Title CENTRE OF MASS & COLLISION.pdf ✅

 Digital www.allendigital.in [ 47 ] Introduction of Centre of Mass Centre of mass is a point where we can concentrate whole mass of the body and it behave in similar manner as a point object behave under same circumstances. OR The point in a system at which whole mass of the system may be assumed to be concentrated for all translational effects of system is called Centre of Mass (COM). Properties of Centre of Mass (i) For symmetrical bodies having uniform distribution of mass, it coincides with centre of symmetry or geometrical centre. (ii) For a given shape it depends on the distribution of mass within the body and is closer to massive part. (iii) There may or may not be any mass present physically at centre of mass and it may be within or outside the body. Illustration 1: The centre of mass of rigid body always lie inside the body, is this statement true or false? Solution: False. Illustration 2: The centre of mass always lies on the axis of symmetry if it exists. Is this statement true or false? Solution: True. CM CM CM CM CM CM Irregular mass distribution CM CM CM 02 Centre of Mass & Collisions
NEET : Physics [ 48 ] www.allendigital.in  Digital Centre of Mass of Discrete mass System Assume there are n-discrete particles with position vector 1 2 n r ,r ,.......,r respectively as shown in the figure. From figure, we can say that 1 1 1 1 ˆ ˆ ˆ r x i y j z k = + + From definition of COM, Position vector of centre of mass of all n-particles i i cm i m r r m =   1 1 2 2 3 3 n n 1 2 3 n m r m r m r ......... m r m m m ....... m + + + + = + + + + rcm = 1 1 1 1 2 2 2 2 3 3 3 3 n n n n ( ) ( ) 1 2 3 n m (x i y j z k) m (x i y j z k) m x i y j z k ........... m x i y ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ j z k m m m ........................ m + + + + + + + + + + + + + + + + rcm = 1 1 2 2 n n 1 1 2 2 n n 1 1 2 2 n n 1 2 3 n 1 2 3 n 1 2 3 n ˆ ˆ ˆ (m x m x ........ m x )i (m y m y ....... m y )j (m z m z ........ m z )k m m m ....... m m m m ....... m m m m ....... m + + + + + + + + + + + + + + + + + + + + + + rcm = cm cm cm ˆ ˆ ˆ x i y j z k + + Where; 1 1 2 2 n n cm l 2 n m x m x ....... m x x m m ....... m + + + = + + + ; 1 1 2 2 n n cm l 2 n m y m y ....... m y y m m ....... m + + + = + + + ; 1 1 2 2 n n cm l 2 n m z m z ....... m z z m m ....... m + + + = + + + Note : If COM is at origin, r 0 cm =  i i i m r 0 m =    i m r 0 i = m r m r m r ..... 0 1 2 3 1 2 + + + = 3 Illustration 3: Find position of COM? Solution: cm 1(0) 2(2) 3(0) 2 x 6 3 + + = = and cm 1(0) 2(0) 3(3) 3 y 6 2 + + = = So position of COM 2 3 , 3 2       . Illustration 4: Find position of COM? Solution: cm 1(1) 2(2) 3(3).........n(n) x 1 2 3 ......... + + = + + + 2 2 2 2 1 2 3 ........ n 1 2 3 ......... n + + + + = + + + + = n(n 1)(2n 1) 6 n(n 1) 2   + +       +     = 2n 1 3 + Illustration 5: The position vector of three particles of masses m1 = 1 kg, m2 = 2 kg and m3 = 3 kg are 1 ˆ ˆ ˆ r (i 4j k)m, = + + 2 ˆ ˆ ˆ r (i j k)m = + + and 3 ˆ ˆ ˆ r (2i j 2k)m = − − respectively. Find the position vector of their centre of mass? 3kg (0, 3) 1kg (0, 0) 2kg (2, 0) 1kg 2kg 3kg .................... nkg (1, 0) (2, 0) (3, 0) ................ (n, 0)
Centre of Mass & Collisions  Digital www.allendigital.in [ 49 ] Solution: The position vector of COM of the three particles will be given by 1 1 2 2 3 3 COM 1 2 3 m r m r m r r mmm + + = + + Substituting the values, we get COM ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (1)(i 4j k) (2)(i j k) (3)(2i–j–2k) r 1 2 3 + + + + + + = + + 1 ˆ ˆ ˆ (3i j k)m 2 = + − Centre of Mass of Two Particle System Consider two particles of masses m1 and m2 with position vectors 1 r and 2 r respectively. Let their centre of mass C have position vector. From definition, we have i i c m r r M  = 1 1 2 2 c 1 2 m r m r r m m +  = + From the result obtained above we have, 1 1 2 2 c 1 2 m x m x x m m + = + and 1 1 2 2 c 1 2 m y m y y m m + = + If we assume origin to be at the centre of mass, then the vector c r vanishes and we have m r m r 0 1 1 2 2 + = Since neither of the masses m1 and m2 can be negative, to satisfy the above equation, vectors 1 r and 2 r must have opposite signs. It is geometrically possible only when the centre of mass C lies between the two particles on the line joining them as shown in the figure. If we substitute magnitudes r1 and r2 of vectors 1 r and 2 r in the above equation, we have; m1r1 = m2r2  1 2 2 1 r m = r m ...(i) 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 ...(ii) 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, using eqn (i) and (ii) we can write 2 1 1 2 m r r = m + m and 1 2 1 2 m r r = m + m m2 y m1 C x m2 r r1 r2 m1 C
NEET : Physics [ 50 ] www.allendigital.in  Digital Illustration 6: Two particles of mass 1 kg and 2 kg are located at x = 0 and x = 3 m. Find the position of their centre of mass? Solution: 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 Distance of COM from the particle of mass 1 kg is r1 = x and distance of COM from the particle of mass 2 kg is r2 = (3 – x) Using 1 2 r r = 2 1 m m  x 2 3 x 1 = −  x = 2 m Thus, the COM of the two particles is located at x = 2 m. Illustration 7: Calculate the position of the centre of mass of a system consisting of two particles of masses m1 and m2 separated by a distance L, in relative to m1. Solution: Treating the line joining the two particles as x axis 1 2 2 CM CM CM 1 2 1 2 m 0 m L m L x , y 0, z 0 m m m m  +  = = = = + + Centre of Mass of Continuous Mass Distribution If a system has continuous distribution of mass, treating the mass element dm at position r as a point mass and replacing summation by integration: cm rdm r m =  ; where m dm =  cm xdm xdm x dm m = =     cm ydm ydm y dm m = =    cm zdm zdm z dm m = =    Mass Densities Sometimes mass of an object is not uniform in that case we can conveniently express masses in terms of mass density. Types of Mass Densities (a) Linear Mass Density (For 1-D Objects) If we take an element of mass 'dm' having length 'dx' from the rod shown below then linear mass density of the rod, m L = We can write mass of the element dm dx = Total mass of the rod m dm dx = =    Example:- If linear mass density of ring is . Then mass of ring y z x rԦ dm x=0 m1=1kg COM M2=2kg xCM=x x=3 r1=x r2=(3–x) m1 (0,0) (3, 0) m2 xcm Y X L m1 m2 (0,0) (L,0) dx = dm R M =  × 2R