Q. A lead bullet strikes against a steel plate with a velocity 200 m/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 of the bullet is (specific heat capacity of lead=${125Jkg}^{- 1}$ $K^{- 1}$)
A) 80${^\circ}$C
B) 60${^\circ}$C
C) 40${^\circ}$C
D) 120${^\circ}$C
Explanation
(a) According to the question, $\frac{1}{2} \times \frac{1}{2}\ mv^{2} = m \times s \times \mathrm{\Delta}T$
$\frac{1}{4}\ m \times 4 \times 10^{4} = 125 \times m \times \mathrm{\Delta}T$
$\mathrm{\Delta}T = \ \frac{4 \times 10^{4}}{500} = \ 80℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.