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jamesbiomed
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Homework Statement
A 3000-kg Cessna airplane flying north at v1 = 70 m/s at an altitude of 1750 m over the jungles of Brazil collided with a 7000-kg cargo plane flying at an angle of θ = 34° north of west with speed v2 = 111 m/s. As measured from a point on the ground directly below the collision, the Cessna wreckage was found 1000 m away at an angle of 25° south of west, as shown in the figure. The cargo plane broke into two pieces. Rescuers located a 4000-kg piece 1800 m away from the same point at an angle of 22° east of north. Where should they look for the other piece of the cargo plane? Give a distance and a direction from the point directly below the collision. (Let to the east be the +x-direction and to the north be the +y-direction.)
http://www.webassign.net/bauerphys1/7-p-064a-alt.gif
http://www.webassign.net/bauerphys1/7-p-064b.gif
Homework Equations
P0=P.
Px=7000kg(111 m/s) cos34
Py=3000kg(70m/s)+7000kg(111 m/s)sin34
P=mv
Perhaps kinematics to find distance.
The Attempt at a Solution
px=7000(111)cos34=644,162.2 kg m/s
py=7000(111)sin34+(70*3000)=644,492.9 kg m/s
P=sqrt (px^2+py^2)=911216.8 kg m/s
Now here's where I'm grasping for straws:
Py/Px=1.00. arctan (1.00)==45
Final angles: 22+25+90=137.
Tan(137)=-.933.
1.00x=.933=>x=.933
.933*911216.8=849718.8.
849718.8/ (mass of third piece=3000kg)=283.24 m/s
Tried kinematics to get a final distance traveled, but to no avail.
For the angle: 1.0-.93=.07. arctan.07=4 degrees. which was incorrect.
Thanks!