Posted by Ryan on Wednesday, March 3, 2010 at 11:08pm.
You 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.
i agree with that but how would you find the kinetic energy acquired by the nail after you get these numbers?
got it
v=[(2*m1)/(m1+m2)]Vo
then take your answer and insert into 1/2mv^2 and divide by 1000
got 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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