Question

A 15-g lead bullet is tested by firing it into a fixed block of wood with a mass of 35 kg. The block and imbedded bullet together absorb all the heat generated. After thermal equilibrium has been reached, the system has a temperature rise measured as 0.020 C°. Estimate the bullet's entering speed.

Short answer

The bullet's entering speed is \(4.0 \times {10^2}\;{\rm{m/s}}\).

Step-by-step solution

Given data

The mass of the bullet is\(m = 15\;{\rm{g}} = 0.015\;{\rm{kg}}\).

The mass of the wood is\(M = 35\;{\rm{kg}}\).

The temperature increase is\(\Delta T = {0.020^ \circ }{\rm{C}}\).

Let v be the entering speed of the bullet.

Concepts

When the bullet stops, its kinetic energy helps to increase the temperature.

The bullet's kinetic energy equals the sum of the heat absorbed by the wood and the material of the bullet.

Calculation of the entering speed of the bullet

The initial kinetic energy of the bullet is \(K = \frac{1}{2}m{v^2}\).

The specific heat of lead is \(s = 130\;{\rm{J/kg}}{ \cdot ^{\rm{o}}}{\rm{C}}\).

The specific heat of wood is \(S = 1700\;{\rm{J/kg}}{ \cdot ^{\rm{o}}}{\rm{C}}\).

The heat absorbed by the wood and the bullet is

\(\begin{aligned}{c}Q = ms\Delta T + MS\Delta T\\ = \left( {ms + MS} \right)\Delta T.\end{aligned}\)

Now,

\(\begin{aligned}{c}K = Q\\\frac{1}{2}m{v^2} = \left( {ms + MS} \right)\Delta T\\{v^2} = 2\left( {s + \frac{M}{m}S} \right)\Delta T\\v = \sqrt {2\left( {s + \frac{M}{m}S} \right)\Delta T} .\end{aligned}\)

Put the values in the above equation.

\(\begin{aligned}{c}v = \sqrt {2\left( {130\;{\rm{J/kg}}{ \cdot ^{\rm{o}}}{\rm{C}} + \left( {\frac{{35\;{\rm{kg}}}}{{0.015\;{\rm{kg}}}}} \right)1700\;{\rm{J/kg}}{ \cdot ^{\rm{o}}}{\rm{C}}} \right)\left( {{{0.020}^ \circ }{\rm{C}}} \right)} \\ = 4.0 \times {10^2}\;{\rm{m/s}}\end{aligned}\)

Hence, the bullet's entering speed is \(4.0 \times {10^2}\;{\rm{m/s}}\).