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 Subject: Speed of a bullet. Category: Science > Physics Asked by: alwysnforevr002-ga List Price: \$3.00 Posted: 22 Jul 2006 17:45 PDT Expires: 21 Aug 2006 17:45 PDT Question ID: 748631
 ```A bullet with a mass of 6.00 g is fired through a 1.25 kg block of wood on a frictionless surface. The initial speed of the bullet is 896 m/s, and the speed of the bullet after it exits the block is 435 m/s. At what speed does the block move after the bullet passes through it?```
 ```Seems like a question from school assignment, so just a hint. Apply conservation of linear momentum. Initial momentum of bullet = sum of final momentums of block and bullet```
 ```First step convert masses to equivalents. i.e. 6g bullet becomes 0.06kg Now momentum p=mv where p = momentum m = mass v = velocity There will be conservation of momentum pre and post interaction. Therefore p(bullet before interaction) = 0.06*896 = 53.76 So 53.76 = p(bullet post interaction) + p (block post interaction) 53.76 = (0.06*435) + 1.25v v = 22.128 m/s v = 22.1 m/s to 3 s.f.```