Calculating Time of Contact for Hammer and Nail

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Homework Statement


A 1.2- kg hammer hits a nail at a speed of 15 m/s and rebounds at 60 percent of that speed. The resisting force of the nail is 9000 N. Approximately how long is the hammer in contact with the head of the nail?



Homework Equations



P=mv
Pf-Pi=Ft

The Attempt at a Solution


This is what is was told to do:
Find P of hammer before, then find v after by mult.
by the % (60%= .60). Use that to find P after. add them,
then divide by the force to find time. deltaP=N*t

before: 1.2*15=18
18*.6=10.8
10.8*1.2=12.24
12.24+18=30.24
30.24/9000
answer=0.00336s
i don't understand why this answer is wrong
any help would be greatly appreciated.
 
Last edited:
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the 80% is an example from how i was told to do it my percent is 60% i forgot to fix that is that the problem with line two you are talking about?
i'll edit it right now.
 
ok i see so instead of what i did first it would be:
1.2*15=18
15*.6=9
9*1.2=10.8
10.8+18=28.8
28.8/9000=
0.0032s
 
Cole07 said:
ok i see so instead of what i did first it would be:
1.2*15=18
15*.6=9
9*1.2=10.8
10.8+18=28.8
28.8/9000=
0.0032s
Yes, but you may have got the answer inadvertently, since you did not proceed systematically. Forgeting the units at first, the initial momentum P_i is 18 down, the final momentum P_f is 10.8 up. Using down as the positive direction, the change in momentum is P_f - P_i = -10.8 - (18) =
- 28.8 Kg.m/s (that is up). So the force on the hammer acts up? Did you understand this when you added the 2 momenta??
 
i did realize it but i just didn't pay any attention to it on this problem thank you for pointing that out.
 
okay so you guys figured out the speed which is exellent how much force in pounds on average does it take to extract a nail from wood ? !
i can't figure it out what so ever ? !