Question

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 0.34 c relative to the original nucleus. Find the velocity of piece A as measured by an observer in the laboratory. Do the same for piece B.

Answer 1

Answer with Explanation:

We are given that

Velocity of uranium=v=0.94 c

Speed of piece A relative to the original nucleus=$u'_A=0.43c$

Speed of piece B relative to the original nucleus=$u'_B=0.34c$

Velocity of piece A observed by observer

$u_A=\frac{u'_A+v}{1+\frac{u'_A v}{c^2}}$

Substitute the values

$u_A=\frac{0.43c+0.94c}{1+\frac{0.43c\times 0.94c}{c^2}}$

$u_A=\frac{1.37c}{1+0.4042}=0.98c$

Velocity of piece B observed by observer

$u_B=\frac{0.34c+0.94c}{1+\frac{0.34c\times 0.94c}{c^2}}$

$u_B=\frac{1.28c}{1+0.3196}$

$u_B=0.97 c$

The velocity of piece A and piece B as measured  by an observer in the laboratory are not same.