. Find net force on the ass and apply dv F net ( ass at the particular instant) dt 4. Integrate it with proper liits to find velocity at any tie t. Ro cket pro pu lsion : Let 0 be the ass of the rocket at tie t 0. its ass at any tie t and v its velocity at that oent. Initially, let us suppose that the velocity of the rocket is u. d Further, let dt be the ass of the gas ejected per unit tie and vr the exhaust velocity of the d gases with respect to rocket. Usually and vr are kept constant throughout the journey of dt the rocket. Now, let us write few equations which can be used in the probles of rocket propulsion. At tie t t, d. Thrust force on the rocket F t v r dt (upwards). W eight of the rocket W g (downwards). Net force on the rocket F net F t W (upwards) AIEEE_CENTRE OF MASS - 8

30 d or F net v r dt g 4. Net acceleration of the rocket a F or dv v r d g dt dt v r or dv d g dt v d t or dv v r u g dt Thus, v u gt + v r n...(i) d d Note :. F t v r is upwards, as vr is downwards and is negative. dt dt. If gravity is ignored and initial velocity of the rocket u 0, Eq. (i) reduces to v v r ln 0. Exaple. A rocket, with an initial ass of 000 kg, is launched vertically upwards fro rest under gravity. The rocket burns fuel at the rate of 0 kg per second. The burnt atter is ejected vertically downwards with a speed of 000 s relative to the rocket. If burning stopsafter one inute. Find the axiu velocity of the rocket. (Take g as at 0 s ) Using the velocity equation v u gt + v r ln Here u 0, t 60s, g 0 /s, v r 000 /s, kg and kg 0 We get 000 v ln 400 or v 000 ln The axiu velocity of the rocket is 00(0 ln.5 ).6 s Ans. Exaple. A flat car of ass 0 starts oving to the right due to a constant horizontal force F. Sand spills on the flat car fro a stationary hopper. The rate of loading is constant and equal to k g/s. Find the ti e dependence of the velocity and the acceleration of the flat car in the process of loading. The friction is negligibly sall. 0 F Initial velocity of the flat car is zero. Let v be its velocity at tie t and its ass at that instant. Then AIEEE_CENTRE OF MASS - 9

31 At t 0, v 0 and 0 at t t, v v and 0 + t Here, v r v (backwards) d dt F t v r d v (backwards) dt Net force on the flat car at tie t is F net F F t or dv F v...(i) dt or dv v dv ( 0 + t) F v or dt F 0 v t dt 0 0 t [n (F v)]0 v [n (0 + t)] 0 t F n F v n 0 t 0 F F v 0 t 0 or v Ft 0 t Ans. Fro Eq. (i), dv acceleration of flat car at tie t dt or F v a F Ft t 0 0 t F0 or a ( t) 0 Ans. Exaple 4. A cart loaded with sand oves along a horizontal floor due to a constant force F coinciding in direction with the cart s velocity vector. In the process sand spills through a hole in the botto with a constant rate kg/s. Find the acceleration and velocity of the cart at the oent t, if at the initial oent t 0 the cart with loaded sand had the ass 0 and its velocity was equal to zero. Friction is to be neglected. In this proble the sand spills through a hole in the botto of the cart. Hence, the relative velocity of the sand v r will be zero because it will acquire the sae velocity as that of the cart at the oent. v r 0 d Thus, F t 0 as Ft vr dt and the net force will be F only. F net F v v F dv or dt F...(i) dv But here 0 t ( 0 t) dt F AIEEE_CENTRE OF MASS - 0

32 or v 0 dv t 0 Fdt 0 t F F n(0 t ) or v ln 0 0 t v t 0 Fro eq. (i), acceleration of the cart dv a dt F or a F 0 t Ans. Ans. LINEAR MOMENTUM CONSERVATION IN PRESENCE OF EXTERNAL FORCE. F ext dp dt F ext dt dp dp F ext ) pulsive dt If F ext ) pulsive 0 dp 0 or P is constant Note: Moentu is conserved if the external force present is non-ipulsive. eg. gravitation or spring force Exaple 5. Two balls are oving towards each other on a vertical line collides with each other as shown. Find their velocities just after collision. kg /s 4/s 4kg Let the final velocity of 4 kg ball just after collision be v. Since, external force is gravitational which is non - ipulsive, hence, linear oentu will be conserved. Applying linear oentu conservation: ( ) + 4(4) (4) + 4(v) 4/s kg v or v /s 4kg AIEEE_CENTRE OF MASS -

33 Exaple 6. A bullet of ass 50g is fired fro below into the bob of ass 450g of a long siple pendulu as shown in figure. The bullet reains inside the bob and the bob rises through a height of.8. Find the speed of the bullet. Take g 0 /s. ///////////////////// Let the speed of the bullet be v. Let the coon velocity of the bullet and the bob, after the bullet is ebedded into the bob, is V. By the principle of conservation of the linear oentu, (0.05 kg) v v V 0.45 kg 0.05 kg 0 The string becoes loose and the bob will go up with a deceleration of g 0 /s. As it coes to rest at a height of.8, using the equation v u + ax, (v /0).8 0 / s or, v 60 /s. v Proble. Three particles of asses 0.5 kg,.0 kg and.5 kg are placed at the three corners of a right angled triangle of sides.0 c, 4.0 c and 5.0 c as shown in figure. Locate the centre of ass of the syste. taking x and y axes as shown. coordinates of body A (0,0) coordinates of body B (4,0) coordinates of body C (0,).5 kg y c (C) 5c x - coordinate of c.. Ax A BxB M r A B C C C 0.5 kg (A) 4c.0 kg (B) x c c.c kg siilarly y - wordinates of c So, certre of ass is. c right and.5 c above particle A..5 c Proble. A block A (ass 4M) is placed on the top of a wedge B of base length l (ass 0 M) as shown in figure. When the syste is released fro rest. Find the distance oved by the wedge B till the block A reaches at lowest end of wedge. As sue all surfaces are frictionless. AIEEE_CENTRE OF MASS -

34 Initial position of centre of ass XBM M B B X AM M B A X B.0M.4M 4M Final position of centre of ass ( XB x)0m 4Mx 4M 5(X B x) x 6 5X B 6 since there is no horizontal force on syste centre of ass initially centre of ass finally. 5X B + 5X B + 5x + x 6x x 6 Proble. An isolated particle of ass is oving in a horizontal xy plane, along x-axis. At a certain height above ground, it suddenly explodes into two fragents of asses /4 and /4. An instant later, the saller fragent is at y + 5 c. Find the position of heavier fragent at this instant. As particle is oving along x-axis, so, y-coordinate of COM is zero. Y M M Y M 4 M M + Y M M M 0 M 5 + Y M Y M 4 5c Proble 4. A shell at rest at origin explodes into three fragents of asses kg, kg and kg. The fragents of asses kg and kg fly off with speeds /s along x-axis and 8 /s along y- axis respectively. If kg flies off with speed 40 /s then find the total ass of the shell. As initial velocity 0, Initial oentu ( + + ) 0 0 Finally, let velocity of M V. We know V 40 /s. Initial oentu final oentu. 0 î + 8 ĵ + V V ( î 6ĵ) V ( ) (6) ( ) (6) 40 {given} () (6) 0.5 kg 40 Total ass kg Proble 5. A block oving horizontally on a sooth surface with a speed of 0 /s bursts into two equal parts continuing in the sae direction. If one of the parts oves at 0 /s, with what speed does the second part ove and what is the fractional change in the kinetic energy of the syste. AIEEE_CENTRE OF MASS -

35 Applying oentu conservation ; 0 V + 0 V So, V 0 /s initial kinetic energy (0) 00 final kinetic energy.. (0) + (0) fractional change in kinetic energy ( final K. E) (initial K.E) initial K.E Proble 6. A block at rest explodes into three equal parts. Two parts starts oving along X and Y axes respectively with equal speeds of 0 /s. Find the initial velocity of the third part. Let total ass, initial linear oentu 0 Let velocity of third part V Using conservation of linear oentu : 0 î + 0 ĵ + V 0 So, V ( 0 î 0 ĵ ) /sec. V ( 0) (0) 0, aking angle 5 o below x-axis Proble 7. Blocks A and B have asses 40 kg and 60 kg respectively. They are placed on a sooth surface and the spring connected between the is stretched by.5. If they are released fro rest, deterine the speeds of both blocks at the instant the spring becoes unstretched. Let, both block start oving with velocity V and V as shown in figure Since no horizontal force on syste so, applying oentu conservation 0 40 V 60 V V V...() Now applying energy conservation, Loss in potential energy gain in kinetic energy kx V + V 600 (.5) 40 V + 60 V...() Solving euation () and () we get, V 4.5 /s, V /s. Proble 8. Find the ass of the rocket as a function of tie, if it oves with a constant accleration a, in absence of external forces. The gas escaps with a constant velocity u relative to the rocket and its initial ass was 0. AIEEE_CENTRE OF MASS - 4