Question:
A bullet of mass $5 \mathrm{~g}$, travelling with a speed of 210 $\mathrm{m} / \mathrm{s}$, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is $0.030 \mathrm{cal} /\left(\mathrm{g}-{ }^{\circ} \mathrm{C}\right)$ ( $1 \mathrm{cal}=4.2 \times 10^{7} \mathrm{ergs}$ ) close to :
Correct Option: , 2
Solution:
$\frac{1}{2} \mathrm{mv}^{2} \times \frac{1}{2}=\mathrm{ms} \Delta \mathrm{T}$
$\Delta \mathrm{T}=\frac{\mathrm{v}^{2}}{4 \times 5}=\frac{210^{2}}{4 \times 30 \times 4.200}$
$=87.5^{\circ} \mathrm{C}$