Rocket Q: Solving 2-Part Launch Problem w/ 900kg Rocket

[Naeem]

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.

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:

t1.07143
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

 

[member 553083]

!The correct answer is 22.5 m. The nose cone will travel a distance of 11.25 m horizontally in 1.071 seconds and then fall the same distance vertically to the ground. So the total distance travelled is 11.25 + 11.25 = 22.5 m.

 

[wais]

!

To find the distance from the launch point, we first need to find the time it takes for the nose cone to reach the ground. We can use the equation Δy = v0t + 1/2at^2 to find this time, where Δy is the vertical displacement, v0 is the initial velocity, a is acceleration (in this case, due to gravity), and t is time. Since we know that the initial velocity is 21 m/s and the acceleration is -9.8 m/s^2 (due to gravity), and the final displacement is 0 (since it reaches the ground), we can solve for t:

0 = 21t + 1/2(-9.8)t^2
0 = 21t - 4.9t^2
0 = 4.9t(4.286 - t)
t = 4.286 seconds

So it takes 4.286 seconds for the nose cone to reach the ground. Now, we can use the equation Δx = v0t to find the horizontal distance it travels during this time:

Δx = 21(4.286)
Δx = 90.006 m

Therefore, the nose cone will land approximately 90 meters from the launch point. **ANSWER!**