Calculate KINETIC ENERGY of proton traveling close to speed of light

[LBRRIT2390]

Homework Statement

A proton is traveling with a speed of v = 2.990x10⁸ m/s. Calculate the (relativistically correct) value of the kinetic energy for this proton. Assume that the speed of light is c = 3.000 x 10⁸ m/s and that the mass of the proton is 1.673 x 10^-27 kg.

Note: Do your calculations to 4 significant figures. Then round off to give an answer good to 3 significant figures. Be sure to include the correct abbreviation for the SI unit.

Homework Equations

K = 1/2 mv²

Limit is K = 1/2mC²

The Attempt at a Solution

K=1/2(1.673x10^-27)(2.990x10⁸)
K=2.50x10^-19J

Is this correct? I feel like I'm missing something here...

 

[LBRRIT2390]

O should I be using the equation:

K =( g - 1 )MC²
or
K= (1/2)gMV²
?

What is the correct formula for calculating kinetic energy when a particle's speed approaches the speed of light??

 

[thebigstar25]

you should use the relativistic equations to calculate the kinetic energy .. I suggest you to use :

Ek = gama*m*c^2 - mc^2 , where gama= 1/sqrt(1-(v/c)^2 ) ..

 

[imnitsy]

when a particle moves with very high velocity ( close to the speed of light )
then the moving mass of the particle is given as :

M = m/sqrt(1-(v^2/c^2))

here,
m is ths rest mass of the particle = 1.673X10^(-27)kg
v is the velocity of the particle = 2.990X10^(8)m/s
c is the speed of light = 3.00X10^(8)m/s

using above formula the mass of the proton moving with the speed close to that of light can be calculated :

M = 6.480X10^(-26)kg

then we can obtain the kinectic energy of the proton as:

KE = (1/2)M(v^2)

KE = 2.914X10^(-19) joule

please correct if m wrong!

 

[thebigstar25]

when a particle moves with very high velocity ( close to the speed of light )
then the moving mass of the particle is given as :

M = m/sqrt(1-(v^2/c^2))

here,
m is ths rest mass of the particle = 1.673X10^(-27)kg
v is the velocity of the particle = 2.990X10^(8)m/s
c is the speed of light = 3.00X10^(8)m/s

using above formula the mass of the proton moving with the speed close to that of light can be calculated :

M = 6.480X10^(-26)kg

then we can obtain the kinectic energy of the proton as:

KE = (1/2)M(v^2)

KE = 2.914X10^(-9) joule

please correct if m wrong!

I don't recall doing this when I used to take that course .. I just substitute in the formula I wrote

 

[vela]

when a particle moves with very high velocity ( close to the speed of light )
then the moving mass of the particle is given as :

M = m/sqrt(1-(v^2/c^2))

here,
m is ths rest mass of the particle = 1.673X10^(-27)kg
v is the velocity of the particle = 2.990X10^(8)m/s
c is the speed of light = 3.00X10^(8)m/s

using above formula the mass of the proton moving with the speed close to that of light can be calculated :

M = 6.480X10^(-26)kg

then we can obtain the kinectic energy of the proton as:

KE = (1/2)M(v^2)

KE = 2.914X10^(-19) joule

please correct if m wrong!

That's wrong. You should abandon the notion of relativistic mass because it leads to errors like this.

In special relativity, the energy of an object of rest mass m is given by

$$E=\frac{mc^2}{\sqrt{1-(v/c)^2}}$$

When v=0, you get E=mc². This is the rest energy of the object. The object's kinetic energy is the energy in excess of the rest energy, so it's given by

$$K=E-mc^2$$

If you express the total energy E as a series in (v/c)², you get

$$E=mc^2\left[1+\frac{1}{2}\left(\frac{v}{c}\right)^2+...\right]\cong mc^2+\frac{1}{2}mv^2+...$$

When the speed of the object is much smaller than the speed of light, you can neglect the higher-order terms and recover the Newtonian expression for kinetic energy

$$K=\frac{1}{2}mv^2$$

To get this result, we assumed v<<c, so this formula is only good when the speeds are not a significant fraction of the speed of light. In particular, you can not use it in this problem.

 

[imnitsy]

That's wrong. You should abandon the notion of relativistic mass because it leads to errors like this.

In special relativity, the energy of an object of rest mass m is given by

$$E=\frac{mc^2}{\sqrt{1-(v/c)^2}}$$

When v=0, you get E=mc². This is the rest energy of the object. The object's kinetic energy is the energy in excess of the rest energy, so it's given by

$$K=E-mc^2$$

If you express the total energy E as a series in (v/c)², you get

$$E=mc^2\left[1+\frac{1}{2}\left(\frac{v}{c}\right)^2+...\right]\cong mc^2+\frac{1}{2}mv^2+...$$

When the speed of the object is much smaller than the speed of light, you can neglect the higher-order terms and recover the Newtonian expression for kinetic energy

$$K=\frac{1}{2}mv^2$$

To get this result, we assumed v<<c, so this formula is only good when the speeds are not a significant fraction of the speed of light. In particular, you can not use it in this problem.

then what should be the kinetic energy for such problem!

 

[vela]

I clearly said what it was in my previous post.

 

[imnitsy]

I clearly said what it was in my previous post.

ohh you ..thanx