Question

A bullet of mass $100g$ has a velocity of $210m{s}^{-1}$ in the horizontal direction. It gets embedded in a block of mass $2kg$ which was kept on a rough horizontal surface. Find the distance moved by the block-bullet system (in $m$ ). (Given: coefficient of kinetic friction between the block and surface is $0.5$ & take acceleration due to gravity $=10m{s}^{-2}$ )

Answer 1

Answer: 10

Let mass of bullet be $m$ and mass of block be $M$ .
After collision, the block-bullet system will move with velocity $v_{sys}$
According to conservation of linear momentum,
$mv=(m+M)v_{sys}$
$\therefore 0.1\times 210=2.10\times v_{sy}$
$\therefore v_{sys}=10m/s$
Also, $K.E.$ acquired by system is converted to work done in moving the system,
i.e., according to work-energy thorem,
$\begin{array}{l}\frac{1}{2}(m+M){\left(v_{sys}\right)}^2=F\times s\\ \therefore \frac{1}{2}(m+M)(10=\mu (m+M)g\times s\\ \therefore s=\frac{100}{2\times 0.5\times 10}=10m\end{array}$