Question

A. A meteorite of mass 1500kg moves with a speed of 0.700c . Find the magnitude of its momentum p.Express your answer in kilograms times meters per second to three significant figures.B. What is the total energy E of the meteorite?C. What would the energy of the meteorite be if it were at rest?D.What is the relativistic kinetic energy K of the meteorite when it travels at 0.700c ?

Answer 1

Answers:

The expression for the relativistic energy $E$ is given by:

$E=mc^{2}$   (1)

being $c=3(10)^{8}m/s$

This famous equation includes the relativistic Kinetic energy $K$ and the energy at rest $E_{o}$:

$E=K+E_{o}$   (2)

Where:

$E_{o}=m_{o}c^{2}$   (3)

Being $m_{o}=1500kg$ the mass at rest for the meteorite

$K=E-E_{o}$   (4)

In addition, there is a relation between the relativistic energy and the momentum $p$:

$E=\sqrt{p^{2}c^{2}+m_{o}^{2}c^{4}}$   (5)

Where:

$p=\frac{m_{o}v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$   (6)

Knowing this, let's begin with the answers:

a)Momentum

In order to solve this part, equation (6) will be helpful, since we already know the mass of the meteorite and its speed $v=0.700c$:

$p=\frac{m_{o}v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$p=\frac{(1500kg)(0.7c)}{\sqrt{1-\frac{(0.7c)^{2}}{c^{2}}}}$

$p=\frac{(1500kg)(0.7(3(10)^{8}m/s))}{\sqrt{1-0.49}}$

$p=4.411(10)^{11}kg.m/s$   (7) >>>>This is the meteorite's momentum

b) Total Energy

Remembering equation (5), which relates the total energy with the momentum:

$E=\sqrt{p^{2}c^{2}+m_{o}^{2}c^{4}}$

We can substitute the value of the momentum found on (7):

$E=\sqrt{(4.411(10)^{11}kg.m/s)^{2}(3(10)^{8}m/s)^{2}+(1500kg)^{2}(3(10)^{8}m/s)^{4}}$

Then:

$E=1.890(10)^{20}kg.m^{2}/s^{2}$

Knowing $1kg.m^{2}/s^{2}=1N.m=1J=1 Joule$:

$E=1.890(10)^{20}J$ (8)>>>This is the total energy of the meteorite

c) Energy at rest

Using equation (3):

$E_{o}=m_{o}c^{2}$  

$E_{o}=1500kg(3(10)^{8}m/s)^{2}$  

$E_{o}=1.35(10)^{20}J$  (9) >>>Meteorite's energy at rest

d)  Relativistic kinetic energy

According to equation (4) the  relativistic kinetic energy depends on the total energy and the energy at rest:

$K=E-E_{o}$  

We already know the values of $E$ and $E_{o}$ from (8) and (9). Hence we only have to substitute them on the equation:

$K=1.890(10)^{20}J-1.35(10)^{20}J$  

$K=5.4(10)^{9}J$ >>>Meteorite's Relativistic energy