91Ó°ÊÓ

In this problem, the meteor striking the Earth and coming to rest is an inelastic collision. The frame of reference of Earth will be used when it is at rest before the collision.

Given data

The mass of the whole meteor is\(m = 1.5 \times {10^8}\;{\rm{kg}}\).

The mass of Earth is\(m' = 6 \times {10^{24}}\;{\rm{kg}}\).

The speed of the meteor is \(v = 25\;{\rm{km/s}}\).

The relation to calculate recoil speed can be written as follows:

\(\begin{array}{c}mv = \left( {m + m'} \right)v'\\v' = \frac{{mv}}{{\left( {m + m'} \right)}}\end{array}\)

Here,\(v'\)is the recoil speed of the Earth.

Plugging the values in the above equation,

\(\begin{array}{l}v' = \left[ {\frac{{\left( {1.5 \times {{10}^8}\;{\rm{kg}}} \right)\left( {25\;{\rm{km/s}}} \right)}}{{\left( {\left( {1.5 \times {{10}^8}\;{\rm{kg}}} \right) + \left( {6 \times {{10}^{24}}\;{\rm{kg}}} \right)} \right)}}} \right]\\v' = 6.25 \times {10^{ - 16}}\;{\rm{km/s}}\end{array}\)

Thus, \(v' = 6.25 \times {10^{ - 16}}\;{\rm{km/s}}\) is the required speed of Earth.

The relation to calculate kinetic energy can be written as follows:

\(\frac{{{\rm{KE}}}}{{{\rm{KE'}}}} = \frac{{\frac{1}{2}m'{{v'}^2}}}{{\frac{1}{2}m{v^2}}}\)

Plugging the values in the above equation,

\(\begin{array}{l}\frac{{{\rm{KE}}}}{{{\rm{KE'}}}} = \left[ {\frac{{\frac{1}{2}\left( {6 \times {{10}^{24}}\;{\rm{kg}}} \right)\left( {6.25 \times {{10}^{ - 16}}\;{\rm{km/s}}} \right)}}{{\frac{1}{2}\left( {1.5 \times {{10}^8}\;{\rm{kg}}} \right)\left( {25\;{\rm{km/s}}} \right)}}} \right]\\\frac{{{\rm{KE}}}}{{{\rm{KE'}}}} = 2.5 \times {10^{ - 17}}\end{array}\)

Thus, \(\frac{{{\rm{KE}}}}{{{\rm{KE'}}}} = 2.5 \times {10^{ - 17}}\) is the required fraction of kinetic energy.

The relation to calculate the change in kinetic energy can be written as follows:

\(\Delta {\rm{KE}} = {\rm{K}}{{\rm{E}}_{\rm{f}}} - {\rm{K}}{{\rm{E}}_{\rm{i}}}\)

Here,\({\rm{K}}{{\rm{E}}_{\rm{i}}}\)and\({\rm{K}}{{\rm{E}}_{\rm{f}}}\)are the initial and final kinetic energies, where the value of initial kinetic energy is zero.

Plugging the values in the above equation,

\(\begin{array}{l}\Delta {\rm{KE}} = \frac{1}{2}m'{{v'}^2} - 0\\\Delta {\rm{KE}} = \frac{1}{2}\left( {6 \times {{10}^{24}}\;{\rm{kg}}} \right){\left( {6.25 \times {{10}^{ - 16}}\;{\rm{km/s}} \times \frac{{1000\;{\rm{m/s}}}}{{1\;{\rm{km/s}}}}} \right)^2}\\\Delta {\rm{KE}} = 1.17\;{\rm{J}}\end{array}\)

Thus, \(\Delta {\rm{KE}} = 1.17\;{\rm{J}}\) is the change in kinetic energy.