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=(u'_A+v)/(1+(u'_A v)/(c^2))`

Substitute the values

`u_A=(0.43c+0.94c)/(1+(0.43c* 0.94c)/(c^2))`

`u_A=(1.37c)/(1+0.4042)=0.98c`

Velocity of piece B observed by observer

`u_B=(0.34c+0.94c)/(1+(0.34c* 0.94c)/(c^2))`

`u_B=(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.