Question

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?

Answer 1

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,

v_bx = v_ax - v_b/[1 - v_axv_b/c²]

Now Substitue v_ax = 0.3c and v_b = 0.9c

v_bx = 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 = 1/2mv_bx²

[Note: As mass of the particle is not mentioned in the question.]

Please comment the mass for kinetic energy.

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