Question

A lead bullet at $550.15K$ 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 $10Xm{s}^{-1}$. Find the value of $X$. (Take lead $=5.5cal{g}^{-1},1cal=4.20J)$

Answer 1

Answer: 280

Let mass of the bullet be ${}^{\prime }{m}^{\prime }g$ . Total heat required to melt the bullet is,
$Q_1=m{c}_l\Delta \theta +mL$
$=[m\times 0.03\times (327-277)]+(m\times 5.5)$
$=(1.5\times m)+(m\times 5.5)cal=(7m\times 4.20)J$
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,
$Q_2=\frac{75}{100}\times \frac{1}{2}m{v}^2\times 10^{-3}=\frac{3}{8}m{v}^2\times 10^{-3}J$
The bullet will melt if $Q_2\ge Q_1$
$\therefore \frac{3}{8}m{v}_{nin}^2\times 10^{-3}=7m\times 4.20$
$\therefore v_{min}^2=\frac{7m\times 4.20}{\frac{3}{8}m\times 10^{-3}}=78400$
$\therefore v=280m/s$