Projectile Motion Problem with Momentum

[SprintsMcgee]

Problem: In a military test, a 575 kg unmanned spy plane is traveling north at an altitude of 2700m and a speed of 450m/s . It is intercepted by a 1280kg rocket traveling east at 725m/s. If the rocket and the spy plane become enmeshed in a tangled mess, where, relative to the point of impact, do they hit the ground? Give the direction as an angle east of north.



The relevant equations I used were the momentum equations in the x and y direction which are Px = m1v1 = (m1 + m2)vfx and Py = m2v2 = (m1 + m2)vfy. Then I used the equations to find the mag of Vf and the ø of the equation.


I was able to get the ø to be 74.4 in the east of North direction and then I got the Vf = 519.38. I know that that's the Vi for a projectile motion equation but I cannot figure out what I am doing wrong. I keep getting answers close to 14 km. What mistakes am I making!?

 

[Tojo Joe]

could you provide your work?
i am currently learning this in class and could double check you if i can see your wrok.

 

[SprintsMcgee]

Px = m1v1 = (m1+m2)vfx
= 575(450) = (575+1280)vf => vf = 139.5
Py = m2v2 = (m1+m2)vfy
= -1280(725) = (575+1280)vfy => vf = -500.3

Ptot = √(vfy^2 +vfx^2) = √(139.5^2 + 500.3^2) = 519.38

omega = arctan( y/x) = arctan(-500.3/139.5) = -74.4 degrees.

Thats all I got for now.

 

[Tojo Joe]

We haven't gotten that far yet unfortunately. We just finished up doing problems like two cars crash into each other head on. When i did my work i got an answer that didn't make sense either.

Your answer of fourteen km may be right because the forward momentum of the tangled mess will quickly dispipate and it will fall straight down rather quickly. Wish I could be more help but I am only in high school level physics.

 

[Tojo Joe]

omega = arctan( y/x) = arctan(-500.3/139.5) = -74.4 degrees.

What is omega and arctan mean? Opposite/Adjacent?

 

[SprintsMcgee]

its omega = tan^-1 (y/x) the y component and the x component of the velocity.