246 MOTION OF SYSTEMS OF PARTICLES 



If the center of gravity of the two bodies is at rest, or, what 

 comes to the same thing, if we measure all velocities relatively to 

 the center of gravity, we have 



mu + m'u' == 0, 



,, m'(u u r ) 



so that "' a ^ 



m + m' 



m(u u') 

 v r = e -/ 

 m -f m' 



Using the relation mu = m'u', these become 



v = eu, 

 v'= eu', 



so that the bodies rebound from one another as though they had 

 impinged on a fixed plane of elasticity e. 



The kinetic energy, either before or after the collision, is equal 

 to the kinetic energy of a single particle moving with the center of 

 gravity, together with that of the system relative to the center of 

 gravity. The former remains unchanged by collision, so that the 

 loss in the total kinetic energy produced by collision is equal to 

 the loss in the kinetic energy relative to the center of gravity. 



If the bodies are smooth, this loss of kinetic energy 



= ^ (mu 2 + m'u' 2 mv 2 .raV 2 ) 

 = ^(mu 2 +m'u' 2 )(l-e 2 ). 



Thus the loss of kinetic energy is (1 e 2 ) times the original kinetic 

 energy relative to the center of gravity. If the bodies are perfectly 

 elastic, e = 1, so that there is no loss of energy ; while if e = 0, the 

 original energy relative to the center of gravity is all lost. 



Impact of Two Smooth Spheres 



202. Let us apply the principles just explained to determining 

 the motion, after impact, of two smooth spheres. 



At the moment of impact let A, B be the centers of the two 

 spheres, so that the line AB is the common normal to the surfaces 

 at the point of impact C. 



