Then F = Delta P/Delta T = (-7.2 kgm/s)/(0.002s) = -3600 N

[reincarnate]

1. The head of a 1.0 kg hammer, moving at 3.6 m/s, strikes a nail and rives it into hardwood.

a. The head stays in contact 2.0 ms and rebounds with negligible velocity. What is the average force exerted on the nail?

b. When the same hammer hits a springy nail, it rebounds with its initial speed, 3.6 m/s. The contact time is the same. What force is exerted this time?


For part a:

P(hammer) = (1kg)(3.6m/s) = 3.1 kgm/s Delta P = Pf - Pi = 3.6 kgm/s - 0 kgm/s = 3.6 kgm/s
Delta P = F * Delta T F = Delta P/Delta T = (3.6kgm/s)/(0.002s) = 1800 N

Not sure if i did this part correctly...Anyone like to confirm my answer?

For part b:
This part I'm pretty confused...Not sure what I should start to do..
Do I find the change in momentum of the hammer first?
Delta P = m*delta v
= m*(Vf - Vi)
= (1kg)(-3.6 m/s - 3.6 m/s)
= -7.2 kgm/s

Is that right?

 

[Filip Larsen]

You are correct in part a and on the right track for part b.