262 MISCELLANEOUS METHODS AND APPLICATIONS. [CHAP. XII. 



The reaction between the spheres, being localised in the 

 common normal, has no resolved part at right angles thereto, and 

 thus the momentum of each sphere at right angles to the line of 

 centres is unaltered by the impact. This gives us 

 v = F, and v = V. 



Now we may solve the first two equations for u and u t and we 

 shall obtain 



(m - m e) U + m! (1 + e) U 



U = 7 , 



m +m 

 - me) U + m (1 + e) U 



Thus the motion after impact is determined. 

 The impulsive pressure on m is 



in the sense opposite to that of U, and there is an equal impulsive 

 pressure on m in the opposite sense. Let the magnitude of this 

 pressure be denoted by R. 



To find the kinetic energy lost, we have 

 kinetic energy before impact kinetic energy after impact 

 = \m ( U 2 + F 2 ) + \m ( U * + F 2 ) - {$m ( 

 = mU*- u* + m U z -u * 



since m(U-u) = m (u - U ) = R. 



Hence the kinetic energy lost is ^R(U U )(l e) 



= _ ^_ 



^ m + rn^ 



It is to be noted that the expression 



for the change of kinetic energy produced by the impulse is in 

 accordance with that obtained in the general case of a system of 

 bodies in Article 155. The impulse in the present case is an 

 internal one but it contributes something to the change of kinetic 

 energy, on the other hand the internal impulsive actions between 

 parts of the spheres contribute nothing to the change of kinetic 

 energy, and we shall be able to prove later on that this is always 

 the case for a rigid body. 



