236 MOTION OF SYSTEMS OF PARTICLES 



If the mass is not moving in the same direction as the line of 

 action of the impulse, the foregoing result will obviously be true 

 if u, v are taken to be the components of the velocities along the 

 line of action. 



ILLUSTRATIVE EXAMPLES 



1. A shot of 14 pounds is fired into a target of mass 200 pounds which is 

 suspended by chains so that it is free to start into motion horizontally. If the 

 shot, before impact, was moving with a horizontal velocity of 1000 feet a second, 

 and afterwards remains embedded in the target, find the loss of energy caused 

 by the impact. 



Let V denote the horizontal velocity, measured in feet per second, with 

 which the target and bullet together start into motion after the impact. Then, 

 by the conservation of momentum, equating the momentum before the impact 

 to that after, 



1000 x 14 = V x 214, 



so that F 



The kinetic energy before impact was | 14 (1000) 2 ; that afterwards is 

 \ - 214 F 2 . Thus the loss of energy is 



| (14,000,000 - 214 F 2 ) = 6,540,000 foot poundals, approximately. 



2. A heavy chain, of length I and mass m per unit length, is held with a length 

 h hanging over the edge of a table, and the remainder coiled up at the extreme edge 

 of the table. If the chain is set free, find the velocity at any stage of the motion. 



Suppose that at any stage of the motion a length x is hanging vertically, so 

 that a length I x is coiled up on the table. After an infinitesimal time dt let 

 x be supposed to have increased from x to x + dx. Then if v is the downward 

 velocity of the chain, we clearly have 



dt 



At the beginning of the interval dt, the downward momentum of the chain 

 was that of a mass mx moving with a velocity v. It was accordingly mvx. At 

 the end of the interval, the momentum is that of a mass m (x + dx) moving with 

 a velocity which may be denoted by (v + dv). Thus the gain in momentum is 



m (x + dx) (v + dv) mxv, 



or, neglecting the small quantity of the second order dv dx, the gain is 



m(xdv + vdx). 



