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69 Physics Pre-Medical...

69 Physics Pre-Medical © 2 C 03 Centre of Mass & Collisions h ontents apter 21 01. Centre of mass 71 0- 02 02. Motion of Centre of Mass 78 :2 03. Application of Methods of Impulse and n Momentum to a System of Particles 79 io ss 04. Collision 85 Se NEET SYLLABUS Centre of mass of a two-particle system, momentum conservation and centre of mass motion. Centre of mass of a rigid body; centre of mass of uniform rod. Law of conservation of linear momentum and its applications. elastic and inelastic collisions in one and two dimensions. 70 ARYABHATA Aryabhata was born (476 CE) in Kusumapura (presend day Patna) in Bihar, India. Aryabhata was an acclaimed mathematician – astronomer. His contribution to mathematics, science and astronomy is immense, and yet he has not been accorded the recognition in the world history of science. At the age of 24, he wrote his famed “Aryabhatiya”. He was aware of the concept of zero, as well as the use of large numbers upto 1018. He was the first to calculate the value for ‘pi’ accurately to the fourth decimal point. He devised the formula for calculating areas of triangles and circles. He calculated the circumference of the earth as 62,832 miles, which is an excellent approximation and suggested that the apparent rotation of the heavens was due to the axial rotation of the earth on its axis. He was the first known astronomer to devise a continuous counting of solar days, designating each day with a number. He asserted that the planets shine due to the reflection of sunlight, and that the eclipses ocur due to the shadows of moon and earth. His observations discount the “Flat earth” concept, and lay the foundation for the belief that earth and other planets orbit the sun. Aryabhata’s major work, Aryabhatiya, a compedium of mathematics and astronomy, was extensively referred to in the Indian mathematical literature, and has survived to modern times. The Aryabhatiya covers arithmetic, algebra and trigonometry. 21 0- 02 :2 MEGHNAD SAHA n io Born : October 6, 1893 Died : February 16, 1956. Meghnad Saha was born ss on Octobe 6, 1893 in Sheoratali, a village in the District of Dacca, now in Se Bangladesh. He took admission in the Kishorilal Jubili school and passed the Entrance Examination of the Calcutta University in 1909, standing first among the student from East Bengal obtaining the highest marks in languages (English, Bengali and Sanskrit combined) and in Mathematics. While studying in Presidency College, Meghnad got involved with Anushilan Samiti to take part in freedom fighting movement. Meghnad Saha joined as lecturer at the newly opened University College of Science in Calcutta. He taught Quantum Physics. In 1919, American Astrophysical Journal published – ”On selective Radiation Pressure and it’s application” – a research paper by Meghnad Saha. In 1927, Meghnad Saha was elected as a fellow of London’s Royal Society. Meghnad Saha moved to Allahabad and in 1932 Uttar Pradesh Academy of Science was established. In 1947, he established Institute of Nuclear Physics which later was named after him as Saha Institute of Nuclear Physics.Having seen cyclotrons used for research in nuclear physics abroad, he ordered one to be installed in the institute. In 1950, India had its first cyclotron in operation. In 1952 he stood as an independent candiate for Parliament and was elected by a wide margin. He died on February 16, 1956 due to a heart attack.  ® ALLEN Pre-Medical : Physics 71 CENTRE OF MASS & COLLISION 1. CENTRE OF MASS For a system of particles centre of mass is that point at which its total mass is supposed to be concentrated. The centre of mass of an object is a point that represents the entire body and moves in the same way as a point mass having mass equal to that of the object, when subjected to the same external forces that act on the object. 1.1 Centre of mass of a system of discrete particles l Centre of Mass of a Two Particles System r r Consider two particles of masses m1 and m2 with position vectors r1 and r2 respectively. Let their centre r of mass C have position vector rc. y From definition , we have r r r m1 r Smi ri r m r + m2 r2 rc = Þ rc = 1 1 M m1 + m2 C r1 rC m2 From the result obtained above we have, r2 x 21 m x + m2 x 2 m y + m2 y2 O xc = 1 1 and yc = 1 1 m1 + m2 m1 + m2 z 0- 02 r If we assume origin to be at the centre of mass, then the vector rc vanishes and we have :2 r r r m1 r1 + m2 r2 = 0. n io y Since neither of the masses m1 and m2 can be negative, to satisfy the ss m1 r r Se above equation, vectors r1 and r2 must have opposite signs. It is geometrically Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_02\03-CENTRE OF MASS & COLLISIONS\01-THEORY.P65 r1 C x possible only when the centre of mass C lies between the two particles on r2 the line joining them as shown in the figure. m2 r r If we substitute magnitudes r1 and r2 of vectors r1 and r2 in the above equation, we have r1 m2 m1r1 = m2r2 , or r = m. 2 1 We conclude that the centre of mass of the two particles system lies between the two particles on the line joining them which divides the distance between them in the inverse ratio of their respective masses. Consider two particles of masses m1 and m2 at a distance r from each other. Their centre of mass C must lie in between them on the line joining them. Let the distances of these particles from the centre of mass be r1 and r2. r r1 r2 m1 C m2 E  ® 72 Pre-Medical : Physics ALLEN Since centre of mass of a two particles system lies between the two particles on the line joining them which divides the distance between them in the inverse ratio of masses of the particles, we can write m2 r m1 r r1 = and r2 = m + m m1 + m2 1 2 y l Centre of mass (COM) of several Particles m1(x1, y1, z 1) If the co-ordinates of particles of masses m1, m2,.... are respectively m2(x2, y2, z 2) m3(x3, y3, z3) (x1, y1, z1), (x2, y2, z2).... r1 r3 r2 then position vector of their centre of mass is mn(xn, yn, zn) r rn R CM = xcm î + ycm $j + zcm k̂ x (0,0,0) z = ˆ + m (x ˆi + y ˆj + z k) m1 (x1ˆi + y1ˆj + z1k) 2 2 2 2 3 ( 3 3 3 ) ˆ + m x ˆi + y ˆj + z kˆ +... m1 + m2 + m3 +... (m1x1 + m2x2 +....)iˆ + (m1y1 + m2 y 2...)jˆ + (m1z1 + m2z 2 +..)kˆ = m1 + m2 + m3 +.. æ m1 x1 + m2 x2 +....... ö æ m1 y1 + m2 y2 +....... ö æ m1z1 + m2z2 +.......ö So, xcm = ç , ycm = ç ÷ , zcm = ç è m1 + m2 + m3 +......ø ÷ è m1 + m2 +.......... ø è m1 + m2 +......... ÷ø 21 0- 1.2 Centre of Mass of Continuous Distribution of Mass 02 y If a system has continuous distribution of mass, treating the mass element r :2 dm at position r as a point mass and replacing summation by integration. r 1 r n R CM = ò rdm ; where m = ò dm dm io M r ss 1 1 1 ò Se Mò y dm and z cm = ò z dm Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_02\03-CENTRE OF MASS & COLLISIONS\01-THEORY.P65 So that x cm = x dm , y cm = x M M (0,0,0) z 1.3 Centre of mass of composite bodies y In order to find the centre of mass, the component bodies are assumed to be particles of masses equal to the corresponding bodies located at their O x respective centres of masses. Then we use the equation to find the coordinates of the centre of mass of the composite body. To find the centre of mass of the composite body, we first have to calculate the masses of the bodies, because their mass distribution is given. If we denote the surface mass density (mass per unit area) by s then the masses of the bodies assumed to be uniform are Mass of the disc m d =Mass per unit area ´ Area = s (A d ) Mass of the square plate m s = Mass per unit area ´ Area = s ( A s ) Location of centre of mass of the disc º (xd, yd) Location of centre of mass of the square plate º (xp, yp) E  ® ALLEN Pre-Medical : Physics 73 Using eq. corresponding to centre of mass, we obtain its coordinates (xc, yc) of the composite body. m d x d + m s xs m d y d + ms y s xc = and yc = m d + ms md + ms A d xd + As xs A d y d + A s ys = and = A d + As Ad + As 1.4 Centre of mass of truncated bodies To find the centre of mass of truncated bodies or bodies with cavities we can make use of superposition principle that is, if we restore the removed portion in the same place we obtain the original body. The idea is illustrated in the following figure. y y y O x O x O x The removed portion is added to the truncated body keeping their location unchanged relative to the coordinate frame. If a portion of a body is taken out, the remaining portion may be considered as, [Original mass (M) – mass of the removed part (m)] = {original mass (M)} + { – mass of the removed part (m)} 21 Mx - mx ¢ My - my ¢ Mz - mz ¢ 0- The formula changes to : xcm = ; ycm = ; zcm= M-m M-m M-m 02 Where x', y' and z' represent the coordinates of the centre of mass of the removed part. :2 1.5 Centre of gravity n io Centre of gravity of a body is that point where it is assumed that the gravitational force of earth i.e. weight ss of its body acts on it. Se Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_02\03-CENTRE OF MASS & COLLISIONS\01-THEORY.P65 In normal cases, if the acceleration due to gravity remains the same throughout the mass distribution then centre of gravity coincides with the centre of mass and both in turn coincide with the geometrical centre of the body. GOLDEN KEY POINTS CM l There may or may not be any mass present physically at the centre of mass (See figure A, B, C, D) l Centre of mass may be inside or outside a body (See figure A, B, C, D) l Position of centre of mass depends on the shape of the body. (See figure A, B, C, D) l For a given shape, it depends on the distribution of mass within the body and is closer to massive portion. (See figure A,C) E  ® 74 Pre-Medical : Physics ALLEN l For symmetrical bodies having homogeneous distribution of mass it coincides with the centre of symmetry or the geometrical centre. (See figure B,D). l If we know the centre of mass of parts of the system and their masses, we can get the combined centre of mass by treating the parts as particles placed at their respective centre of masses. l It is independent of the co-ordinate system, e.g., the centre of mass of a ring is at its centre whatever be the co-ordinate system. r r l If the origin of co-ordinate system is at the centre of mass, i.e., R CM = 0 , then by definition, 1 r r Smi ri = 0 Þ Sm i ri = 0. M The sum of the moments of the masses of a system about its centre of mass is always zero. Centre of mass of some uniform symmetric bodies are (i) Semicircular ring of radius R (ii) Semicircular disc CM CM 2R 4R p 3p (iii) Hemispherical shell (iv) 21Solid hemisphere CM CM 0- R 3R 2 8 02 :2 (v) Solid cone (vi) Hollow cone n io ss Se Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_02\03-CENTRE OF MASS & COLLISIONS\01-THEORY.P65 h CM h CM h h 4 3 (vii) Circular arc (viii) Sector of a circular plate R r 2q CM 2q CM X X xc r xc r sin q 2R sin q xc = xc = q 3q Note : Here q is in radians. E  ® ALLEN Pre-Medical : Physics 75 Illustrations Illustration 1. æ a a 3ö a;a 3 Three bodies of equal masses are placed at (0, 0), (a, 0) and at ç 2 , 2 ÷. 2 2 è ø Find out the co-ordinates of centre of mass. Solution (a,0) a a 3 0´m+a´m+ ´m a 0´m+0´m+ ´m a 3 (0,0) x CM = 2 = , y CM = 2 = m+m+m 2 m+m+m 6 Illustration 2. Y Calculate the position of the centre of mass of a system consisting of two xc (0,0) (L,0) particles of masses m1 and m2 separated by a distance L, in relative to m1. X m1 m2 Solution L Treating the line joining the two particles as x axis m1 ´ 0 + m2 ´ L m2 L x CM = = , yCM = 0 zCM = 0 m1 + m2 m1 + m2 Illustration 3. Three rods of the same mass are placed as shown in the figure. 21 Calculate the coordinates of the centre of mass of the system. 0- Solution 02 æa ö æ aö æa aö CM of rod OA is at ç ,0 ÷ , CM of rod OB is at ç 0, ÷ and CM of rod AB is at ç , ÷ :2 è2 ø è 2ø è2 2ø n a a io a a m´ +m´0+m´ m´0+m´ +m´ 2 = a 2 = a ss For the system, xcm = 2 Þ ycm = 2 m+m+m 3 m+m+m 3 Se Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_02\03-CENTRE OF MASS & COLLISIONS\01-THEORY.P65 Illustration 4. If the linear density of a rod of length L varies as l = A + Bx, determine the position of its centre of mass. (where x is the distance from one of its ends) Solution Let the X–axis be along the length of the rod with origin at one of its end as shown in figure. As the rod is along x–axis, so, yCM = 0 and zCM = 0 i.e., centre of mass will be on the rod. Now consider an element of rod of length dx at a distance x from the origin, mass of this element dm = ldx = (A + Bx)dx so, L L dx ò xdm ò x(A + Bx)dx AL2 BL3 + x CM = 0L = 0L = 2 3 = L(3A + 2BL) x BL2 3(2A + BL) ò0 dm ò0 (A + Bx)dx AL + 2 Note : (i) If the rod is of uniform density then l = A = constant & B = 0 then xCM= L/2 (ii) If the density of rod varies linearly with x, then l = Bx and A = 0 then xCM = 2L/3 E  ® 76 Pre-Medical : Physics ALLEN Illustration 5. R A disc of radius R is cut off from a uniform thin sheet of metal. A circular hole of radius is now cut out 2 from the disc, with the hole being tangent to the rim of the disc. Find the distance of the centre of mass from the centre of the original disc. Solution We treat the hole as a 'negative mass' object that is combined with the original uncut disc. (When the two are overlapped together, the hole region then has zero mass). By symmetry, the CM lies along the +y–axis in figure, so xCM = 0. With the origin at the centre of the original circle whose mass is assumed to be m. Mass of original uncut circle m1 = m & Location of CM = (0,0) m æ Rö y Mass of hole of negative mass : m2 = ; Location of CM = ç 0, ÷ 4 è 2ø R 2 æ möR x m(0) + ç - ÷ 0 m1 y1 + m2 y2 è 4 ø2 =-R R Thus y CM = = m1 + m2 æ mö 6 m + ç- ÷ è 4 ø So the centre of mass is at the point æ 0, - R ö. çè ÷ 6ø Thus, the required distance is R/6. 21 Illustration 6. 0- Find the position vector of centre of mass of a system of three particles of masses 1 kg, 2 kg and 3 kg located 02 r r r at position vectors r1 = ( 4iˆ + 2ˆj - 3kˆ ) m, r2 = (ˆi - 4ˆj + 2kˆ ) m and r3 = ( 2iˆ - 2ˆj + kˆ ) m respectively. :2 n Solution. io From eq. corresponding to CM, we have ss r Sm rr Se Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_02\03-CENTRE OF MASS & COLLISIONS\01-THEORY.P65 rc = i i M r 1(4iˆ + 2jˆ - 3k) ˆ + 2(iˆ - 4jˆ + 2k) ˆ + 3(2iˆ - 2jˆ + k) ˆ æ 2 ö rc = = ç 2iˆ - 2jˆ + kˆ ÷ m 1+2+3 è 3 ø Illustration 7. Find coordinates of center of mass of a quarter ring of radius r placed in the first quadrant of a Cartesian coordinate system, with centre at origin. Solution. y Making use of the result of circular arc, distance OC of the center of mass from yc C r sin ( p/ 4 ) 2 2r the center is OC = =. Coordinates of the center of mass p/4 p p/4 O xc x æ 2r 2r ö (xc, yc) are ç , ÷ è p pø E  ® ALLEN Pre-Medical : Physics 77 Illustration 8. Find coordinates of center of mass of a semicircular ring of radius r placed symmetric to the y-axis of a Cartesian coordinate system. Solution. The y-axis is the line of symmetry, therefore center of mass of the ring y lies on it making x-coordinate zero. Distance OC of center of mass from center is given by the result obtained C for circular arc yc p/2 r sin q r sin ( p / 2 ) 2r æ 2r ö O x OC = Þ yc = = , So coordinates are ç 0, ÷ q p/2 p è pø Illustration 9. Find coordinates of center of mass of a quarter sector of a uniform disk of radius r placed in the first quadrant of a Cartesian coordinate system with centre at origin. Solution. From the result obtained for sector of circular plate distance OC of the center of mass form the center is y 2r sin ( p / 4 ) 4 2r OC = = 3p / 4 3p yc C æ 4r 4r ö p/4 21 Coordinates of the center of mass (xc, yc) are ç , ÷ O xc x è 3p 3p ø 0- 02 BEGINNER'S BOX-1 :2 1. What are the co–ordinates of the centre of mass of the three particles system shown in figure? n y (m) io ss Se Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_02\03-CENTRE OF MASS & COLLISIONS\01-THEORY.P65 2 1 3 kg x (m) 1 2 3 2. Four particles of masses m, 2m, 3m, 4m are placed at the corners of a square of side 'a' as shown in fig. Find out the co-ordinates of centre of mass. Y X 0 3. A rigid body consists of a 3 kg mass connected to a 2 kg mass by a massless rod. The 3 kg mass is located r r at r1 = (2$i + 5$j ) m and the 2 kg mass at r2 = (4$i + 2$j ) m. Find the position and coordinates of the centre of mass. E  ® 78 Pre-Medical : Physics ALLEN 4. Fig. shows a uniform square plate from which one or more of the four identical squares at the corners will be removed. (a) Where is the centre of mass of the plate originally. (b) Where is the C.M. after square 1 is removed. (c) Where is the C.M. after squares 1 and 2 removed. (d) Where is the C.M. after squares 1 and 3 are removed. (e) Where is the C.M. after squares 1, 2 and 3 are removed. (f) Where is the C.M. after all the four squares are removed. Give your answers in terms of quadrants and axis. 5. Find the centre of mass of a uniform disc of radius 'a' from which a circular section of radius 'b' has been removed. The centre of the hole is at a distance c from the centre of the disc. 2. MOTION OF CENTRE OF MASS 2.1 Motion of Centre of Mass 21 r r r r 0- m r + m2 r2 + m3 r3 +... As for a system of particles, position of centre of mass is give by R CM = 1 1 02 m1 + m2 + m3 +... :2 r r r dr1 dr2 dr3 r r r r m + m + m +... r dR CM m1 v1 + m2 v 2 +... ( ) n d 1 dt 2 dt 3 dt v = = So R CM = Þ velocity of centre of mass CM io dt m1 + m2 + m3 +... dt m1 + m2 +.... ss Se Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_02\03-CENTRE OF MASS & COLLISIONS\01-THEORY.P65 r r r d r m1 a1 + m2 a2 +... Similarly acceleration a CM = ( v CM ) = m + m +.... dt 1 2 r r r r r r r r We can write Mv CM = m1 v1 + m2 v 2 +... = p1 + p2 + p3 +.... [ Q p = mv ] r r r r Mv CM = p CM [ Q Sp i = p CM ] Linear momentum of a system of particles is equal to the product of mass of the system with velocity of its r r d ( Mv CM ) centre of mass. From Newton's second law Fext. = dt r r r If Fext. = 0 then v CM = constant If no external force acts on a system the velocity of its centre of mass remains constant, i.e., velocity of centre of mass is unaffected by internal forces. E  ® ALLEN Pre-Medical : Physics 79 3. APPLICATION OF METHODS OF IMPULSE AND MOMENTUM TO A SYSTEM OF PARTICLES In a phenomenon, when a system changes its configuration, some or all of its particles change their respective locations and momenta. Sum of linear momenta of all the particles equals to the linear momentum due to translation of centre of mass. Impulse momentum theorem : Impulse = Change in momentum r r r r i.e., ò F dt = Dp = p final – p initial 3.1 Conservation of Linear momentum Total linear momentum of a system of particles remains conserved in a time interval in which impulse of external forces is zero. Total momentum of a system of particles cannot change under the action of internal forces and if net impulse of the external forces in a time interval is zero, the total momentum of the system in that time interval will remain conserved. r r p final = p initial The above statement is known as the principle of conservation of momentum. Since force, impulse and momentum are vectors, component of momentum of a system in a particular direction is conserved, if net impulse of all external forces in that direction vanishes. No external force Þ Stationary mass relative to an inertial frame remains at rest Example : Firing a Bullet from a Gun : V M m 21 v If the bullet is the system, the force exerted by trigger will be Bullet external and so the linear momentum of the bullet will change 0- Gun from 0 to mv. This is not the violation of the law of conservation 02 of linear momentum as linear momentum is conserved only in :2 the absence of external force. If the bullet and gun is the system, then the force exerted by trigger will be internal so. n r r r io total momentum of the system ps = p B + pG = constant.....(i) ss ® ® r Now, as initially both bullet and gun are at rest so pB + pG = 0. From this it is evident that : Se Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_02\03-CENTRE OF MASS & COLLISIONS\01-THEORY.P65 r r pG = -pB , i.e., if bullet acquires forward momentum, the gun will acquire equal and opposite (backward) momentum. r r r r mr From (i) mv + MV = 0, i.e, V = - v i.e, if the bullet moves forward, the gun 'recoils' or 'kicks backwards'. M Heavier the gun lesser will be the recoil velocity V. p2 r r p2 Kinetic energy K = and pB = p G = p. Kinetic energy of gun K G = , 2m 2M p2 K m Kinetic energy of bullet K B = \ G = < 1 (Q M >> m). Thus kinetic energy of gun is lesser than 2m KB M that of bullet i.e., kinetic energy of bullet and gun will not be equal. Initial kinetic energy of the system is zero as both are at rest. Final kinetic energy of the system is greater than zero. So, here kinetic energy of the system is not constant but increases. If PE is assumed to be constant then Mechanical energy = (kinetic energy + potential energy) will also increase. However, energy is always conserved. Here chemical energy of gun powder is converted into KE. E  ® 80 Pre-Medical : Physics ALLEN Example : Block-Bullet System : (a) When bullet remains embedded in the block Conserving momentum of bullet and block mv + 0= (M+m) V mv q Velocity of block V =...(i) M+m L Lcosq By conservation of mechanical energy m+M 1 (M + m)V 2 = (M + m)gh Þ V= 2gh....(ii) 2 h m v mv From eqn. (i) and eqn. (ii) = 2gh M+m M (M + m) 2gh Speed of bullet v = , m V2 m2 v 2 Maximum height gained by block h = = 2g 2g(M + m)2 h æ hö Q h = L – L cosq \ cosq = 1 - Þ q = cos–1 çè 1 - L ÷ø L (b) If bullet emerges out of the block Conserving momentum mv + 0 = mv1 + Mv2 m (v – v1) = Mv2........(i) v2 m v m v1 M 21 1 2 Conserving energy Mv2 = Mgh Þ v 2 = 2gh........(ii) 2 0- 02 m2 (v - v1 )2 From eqn. (i) & eqn. (ii) m(v – v1) = M 2gh Þ h = :2 2gM2 Example : Explosion of a Bomb at rest n p2 Conserving momentum io ss r r r r r r r r r 2 p3= p 1 +p22 p1 + p2 + p3 = 0 Þ p3 = - ( p1 + p2 ) Þ p3 = p12 + p22 as p1 ^ p2 Se Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_02\03-CENTRE OF MASS & COLLISIONS\01-THEORY.P65 r r p1 Angle made by p3 with p1 = p + q q p3 r r p Angle made by p3 with p2 = +q 2 æp ö where q = tan–1 ç 2 ÷. è p1 ø p12 p22 p 32 p12 p22 p23 Energy released in explosion = Kf–Ki = 2m + + + + 1 2m 2 2m3 – 0 = 2m1 2m 2 2m 3. Example : Motion of Two Masses Connected to a Spring Consider two blocks, resting on a frictionless surface and connected by a massless spring as shown in figure. If the spring is stretched (or compressed) and then released from rest, r r r v2 v1 Then Fext=0 so p s = p1 + p2 = constant r r r m2 m1 However, initially both the blocks were at rest so, p1 + p2 = 0 E  ® ALLEN Pre-Medical : Physics 81 It is clear that : r r l p2 = -p1 , i.e., at any instant the two blocks will have momentum equal in magnitude but opposite in direction (though they have different values of momentum at different positions). r r r r r r æm ö r l As momentum p = mv , m1 v1 + m2 v 2 = 0 Þ v 2 = - ç 1 ÷ v1 è m2 ø The two blocks always move in opposite directions with lighter block moving faster. p2 r r KE1 m2 l Kinetic energy KE = and p1 = p2 , KE = m or the kinetic energy of two blocks will not be 2m 2 1 equal but in the inverse ratio of their masses and so lighter block will have greater kinetic energy. l Initially kinetic energy of the blocks is zero (as both are at rest) but after some time kinetic energy of the blocks is not zero (as both are in motion). So, kinetic energy is not constant but changes. Here during the motion of the blocks KE is converted into elastic potential energy of the spring and vice– versa but total mechanical energy of the system remain constant. Kinetic energy + Potential energy = Mechanical Energy = Constant GOLDEN KEY POINTS l For an isolated system, initial momentum of the system is equal to the final momentum of the system. If the system consists of n bodies having momenta 21 r r r r r r r r p1 , p2 , p 3 ,.......... p n , then p1 + p2 + p3 +.......... + p n = constant 0- 02 l As linear momentum depends on frame of reference, observers in different frames would find different values of linear momenta of a given system but each would agree that his own value of linear momentum does :2 not change with time. But the system should be isolated and closed, i.e., law of conservation of linear momentum n is independent of frame of reference though linear momentum depends on the frame of reference. io l Conservation of linear momentum is equivalent to Newton's III law of motion for a system of two particles. ss In the absence of external force from law of conservation of linear momentum, Se Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_02\03-CENTRE OF MASS & COLLISIONS\01-THEORY.P65 r r r r Þ p1 + p2 = constant i.e. m1 v1 + m2 v 2 = constant r r r dv 1 dv Differentiating the above expression with respect to time m1 + m2 2 = 0 [as m is constant] dt dt r r r r dv1 r r r r r r r Þ m1a1 + m2 a2 = 0 [Q =a] Þ F1 + F2 = 0 [Q F = mar ] Þ F1 = -F2 dt i.e., for every action there is equal and opposite reaction which is Newton's III law of motion. l This law is universal, i.e., it applies to macroscopic as well as microscopic systems. r r l Sum of mass moments in centroidal frame (i.e. centre of mass frame) become zero. It implies Sm n rn = 0 r r r r or m1 r1 + m2 r2 +............... + mn rn = 0. r r l Total linear momentum of the system in centroidal frame is zero. It implies Smn v n = 0 or r r r r m1 v1 + m2 v 2 +............... + mn v n = 0. E  ® 82 Pre-Medical : Physics ALLEN Illustrations Illustration 10. Two particles of masses 1 kg and 0.5 kg are moving in the same direction with speeds of 2 m/s and 6 m/s, respectively, on a smooth horizontal surface. Find the speed of the centre of mass of the system. Solution r r r m1 v 1 + m2 v 2 Velocity of centre of mass of the system v cm = m + m. Since the two particles are moving in same 1 2 r r direction, m1 v1 and m2 v 2 are parallel. r r Þ m1 v 1 + m2 v 2 = m1 v1 + m2 v 2. r r m1 v1 + m2 v 2 m1 v 1 + m 2 v 2 (1)(2) + æçè 12 ö÷ø (6) Therefore, v cm = = = = 3.33 m/s. m1 + m2 m1 + m2 æ 1ö çè 1 + 2 ÷ø Illustration 11. Two particles of masses 2 kg and 4 kg are approaching towards each other with accelerations of 1 m/s 2 and 2 m/s2 respectively, on a smooth horizontal surface. Find the acceleration of centre of mass of the system. Solution 21 r r r r 0- r m1a1 + m2a2 m1a1 + m2a2 The acceleration of centre of mass of the system a = Þ a = 02 cm cm m1 + m2 m1 + m2 :2 r r m1 a1 - m2 a 2 (2)(1) - ( 4)(2) Since a1 and a2 are anti–parallel, so a cm = m + m = 1 m/s2. n = 2+4 io 1 2 ss Since m2a2>m1a1 so the direction of acceleration of centre of mass is along in the direction of a2. Se Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_02\03-CENTRE OF MASS & COLLISIONS\01-THEORY.P65 Illustration 12. A block of mass M is placed on the top of a bigger block of mass 10M as shown in figure. All the surfaces are frictionless. The system is released from rest. 2.2 m Find the distance moved by the bigger block at the instant when the smaller block reaches the ground. Solution If the bigger block moves toward right by a distance (x) then the smaller block will move toward left by a distance (2.2 – x). Now considering both the blocks together as a system, horizontal position of CM remains same. As the sum of mass moments about centre of mass is zero i.e. åm x i i / cm = 0. M(2.2 – x) = 10 Mx Þ x = 0.2 m. E  ® ALLEN Pre-Medical : Physics 83 Illustration 13. Two blocks A and B are joined together with a compressed spring. When the system is released, the two blocks appear to be moving with unequal speeds in opposite directions as shown in figure. Select the correct statement : 10m/s 10 m/s 15m/s 15 m/s -1 K=500Nm K = 500 Nm –1 A B (A) The centre of mass of the system will remain stationary. (B) Mass of block A is equal to that of block B. (C) The centre of mass of the system will move towards right. (D) It is an impossible physical situation. Solution Ans. (A) As net force on the system = 0 (after being released) So centre of mass of the system remains stationary. Illustration 14. A man of mass 80 kg stands on a plank of mass 40 kg. The plank is lying on a smooth horizontal floor. Initially both are at rest. The man starts walking on the plank towards north and stops after moving a distance of 6 m on the plank. Then 21 (A) the centre of mass of plank-man system remains stationary. 0- (B) the plank will slide to the north by a distance of 4 m 02 (C) the plank will slide to the south by a distance of 4 m :2 (D) the plank will slide to the south by a distance of 12 m n Solution Ans. (A,C) io ss Since net force is zero so centre of mass remains stationary 6m Se Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_02\03-CENTRE OF MASS & COLLISIONS\01-THEORY.P65 Let x be the displacement of the plank. Since CM of the system remains stationary south x north so 80 (6–x) = 40 x Þ 12 – 2x = x Þ x = 4 m. Illustration 15. Two bodies of masses m1 and m2 ( KEafter collision (c) Perfect inelastic collision : Both the bodies stick together after collision. momentum remains conserved in all types of collisions. 21 0- 02 :2 n io ss Se Z:\NODE02\B0B0-BA\TARGET\PHY\ENG\MODULE_02\03-CENTRE OF MASS & COLLISIONS\01-THEORY.P65 Head on Elastic collision The head on elastic collision is one in which the colliding bodies move along the same straight line path before and after the collision. u1 u2 v1 v2 A B A B A B m1 m2 m1 m2