Solution 16.1
From Mechanics
(Difference between revisions)
(New page: Using <math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math> and <math>{{u}_{A}}={{u}_{B}}</math> <math>\begin{align} & 1000\times 20+4000\times 10=(1...) |
(New page: Using <math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math> and <math>{{u}_{A}}={{u}_{B}}</math> <math>\begin{align} & 1000\times 20+4000\times 10=(1...) |
Current revision
Using
\displaystyle {{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}
and
\displaystyle {{u}_{A}}={{u}_{B}}
\displaystyle \begin{align} & 1000\times 20+4000\times 10=(1000+4000)v \\ & v=\frac{60000}{5000}=12\text{ m}{{\text{s}}^{\text{-1}}} \\ \end{align}