Question

A bullet of mass 5g moving with a speed of 210 m/s hit a stationary plank and comes to rest. If halt of its kinetic energy Ps absorbed by the bullet itself and the remaining half kinetic energy by plank. then what is the size in the temperature of the bulletin °C : (Given specific heat of bullet is 6125 J/g°C) .

Answer 1

Here the final kinetic energy of the bullet is half its initial kinetic energy.

Thus change in kinetic energy,

$\sf{\longrightarrow \Delta K=\dfrac{1}{4}\,mu^2-\dfrac{1}{2}\,mu^2}$

$\sf{\longrightarrow \Delta K=-\dfrac{1}{4}\,mu^2}$

The lost kinetic energy is evolved as heat. Thus we have,

$\sf{\longrightarrow\Delta Q=\Delta K}$

$\sf{\longrightarrow mc\Delta T=-\dfrac{1}{4}\,mu^2}$

Then, change in temperature,

$\sf{\longrightarrow\Delta T=\dfrac{u^2}{4c}}$

$\sf{\longrightarrow\Delta T=\dfrac{210^2}{4\times6125}}$

$\sf{\longrightarrow\underline{\underline{\Delta T=1.8^oC}}}$

Hence (4) is the answer.