Q147 / 155 · WBJEE 2017 · Physics
A bullet of mass $4.2 \times 10^{-2} \mathrm{kg},$ moving at a speed of $300 \mathrm{ms}^{-1}$, gets stuck into a block with a mass 9 times that of the bullet. If the block is free to move without any kind of friction, the heat generated in the process will be
- A$45 \mathrm{cal}$
- Correct answerB$405 \mathrm{cal}$
- C$450 \mathrm{cal}$
- D$1701 \mathrm{cal}$
Solution
Let mass of bullet = $m$
Mass of block $=\mathrm{M}$
Velocity of bullet $=v=300 \mathrm{m} / \mathrm{s}$
Velocity of combined system $M+m=V$
Here, fhom momentum conservation $\frac{M+m}{m} V=v$
$v=\frac{300 \times 42 \times 10^{-4}}{4.2 \times 10^{-4}+9\left(4.2 \times 10^{-4}\right)}$
$30 \mathrm{ms}$
Now, heat produced = Loss in kinetic energy of bullet
$$
\begin{array}{l}
=\frac{1}{2} m^{2}-\frac{1}{2}(M+m) V^{2} \\
=\frac{1}{2} \times 4.2 \times 10^{-4}(300)^{2}-\frac{1}{2}\left(4.2 \times 10^{-2}\right. \\
+9 \times 4.2 \times 1)(30)^{2} \\
=63 \times 270 \\
=1701 \mathrm{J} \\
=\frac{1701}{4.2} \mathrm{Cal}
\end{array}
$$
$=405 \mathrm{Cal}$
Hence, the correct answer is option: 'B'.