Question

A lead bullet strikes against a steel plate with a velocity $200m/s$. If the impact is perfectly inelastic and the heat produced is equally shared between the bullet and the target, then the rise in temperature (in ${}^{\circ }C$) of the bullet is (specific heat capacity of lead $s=125J/kgK$)

Answer 1

We know that: $Q=ms\Delta \theta$

Given: $s=125J/kgK,v=200m/s$

Half of kinetic energy is getting converted into heat to increase temperature of lead bullet,

$\frac{1}{2}m{v}^2\times \frac{1}{2}=ms\Delta \theta$

$\frac{(200{)}^2}{4}=125\Delta \theta$

$\Delta \theta ={80}^{\circ }C$

Final answer: ${80}^{\circ }C$