A space vehicle is traveling at 4300 km / h relative to Earth when the exhausted rocket motor (m

A space vehicle is traveling at 4300 km / h relative to...

A space vehicle is traveling at 4300 $\mathrm{km} / \mathrm{h}$ relative to Earth when the exhausted rocket motor (mass 4$m )$ is disengaged and sent backward with a speed of 82 $\mathrm{km} / \mathrm{h}$ relative to the command module (mass $m ) .$ What is the speed of the command module rel- ative to Earth just after the sparation?

A space vehicle is traveling at 4300 $\mathrm{km} / \mathrm{h}$ relative to Earth when the exhausted rocket motor (mass 4$m )$ is disengaged and
sent backward with a speed of 82 $\mathrm{km} / \mathrm{h}$ relative to the command
module (mass $m ) .$ What is the speed of the command module rel-
ative to Earth just after the sparation?

Key Concepts

Conservation of Momentum

This principle states that in the absence of external forces, the total momentum of a system remains constant during any interactions within the system. In this problem, the momentum of the space vehicle before separation is equal to the sum of the momenta of the command module and the exhausted rocket motor after separation.

Relative Velocity Transformation

This concept involves converting velocities between different frames of reference. It is essential when determining the actual speed of one part of the system (the command module) relative to an external observer (Earth) given its speed relative to another component (the rocket motor) within the moving system.

Mass Distribution in Multi-body Systems

Understanding how the masses of different components affect the dynamics of a system is crucial. In this case, the different mass values for the command module and the rocket motor determine how the velocity is partitioned between them after the separation, ensuring that the overall momentum remains conserved.

Transcript

00:01 Our question says that we have a space vehicle traveling at 4 ,300 kilometers per hour.

00:07 So i write the initial velocity v0 as 4300 kilometers per hour relative to earth.

00:12 When the exhaust rocket motor, which has mass 4m, so i write the exhaust rocket motor mass as m subr is equal to 4m, and it's disengaged and sent backwards with the speed of 82 kilometers per hour relative to the command module, which has mass m.

00:28 So i write the mass of the command module as m sub c, and that's just equal.

00:31 To m and the speed of the rocket motor relative to the command module i write as v sub r and that's 82 kilometers per hour here okay so we can use a conservation of momentum to then find what we're asked to find which is the relative speed of the command module compared to earth just after separation so we can use the fact that the initial momentum is equal to the final momentum since momentum is conserved in this case.

01:08 So the initial momentum is just the mass of the rocket plus the mass of the command module, because that's the total mass of the object before disengaging, multiplied by the speed at which the object is traveling, which is v of zero.

01:22 And this is equal to the final momentum, which is the momentum of the command module, which is the mass of the command module times the speed of the command module plus the momentum of the rocket.

01:38 And the momentum of the rocket is the mass of the rocket times the speed of the rocket.

01:44 Well, the speed of the rocket is just the speed of the command module minus the relative speed of the rocket to earth, which is v sub r.

01:52 Okay, so we want to solve for v .c.

01:55 So let's put everything that has v .c in it on the left side of the equation.

01:58 So we have mc times v subc plus mr times v subc is equal to mc or at mr plus mc times mc times v0 plus mr times vr...

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