Final answer:
The derivative of relativistic kinetic energy with respect to momentum is the velocity of the particle, as derived from the relativistic energy-momentum relationship, making the correct answer de/dp = v.
Step-by-step explanation:
The question you're asking relates to the derivative of relativistic kinetic energy (KErel) with respect to momentum (p), which is derived from the relativistic energy-momentum relationship.
According to the theory of relativity, the total energy (E) of a particle is related to its momentum (p) and rest mass (m) by the equation E² = (pc)² + (mc²)². When we talk about relativistic kinetic energy, it's the portion of the total energy excluding the rest mass energy (mc²), and is dependent on the particle's velocity (u).
From the given relationship E² = (pc)² + (mc²)², we can differentiate both sides with respect to p to find de/dp. For highly relativistic velocities (speeds close to the speed of light), we find that the derivative of the relativistic kinetic energy with respect to momentum is equivalent to the velocity of the particle,
The student is asked to show that the equation de/dp = v remains valid when e represents the relativistic kinetic energy of a particle. In order to do this, we can use the work-energy theorem, which states that the net work done on a particle is equal to its final kinetic energy.
Since the force is given by F = dp/dt, we can substitute this into the work-energy theorem and rearrange to get de/dp = v, which shows that the equation remains valid.
hence the correct answer to your question is de/dp = v.