Hi,
Answer 2c presented in the first post.
I improved 2b according to your instruction, use conservation of energy.
2b) W=PE PE=KE KE=Tm*v3^2/2
Tm=4400kg
v3=2.7059m/s
W=Tm*v3^2/2=4400*2.7059^2/2=16108.17J
Same answer I received in momentum 1b. Hope is ok.
2d)
m1=1700kg (hammer)
m2=2700kg...
Hi,
This is whole exercise. Thanks KURUMAN for your tips. I found resolution for 2b, 2d but still strugling with 1d. I do not know how to find initial energy in conservation of momentum.
Homework Statement
Hi,
I do not know how to resolve question 1d and 2d at all. I hope the rest is well resolved, not sure about 1c and 2c. Can someone check my work and help me resolve problem with this questions, please?
Task
A hammer of mass 1700kg is dropped 2.5m under its own in fluency of...