Q. A 900 kg rocket with an initial velocity of 21 m/sec is launched with at an angle of 60o to the horizontal as shown to the left. At the top of it, the nose-cone of the rocket separates from the first stage of the rocket. Both the first stage and the nose cone are of equal mass and the separation happens in such a way the the first stage falls vertically from the point of separation.(adsbygoogle = window.adsbygoogle || []).push({});

First of all we know that because the rocket separates at the very top, the vertical component of the velocity is 0.

Define some parameters:

To get the horizontal component,

Define . The horizontal component is =10.5 m/s

Hence, the velocity is 10.5 m/s **ANSWER!**

vCOM= ____________________ m/sec

________________________________________

b) What is the velocity of the COM of the two part sytstem immediately after the separation?

vCOM= ____________________ m/sec

Because there are no external forces during the separation, the momentum does not change.

Conservation of x-component:

Conservation of y-component:

= 10.5m/s

= 0. m/s

Therefore, the total velocity is 10.5 m/s. **ANSWER!**

Note that it does not change!

________________________________________

c) What is the velocity of the COM of the two part system 1.4 seconds after the separation?

vCOM= ____________________ m/sec

The only external force acting is gravity, so the x component remains the same.

So =21 m/s

=0 m/s

After 1.4 seconds, the vertical component of the velocity of the nose is:

= -13.72 m/s

and that of the first stage is:

= -13.72 m/s

So the = 10.5 m/s

= -13.72

And = 17.276 m/s **ANSWER!**

________________________________________

d) What was the impulse delivered to the first stage of the rocket during the separation?

The impulse is change in momentum.

Before the separation, the momentum was

= 4725 kg*m/s

=0 kg*m/s

After the separation, its momentum was:

=0

=0

since basically it stops and then just falls to the ground.

Therefore, the total change in momentum, or its impulse is 4725 kg*m/s **ANSWER!**

(in the x-direction)

vCOM (you probably mean the impulse, not vCOM) ____________________ kg-m/sec

________________________________________

e) How far from the launch point will the nose-cone of the rocket land?

Let’s find where the collision takes place:

t1.07143

So the collision takes place after 1.071 seconds, at which the rocket has traveled =11.25m

From this point, the nose has a horizontal velocity of =21 m/s. It will take the same amount of time to fall to the ground, at which point it has traveled =22.5m **ANSWER!**, which is wrong!!!!!!!!, don't know if my reasoning is correct or not.

Δx= ____________________

Please help on this last part folks!!!!!!!!!

Thanks a lot

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# Homework Help: Two Part Rocket

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