Question

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