Question

An atom A, moving relative to the observer, with velocity 2 X 10^8m/s emits a particle B which moves with a velocity of 2.8 X 10^8m/s with respect to the atom. The velocity of the emitter particle relative to the scientist is _____________

(a) 0.8 X 10^8m/s

(b) 2.4 X 10^8m/s

(c) 3 x 10^8m/s

(d) 2.95 X 10^8m/s

This question was posed to me in an international level competition.

Asked question is from Law of Addition of Velocities in section Ultrasonics & Relativity of Engineering Physics I

Answer 1

Correct answer is (d) 2.95 X 10^8m/s

To explain I would say: Now we know, ux = \(\frac{u^{‘}+v}{1+\frac{u^{‘}v}{c^2}}\)

Here, u’ = 2.8 X 10^8m/s

u = 2 X 10^8m/x

ux = ?

ux = 4.8/ (1 + 5.6/3) X 10^8m/s

= 2.95 X 10^8m/s.