Question

A lead bullet pertrates into a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, if the initial speed (in m/s) of the bullet is 100x100x J/Kg and specific heat capacity of lead = 125 J/kgK.

Answer 1

Given - $t_i={27}^{o}C,t_m={327}^{o}C,L=2.5\times {10}^{4}J/kg,c=125J/kg-K=125J/kg{-}^{o}C$ ,

change in temperature $\Delta t=327-27={300}^{o}C$ ,

let $v$ be the initial velocity of bullet and $m$ be the mass of bullet ,

the heat gained by bullet from kinetic energy , will first raise the temperature of bullet upto its melting point and then rest of heat will be used to melt it , without increasing its temperature , this can be represented by the following equation ,

$\frac{1}{2}\left(\frac{1}{2}m{v}^2\right)=mc\Delta t+mL$ ,

or $\frac{1}{4}{v}^2=c\Delta t+L$ ,

or $v^2=4c\Delta t+4L$ ,

or $v^2=4\times 125\times 300+4\times 2.5\times {10}^4$ ,

or $v^2=15\times {10}^4+10\times {10}^4=25\times {10}^4$ ,

or $v=\sqrt{25\times {10}^4}=500m/s$

Therefore, $x=5$.