Question

A space vehicle is traveling at 4720 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The relative speed between the motor and the command module is then 93 km/h. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth just after the separation?

Answer 1

4794.4 km/h

Step-by-step explanation:

Given:

The initial velocity v₁ = 4720 km/h

velocity of the motor,
`v_b-v_a` = 93 km/h (relative to the module)

`v_a=v_b- 93\ km/h`

where,

`v_a` is the velocity of the motor

`v_b` 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

`(5m)v_1 = 4m* v_a + m* v_b`

on substituting the values in the above equation, we get

`(5m)* 4720 = 4m* (v_b-93) + m* v_b`

or

`23600 = 4* v_b-4*93 + v_b`

or

`23600 = 5* v_b-372`

or

`v_b` = 4794.4 km/h