A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two pieces. Piece A is propelled in the forward direction with a speed of 0.43 c relative to the original nucleus. Piece B is sent backward at 0.35 c relative to the original nucleus. Part A Find the velocity of piece A as measured by an observer in the laboratory. Do the same for piece B.

A u = 0.36c B u = 0.961c

Explanation:

In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains

u ' = (u-v) / (1 - uv / c²)

Where u' is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory

The data give is u ' = 0.43c and the initial core velocity v = 0.94c

Let's clear the speed with respect to the observer (u)

u' (1 - u v / c²) = u - v

u + u 'uv / c² = v - u'

u (1 + u 'v / c²) = v - u'

u = (v-u ') / (1 + u' v / c²)

Let's calculate

u = (0.94 c - 0.43c) / (1 + 0.43c 0.94 c / c²)

u = 0.51c / (1 + 0.4042)

u = 0.36c

We repeat the calculation for the other piece

In this case u ' = - 0.35c

We calculate

u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)

u = 1.29c / (1 - 0.329)

u = 0.961c