Question

A bullet of mass $4.2\times 10^{-2}$ kg, moving at a speed of $300m{s}^{-1}$, 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 cal
• B. 405 cal
• C. 450 cal
• D. 1701 cal

Answer 1

Answer: (B)

Let mass of bullet $=m$
Mass of block $=M$
Velocity of bullet $=v=300m/s$
Velocity of combined system $M+m=V$
Here, from momentum conservation
$\frac{M+m}{m}V=v$
$\Rightarrow V=\frac{300\times 42\times 10^{-2}}{4.2\times 10^{-2}+9\left(4.2\times 10^{-2}\right)}$
$=30m/s$
Now, heat produced = Loss in kinetic energy of builet
$=\frac{1}{2}m{v}^2-\frac{1}{2}(M+m)V^2$
$=\frac{1}{2}\times 4.2\times 10^{-2}(300{)}^2-\frac{1}{2}(4.2\times 10^{-2}$
$+9\times 4.2\times 1)(30{)}^2$
$=6:3\times 270$
$=1701J$
$=\frac{1701}{4.2}Cal$
$=405Cal$