In: Physics
Particle A moves along an axis in the laboratory with velocity V = 0.3c. Particle b moves with velocity of V = .9c along the direction of motion of particle A.
What kinetic energy does the particle b measure for the particle A?
Solution:
Using the Lorentz velocity transformation in lab frame,
First, find the velocity of A observed by B,
Using the relative velocity of object moving same directions,
vbx = vax - vb/[1 - vaxvb/c2]
Now Substitue vax = 0.3c and vb = 0.9c
vbx = 0.3c - 0.9c /[1 - (0.3*0.9)/c2]
= - 0.6c/[1-0.27]
= - 0.822c.
We go the velocity of A as measured by B,
Now, just calcualte the kinetic energy,
K.E = 1/2mvbx2
[Note: As mass of the particle is not mentioned in the question.]
Please comment the mass for kinetic energy.
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