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April 25, 2014

April 25, 2014

Posted by **Ryan** on Wednesday, March 3, 2010 at 11:08pm.

- Physics -
**drwls**, Thursday, March 4, 2010 at 12:27amYou might consider the collision as nearly elastic for the brief instant before the nail moves into the board. You need to consider conservation of momentum also, to figure out what fraction of the hammer's kinetic energy is transferred to the nail.

By assuming both momentum and total kinetic energy conservation, you should obtain

V = Vo (M-m)/(M+m) = (535/565)*5.1 m/s = 4.83 m/s for the hammer velocity after impact.

For the nail velocity v,

v = (M/m)(Vo -V) = 36.67*.27 m/s

= 9.9 m/s

Try the calculations yourself and see if you agree.

- Physics -
**Kelly**, Tuesday, April 27, 2010 at 10:44pmi agree with that but how would you find the kinetic energy acquired by the nail after you get these numbers?

- Physics -
**Kelly**, Tuesday, April 27, 2010 at 11:20pmgot it

v=[(2*m1)/(m1+m2)]Vo

then take your answer and insert into 1/2mv^2 and divide by 1000

- Physics -
**Kelly**, Tuesday, April 27, 2010 at 11:20pmgot it

v=[(2*m1)/(m1+m2)]Vo

then take your answer and insert into 1/2mv^2 and divide by 1000

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