A) 0.4 c
B) 0.8 c
C) 0.9 c
D) 1.2 c
Correct Answer: C
Solution :
The particles are moving with velocities \[0.8\text{ }c\] and \[-0.4\text{ }c\]in the laboratory frame, say S frame. Let S be a reference frame in which the particle with velocity \[-0.4\text{ }c\] is at nest. Then, the velocity of S (laboratory) relative to S is \[v=0.4\text{ }c\]. Therefore, the particle which in S has velocity \[u=+0.8\text{ }c\]has a velocity in S given by \[u=\frac{u+v}{1+\frac{uv}{{{c}^{2}}}}\] \[=\frac{0.8c+0.4c}{1+\frac{(0.8c)\,(0.4c)}{{{c}^{2}}}}\] \[=\frac{1.2c}{1+0.32}=0.9c\]
You need to login to perform this action.
You will be redirected in
3 sec
