In the Figure below, a $3\mathrm{.}50$ g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block $1($mass $1\mathrm{.}39$ kg $)$ and embeds itself in block $2($mass $1\mathrm{.}89$ kg $).$ The blocks $1$ and $2$ end up with speeds $\backslash ($ v$\_\{1\}=0\mathrm{.}600\backslash$mathrm$\{$~m$\}/\backslash$mathrm$\{$s$\}\backslash )$ and $\backslash ($ v$\_\{2\}=1\mathrm{.}44\backslash$mathrm$\{$~m$\}/\backslash$mathrm$\{$s$\}\backslash )$ respectively.

$-$ We assume direction is to the right.

$-$ Neglect the material removed from block $1$ by the bullet.

$-$ Whenever needed, round your final answers to nearest whole numbers.

Bullet is fired horizontally at two blocks at rest on a frictionless table.

PS: Indicate which equation $($law$)$ you are using, brief necessary calculation steps, and final answers must be SI units in the answer box below.

A$)$ By applying the law of conservation of momentum, find the velocity $\backslash (\backslash$overrightarrow$\{$v$\_\{$f$\}\}\backslash )$ of the bullet as it leaves block $1.$

B$)$ Calculate the velocity $\backslash (\backslash$overrightarrow$\{$v$\_\{$i$\}\}\backslash )$ of the bullet as it enters block $1.$

C$)$ What's the type of collision as the bullet embeds itself in block $2?$ Explain why?

D$)$ Compare the Kinetic energy, at the beginning of the motion, as the bullet passes through block $1.$ Can we say the law of conservation of kinetic energy has been applied?