Question

A 50kg meteorite moving at 1000 m/s strikes Earth. Assume the velocity is along the line joining Earth's center of mass and the meteor's center of mass. What is the gain in Earth's Kinetic Energy?

Answer 1

The gain in the kinetic energy of earth after the meteorite strikes earth is $\fbox{\begin\\2.5 \times {10^7}\,{\text{J}}\end{minispace}}$.

Further Explanation:

When the collision between two objects takes place the total momentum of objects before collision is equal to the total momentum of objects after collision in an isolated system. This is known as the law of conservation of momentum.

Given:

The mass of the meteorite is $50\,{\text{kg}}$.

The velocity of the meteorite $1000\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$.

The mass of the earth is $5.972 \times {10^{24}}\,{\text{kg}}$.

Concept:

The velocity of the earth is zero because the earth is at rest. After the collision of meteorite with earth, the meteorite will get merged with the earth and the meteorite and earth will acquire same velocity. The final mass of the meteorite and earth system is equal to the sum of mass of earth and mass of the meteorite.

The law of conservation of momentum for meteorite and earth is:

${m_1}{v_1} + {m_2}{v_2} = \left( {{m_1} + {m_2}} \right)v$                                               …… (1)

Here, ${m_1}$ is the mass of the meteorite, ${v_1}$ is the velocity with which meteorite is moving, ${m_2}$ is the mass of earth, ${v_2}$ is the velocity of the earth and v is the velocity with which earth and meteorite system is moving.

Substitute $0$ for ${v_2}$ in equation (1) and rearrange the expression.

$\fbox{\begin\\v=\dfrac{{m_1}{v_1}}{{m_1}+{m_2}} \end{minispace}}$                                                  …… (2)

Substitute the values in equation (2).

$\begin{aligned}v&=\frac{{\left( {50\,{\text{kg}}} \right)\left( {1000\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}} \right)}}{{\left( {50\,{\text{kg}} + 5.972 \times {{10}^{24}}\,{\text{kg}}} \right)}} \\&=8.372 \times {10^{ - 21}}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}$

The amount of kinetic energy lost by the meteorite is equal to the amount of kinetic energy gain by the earth after collision. The gain in kinetic energy of the earth is equal to the difference between the kinetic energy of the meteorite earth system before collision and the kinetic energy of the meteorite earth after collision.

The gain in the kinetic energy of the earth is:

$\Delta k=\left| {{k_f} - {k_i}} \right|$                                                                            …… (3)

Here, ${k_f}$ is the kinetic energy of the meteorite earth system after collision, ${k_i}$ is the kinetic energy of the meteorite earth system before collision and $\Delta k$ is the gain in the kinetic energy of the earth.

The kinetic energy of the meteorite earth system before collision is:

${k_i}=\frac{1}{2}{m_1}v_1^2$                                                                              …… (4)

The kinetic energy of the meteorite earth system after collision is:

${k_f}=\frac{1}{2}\left( {{m_1} + {m_2}} \right){v^2}$                                            …… (5)

Substitute the values in equation (3).

$\fbox{\begin\Delta k=\left| {\frac{1}{2}\left( {{m_1} + {m_2}} \right){v^2} - \frac{1}{2}{m_1}v_1^2} \right|\end{minispace}}$                                               ……. (6)

Substitute the value in equation (6).

$\begin{aligned}\\{\Delta k&=\frac{1}{2}( 50+5.972\times10^{24})(8.372\times 10^{-21})^{2}-\frac{1}{2}{50}(1000)^{2}\\\\&=2.5\times10^{7}\text{J}\end{aligned}}$

Thus, the gain in the kinetic energy of earth after the meteorite strikes earth is $\fbox{\begin\\2.5 \times {10^7}\,{\text{J}}\end{minispace}}$.

Learn more:

1. Conservation of energy brainly.com/question/3943029

2. Average kinetic energy  brainly.com/question/9078768

3. Wind and solar energy brainly.com/question/1062501

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Kinetic energy, conservation of energy, meteorite, earth, inelastic collision, meteorites colliding with earth, kinetic energy gained by earth, energy lost by meteorite, collision, earth is at stationary position.

Answer 2

As per the question, the mass of meteorite [ m]= 50 kg

                       The velocity of the meteorite [v] = 1000 m/s

When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.

We are asked to calculate the gain in kinetic energy of earth.

The kinetic energy of meteorite is calculated as -

                                       $Kinetic\ energy\ [K.E]\ =\frac{1}{2} mv^2$

                                                             $=\frac{1}{2}50kg*[1000\ m/s]^2$

                                                               $=\frac{1}{2}50* 10^{6}\ J$

                                                               $=25*10^6\ J$    

Here, J stands for Joule which is the S.I unit of energy.

$Hence,\ the\ kinetic\ energy\ gained\ by\ earth\ is\ 25*10^6\ J$