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-(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-(v^(2))/(c^(2))}}`

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

`p=((1500kg)(0.7(3(10)^(8)m/s)))/(√(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