Answer:
The speed of the proton relative to the laboratory frame is 0.981c
Step-by-step explanation:
Given that,
Speed of electron v= 0.90c
Speed of proton u= 0.70c
We need to calculate the speed of the proton relative to the laboratory frame
Using formula of speed
![u'=(u+v)/(1+(uv)/(c^2))](https://img.qammunity.org/2021/formulas/physics/college/feunggpggaafporz2tnefm7hiq0sfb2z3g.png)
Where, u = speed of the proton relative to the electron
v = speed of electron relative to the laboratory frame
Put the value into the formula
![u'=(0.70+0.90)/(1+(0.70*0.90* c^2)/(c^2))](https://img.qammunity.org/2021/formulas/physics/college/bq1jl6e5q9j788fqcznzk4yw1x2it2hlzp.png)
![u'=c(0.70+0.90)/(1+(0.70*0.90))](https://img.qammunity.org/2021/formulas/physics/college/neuaxhzus3ncqatu0hn9oujykn9kx15wfy.png)
![u'=0.981c](https://img.qammunity.org/2021/formulas/physics/college/6990xbv204yic4v6yeb1cduf17qxgl2ud3.png)
Hence, The speed of the proton relative to the laboratory frame is 0.981c