Maths問題請教 , 各位知識朋友 , 入黎幫我手,要有步驟,唔要得個答案-.-

a)By conservation of energy,
When the hand tool drops from 50m,it's intial speed is zero,that means it's kE is zero,but when it just reaches the ground,it's potential energy is zero,but it has greater vertical speed.
So Gain in KE=Gain in PE
0.5mv^2=mgh
v=√2gh
v=31.6ms-1
b)The KE of the hand tool before collision:
(0.5)(1)(√2x10x50)^2
=500J
The KE of the hand tool after collision:
(0.5)(1)(100^2)
=5000J
Note that for the KE of the hand tool after collision,the mass in the equation should be the mass of the hand tool.

2008-07-17 18:50:39 補充：
When the hand tool drops from 50m,it's KE is zero because it's velocty is zerobut it has greatest potential energy ,the toal mechanical energy at that moment is the potential energy ,i.e.mgh=1x10x50

2008-07-17 18:50:42 補充：
When the hand tool just reaches the ground,it's PE is zero(because it's 0m from the ground),but it has greatest KE,the total mechanical energy at that moment is the Kinetic energy i.e.0.5mv^2=0.5v^2
since energy is conserved,so mgh=0.5mv^2

A) K.E.=P.E.
=mgs
=1(10)(50)

K.E.=1/2 mv'2


B) m1u1+m2u2=m1v1+m2v2
=(m1+m2)V because they move together