A particle has a kinetic energy equal to its rest energy.
(a) What is the speed of this particle?
(b) If the kinetic energy of this particle is doubled, does its speed increase by more than, less than, or exactly a factor of 2?
(c) Explain
(d) Calculate the speed of a particle whose kinetic energy is twice its rest energy.
I got (a) and (d)
(a) .866c and (d) .943c
but I'm stuck on (b) and (c) I think it might be less than 2 but I have no idea why that would be the case, can someone please explain it to me!? Thank you!
(a)
KE = m(o) •c^2•[1/sqrt(1-β^2) -1],
E(o) = m(o) •c^2,
KE = E(o),
m(o) •c^2•[1/sqrt(1-β^2) -1] =
= m(o) •c^2,
1/sqrt(1-β^2) =2,
β = v/c = 0.866.
v = 0.866•c =0.866•3•10^8 =2.6•10^8 m/s.
(b),(c),(d)
KE = 2 •E(o)
m(o) •c^2•[1/sqrt(1-β^2) -1] =
=2•m(o) •c^2,
1/sqrt(1-β^2) =3,
β = v/c = 0.94.
v =0.94•c =2.82•10^8 m/s.
To answer part (b), let's analyze the relationship between kinetic energy and speed in the context of special relativity.
According to the theory of special relativity, the total energy (E) of a particle is given by the equation:
E^2 = (mc^2)^2 + (pc)^2
where m is the rest mass of the particle, c is the speed of light, and p is the momentum of the particle.
The rest energy (E_rest) of the particle is simply its energy at rest, which is given by:
E_rest = mc^2.
Given that the kinetic energy (KE) of the particle is equal to its rest energy, we can write:
KE = E - E_rest.
Since E_rest = mc^2 and E^2 = (mc^2)^2 + (pc)^2, we have:
KE = sqrt((mc^2)^2 + (pc)^2) - mc^2.
Now, let's consider the case when the kinetic energy is doubled. We'll call this new value KE_double.
KE_double = 2 * KE.
Substituting the values above, we get:
2 * sqrt((mc^2)^2 + (pc)^2) - 2 * mc^2 = sqrt((mc^2)^2 + (pc)^2) - mc^2.
Let's simplify this expression:
sqrt((mc^2)^2 + (pc)^2) - mc^2 = 0.
Now, let's solve for the momentum (p):
(sqrt((mc^2)^2 + (pc)^2) - mc^2)^2 = 0.
Expanding and simplifying:
(mc^2)^2 + (pc)^2 - 2 * mc^2 * sqrt((mc^2)^2 + (pc)^2) + (mc^2)^2 = 0.
Simplifying further:
(pc)^2 - 2 * mc^2 * sqrt((mc^2)^2 + (pc)^2) = 0.
(pc)^2 = 2 * mc^2 * sqrt((mc^2)^2 + (pc)^2).
Squaring both sides:
(p^2)(c^2) = 4 * (m^2)(c^4) * [(m^2)(c^2) + (p^2)(c^2)].
p^2 = 4 * (m^2)(c^2) * [(m^2)(c^2) + (p^2)(c^2)].
Dividing both sides by (m^2)(c^2):
p^2 / (m^2)(c^2) = 4 * [(m^2)(c^2) + (p^2)(c^2)].
p^2 / (m^2)(c^2) = 4 * (c^2) * [m^2 + p^2].
p^2 / m^2 = 4 * (m^2 + p^2).
Dividing both sides by m^2:
(p^2 / m^2) = 4 * (1 + p^2 / m^2).
Let's simplify this equation:
p^2 / m^2 = 4 + 4 * (p^2 / m^2).
Rearranging terms:
3 * (p^2 / m^2) = 4.
Dividing both sides by 3:
(p^2 / m^2) = 4/3.
Now, we can substitute the expression for momentum (p) and rest mass (m) in terms of speed (v):
[(m * v) ^ 2 / m^2] = 4/3.
Cancelling out the m^2 term:
v^2 = 4/3.
v = sqrt(4/3).
Therefore, the speed of the particle whose kinetic energy is twice its rest energy is given by sqrt(4/3), approximately equal to 1.155 times the speed of light (c).
Hence, the speed increase is less than a factor of 2.
To summarize, the speed increase is less than a factor of 2 because, as the kinetic energy increases, the speed of the particle approaches but never reaches the speed of light. According to special relativity, as an object with mass accelerates, its energy and momentum increase, but its speed cannot exceed the speed of light in a vacuum.
To understand why the speed of the particle does not increase by exactly a factor of 2 when its kinetic energy is doubled, we need to consider the relationship between kinetic energy (KE) and relativistic mass (m).
The relativistic mass of an object is given by the equation:
m = m₀ / √(1 - (v²/c²))
where m₀ is the rest mass of the object, v is its velocity, and c is the speed of light.
Now, the kinetic energy of an object can be expressed as:
KE = (γ - 1) * m₀ * c²
where γ is the Lorentz factor, given by:
γ = 1 / √(1 - (v²/c²))
In the given scenario, the kinetic energy of the particle is equal to its rest energy, so we can equate the two equations:
(γ - 1) * m₀ * c² = m₀ * c²
Canceling out the m₀ * c² terms and simplifying, we get:
γ - 1 = 1
γ = 2
To find the velocity of the particle (v), we substitute the value of γ into the equation for the Lorentz factor:
2 = 1 / √(1 - (v²/c²))
Squaring both sides of the equation, we get:
4 = 1 - (v²/c²)
Rearranging the equation, we have:
v²/c² = 1 - 4
v²/c² = -3
This implies that the speed of the particle squared is negative, which is not possible. Therefore, the speed must be imaginary, and in accordance with the laws of physics, it means that it cannot travel faster than the speed of light.
Hence, in the relativistic regime, when the kinetic energy of a particle is doubled, its speed does not increase by more than, less than, or exactly a factor of 2. The speed remains limited to the speed of light (c).