Question

A lead bullet penetrates into a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, calculate the initial speed of the bullet. The initial temperature of the bullet is ${27}^{\circ }C$ and its melting point is ${327}^{\circ }C$. Latent heat of fusion of lead = $2.5\times {10}^{4}Jk{g}^{-1}$ and specific heat capacity of lead = $125Jk{g}^{-1}{K}^{-1}$.

• A. 480 m/s
• B. 5000 m/s
• C. 100 m/s
• D. 1000m/s

Answer 1

The correct option is A

480 m/s

Let the mass of the bullet = $m$.

Heat required to take the bullet from ${27}^{\circ }C$ to ${327}^{\circ }C$.

= $m\times \left(125Jk{g}^{-1}{K}^{-1}\right)\left(300K\right)$

= $m\times (3.75\times {10}^{4}Jk{g}^{-1})$.

Heat required to melt the bullet

= $m\times (2.5\times {10}^{4}Jk{g}^{-1})$

If the initial speed by v, the kinetic energy is $\frac{1}{2}m{v}^2$ and hence the heat developed is

$\frac{1}{2}\left(\frac{1}{2}m{v}^2\right)$ = $\frac{1}{4}m{v}^2$. Thus, $\frac{1}{4}m{v}^2$ = $m(3.75+2.5)\times {10}^{4}Jk{g}^{-1}$

or, $v$ = $480m{s}^{-1}$.