Aspace 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?

4794.4 km/h

Explanation:

Given:

The initial velocity v₁ = 4720 km/h

velocity of the motor, = 93 km/h (relative to the module)

where,

is the velocity of the motor

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

on substituting the values in the above equation, we get

or

or

or

= 4794.4 km/h

a and c

c is 100% but i am confused in a

well, you would divide 135 by 33 to get your answer. then that is how many seconds it would take. boardman should fly about 132 m or 4.090909090909091 seconds.