Let:

\(\displaystyle{m}={3.6}{k}{g}\)

\(\displaystyle{v}_{{i}}={9.3}\frac{{m}}{{s}}\)

\(\displaystyle{v}_{{f}}=-{2.7}\frac{{m}}{{s}}\)

V=to be found

M=to get rid of

Conservation of momentun:

p initial=p final:

\(\displaystyle{m}{v}_{{i}}={m}{v}_{{f}}+{M}{V}\)

Solving for M:

\(\displaystyle{M}{V}={m}{v}_{{i}}-{m}{v}_{{f}}\)

\(\displaystyle{M}={\frac{{{m}{v}_{{i}}-{m}{v}_{{f}}}}{{{V}}}}={\frac{{{m}{\left({v}_{{i}}-{v}_{{f}}\right)}}}{{{V}}}}\)

Conservation of kinetic energy:

KE initial=KE final:

\(\displaystyle{.5}{{v}_{{i}}^{{2}}}={.5}{m}{{v}_{{f}}^{{2}}}+{.5}{M}{V}^{{2}}\)

Getting rid of the .5

\(\displaystyle{m}{{v}_{{i}}^{{2}}}={m}{{v}_{{f}}^{{2}}}+{M}{V}^{{2}}\)

Solving for M:

\(\displaystyle{M}{V}^{{2}}={m}{{v}_{{i}}^{{2}}}-{m}{{v}_{{f}}^{{2}}}\)

\(\displaystyle{M}={\frac{{{m}{{v}_{{i}}^{{2}}}-{m}{{v}_{{f}}^{{2}}}}}{{{V}}}}={\frac{{{m}{\left({{v}_{{i}}^{{2}}}-{{v}_{{f}}^{{2}}}\right\rbrace}{\left\lbrace{V}\right\rbrace}}}{}}\)

Setting both expressions for M equal:

\(\displaystyle{\frac{{{m}{\left({v}_{{i}}-{v}_{{f}}\right)}}}{{{V}}}}={\frac{{{m}{\left({{v}_{{i}}^{{2}}}-{{v}_{{f}}^{{2}}}\right)}}}{{{V}^{{2}}}}}\)

Canceling out m and knowing that \(\displaystyle{\left({{v}_{{i}}^{{2}}}-{{v}_{{f}}^{{2}}}\right)}\) is the difference of two squares:

\(\displaystyle{\frac{{{v}_{{i}}-{v}_{{f}}}}{{{V}}}}={\frac{{{\left({v}_{{i}}-{v}_{{f}}\right)}{\left({v}_{{i}}+{v}_{{f}}\right)}}}{{{V}^{{2}}}}}\)

Solving for V:

\(\displaystyle{V}={v}_{{i}}+{v}_{{f}}={\left({9.3}\frac{{m}}{{s}}\right)}+{\left(-{2.7}\frac{{m}}{{s}}\right)}={6.6}\frac{{m}}{{s}}\)

\(\displaystyle{m}={3.6}{k}{g}\)

\(\displaystyle{v}_{{i}}={9.3}\frac{{m}}{{s}}\)

\(\displaystyle{v}_{{f}}=-{2.7}\frac{{m}}{{s}}\)

V=to be found

M=to get rid of

Conservation of momentun:

p initial=p final:

\(\displaystyle{m}{v}_{{i}}={m}{v}_{{f}}+{M}{V}\)

Solving for M:

\(\displaystyle{M}{V}={m}{v}_{{i}}-{m}{v}_{{f}}\)

\(\displaystyle{M}={\frac{{{m}{v}_{{i}}-{m}{v}_{{f}}}}{{{V}}}}={\frac{{{m}{\left({v}_{{i}}-{v}_{{f}}\right)}}}{{{V}}}}\)

Conservation of kinetic energy:

KE initial=KE final:

\(\displaystyle{.5}{{v}_{{i}}^{{2}}}={.5}{m}{{v}_{{f}}^{{2}}}+{.5}{M}{V}^{{2}}\)

Getting rid of the .5

\(\displaystyle{m}{{v}_{{i}}^{{2}}}={m}{{v}_{{f}}^{{2}}}+{M}{V}^{{2}}\)

Solving for M:

\(\displaystyle{M}{V}^{{2}}={m}{{v}_{{i}}^{{2}}}-{m}{{v}_{{f}}^{{2}}}\)

\(\displaystyle{M}={\frac{{{m}{{v}_{{i}}^{{2}}}-{m}{{v}_{{f}}^{{2}}}}}{{{V}}}}={\frac{{{m}{\left({{v}_{{i}}^{{2}}}-{{v}_{{f}}^{{2}}}\right\rbrace}{\left\lbrace{V}\right\rbrace}}}{}}\)

Setting both expressions for M equal:

\(\displaystyle{\frac{{{m}{\left({v}_{{i}}-{v}_{{f}}\right)}}}{{{V}}}}={\frac{{{m}{\left({{v}_{{i}}^{{2}}}-{{v}_{{f}}^{{2}}}\right)}}}{{{V}^{{2}}}}}\)

Canceling out m and knowing that \(\displaystyle{\left({{v}_{{i}}^{{2}}}-{{v}_{{f}}^{{2}}}\right)}\) is the difference of two squares:

\(\displaystyle{\frac{{{v}_{{i}}-{v}_{{f}}}}{{{V}}}}={\frac{{{\left({v}_{{i}}-{v}_{{f}}\right)}{\left({v}_{{i}}+{v}_{{f}}\right)}}}{{{V}^{{2}}}}}\)

Solving for V:

\(\displaystyle{V}={v}_{{i}}+{v}_{{f}}={\left({9.3}\frac{{m}}{{s}}\right)}+{\left(-{2.7}\frac{{m}}{{s}}\right)}={6.6}\frac{{m}}{{s}}\)