Physics

Physics

Problem

A bullet of mass 5 g, travelling with a speed of 210 m/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 cal/(g⋅∘C)0.030 \text{ cal}/(\text{g}\cdot^\circ\text{C}) ( 1 cal=4.2×107 ergs1 \text{ cal} = 4.2 \times 10^7 \text{ ergs} ) close to : a. 87.5∘C87.5^\circ\text{C} b. 83.3∘C83.3^\circ\text{C} c. 119.2∘C119.2^\circ\text{C} d. 38.4∘C38.4^\circ\text{C}
Intuitive explanation

Energy conservation is your key here: the bullet's kinetic energy gets split equally between heating itself and the wood, so you only need to track half the original KEKE going into the bullet. Convert that energy into calories using the given conversion factor, then use Q=mcΔTQ = mc\Delta T to find the temperature rise—the specific heat tells you how much energy is needed per gram per degree.

Intuitive explanation

Energy conservation is your key here: the bullet's kinetic energy gets split equally between heating itself and the wood, so you only need to track half the original KEKE going into the bullet. Convert that energy into calories using the given conversion factor, then use Q=mcΔTQ = mc\Delta T to find the temperature rise—the specific heat tells you how much energy is needed per gram per degree.

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