Question

A lead bullet at $550.15 K$ just melts when stopped by an obstacle. Assuming that the obstacle absorbs $25 \%$ of heat, if the minimum velocity of the bullet at the time of strike is $10 X m s ^{-1}$. Find the value of $X$. (Take lead $\left.=5.5 cal g ^{-1}, 1 cal =4.20 J \right)$

Answer 1

Let mass of the bullet be $'m'g$ . Total heat required to melt the bullet is,
$Q_{1}=mc_{l}Δ\theta +mL$
$=\left[\right.m\times 0.03\times \left(\right.327-277\left.\right)\left]\right.+\left(\right.m\times 5.5\left.\right)$
$=\left(\right.1.5\times m\left.\right)+\left(\right.m\times 5.5\left.\right)cal=\left(\right.7m\times 4.20\left.\right)J$
Loss in mechanical energy of the bullet due to obstacle
$=\frac{1}{2}\left( m \times 10^{-3}\right) v ^{2} J$
Now, $25\%$ of this energy is absorbed by the obstacle.
$\therefore $ Energy absorbed by the bullet,
$Q_{2}=\frac{75}{100}\times \frac{1}{2}mv^{2}\times 10^{- 3}=\frac{3}{8}mv^{2}\times 10^{- 3}J$
The bullet will melt if $Q_{2}\geq Q_{1}$
$\therefore \frac{3}{8}mv_{nin}^{2}\times 10^{- 3}=7m\times 4.20$
$\therefore v_{min}^{2}=\frac{7 m \times 4 . 20}{\frac{3}{8} m \times 10^{- 3}}=78400$
$\therefore v=280m/s$