Question

A bullet of mass $100g$ has a velocity of $210ms^{- 1}$ 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 $=10ms^{- 2}$ )

Answer 1

Let mass of bullet be $m$ and mass of block be $M$ .
After collision, the block-bullet system will move with velocity $v_{s y s}$
According to conservation of linear momentum,
$mv=\left(\right.m+M\left.\right)v_{sys}$
$\therefore 0.1\times 210=2.10\times v_{sy}$
$\therefore v_{s y s}=10m/s$
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_{s y s}\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}=10 m \end{array} $