Question

A free ${}^{238}U$ nucleus kept in a train emits an alpha particle. When the train is stationary, a nucleus decays and a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes x at time t after the decay. If the decay takes place while the train is moving at a uniform velocity v, the distance between the alpha particle and the recoiling nucleus at a time t after the decay, as measured by the passenger, is
(a) x + v t
(b) x − v t
(c) x
(d) depends on the direction of the train

Answer 1

(c) x

The moving train does not put any extra force on the alpha particle and the recoiling nucleus. So, the distance between the alpha particle and the recoiling nucleus at a time t after the decay, as measured by the passenger, will be same as before, i.e. x.