Mass $\backslash (\backslash$mathrm$\{$m$\}=0\mathrm{.}1\backslash$mathrm$\{$~kg$\}\backslash )$ moves to the right with speed $\backslash (\backslash$mathrm$\{$v$\}=0\mathrm{.}62\backslash$mathrm$\{$~m$\}/\backslash$mathrm$\{$s$\}\backslash )$ and collides with an equal mass initially at rest. After this inelastic collision the system retains a fraction $\backslash (=0\mathrm{.}84\backslash )$ of its original kinetic energy. What is the speed $\backslash (\backslash$mathrm$\{$V$\}\_\{\backslash$mathrm$\{$R$\}\}\backslash )($in $\backslash (\backslash$mathrm$\{$m$\}/\backslash$mathrm$\{$s$\}\backslash ))$ of the mass which was originally at rest before the collision?

Hints: All motion is in $1$D$.$ Ignore friction between the masses and the horizontal surface. You will probably need to use the quadratic formula to solve the resulting equations. $\backslash (\backslash$mathrm$\{$V$\}\_\{\backslash$mathrm$\{$R$\}\}\backslash )$ must be greater than $\backslash (\backslash$mathrm$\{$V$\}\_\{\backslash$mathrm$\{$L$\}\}\backslash )$ since the masses can't pass through each other!

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