98 Mr. D. L. Chapman on the 



a uniform velocity. Therefore during the explosion momentum 

 is generated by the moving piston. In an actual explosion in 

 a tube not provided with a piston the whole mass of gas cannot 

 move forward with this uniform velocity, for there would then 

 be a vacuum at the end of the tube where the explosion started, 

 and the pressure at that end would be zero, making it im- 

 possible to account for the generation of momentum. There 

 is, however, no need to assume that the whole exploded gas 

 acquires a uniform velocity. In fact the velocity of the wave 

 would be the same if it were followed by a layer of exploded 

 gas of uniform density and velocity, and would be un affected 

 by any subsequent disturbance which must take place behind 

 the explosive wave. 



It is therefore necessary to prove that behind the explosive 

 wave there is a layer of homogeneous gas. This evidently 

 must he if any disturbance behind the wave can only move 

 forward with a velocity less than that of the wave itself. 



The forward velocity of any disturbance in the exploded 

 gas will be given by the sum of the velocity of the gas and 

 the velocity of sound in the gas. 



The velocity of the gas 



=V-u=Y(l-?-) 



v Vc„+cJ 



'V 



The velocity of sound 



_ /m&t Up 



In the complete expression for V 2 the first term may be 

 here neglected. Also in the complete expression for t 

 (equation (3)) the last three terms are small. We may 

 therefore write 



and 



^Y 2 



h +2^(v-Vo) 



Employing these values, the velocity of the gas becomes 



V— fere) 



