Solution 16.4
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(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} & 0\textrm{.}2\times 1\textrm{....)
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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} & 0\textrm{.}2\times 1\textrm{.}8+0\textrm{.}3\times 0=(0\textrm{.}2+0\textrm{.}3)v \\ & 0\textrm{.}36=0\textrm{.}5v \\ & v=0\textrm{.}72\text{ m}{{\text{s}}^{\text{-1}}} \\ \end{align}
