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April 17, 2014

Homework Help: physics

Posted by physics on Wednesday, March 11, 2009 at 6:38pm.

here is my question: 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 through it.

here is my answer: mass of bullet = 6gm
= 6/1000 kg = .800kg
let 'v' be speed of block. apply law of conservation of energy
initial energy of (block+bullet)
= final energy of (block+bullet)
kinetic energy=1/2 mV^2 (formula)
solve for v: 1/2x.006*(896)^2=1/2.006*(435)^2 +1/2(1.25)(v)^2

1.25v^2=.006(896^2-435^2)
v=54.27m/s

please show me if there is a simpler way to write this out. please write out the equaiton and fill in the appropriate data.
i think what i have is too confusing

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