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

Answer:

4794.4 km/h

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

$(5m)v_1 = 4m\times v_a + m\times v_b$

on substituting the values in the above equation, we get

$(5m)\times 4720 = 4m\times (v_b-93) + m\times v_b$

or

$23600 = 4\times v_b-4\times93 + v_b$

or

$23600 = 5\times v_b-372$

or

$v_b$ = 4794.4 km/h