
By the conservation of linear momentum,
m1v=(m+m1)v1
⇒v1=m+m1m1v...(1)
After collision by work energy theorem, we have
Wfriction=Δk
⇒−μ(m+m1)gx=−21(m+m1)v12
⇒μgx=21v12
⇒0.05×10×2=21(m+m1m1v)2
⇒1005×10×4=(100050)2×(10+100050)2v2
⇒2=4001(201)2v2×400
v2=(201)22
v=2×201
For a freely falling body,
v′=2gH
10v=2gH
102×201=2×gH
100(201)2=10H
H=1000(201)2=40m
=0.04km