Centre of Mass & Collisions Tips and Formulas

1. Impulse-Momentum theorem r r Fdt   p t2 Impulse of a force is equal to the change of momentum force time graph area gives change in momentum. 2. t1 Collision of bodies The event of the process, in which two bodies either coming in contact with each other or due to mutual interaction at distance apart, affect each others motion (velocity, momentum, energy or direction of motion) is defined as a collision. 3. In collision The particles come closer before collision and after collision they either stick together or move away from each other. The particles need not come in contact with each other for a collision. The law of conservation of linear momentum is necessarily applicable in a collision, whereas the law of conservation of mechanical energy is not. 4. Coefficient of restitution (Newton`s law) e velocity of separation along line of impact v 2  v1  velocity of approach along line of impact u1  u 2 Value of e is 1 for elastic collision, 0 for perfectly inelastic collision and 0 < e < 1 for inelastic collision. 5. Head on collision A 6. u1 B u2 A m2 m1 Before Collision B Collision Head on inelastic collision of two particles A v1 m1 m2 After Collision Let the coefficient of restitution for collision is e (a) Momentum is conserved m1u1 + m2u2 = m1v1 +m2v2 (b) Kinetic energy is not conserved v v 2 1 (c) According to Newton`s law e  u  u 1 2 ...(i) ....(ii) By solving eq. (i) and (ii) : m1u1  m2 u 2  m2 e  u1  u 2   1  e  m2   m  em2  v1   1 u 2   u1   m1  m2  m1  m2   m1  m2  m1u1  m 2 u 2  m1e  u 2  u1   1  e  m1   m  em1  v2   2  u1   u2   m1  m 2  m1  m 2   m1  m 2  7. B Elastic Collision (e = 1) If the two bodies are of equal masses : m1 = m2 = m, v1 = u2 and v2 = u1 Thus, if two bodies of equal masses undergo elastic collision in one dimension, then after the collision, the bodies will exchange their velocities. v2 x CENTRE OF MASS & COLLISIONS If the mass of a body is negligible as compared to other. If m1 >> m2 and u2 = 0 then v1 = u1, v2 = 2u1 when a heavy body A collides against a light body B at rest, the body A should keep on moving with same velocity and the body B will move with velocity double that of A. If m2 >> m1 and u2 = 0 then v2 = 0, v1 = – u1 When light body A collides against a heavy body B at rest, the body A should starts moving with same velocity just in opposite direction while the body B should practically remains at rest. 8. Loss in kinetic energy in inelastic collision K  9. m1m 2 2 1  e 2   u1  u 2   2  m1  m 2  Oblique Collision Conserving the momentum of system in directions along normal (x-axis in our case) and tangential (y axis our case) m1u1 cos1 + m2u2cos2 = m1v1cos 2 – m1v1sin 1 m2u2 sin2 – m1u1sin1 = m2v2sin 2 – m1v1sin 1 y u2 m2 m1 v2 m2 m2 x u1 Before Collision v1 After Collision Since no force is acting on m1 and m2 along the tangent (i.e. y-axis) the individual momentum of m1 and m2 remains conserved. m1u1sin1 = m1v1sin 1 & m2u2 sin2 m2v2sin 2 By using Newton`s experimental law along the line of impact e 10. v 2 cos 2  v1 cos 1 u1 cos 1  u 2 cos  2 Centre of mass For a system of particles centre of mass is that point at which its total mass is supposed to be concentrated. 11. Centre of mass of system of discrete particles Total mass of the body : M = m1 + m2 + .... + mn then r r r r m1 r1  m 2 r2  m3 r3  .... 1 r R CM    m1 r1 m1  m 2  m 3  ... M co-ordinates of centre of mass : x cm  12. 1 1 1  mi x i , ycm  M  mi yi and z cm  M  mi zi M Centre of mass of continuous distribution of particles y r 1 r R cm   rdm M x cm  1 1 x dm , ycm   y dm  M M 1 z dm M x, y, z are the co-ordinate of the COM of the dm mass. r and z cm  13. dm (0,0,0) x z The centre of mass after removal of a part of a body Original mass (M) – mass of the removed part (m) = {original mass(M)} + {– mass of the removed part (m)} The formula changes to : x cm  14. Mx  mx  My  my Mz  mz  ; y cm  ; ycm  Mm Mm Mm MOTION OF CENTRE OF MASS For a system of particles, r r r dR CM m1 v1  m 2 v 2  ..... r  velocity of centre of mass v CM  dt m1  m 2  .... r r m1a1  m 2 a 2  .... r d r Similarly acceleration a CM    vCM  dt m1  m 2  .... 15. Law of conservation of linear momentum Linear momentum of a system of particles is equal to the product of mass of the system with velocity of its centre of mass. r r d  MvCM  From Newton`s second law Fext.  dt r r r If Fext. = 0 then Mv CM = constant 16. 17. 18. 19. 20. 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. r r Sum of mass moments about centre of mass is zero. i.e.  mi ri /cm  0 A quick collision between two bodies is more violent then slow collision, even when initial and final velocities are equal because the rate of change of momentum determines that the impulsive force small or large. Heavy water is used as moderator in nuclear reactors as energy transfer is maximum if m1 = m2. Impulse momentum theorem is equivalent to Newton`s second law of motion For a system, conservation of linear momentum is equivalent to Newton`s third law of motion.