Question

A space vehicle is traveling at 4000 km/h relative to the Earth when the exhausted rocket motor (mass 4m) is disengaged and sent backward with a speed of 85 km/h relative to the command module (mass m). What is the speed of the command module relative to Earth just after the separation?

Answer 1

the velocity of command module with respect to earth is equal to 3932 km/h

Step-by-step explanation:

given,

velocity of space vehicle = 4000 km/h

velocity of motor relative to module ( vb - v a ) = 85 Km/h

`v_a =v_b - 85`

mass of rocket motor = 4 m

mass of command module = m

the mass of the vehicle before disengaged = 4 m + m = 5 m

from law of conservation of momentum

`5 m v_i = 4 m v_a + m(v_b)`

`5 m v_i = 4 m (v_b - 85) + m(v_b)`

`5* 4000 = 4(v_b -85)+ v_b`

`20000\ = 5 v_b - 340\ km/h`

`v_b = 3932\ km/h`

hence, the velocity of command module with respect to earth is equal to 3932 km/h

Answer 2

4068 Km/h

Step-by-step explanation:

Data provided:

Initial Speed of space vehicle relative to Earth, V = 4000 km/h

Rocket motor mass = 4m

speed of the rocket motor, v' = 85 km/h relative to command module

Mass of the command module = m

Now,

let us assume the speed of command module relative to earth be 'v'

applying the concept of conservation of momentum,

( 4m + m ) × V = m × v + (4m × (v - v'))

or

5m × 4000 = mv + 4mv - 4mv'

or

20000 = 5v - ( 4 × 85 )

or

20000 = 5v - 340

or

v = 4068 Km/h