# A block of mass m=3.6 kg, moving on africtionless surface with a speed v_1=9.3 m/s makes a perfectly elastic collision with a block of mass Mat rest.

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A block of mass m=3.6 kg, moving on africtionless surface with a speed $$\displaystyle{v}_{{1}}={9.3}$$ m/s makes a perfectly elastic collision with a block of mass Mat rest. After the collision, the 3.6 kg block recoils with a speed of $$\displaystyle{v}_{{1}}={2.7}$$ m/s in figure, the speed of the vlock of mass M after the collision is closest to:
a. 9.3 m/s
b. 6.6 m/s
c. 8.0 m/s
d. 10.7 m/s
e. 12.0 m/s

2021-02-25
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:
$$\displaystyle{V}={v}_{{i}}+{v}_{{f}}={\left({9.3}\frac{{m}}{{s}}\right)}+{\left(-{2.7}\frac{{m}}{{s}}\right)}={6.6}\frac{{m}}{{s}}$$