228 ANSWERS. 



Pages 24, 25. 



(1) (b) 250 pounds. 



(2) (a) 8 ft. per second; (b) 20 ft. 



(3) ( a ) 825 pounds; (b) i\ miles; (c) about 1000 pounds. 



(4) 4-9 sec. 



(5) It would be greater by -^ oz. 



(6) j == (OT! sin 0! m 2 sin # 2 /x^ cos 0! /x 2 w 2 cos 

 7"= (sin 0j -f sin 2 /A! cos 0j + /x, 2 cos 2 ) niim. 2 g/(m^ + 



(7) /= 5.4 ft. per second ; T= f pound. 



(9) 0-0363. (n) (0) 1528 pounds; (b) 1910 pounds. 



(10) 0-025. ( I2 ) 5**9 ft- 



Pages 33, 34. 



(1) (a) 15 270 foot-pounds; (b) 30^ ft. 



(2) (a) 917 pounds; (b) 1557 pounds; (c) 640 pounds. 



(3) (a) 1267 foot-tons; (b) 4435 foot-tons; (c) 5 : 2. 



(4) 2016 foot-tons. 



(5) 864 ft. per second. 



(7) () ^ = 



(8) * = 



(9) At the time /, let s be the distance of mi and m 2 ; s lt J a 

 their distances from their initial positions, so that s l + s + s 2 = s . 



Then we find s 1= ^ (s -s), s 2 = ^ (SQ-S), and 



^i + ^2 



f ) = 2 K (^! -f w 2 ) [- -). To integrate, put j- = j- cos 2 i<#> ; then 

 \dt) \s sj 



we find /=-\/ - - - (< + sin<f>). The particles meet at the 

 ^ - 



-f 



2 



(ii) (-) . W 



