Answer:
4794.4 km/h
Explanation:
Given:
The initial velocity v₁ = 4720 km/h
velocity of the motor,[tex]v_b-v_a[/tex] = 93 km/h (relative to the module)
[tex]v_a=v_b- 93\ km/h [/tex]
where,
[tex]v_a[/tex] is the velocity of the motor
[tex]v_b[/tex] is the velocity of the command module
let, the mass of the command module be m
thus, the mass of the motor will be '4m'
Now. the mass of the vehicle before disengaged = 4m + m = 5m
using the concept conservation of momentum ,
we have
[tex](5m)v_1 = 4m\times v_a + m\times v_b[/tex]
on substituting the values in the above equation, we get
[tex](5m)\times 4720 = 4m\times (v_b-93) + m\times v_b[/tex]
or
[tex]23600 = 4\times v_b-4\times93 + v_b[/tex]
or
[tex]23600 = 5\times v_b-372 [/tex]
or
[tex]v_b[/tex] = 4794.4 km/h