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?

Respuesta :

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

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